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University Physics Volume 2 SENIOR CONTRIBUTING AUTHORS SAMUEL J. LING, TRUMAN STATE UNIVERSITY JEFF SANNY, LOYOLA MARYMOUNT UNIVERSITY WILLIAM MOEBS, PHD




 

 


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Table of ContentsPreface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1Unit 1. ThermodynamicsChapter 1: Temperature and Heat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.1 Temperature and Thermal Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.2 Thermometers and Temperature Scales . . . . . . . . . . . . . . . . . . . . . . . . . . 91.3 Thermal Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.4 Heat Transfer, Specific Heat, and Calorimetry . . . . . . . . . . . . . . . . . . . . . . 191.5 Phase Changes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261.6 Mechanisms of Heat Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35Chapter 2: The Kinetic Theory of Gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 672.1 Molecular Model of an Ideal Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 682.2 Pressure, Temperature, and RMS Speed . . . . . . . . . . . . . . . . . . . . . . . . . 782.3 Heat Capacity and Equipartition of Energy . . . . . . . . . . . . . . . . . . . . . . . . 882.4 Distribution of Molecular Speeds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93Chapter 3: The First Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . 1093.1 Thermodynamic Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1103.2 Work, Heat, and Internal Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1123.3 First Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1163.4 Thermodynamic Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1223.5 Heat Capacities of an Ideal Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1263.6 Adiabatic Processes for an Ideal Gas . . . . . . . . . . . . . . . . . . . . . . . . . . . 128Chapter 4: The Second Law of Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . 1454.1 Reversible and Irreversible Processes . . . . . . . . . . . . . . . . . . . . . . . . . . 1464.2 Heat Engines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1484.3 Refrigerators and Heat Pumps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1504.4 Statements of the Second Law of Thermodynamics . . . . . . . . . . . . . . . . . . . 1534.5 The Carnot Cycle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1554.6 Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1604.7 Entropy on a Microscopic Scale . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166Unit 2. Electricity and MagnetismChapter 5: Electric Charges and Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1815.1 Electric Charge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1825.2 Conductors, Insulators, and Charging by Induction . . . . . . . . . . . . . . . . . . . . 1885.3 Coulomb's Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1925.4 Electric Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1975.5 Calculating Electric Fields of Charge Distributions . . . . . . . . . . . . . . . . . . . . 2045.6 Electric Field Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2135.7 Electric Dipoles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217Chapter 6: Gauss's Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2356.1 Electric Flux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2366.2 Explaining Gauss’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2456.3 Applying Gauss’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2516.4 Conductors in Electrostatic Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . 265Chapter 7: Electric Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2857.1 Electric Potential Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2867.2 Electric Potential and Potential Difference . . . . . . . . . . . . . . . . . . . . . . . . 2937.3 Calculations of Electric Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3057.4 Determining Field from Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3167.5 Equipotential Surfaces and Conductors . . . . . . . . . . . . . . . . . . . . . . . . . . 3197.6 Applications of Electrostatics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 328Chapter 8: Capacitance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3458.1 Capacitors and Capacitance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3458.2 Capacitors in Series and in Parallel . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3558.3 Energy Stored in a Capacitor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3618.4 Capacitor with a Dielectric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3658.5 Molecular Model of a Dielectric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 368Chapter 9: Current and Resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385




9.1 Electrical Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3869.2 Model of Conduction in Metals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3919.3 Resistivity and Resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3979.4 Ohm's Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4069.5 Electrical Energy and Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4099.6 Superconductors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415Chapter 10: Direct-Current Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43110.1 Electromotive Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43210.2 Resistors in Series and Parallel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44010.3 Kirchhoff's Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45310.4 Electrical Measuring Instruments . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46610.5 RC Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46910.6 Household Wiring and Electrical Safety . . . . . . . . . . . . . . . . . . . . . . . . . 475Chapter 11: Magnetic Forces and Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49311.1 Magnetism and Its Historical Discoveries . . . . . . . . . . . . . . . . . . . . . . . . 49411.2 Magnetic Fields and Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49611.3 Motion of a Charged Particle in a Magnetic Field . . . . . . . . . . . . . . . . . . . . 50111.4 Magnetic Force on a Current-Carrying Conductor . . . . . . . . . . . . . . . . . . . . 50611.5 Force and Torque on a Current Loop . . . . . . . . . . . . . . . . . . . . . . . . . . 51111.6 The Hall Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51411.7 Applications of Magnetic Forces and Fields . . . . . . . . . . . . . . . . . . . . . . . 517Chapter 12: Sources of Magnetic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53512.1 The Biot-Savart Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53612.2 Magnetic Field Due to a Thin Straight Wire . . . . . . . . . . . . . . . . . . . . . . . 54012.3 Magnetic Force between Two Parallel Currents . . . . . . . . . . . . . . . . . . . . . 54312.4 Magnetic Field of a Current Loop . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54612.5 Ampère’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54912.6 Solenoids and Toroids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55512.7 Magnetism in Matter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 560Chapter 13: Electromagnetic Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58113.1 Faraday’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58213.2 Lenz's Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58613.3 Motional Emf . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59113.4 Induced Electric Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59813.5 Eddy Currents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60213.6 Electric Generators and Back Emf . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60613.7 Applications of Electromagnetic Induction . . . . . . . . . . . . . . . . . . . . . . . . 613Chapter 14: Inductance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62714.1 Mutual Inductance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62814.2 Self-Inductance and Inductors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63114.3 Energy in a Magnetic Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63714.4 RL Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63914.5 Oscillations in an LC Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64514.6 RLC Series Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 648Chapter 15: Alternating-Current Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66115.1 AC Sources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66215.2 Simple AC Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66315.3 RLC Series Circuits with AC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67015.4 Power in an AC Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67515.5 Resonance in an AC Circuit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67915.6 Transformers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 684Chapter 16: Electromagnetic Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69916.1 Maxwell’s Equations and Electromagnetic Waves . . . . . . . . . . . . . . . . . . . . 70016.2 Plane Electromagnetic Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70616.3 Energy Carried by Electromagnetic Waves . . . . . . . . . . . . . . . . . . . . . . . 71216.4 Momentum and Radiation Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . 71716.5 The Electromagnetic Spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 722Appendix A: Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 741Appendix B: Conversion Factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 745


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Appendix C: Fundamental Constants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 749Appendix D: Astronomical Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 751Appendix E: Mathematical Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 753Appendix F: Chemistry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 757Appendix G: The Greek Alphabet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 759Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 807




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PREFACE
Welcome to University Physics, an OpenStax resource. This textbook was written to increase student access to high-quality learning materials, maintaining highest standards of academic rigor at little to no cost.
About OpenStax
OpenStax is a nonprofit based at Rice University, and it’s our mission to improve student access to education. Our first openly licensed college textbook was published in 2012 and our library has since scaled to over 20 books used by hundreds of thousands of students across the globe. Our adaptive learning technology, designed to improve learning outcomes through personalized educational paths, is currently being piloted for K–12 and college. The OpenStax mission is made possible through the generous support of philanthropic foundations. Through these partnerships and with the help of additional low-cost resources from our OpenStax partners, OpenStax is breaking down the most common barriers to learning and empowering students and instructors to succeed.
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Errata
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Format
You can access this textbook for free in web view or PDF through openstax.org, and for a low cost in print.
About University Physics
University Physics is designed for the two- or three-semester calculus-based physics course. The text has been developed to meet the scope and sequence of most university physics courses and provides a foundation for a career in mathematics, science, or engineering. The book provides an important opportunity for students to learn the core concepts of physics and understand how those concepts apply to their lives and to the world around them.
Due to the comprehensive nature of the material, we are offering the book in three volumes for flexibility and efficiency.
Coverage and Scope
Our University Physics textbook adheres to the scope and sequence of most two- and three-semester physics courses nationwide. We have worked to make physics interesting and accessible to students while maintaining the mathematical rigor inherent in the subject. With this objective in mind, the content of this textbook has been developed and arranged to provide a logical progression from fundamental to more advanced concepts, building upon what students have already learned and emphasizing connections between topics and between theory and applications. The goal of each section is to enable students not just to recognize concepts, but to work with them in ways that will be useful in later courses and future careers. The organization and pedagogical features were developed and vetted with feedback from science educators dedicated to the project.


Preface 1




VOLUME I
Unit 1: Mechanics


Chapter 1: Units and Measurement
Chapter 2: Vectors
Chapter 3: Motion Along a Straight Line
Chapter 4: Motion in Two and Three Dimensions
Chapter 5: Newton’s Laws of Motion
Chapter 6: Applications of Newton’s Laws
Chapter 7: Work and Kinetic Energy
Chapter 8: Potential Energy and Conservation of Energy
Chapter 9: Linear Momentum and Collisions
Chapter 10: Fixed-Axis Rotation
Chapter 11: Angular Momentum
Chapter 12: Static Equilibrium and Elasticity
Chapter 13: Gravitation
Chapter 14: Fluid Mechanics


Unit 2: Waves and Acoustics
Chapter 15: Oscillations
Chapter 16: Waves
Chapter 17: Sound


VOLUME II
Unit 1: Thermodynamics


Chapter 1: Temperature and Heat
Chapter 2: The Kinetic Theory of Gases
Chapter 3: The First Law of Thermodynamics
Chapter 4: The Second Law of Thermodynamics


Unit 2: Electricity and Magnetism
Chapter 5: Electric Charges and Fields
Chapter 6: Gauss’s Law
Chapter 7: Electric Potential
Chapter 8: Capacitance
Chapter 9: Current and Resistance
Chapter 10: Direct-Current Circuits
Chapter 11: Magnetic Forces and Fields
Chapter 12: Sources of Magnetic Fields
Chapter 13: Electromagnetic Induction
Chapter 14: Inductance
Chapter 15: Alternating-Current Circuits
Chapter 16: Electromagnetic Waves


VOLUME III
Unit 1: Optics


Chapter 1: The Nature of Light


2 Preface


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Chapter 2: Geometric Optics and Image Formation
Chapter 3: Interference
Chapter 4: Diffraction


Unit 2: Modern Physics
Chapter 5: Relativity
Chapter 6: Photons and Matter Waves
Chapter 7: Quantum Mechanics
Chapter 8: Atomic Structure
Chapter 9: Condensed Matter Physics
Chapter 10: Nuclear Physics
Chapter 11: Particle Physics and Cosmology


Pedagogical Foundation
Throughout University Physics you will find derivations of concepts that present classical ideas and techniques, as wellas modern applications and methods. Most chapters start with observations or experiments that place the material in acontext of physical experience. Presentations and explanations rely on years of classroom experience on the part of long-time physics professors, striving for a balance of clarity and rigor that has proven successful with their students. Throughoutthe text, links enable students to review earlier material and then return to the present discussion, reinforcing connectionsbetween topics. Key historical figures and experiments are discussed in the main text (rather than in boxes or sidebars),maintaining a focus on the development of physical intuition. Key ideas, definitions, and equations are highlighted inthe text and listed in summary form at the end of each chapter. Examples and chapter-opening images often includecontemporary applications from daily life or modern science and engineering that students can relate to, from smart phonesto the internet to GPS devices.
Assessments That Reinforce Key Concepts
In-chapter Examples generally follow a three-part format of Strategy, Solution, and Significance to emphasize how toapproach a problem, how to work with the equations, and how to check and generalize the result. Examples are oftenfollowed by Check Your Understanding questions and answers to help reinforce for students the important ideas of theexamples. Problem-Solving Strategies in each chapter break down methods of approaching various types of problems intosteps students can follow for guidance. The book also includes exercises at the end of each chapter so students can practicewhat they’ve learned.


Conceptual questions do not require calculation but test student learning of the key concepts.
Problems categorized by section test student problem-solving skills and the ability to apply ideas to practicalsituations.
Additional Problems apply knowledge across the chapter, forcing students to identify what concepts and equationsare appropriate for solving given problems. Randomly located throughout the problems are Unreasonable Resultsexercises that ask students to evaluate the answer to a problem and explain why it is not reasonable and whatassumptions made might not be correct.
Challenge Problems extend text ideas to interesting but difficult situations.


Answers for selected exercises are available in an Answer Key at the end of the book.
Additional ResourcesStudent and Instructor Resources
We’ve compiled additional resources for both students and instructors, including Getting Started Guides, PowerPoint slides,and answer and solution guides for instructors and students. Instructor resources require a verified instructor account, whichcan be requested on your openstax.org log-in. Take advantage of these resources to supplement your OpenStax book.
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Preface 3




About the AuthorsSenior Contributing Authors
Samuel J. Ling, Truman State UniversityDr. Samuel Ling has taught introductory and advanced physics for over 25 years at Truman State University, where he iscurrently Professor of Physics and the Department Chair. Dr. Ling has two PhDs from Boston University, one in Chemistryand the other in Physics, and he was a Research Fellow at the Indian Institute of Science, Bangalore, before joining Truman.Dr. Ling is also an author of A First Course in Vibrations and Waves, published by Oxford University Press. Dr. Ling hasconsiderable experience with research in Physics Education and has published research on collaborative learning methods inphysics teaching. He was awarded a Truman Fellow and a Jepson fellow in recognition of his innovative teaching methods.Dr. Ling’s research publications have spanned Cosmology, Solid State Physics, and Nonlinear Optics.
Jeff Sanny, Loyola Marymount UniversityDr. Jeff Sanny earned a BS in Physics from Harvey Mudd College in 1974 and a PhD in Solid State Physics from theUniversity of California–Los Angeles in 1980. He joined the faculty at Loyola Marymount University in the fall of 1980.During his tenure, he has served as department Chair as well as Associate Dean. Dr. Sanny enjoys teaching introductoryphysics in particular. He is also passionate about providing students with research experience and has directed an activeundergraduate student research group in space physics for many years.
Bill Moebs, PhDDr. William Moebs earned a BS and PhD (1959 and 1965) from the University of Michigan. He then joined their staffas a Research Associate for one year, where he continued his doctoral research in particle physics. In 1966, he acceptedan appointment to the Physics Department of Indiana Purdue Fort Wayne (IPFW), where he served as Department Chairfrom 1971 to 1979. In 1979, he moved to Loyola Marymount University (LMU), where he served as Chair of the PhysicsDepartment from 1979 to 1986. He retired from LMU in 2000. He has published research in particle physics, chemicalkinetics, cell division, atomic physics, and physics teaching.
Contributing Authors
David Anderson, Albion CollegeDaniel Bowman, Ferrum CollegeDedra Demaree, Georgetown UniversityGerald Friedman, Santa Fe Community CollegeLev Gasparov, University of North FloridaEdw. S. Ginsberg, University of MassachusettsAlice Kolakowska, University of MemphisLee LaRue, Paris Junior CollegeMark Lattery, University of WisconsinRichard Ludlow, Daniel Webster CollegePatrick Motl, Indiana University–KokomoTao Pang, University of Nevada–Las VegasKenneth Podolak, Plattsburgh State UniversityTakashi Sato, Kwantlen Polytechnic UniversityDavid Smith, University of the Virgin IslandsJoseph Trout, Richard Stockton CollegeKevin Wheelock, Bellevue College
Reviewers
Salameh Ahmad, Rochester Institute of Technology–DubaiJohn Aiken, University of Colorado–BoulderAnand Batra, Howard UniversityRaymond Benge, Terrant County CollegeGavin Buxton, Robert Morris UniversityErik Christensen, South Florida State CollegeClifton Clark, Fort Hays State UniversityNelson Coates, California Maritime AcademyHerve Collin, Kapi’olani Community CollegeCarl Covatto, Arizona State UniversityAlexander Cozzani, Imperial Valley CollegeDanielle Dalafave, The College of New JerseyNicholas Darnton, Georgia Institute of Technology


4 Preface


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Robert Edmonds, Tarrant County CollegeWilliam Falls, Erie Community CollegeStanley Forrester, Broward CollegeUmesh Garg, University of Notre DameMaurizio Giannotti, Barry UniversityBryan Gibbs, Dallas County Community CollegeMark Giroux, East Tennessee State UniversityMatthew Griffiths, University of New HavenAlfonso Hinojosa, University of Texas–ArlingtonSteuard Jensen, Alma CollegeDavid Kagan, University of MassachusettsJill Leggett, Florida State College–JacksonvilleSergei Katsev, University of Minnesota–DuluthAlfredo Louro, University of CalgaryJames Maclaren, Tulane UniversityPonn Maheswaranathan, Winthrop UniversitySeth Major, Hamilton CollegeOleg Maksimov, Excelsior CollegeAristides Marcano, Delaware State UniversityMarles McCurdy, Tarrant County CollegeJames McDonald, University of HartfordRalph McGrew, SUNY–Broome Community CollegePaul Miller, West Virginia UniversityTamar More, University of PortlandFarzaneh Najmabadi, University of PhoenixRichard Olenick, The University of DallasChristopher Porter, Ohio State UniversityLiza Pujji, Manakau Institute of TechnologyBaishali Ray, Young Harris UniversityAndrew Robinson, Carleton UniversityAruvana Roy, Young Harris UniversityAbhijit Sarkar, The Catholic University of AmericaGajendra Tulsian, Daytona State CollegeAdria Updike, Roger Williams UniversityClark Vangilder, Central Arizona UniversitySteven Wolf, Texas State UniversityAlexander Wurm, Western New England UniversityLei Zhang, Winston Salem State UniversityUlrich Zurcher, Cleveland State University


Preface 5




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1 | TEMPERATURE ANDHEAT


Figure 1.1 These snowshoers on Mount Hood in Oregon are enjoying the heat flow and light caused by high temperature. Allthree mechanisms of heat transfer are relevant to this picture. The heat flowing out of the fire also turns the solid snow to liquidwater and vapor. (credit: “Mt. Hood Territory”/Flickr)


Chapter Outline
1.1 Temperature and Thermal Equilibrium
1.2 Thermometers and Temperature Scales
1.3 Thermal Expansion
1.4 Heat Transfer, Specific Heat, and Calorimetry
1.5 Phase Changes
1.6 Mechanisms of Heat Transfer


Introduction
Heat and temperature are important concepts for each of us, every day. How we dress in the morning depends on whetherthe day is hot or cold, and most of what we do requires energy that ultimately comes from the Sun. The study of heatand temperature is part of an area of physics known as thermodynamics. The laws of thermodynamics govern the flowof energy throughout the universe. They are studied in all areas of science and engineering, from chemistry to biology toenvironmental science.
In this chapter, we explore heat and temperature. It is not always easy to distinguish these terms. Heat is the flow of energyfrom one object to another. This flow of energy is caused by a difference in temperature. The transfer of heat can changetemperature, as can work, another kind of energy transfer that is central to thermodynamics. We return to these basic ideasseveral times throughout the next four chapters, and you will see that they affect everything from the behavior of atoms andmolecules to cooking to our weather on Earth to the life cycles of stars.


Chapter 1 | Temperature and Heat 7




1.1 | Temperature and Thermal Equilibrium
Learning Objectives


By the end of this section, you will be able to:
• Define temperature and describe it qualitatively
• Explain thermal equilibrium
• Explain the zeroth law of thermodynamics


Heat is familiar to all of us. We can feel heat entering our bodies from the summer Sun or from hot coffee or tea aftera winter stroll. We can also feel heat leaving our bodies as we feel the chill of night or the cooling effect of sweat afterexercise.
What is heat? How do we define it and how is it related to temperature? What are the effects of heat and how does itflow from place to place? We will find that, in spite of the richness of the phenomena, a small set of underlying physicalprinciples unites these subjects and ties them to other fields. We start by examining temperature and how to define andmeasure it.
Temperature
The concept of temperature has evolved from the common concepts of hot and cold. The scientific definition of temperatureexplains more than our senses of hot and cold. As you may have already learned, many physical quantities are definedsolely in terms of how they are observed or measured, that is, they are defined operationally. Temperature is operationallydefined as the quantity of what we measure with a thermometer. As we will see in detail in a later chapter on the kinetictheory of gases, temperature is proportional to the average kinetic energy of translation, a fact that provides a more physicaldefinition. Differences in temperature maintain the transfer of heat, or heat transfer, throughout the universe. Heat transferis the movement of energy from one place or material to another as a result of a difference in temperature. (You will learnmore about heat transfer later in this chapter.)
Thermal Equilibrium
An important concept related to temperature is thermal equilibrium. Two objects are in thermal equilibrium if they are inclose contact that allows either to gain energy from the other, but nevertheless, no net energy is transferred between them.Even when not in contact, they are in thermal equilibrium if, when they are placed in contact, no net energy is transferredbetween them. If two objects remain in contact for a long time, they typically come to equilibrium. In other words, twoobjects in thermal equilibrium do not exchange energy.
Experimentally, if object A is in equilibrium with object B, and object B is in equilibrium with object C, then (as you mayhave already guessed) object A is in equilibrium with object C. That statement of transitivity is called the zeroth law ofthermodynamics. (The number “zeroth” was suggested by British physicist Ralph Fowler in the 1930s. The first, second,and third laws of thermodynamics were already named and numbered then. The zeroth law had seldom been stated, but itneeds to be discussed before the others, so Fowler gave it a smaller number.) Consider the case where A is a thermometer.The zeroth law tells us that if A reads a certain temperature when in equilibrium with B, and it is then placed in contact withC, it will not exchange energy with C; therefore, its temperature reading will remain the same (Figure 1.2). In other words,if two objects are in thermal equilibrium, they have the same temperature.


Figure 1.2 If thermometer A is in thermal equilibrium withobject B, and B is in thermal equilibrium with C, then A is inthermal equilibrium with C. Therefore, the reading on A staysthe same when A is moved over to make contact with C.


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A thermometer measures its own temperature. It is through the concepts of thermal equilibrium and the zeroth law ofthermodynamics that we can say that a thermometer measures the temperature of something else, and to make sense of thestatement that two objects are at the same temperature.
In the rest of this chapter, we will often refer to “systems” instead of “objects.” As in the chapter on linear momentum andcollisions, a system consists of one or more objects—but in thermodynamics, we require a system to be macroscopic, that
is, to consist of a huge number (such as 1023 ) of molecules. Then we can say that a system is in thermal equilibrium with
itself if all parts of it are at the same temperature. (We will return to the definition of a thermodynamic system in the chapteron the first law of thermodynamics.)
1.2 | Thermometers and Temperature Scales


Learning Objectives
By the end of this section, you will be able to:
• Describe several different types of thermometers
• Convert temperatures between the Celsius, Fahrenheit, and Kelvin scales


Any physical property that depends consistently and reproducibly on temperature can be used as the basis of a thermometer.For example, volume increases with temperature for most substances. This property is the basis for the common alcoholthermometer and the original mercury thermometers. Other properties used to measure temperature include electricalresistance, color, and the emission of infrared radiation (Figure 1.3).


Chapter 1 | Temperature and Heat 9




Figure 1.3 Because many physical properties depend on temperature, the variety of thermometers is remarkable. (a) In thiscommon type of thermometer, the alcohol, containing a red dye, expands more rapidly than the glass encasing it. When thethermometer’s temperature increases, the liquid from the bulb is forced into the narrow tube, producing a large change in thelength of the column for a small change in temperature. (b) Each of the six squares on this plastic (liquid crystal) thermometercontains a film of a different heat-sensitive liquid crystal material. Below 95 °F , all six squares are black. When the plastic
thermometer is exposed to a temperature of 95 °F , the first liquid crystal square changes color. When the temperature reaches
above 96.8 °F , the second liquid crystal square also changes color, and so forth. (c) A firefighter uses a pyrometer to check the
temperature of an aircraft carrier’s ventilation system. The pyrometer measures infrared radiation (whose emission varies withtemperature) from the vent and quickly produces a temperature readout. Infrared thermometers are also frequently used tomeasure body temperature by gently placing them in the ear canal. Such thermometers are more accurate than the alcoholthermometers placed under the tongue or in the armpit. (credit b: modification of work by Tess Watson; credit c: modification ofwork by Lamel J. Hinton)


Thermometers measure temperature according to well-defined scales of measurement. The three most common temperaturescales are Fahrenheit, Celsius, and Kelvin. Temperature scales are created by identifying two reproducible temperatures.The freezing and boiling temperatures of water at standard atmospheric pressure are commonly used.
On the Celsius scale, the freezing point of water is 0 °C and the boiling point is 100 °C. The unit of temperature on this
scale is the degree Celsius (°C) . The Fahrenheit scale (still the most frequently used for common purposes in the United
States) has the freezing point of water at 32 °F and the boiling point at 212 °F. Its unit is the degree Fahrenheit ( °F ).
You can see that 100 Celsius degrees span the same range as 180 Fahrenheit degrees. Thus, a temperature difference of one
degree on the Celsius scale is 1.8 times as large as a difference of one degree on the Fahrenheit scale, or ΔTF = 95ΔTC.
The definition of temperature in terms of molecular motion suggests that there should be a lowest possible temperature,where the average kinetic energy of molecules is zero (or the minimum allowed by quantum mechanics). Experimentsconfirm the existence of such a temperature, called absolute zero. An absolute temperature scale is one whose zero pointis absolute zero. Such scales are convenient in science because several physical quantities, such as the volume of an idealgas, are directly related to absolute temperature.
The Kelvin scale is the absolute temperature scale that is commonly used in science. The SI temperature unit is the kelvin,which is abbreviated K (not accompanied by a degree sign). Thus 0 K is absolute zero. The freezing and boiling points


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of water are 273.15 K and 373.15 K, respectively. Therefore, temperature differences are the same in units of kelvins anddegrees Celsius, or ΔTC = ΔTK.
The relationships between the three common temperature scales are shown in Figure 1.4. Temperatures on these scalescan be converted using the equations in Table 1.1.


Figure 1.4 Relationships between the Fahrenheit, Celsius, and Kelvin temperature scales are shown. The relative sizesof the scales are also shown.
To convert from… Use this equation…
Celsius to Fahrenheit TF = 95TC + 32
Fahrenheit to Celsius TC = 59⎛⎝TF − 32⎞⎠
Celsius to Kelvin TK = TC + 273.15
Kelvin to Celsius TC = TK − 273.15
Fahrenheit to Kelvin TK = 59⎛⎝TF − 32⎞⎠+ 273.15
Kelvin to Fahrenheit TF = 95⎛⎝TK − 273.15⎞⎠+ 32


Table 1.1 Temperature Conversions
To convert between Fahrenheit and Kelvin, convert to Celsius as an intermediate step.
Example 1.1


Converting between Temperature Scales: Room Temperature
“Room temperature” is generally defined in physics to be 25 °C . (a) What is room temperature in °F ? (b) What
is it in K?
Strategy
To answer these questions, all we need to do is choose the correct conversion equations and substitute the knownvalues.
Solution
To convert from °C to °F , use the equation


Chapter 1 | Temperature and Heat 11




TF =
9
5
TC + 32.


Substitute the known value into the equation and solve:
TF =


9
5
(25 °C) + 32 = 77 °F.


Similarly, we find that TK = TC + 273.15 = 298 K .


The Kelvin scale is part of the SI system of units, so its actual definition is more complicated than the one given above.First, it is not defined in terms of the freezing and boiling points of water, but in terms of the triple point. The triple pointis the unique combination of temperature and pressure at which ice, liquid water, and water vapor can coexist stably. Aswill be discussed in the section on phase changes, the coexistence is achieved by lowering the pressure and consequentlythe boiling point to reach the freezing point. The triple-point temperature is defined as 273.16 K. This definition has theadvantage that although the freezing temperature and boiling temperature of water depend on pressure, there is only onetriple-point temperature.
Second, even with two points on the scale defined, different thermometers give somewhat different results for othertemperatures. Therefore, a standard thermometer is required. Metrologists (experts in the science of measurement) havechosen the constant-volume gas thermometer for this purpose. A vessel of constant volume filled with gas is subjected totemperature changes, and the measured temperature is proportional to the change in pressure. Using “TP” to represent thetriple point,


T =
p
pTP


TTP.


The results depend somewhat on the choice of gas, but the less dense the gas in the bulb, the better the results for differentgases agree. If the results are extrapolated to zero density, the results agree quite well, with zero pressure corresponding toa temperature of absolute zero.
Constant-volume gas thermometers are big and come to equilibrium slowly, so they are used mostly as standards to calibrateother thermometers.


Visit this site (https://openstaxcollege.org/l/21consvolgasth) to learn more about the constant-volume gasthermometer.


1.3 | Thermal Expansion
Learning Objectives


By the end of this section, you will be able to:
• Answer qualitative questions about the effects of thermal expansion
• Solve problems involving thermal expansion, including those involving thermal stress


The expansion of alcohol in a thermometer is one of many commonly encountered examples of thermal expansion, whichis the change in size or volume of a given system as its temperature changes. The most visible example is the expansion ofhot air. When air is heated, it expands and becomes less dense than the surrounding air, which then exerts an (upward) forceon the hot air and makes steam and smoke rise, hot air balloons float, and so forth. The same behavior happens in all liquidsand gases, driving natural heat transfer upward in homes, oceans, and weather systems, as we will discuss in an upcomingsection. Solids also undergo thermal expansion. Railroad tracks and bridges, for example, have expansion joints to allowthem to freely expand and contract with temperature changes, as shown in Figure 1.5.


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Figure 1.5 (a) Thermal expansion joints like these in the (b) Auckland Harbour Bridge in New Zealand allow bridges tochange length without buckling. (credit: “ŠJů”/Wikimedia Commons)


What is the underlying cause of thermal expansion? As previously mentioned, an increase in temperature means anincrease in the kinetic energy of individual atoms. In a solid, unlike in a gas, the molecules are held in place by forcesfrom neighboring molecules; as we saw in Oscillations (http://cnx.org/content/m58360/latest/) , the forces canbe modeled as in harmonic springs described by the Lennard-Jones potential. Energy in Simple Harmonic Motion(http://cnx.org/content/m58362/latest/#CNX_UPhysics_15_02_LennaJones) shows that such potentials areasymmetrical in that the potential energy increases more steeply when the molecules get closer to each other than when theyget farther away. Thus, at a given kinetic energy, the distance moved is greater when neighbors move away from each otherthan when they move toward each other. The result is that increased kinetic energy (increased temperature) increases theaverage distance between molecules—the substance expands.
For most substances under ordinary conditions, it is an excellent approximation that there is no preferred direction (that is,the solid is “isotropic”), and an increase in temperature increases the solid’s size by a certain fraction in each dimension.Therefore, if the solid is free to expand or contract, its proportions stay the same; only its overall size changes.


Linear Thermal Expansion
According to experiments, the dependence of thermal expansion on temperature, substance, and original length issummarized in the equation


(1.1)dL
dT


= αL


where L is the original length, dL
dT


is the change in length with respect to temperature, and α is the coefficient of
linear expansion, a material property that varies slightly with temperature. As α is nearly constant and also very
small, for practical purposes, we use the linear approximation:


(1.2)ΔL = αLΔT .
Table 1.2 lists representative values of the coefficient of linear expansion. As noted earlier, ΔT is the same whether it
is expressed in units of degrees Celsius or kelvins; thus, α may have units of 1/°C or 1/K with the same value in either
case. Approximating α as a constant is quite accurate for small changes in temperature and sufficient for most practical
purposes, even for large changes in temperature. We examine this approximation more closely in the next example.


Chapter 1 | Temperature and Heat 13




Material Coefficient of LinearExpansion α(1/°C) Coefficient of VolumeExpansion β(1/°C)
Solids
Aluminum 25 × 10−6 75 × 10−6
Brass 19 × 10−6 56 × 10−6
Copper 17 × 10−6 51 × 10−6
Gold 14 × 10−6 42 × 10−6
Iron or steel 12 × 10−6 35 × 10−6
Invar (nickel-iron alloy) 0.9 × 10−6 2.7 × 10−6
Lead 29 × 10−6 87 × 10−6
Silver 18 × 10−6 54 × 10−6
Glass (ordinary) 9 × 10−6 27 × 10−6
Glass (Pyrex®) 3 × 10−6 9 × 10−6
Quartz 0.4 × 10−6 1 × 10−6
Concrete, brick ~12 × 10−6 ~36 × 10−6
Marble (average) 2.5 × 10−6 7.5 × 10−6
Liquids
Ether 1650 × 10−6
Ethyl alcohol 1100 × 10−6
Gasoline 950 × 10−6
Glycerin 500 × 10−6
Mercury 180 × 10−6
Water 210 × 10−6
Gases
Air and most other gases atatmospheric pressure 3400 × 10−6


Table 1.2 Thermal Expansion Coefficients
Thermal expansion is exploited in the bimetallic strip (Figure 1.6). This device can be used as a thermometer if the curvingstrip is attached to a pointer on a scale. It can also be used to automatically close or open a switch at a certain temperature,as in older or analog thermostats.


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Figure 1.6 The curvature of a bimetallic strip depends ontemperature. (a) The strip is straight at the starting temperature,where its two components have the same length. (b) At a highertemperature, this strip bends to the right, because the metal onthe left has expanded more than the metal on the right. At alower temperature, the strip would bend to the left.


Example 1.2
Calculating Linear Thermal Expansion
The main span of San Francisco’s Golden Gate Bridge is 1275 m long at its coldest. The bridge is exposed totemperatures ranging from – 15 °C to 40 °C . What is its change in length between these temperatures? Assume
that the bridge is made entirely of steel.
Strategy
Use the equation for linear thermal expansion ΔL = αLΔT to calculate the change in length, ΔL . Use the
coefficient of linear expansion α for steel from Table 1.2, and note that the change in temperature ΔT is
55 °C.


Solution
Substitute all of the known values into the equation to solve for ΔL :


ΔL = αLΔT =


12 × 10−6


°C

⎠(1275 m)(55 °C) = 0.84 m.


Significance
Although not large compared with the length of the bridge, this change in length is observable. It is generallyspread over many expansion joints so that the expansion at each joint is small.


Thermal Expansion in Two and Three Dimensions
Unconstrained objects expand in all dimensions, as illustrated in Figure 1.7. That is, their areas and volumes, as well astheir lengths, increase with temperature. Because the proportions stay the same, holes and container volumes also get largerwith temperature. If you cut a hole in a metal plate, the remaining material will expand exactly as it would if the piece youremoved were still in place. The piece would get bigger, so the hole must get bigger too.


Thermal Expansion in Two Dimensions
For small temperature changes, the change in area ΔA is given by


(1.3)ΔA = 2αAΔT
where ΔA is the change in area A, ΔT is the change in temperature, and α is the coefficient of linear expansion,


Chapter 1 | Temperature and Heat 15




which varies slightly with temperature.


Figure 1.7 In general, objects expand in all directions as temperature increases. In these drawings, the originalboundaries of the objects are shown with solid lines, and the expanded boundaries with dashed lines. (a) Area increasesbecause both length and width increase. The area of a circular plug also increases. (b) If the plug is removed, the hole itleaves becomes larger with increasing temperature, just as if the expanding plug were still in place. (c) Volume alsoincreases, because all three dimensions increase.


Thermal Expansion in Three Dimensions
The relationship between volume and temperature dV


dT
is given by dV


dT
= βVΔT , where β is the coefficient of


volume expansion. As you can show in Exercise 1.60, β = 3α . This equation is usually written as
(1.4)ΔV = βVΔT .


Note that the values of β in Table 1.2 are equal to 3α except for rounding.


Volume expansion is defined for liquids, but linear and area expansion are not, as a liquid’s changes in linear dimensionsand area depend on the shape of its container. Thus, Table 1.2 shows liquids’ values of β but not α .
In general, objects expand with increasing temperature. Water is the most important exception to this rule. Water doesexpand with increasing temperature (its density decreases) at temperatures greater than 4 °C (40 °F) . However, it is densest
at +4 °C and expands with decreasing temperature between +4 °C and 0 °C ( 40 °F to 32 °F ), as shown in Figure 1.8.
A striking effect of this phenomenon is the freezing of water in a pond. When water near the surface cools down to 4 °C,
it is denser than the remaining water and thus sinks to the bottom. This “turnover” leaves a layer of warmer water near thesurface, which is then cooled. However, if the temperature in the surface layer drops below 4 °C , that water is less dense
than the water below, and thus stays near the top. As a result, the pond surface can freeze over. The layer of ice insulates theliquid water below it from low air temperatures. Fish and other aquatic life can survive in 4 °C water beneath ice, due to
this unusual characteristic of water.


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Figure 1.8 This curve shows the density of water as a function of temperature. Note that thethermal expansion at low temperatures is very small. The maximum density at 4 °C is only
0.0075% greater than the density at 2 °C , and 0.012% greater than that at 0 °C . The
decrease of density below 4 °C occurs because the liquid water approachs the solid crystal
form of ice, which contains more empty space than the liquid.


Example 1.3
Calculating Thermal Expansion
Suppose your 60.0-L (15.9 -gal -gal) steel gasoline tank is full of gas that is cool because it has just been pumped
from an underground reservoir. Now, both the tank and the gasoline have a temperature of 15.0 °C. How much
gasoline has spilled by the time they warm to 35.0 °C ?
Strategy
The tank and gasoline increase in volume, but the gasoline increases more, so the amount spilled is the differencein their volume changes. We can use the equation for volume expansion to calculate the change in volume of thegasoline and of the tank. (The gasoline tank can be treated as solid steel.)
Solution1. Use the equation for volume expansion to calculate the increase in volume of the steel tank:


ΔVs = βsVsΔT .


2. The increase in volume of the gasoline is given by this equation:
ΔVgas = βgasVgasΔT .


3. Find the difference in volume to determine the amount spilled as
Vspill = ΔVgas − ΔVs.


Alternatively, we can combine these three equations into a single equation. (Note that the original volumes areequal.)
Vspill =



⎝βga s − βs



⎠VΔT


= ⎡⎣(950 − 35) × 10
−6 /°C⎤⎦(60.0 L)(20.0 °C)


= 1.10 L.


Significance
This amount is significant, particularly for a 60.0-L tank. The effect is so striking because the gasoline and steel


Chapter 1 | Temperature and Heat 17




1.1


expand quickly. The rate of change in thermal properties is discussed later in this chapter.
If you try to cap the tank tightly to prevent overflow, you will find that it leaks anyway, either around the cap orby bursting the tank. Tightly constricting the expanding gas is equivalent to compressing it, and both liquids andsolids resist compression with extremely large forces. To avoid rupturing rigid containers, these containers haveair gaps, which allow them to expand and contract without stressing them.


Check Your Understanding Does a given reading on a gasoline gauge indicate more gasoline in coldweather or in hot weather, or does the temperature not matter?


Thermal Stress
If you change the temperature of an object while preventing it from expanding or contracting, the object is subjected tostress that is compressive if the object would expand in the absence of constraint and tensile if it would contract. This stressresulting from temperature changes is known as thermal stress. It can be quite large and can cause damage.
To avoid this stress, engineers may design components so they can expand and contract freely. For instance, in highways,gaps are deliberately left between blocks to prevent thermal stress from developing. When no gaps can be left, engineersmust consider thermal stress in their designs. Thus, the reinforcing rods in concrete are made of steel because steel’scoefficient of linear expansion is nearly equal to that of concrete.
To calculate the thermal stress in a rod whose ends are both fixed rigidly, we can think of the stress as developing in twosteps. First, let the ends be free to expand (or contract) and find the expansion (or contraction). Second, find the stressnecessary to compress (or extend) the rod to its original length by the methods you studied in Static Equilibrium andElasticity (http://cnx.org/content/m58339/latest/) on static equilibrium and elasticity. In other words, the ΔL of the
thermal expansion equals the ΔL of the elastic distortion (except that the signs are opposite).
Example 1.4


Calculating Thermal Stress
Concrete blocks are laid out next to each other on a highway without any space between them, so they cannotexpand. The construction crew did the work on a winter day when the temperature was 5 °C . Find the stress in
the blocks on a hot summer day when the temperature is 38 °C . The compressive Young’s modulus of concrete
is Y = 20 × 109 N/m2 .
Strategy
According to the chapter on static equilibrium and elasticity, the stress F/A is given by


F
A


= YΔL
L0


,


where Y is the Young’s modulus of the material—concrete, in this case. In thermal expansion, ΔL = αL0ΔT .
We combine these two equations by noting that the two ΔL’s are equal, as stated above. Because we are not
given L0 or A, we can obtain a numerical answer only if they both cancel out.
Solution
We substitute the thermal-expansion equation into the elasticity equation to get


F
A


= Y
αL0ΔT


L0
= YαΔT ,


and as we hoped, L0 has canceled and A appears only in F/A, the notation for the quantity we are calculating.
Now we need only insert the numbers:


F
A


= ⎛⎝20 × 10
9 N/m2⎞⎠



⎝12 × 10


−6 /°C⎞⎠(38 °C − 5 °C) = 7.9 × 10
6 N/m2.


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1.2


Significance
The ultimate compressive strength of concrete is 20 × 106 N/m2, so the blocks are unlikely to break. However,
the ultimate shear strength of concrete is only 2 × 106 N/m2, so some might chip off.


Check Your Understanding Two objects A and B have the same dimensions and are constrainedidentically. A is made of a material with a higher thermal expansion coefficient than B. If the objects are heatedidentically, will A feel a greater stress than B?


1.4 | Heat Transfer, Specific Heat, and Calorimetry
Learning Objectives


By the end of this section, you will be able to:
• Explain phenomena involving heat as a form of energy transfer
• Solve problems involving heat transfer


We have seen in previous chapters that energy is one of the fundamental concepts of physics. Heat is a type of energytransfer that is caused by a temperature difference, and it can change the temperature of an object. As we learned earlierin this chapter, heat transfer is the movement of energy from one place or material to another as a result of a differencein temperature. Heat transfer is fundamental to such everyday activities as home heating and cooking, as well as manyindustrial processes. It also forms a basis for the topics in the remainder of this chapter.
We also introduce the concept of internal energy, which can be increased or decreased by heat transfer. We discussanother way to change the internal energy of a system, namely doing work on it. Thus, we are beginning the study of therelationship of heat and work, which is the basis of engines and refrigerators and the central topic (and origin of the name)of thermodynamics.
Internal Energy and Heat
A thermal system has internal energy (also called thermal energy), which is the sum of the mechanical energies of itsmolecules. A system’s internal energy is proportional to its temperature. As we saw earlier in this chapter, if two objects atdifferent temperatures are brought into contact with each other, energy is transferred from the hotter to the colder object untilthe bodies reach thermal equilibrium (that is, they are at the same temperature). No work is done by either object becauseno force acts through a distance (as we discussed in Work and Kinetic Energy (http://cnx.org/content/m58307/latest/) ). These observations reveal that heat is energy transferred spontaneously due to a temperature difference. Figure1.9 shows an example of heat transfer.


Figure 1.9 (a) Here, the soft drink has a higher temperature than the ice, so they are not in thermal equilibrium. (b) When thesoft drink and ice are allowed to interact, heat is transferred from the drink to the ice due to the difference in temperatures untilthey reach the same temperature, T′ , achieving equilibrium. In fact, since the soft drink and ice are both in contact with the
surrounding air and the bench, the ultimate equilibrium temperature will be the same as that of the surroundings.


Chapter 1 | Temperature and Heat 19




The meaning of “heat” in physics is different from its ordinary meaning. For example, in conversation, we may say “theheat was unbearable,” but in physics, we would say that the temperature was high. Heat is a form of energy flow, whereastemperature is not. Incidentally, humans are sensitive to heat flow rather than to temperature.
Since heat is a form of energy, its SI unit is the joule (J). Another common unit of energy often used for heat is the calorie(cal), defined as the energy needed to change the temperature of 1.00 g of water by 1.00 °C —specifically, between
14.5 °C and 15.5 °C , since there is a slight temperature dependence. Also commonly used is the kilocalorie (kcal), which
is the energy needed to change the temperature of 1.00 kg of water by 1.00 °C . Since mass is most often specified in
kilograms, the kilocalorie is convenient. Confusingly, food calories (sometimes called “big calories,” abbreviated Cal) areactually kilocalories, a fact not easily determined from package labeling.
Mechanical Equivalent of Heat
It is also possible to change the temperature of a substance by doing work, which transfers energy into or out of asystem. This realization helped establish that heat is a form of energy. James Prescott Joule (1818–1889) performed manyexperiments to establish the mechanical equivalent of heat—the work needed to produce the same effects as heat transfer.In the units used for these two quantities, the value for this equivalence is


1.000 kcal = 4186 J.


We consider this equation to represent the conversion between two units of energy. (Other numbers that you may see referto calories defined for temperature ranges other than 14.5 °C to 15.5 °C .)
Figure 1.10 shows one of Joule’s most famous experimental setups for demonstrating that work and heat can produce thesame effects and measuring the mechanical equivalent of heat. It helped establish the principle of conservation of energy.Gravitational potential energy (U) was converted into kinetic energy (K), and then randomized by viscosity and turbulenceinto increased average kinetic energy of atoms and molecules in the system, producing a temperature increase. Joule’scontributions to thermodynamics were so significant that the SI unit of energy was named after him.


Figure 1.10 Joule’s experiment established the equivalence of heat andwork. As the masses descended, they caused the paddles to do work,
W = mgh , on the water. The result was a temperature increase, ΔT ,
measured by the thermometer. Joule found that ΔT was proportional to W
and thus determined the mechanical equivalent of heat.


Increasing internal energy by heat transfer gives the same result as increasing it by doing work. Therefore, although asystem has a well-defined internal energy, we cannot say that it has a certain “heat content” or “work content.” A well-


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defined quantity that depends only on the current state of the system, rather than on the history of that system, is known asa state variable. Temperature and internal energy are state variables. To sum up this paragraph, heat and work are not statevariables.
Incidentally, increasing the internal energy of a system does not necessarily increase its temperature. As we’ll see in the nextsection, the temperature does not change when a substance changes from one phase to another. An example is the meltingof ice, which can be accomplished by adding heat or by doing frictional work, as when an ice cube is rubbed against a roughsurface.
Temperature Change and Heat Capacity
We have noted that heat transfer often causes temperature change. Experiments show that with no phase change and nowork done on or by the system, the transferred heat is typically directly proportional to the change in temperature and tothe mass of the system, to a good approximation. (Below we show how to handle situations where the approximation isnot valid.) The constant of proportionality depends on the substance and its phase, which may be gas, liquid, or solid. Weomit discussion of the fourth phase, plasma, because although it is the most common phase in the universe, it is rare andshort-lived on Earth.
We can understand the experimental facts by noting that the transferred heat is the change in the internal energy, whichis the total energy of the molecules. Under typical conditions, the total kinetic energy of the molecules Ktotal is a
constant fraction of the internal energy (for reasons and with exceptions that we’ll see in the next chapter). The averagekinetic energy of a molecule Kave is proportional to the absolute temperature. Therefore, the change in internal energy
of a system is typically proportional to the change in temperature and to the number of molecules, N. Mathematically,
ΔU ∝ ΔKtotal = NKave ∝ NΔT The dependence on the substance results in large part from the different masses of atoms
and molecules. We are considering its heat capacity in terms of its mass, but as we will see in the next chapter, in somecases, heat capacities per molecule are similar for different substances. The dependence on substance and phase also resultsfrom differences in the potential energy associated with interactions between atoms and molecules.


Heat Transfer and Temperature Change
A practical approximation for the relationship between heat transfer and temperature change is:


(1.5)Q = mcΔT ,
where Q is the symbol for heat transfer (“quantity of heat”), m is the mass of the substance, and ΔT is the change
in temperature. The symbol c stands for the specific heat (also called “specific heat capacity”) and depends on thematerial and phase. The specific heat is numerically equal to the amount of heat necessary to change the temperature of
1.00 kg of mass by 1.00 °C . The SI unit for specific heat is J/(kg × K) or J/(kg × °C) . (Recall that the temperature
change ΔT is the same in units of kelvin and degrees Celsius.)


Values of specific heat must generally be measured, because there is no simple way to calculate them precisely. Table 1.3lists representative values of specific heat for various substances. We see from this table that the specific heat of water isfive times that of glass and 10 times that of iron, which means that it takes five times as much heat to raise the temperatureof water a given amount as for glass, and 10 times as much as for iron. In fact, water has one of the largest specific heats ofany material, which is important for sustaining life on Earth.
The specific heats of gases depend on what is maintained constant during the heating—typically either the volume orthe pressure. In the table, the first specific heat value for each gas is measured at constant volume, and the second (inparentheses) is measured at constant pressure. We will return to this topic in the chapter on the kinetic theory of gases.


Chapter 1 | Temperature and Heat 21




Substances Specific Heat (c)
Solids J/kg · °C kcal/kg · °C[2]
Aluminum 900 0.215
Asbestos 800 0.19
Concrete, granite (average) 840 0.20
Copper 387 0.0924
Glass 840 0.20
Gold 129 0.0308
Human body (average at 37 °C ) 3500 0.83
Ice (average, −50 °C to 0 °C ) 2090 0.50
Iron, steel 452 0.108
Lead 128 0.0305
Silver 235 0.0562
Wood 1700 0.40
Liquids
Benzene 1740 0.415
Ethanol 2450 0.586
Glycerin 2410 0.576
Mercury 139 0.0333
Water (15.0 °C) 4186 1.000
Gases[3]
Air (dry) 721 (1015) 0.172 (0.242)
Ammonia 1670 (2190) 0.399 (0.523)
Carbon dioxide 638 (833) 0.152 (0.199)
Nitrogen 739 (1040) 0.177 (0.248)
Oxygen 651 (913) 0.156 (0.218)
Steam (100 °C) 1520 (2020) 0.363 (0.482)


Table 1.3 Specific Heats of Various Substances[1] [1]The values forsolids and liquids are at constant volume and 25 °C , except as noted.[2]These values are identical in units of cal/g · °C. [3]Specific heats at
constant volume and at 20.0 °C except as noted, and at 1.00 atm
pressure. Values in parentheses are specific heats at a constantpressure of 1.00 atm.


In general, specific heat also depends on temperature. Thus, a precise definition of c for a substance must be given in terms
of an infinitesimal change in temperature. To do this, we note that c = 1m ΔQΔT and replace Δ with d:


c = 1m
dQ
dT


.


Except for gases, the temperature and volume dependence of the specific heat of most substances is weak at normal


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temperatures. Therefore, we will generally take specific heats to be constant at the values given in the table.
Example 1.5


Calculating the Required Heat
A 0.500-kg aluminum pan on a stove and 0.250 L of water in it are heated from 20.0 °C to 80.0 °C . (a) How
much heat is required? What percentage of the heat is used to raise the temperature of (b) the pan and (c) thewater?
Strategy
We can assume that the pan and the water are always at the same temperature. When you put the pan on the stove,the temperature of the water and that of the pan are increased by the same amount. We use the equation for theheat transfer for the given temperature change and mass of water and aluminum. The specific heat values forwater and aluminum are given in Table 1.3.
Solution1. Calculate the temperature difference:


ΔT = Tf − Ti = 60.0 °C.


2. Calculate the mass of water. Because the density of water is 1000 kg/m3 , 1 L of water has a mass of 1
kg, and the mass of 0.250 L of water is mw = 0.250 kg .


3. Calculate the heat transferred to the water. Use the specific heat of water in Table 1.3:
Qw = mw cwΔT =



⎝0.250 kg⎞⎠⎛⎝4186 J/kg °C⎞⎠(60.0 °C) = 62.8 kJ.


4. Calculate the heat transferred to the aluminum. Use the specific heat for aluminum in Table 1.3:
QAl = mA1 cA1ΔT =



⎝0.500 kg⎞⎠⎛⎝900 J/kg °C⎞⎠(60.0 °C) = 27.0 kJ.


5. Find the total transferred heat:
QTotal = QW + QAl = 89.8 kJ.


Significance
In this example, the heat transferred to the container is a significant fraction of the total transferred heat. Althoughthe mass of the pan is twice that of the water, the specific heat of water is over four times that of aluminum.Therefore, it takes a bit more than twice as much heat to achieve the given temperature change for the water asfor the aluminum pan.


Example 1.6 illustrates a temperature rise caused by doing work. (The result is the same as if the same amount of energyhad been added with a blowtorch instead of mechanically.)
Example 1.6


Calculating the Temperature Increase from the Work Done on a Substance
Truck brakes used to control speed on a downhill run do work, converting gravitational potential energy intoincreased internal energy (higher temperature) of the brake material (Figure 1.11). This conversion prevents thegravitational potential energy from being converted into kinetic energy of the truck. Since the mass of the truckis much greater than that of the brake material absorbing the energy, the temperature increase may occur too fastfor sufficient heat to transfer from the brakes to the environment; in other words, the brakes may overheat.


Chapter 1 | Temperature and Heat 23




Figure 1.11 The smoking brakes on a braking truck are visible evidence of the mechanical equivalent of heat.


Calculate the temperature increase of 10 kg of brake material with an average specific heat of 800 J/kg · °C if
the material retains 10% of the energy from a 10,000-kg truck descending 75.0 m (in vertical displacement) at aconstant speed.
Strategy
We calculate the gravitational potential energy (Mgh) that the entire truck loses in its descent, equate it to theincrease in the brakes’ internal energy, and then find the temperature increase produced in the brake materialalone.
Solution
First we calculate the change in gravitational potential energy as the truck goes downhill:


Mgh = ⎛⎝10,000 kg⎞⎠⎛⎝9.80 m/s
2⎞
⎠(75.0 m) = 7.35 × 10


6 J.


Because the kinetic energy of the truck does not change, conservation of energy tells us the lost potential energyis dissipated, and we assume that 10% of it is transferred to internal energy of the brakes, so take Q = Mgh/10 .
Then we calculate the temperature change from the heat transferred, using


ΔT = Qmc,


where m is the mass of the brake material. Insert the given values to find
ΔT = 7.35 × 10


5 J
(10 kg)(800 J/kg °C)


= 92 °C.


Significance
If the truck had been traveling for some time, then just before the descent, the brake temperature wouldprobably be higher than the ambient temperature. The temperature increase in the descent would likely raise thetemperature of the brake material very high, so this technique is not practical. Instead, the truck would use thetechnique of engine braking. A different idea underlies the recent technology of hybrid and electric cars, wheremechanical energy (kinetic and gravitational potential energy) is converted by the brakes into electrical energy inthe battery, a process called regenerative braking.


In a common kind of problem, objects at different temperatures are placed in contact with each other but isolated fromeverything else, and they are allowed to come into equilibrium. A container that prevents heat transfer in or out is calleda calorimeter, and the use of a calorimeter to make measurements (typically of heat or specific heat capacity) is calledcalorimetry.
We will use the term “calorimetry problem” to refer to any problem in which the objects concerned are thermally isolated


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from their surroundings. An important idea in solving calorimetry problems is that during a heat transfer between objectsisolated from their surroundings, the heat gained by the colder object must equal the heat lost by the hotter object, due toconservation of energy:


(1.6)Qcold + Qhot = 0.


We express this idea by writing that the sum of the heats equals zero because the heat gained is usually considered positive;the heat lost, negative.
Example 1.7


Calculating the Final Temperature in Calorimetry
Suppose you pour 0.250 kg of 20.0-°C water (about a cup) into a 0.500-kg aluminum pan off the stove with a
temperature of 150 °C . Assume no heat transfer takes place to anything else: The pan is placed on an insulated
pad, and heat transfer to the air is neglected in the short time needed to reach equilibrium. Thus, this is acalorimetry problem, even though no isolating container is specified. Also assume that a negligible amount ofwater boils off. What is the temperature when the water and pan reach thermal equilibrium?
Strategy
Originally, the pan and water are not in thermal equilibrium: The pan is at a higher temperature than the water.Heat transfer restores thermal equilibrium once the water and pan are in contact; it stops once thermal equilibriumbetween the pan and the water is achieved. The heat lost by the pan is equal to the heat gained by the water—thatis the basic principle of calorimetry.
Solution1. Use the equation for heat transfer Q = mcΔT to express the heat lost by the aluminum pan in terms


of the mass of the pan, the specific heat of aluminum, the initial temperature of the pan, and the finaltemperature:
Qhot = mA1 cA1



⎝Tf − 150 °C



⎠.


2. Express the heat gained by the water in terms of the mass of the water, the specific heat of water, theinitial temperature of the water, and the final temperature:
Qcold = mw cw



⎝Tf − 20.0 °C



⎠.


3. Note that Qhot < 0 and Qcold > 0 and that as stated above, they must sum to zero:
Qcold + Qhot = 0


Qcold = −Qhot
mw cw



⎝Tf − 20.0 °C



⎠ = −mA1 cA1



⎝Tf − 150 °C



⎠.


4. This a linear equation for the unknown final temperature, Tf. Solving for Tf,
Tf =


mA1 cA1 (150 °C) + mw cw (20.0 °C)
mA1 cA1 + mw cw


,


and insert the numerical values:
Tf =



⎝0.500 kg⎞⎠⎛⎝900 J/kg °C⎞⎠(150 °C) + ⎛⎝0.250 kg⎞⎠⎛⎝4186 J/kg °C⎞⎠(20.0 °C)



⎝0.500 kg⎞⎠⎛⎝900 J/kg °C⎞⎠+ ⎛⎝0.250 kg⎞⎠⎛⎝4186 J/kg °C⎞⎠


= 59.1 °C.


Significance
Why is the final temperature so much closer to 20.0 °C than to 150 °C ? The reason is that water has a greater
specific heat than most common substances and thus undergoes a smaller temperature change for a given heattransfer. A large body of water, such as a lake, requires a large amount of heat to increase its temperature


Chapter 1 | Temperature and Heat 25




1.3


appreciably. This explains why the temperature of a lake stays relatively constant during the day even whenthe temperature change of the air is large. However, the water temperature does change over longer times (e.g.,summer to winter).
Check Your Understanding If 25 kJ is necessary to raise the temperature of a rock from


25 °C to 30 °C, how much heat is necessary to heat the rock from 45 °C to 50 °C ?


Example 1.8
Temperature-Dependent Heat Capacity
At low temperatures, the specific heats of solids are typically proportional to T 3 . The first understanding of this
behavior was due to the Dutch physicist Peter Debye, who in 1912, treated atomic oscillations with the quantumtheory that Max Planck had recently used for radiation. For instance, a good approximation for the specific heat
of salt, NaCl, is c = 3.33 × 104 J


kg · k



T
321 K





3
. The constant 321 K is called the Debye temperature of NaCl,


ΘD, and the formula works well when T < 0.04ΘD. Using this formula, how much heat is required to raise
the temperature of 24.0 g of NaCl from 5 K to 15 K?
Solution
Because the heat capacity depends on the temperature, we need to use the equation


c = 1m
dQ
dT


.


We solve this equation for Q by integrating both sides: Q = m∫
T1


T2
cdT .


Then we substitute the given values in and evaluate the integral:
Q = (0.024 kg)∫


T1


T2
333 × 104 J


kg · K



T
321 K





3
dT =

⎝6.04 × 10


−4 J
K4

⎠T


4|5 K
15 K


= 30.2 J.


Significance
If we had used the equation Q = mcΔT and the room-temperature specific heat of salt, 880 J/kg · K, we would
have gotten a very different value.


1.5 | Phase Changes
Learning Objectives


By the end of this section, you will be able to:
• Describe phase transitions and equilibrium between phases
• Solve problems involving latent heat
• Solve calorimetry problems involving phase changes


Phase transitions play an important theoretical and practical role in the study of heat flow. In melting (or “fusion”), asolid turns into a liquid; the opposite process is freezing. In evaporation, a liquid turns into a gas; the opposite process iscondensation.
A substance melts or freezes at a temperature called its melting point, and boils (evaporates rapidly) or condenses at its


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boiling point. These temperatures depend on pressure. High pressure favors the denser form, so typically, high pressureraises the melting point and boiling point, and low pressure lowers them. For example, the boiling point of water is 100 °C
at 1.00 atm. At higher pressure, the boiling point is higher, and at lower pressure, it is lower. The main exception is themelting and freezing of water, discussed in the next section.
Phase Diagrams
The phase of a given substance depends on the pressure and temperature. Thus, plots of pressure versus temperatureshowing the phase in each region provide considerable insight into thermal properties of substances. Such a pT graph iscalled a phase diagram.
Figure 1.12 shows the phase diagram for water. Using the graph, if you know the pressure and temperature, youcan determine the phase of water. The solid curves—boundaries between phases—indicate phase transitions, that is,temperatures and pressures at which the phases coexist. For example, the boiling point of water is 100 °C at 1.00 atm.
As the pressure increases, the boiling temperature rises gradually to 374 °C at a pressure of 218 atm. A pressure cooker
(or even a covered pot) cooks food faster than an open pot, because the water can exist as a liquid at temperatures greaterthan 100 °C without all boiling away. (As we’ll see in the next section, liquid water conducts heat better than steam or
hot air.) The boiling point curve ends at a certain point called the critical point—that is, a critical temperature, abovewhich the liquid and gas phases cannot be distinguished; the substance is called a supercritical fluid. At sufficiently highpressure above the critical point, the gas has the density of a liquid but does not condense. Carbon dioxide, for example, issupercritical at all temperatures above 31.0 °C . Critical pressure is the pressure of the critical point.


Figure 1.12 The phase diagram (pT graph) for water showssolid (s), liquid (l), and vapor (v) phases. At temperatures andpressure above those of the critical point, there is no distinctionbetween liquid and vapor. Note that the axes are nonlinear andthe graph is not to scale. This graph is simplified—it omitsseveral exotic phases of ice at higher pressures. The phasediagram of water is unusual because the melting-point curve hasa negative slope, showing that you can melt ice by increasingthe pressure.


Similarly, the curve between the solid and liquid regions in Figure 1.12 gives the melting temperature at various pressures.For example, the melting point is 0 °C at 1.00 atm, as expected. Water has the unusual property that ice is less dense than
liquid water at the melting point, so at a fixed temperature, you can change the phase from solid (ice) to liquid (water) byincreasing the pressure. That is, the melting temperature of ice falls with increased pressure, as the phase diagram shows.For example, when a car is driven over snow, the increased pressure from the tires melts the snowflakes; afterwards, thewater refreezes and forms an ice layer.
As you learned in the earlier section on thermometers and temperature scales, the triple point is the combination of


Chapter 1 | Temperature and Heat 27




temperature and pressure at which ice, liquid water, and water vapor can coexist stably—that is, all three phases exist inequilibrium. For water, the triple point occurs at 273.16 K (0.01 °C) and 611.2 Pa; that is a more accurate calibration
temperature than the melting point of water at 1.00 atm, or 273.15 K (0.0 °C) .


View this video (https://openstaxcollege.org/l/21triplepoint) to see a substance at its triple point.


At pressures below that of the triple point, there is no liquid phase; the substance can exist as either gas or solid. For water,there is no liquid phase at pressures below 0.00600 atm. The phase change from solid to gas is called sublimation. You mayhave noticed that snow can disappear into thin air without a trace of liquid water, or that ice cubes can disappear in a freezer.Both are examples of sublimation. The reverse also happens: Frost can form on very cold windows without going throughthe liquid stage. Figure 1.13 shows the result, as well as showing a familiar example of sublimation. Carbon dioxidehas no liquid phase at atmospheric pressure. Solid CO2 is known as dry ice because instead of melting, it sublimes. Its
sublimation temperature at atmospheric pressure is −78 °C . Certain air fresheners use the sublimation of a solid to spread
a perfume around a room. Some solids, such as osmium tetroxide, are so toxic that they must be kept in sealed containers toprevent human exposure to their sublimation-produced vapors.


Figure 1.13 Direct transitions between solid and vapor are common, sometimes useful, and even beautiful. (a) Dry icesublimes directly to carbon dioxide gas. The visible “smoke” consists of water droplets that condensed in the air cooled by thedry ice. (b) Frost forms patterns on a very cold window, an example of a solid formed directly from a vapor. (credit a:modification of work by Windell Oskay; credit b: modification of work by Liz West)
Equilibrium
At the melting temperature, the solid and liquid phases are in equilibrium. If heat is added, some of the solid will melt,and if heat is removed, some of the liquid will freeze. The situation is somewhat more complex for liquid-gas equilibrium.Generally, liquid and gas are in equilibrium at any temperature. We call the gas phase a vaporwhen it exists at a temperaturebelow the boiling temperature, as it does for water at 20.0 °C . Liquid in a closed container at a fixed temperature
evaporates until the pressure of the gas reaches a certain value, called the vapor pressure, which depends on the gas andthe temperature. At this equilibrium, if heat is added, some of the liquid will evaporate, and if heat is removed, some of thegas will condense; molecules either join the liquid or form suspended droplets. If there is not enough liquid for the gas toreach the vapor pressure in the container, all the liquid eventually evaporates.
If the vapor pressure of the liquid is greater than the total ambient pressure, including that of any air (or other gas), the liquidevaporates rapidly; in other words, it boils. Thus, the boiling point of a liquid at a given pressure is the temperature at whichits vapor pressure equals the ambient pressure. Liquid and gas phases are in equilibrium at the boiling temperature (Figure1.14). If a substance is in a closed container at the boiling point, then the liquid is boiling and the gas is condensing at thesame rate without net change in their amounts.


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1.4


Figure 1.14 Equilibrium between liquid and gas at two different boiling points inside aclosed container. (a) The rates of boiling and condensation are equal at this combination oftemperature and pressure, so the liquid and gas phases are in equilibrium. (b) At a highertemperature, the boiling rate is faster, that is, the rate at which molecules leave the liquid andenter the gas is faster. This increases the number of molecules in the gas, which increases thegas pressure, which in turn increases the rate at which gas molecules condense and enter theliquid. The pressure stops increasing when it reaches the point where the boiling rate and thecondensation rate are equal. The gas and liquid are in equilibrium again at this highertemperature and pressure.


For water, 100 °C is the boiling point at 1.00 atm, so water and steam should exist in equilibrium under these conditions.
Why does an open pot of water at 100 °C boil completely away? The gas surrounding an open pot is not pure water: it is
mixed with air. If pure water and steam are in a closed container at 100 °C and 1.00 atm, they will coexist—but with air
over the pot, there are fewer water molecules to condense, and water boils away. Another way to see this is that at the boilingpoint, the vapor pressure equals the ambient pressure. However, part of the ambient pressure is due to air, so the pressure ofthe steam is less than the vapor pressure at that temperature, and evaporation continues. Incidentally, the equilibrium vaporpressure of solids is not zero, a fact that accounts for sublimation.


Check Your Understanding Explain why a cup of water (or soda) with ice cubes stays at 0 °C, even on
a hot summer day.


Phase Change and Latent Heat
So far, we have discussed heat transfers that cause temperature change. However, in a phase transition, heat transfer doesnot cause any temperature change.
For an example of phase changes, consider the addition of heat to a sample of ice at −20 °C (Figure 1.15) and
atmospheric pressure. The temperature of the ice rises linearly, absorbing heat at a constant rate of 2090 J/kg · ºC until it
reaches 0 °C. Once at this temperature, the ice begins to melt and continues until it has all melted, absorbing 333 kJ/kg
of heat. The temperature remains constant at 0 °C during this phase change. Once all the ice has melted, the temperature
of the liquid water rises, absorbing heat at a new constant rate of 4186 J/kg · ºC. At 100 °C, the water begins to boil.
The temperature again remains constant during this phase change while the water absorbs 2256 kJ/kg of heat and turns intosteam. When all the liquid has become steam, the temperature rises again, absorbing heat at a rate of 2020 J/kg · ºC . If we
started with steam and cooled it to make it condense into liquid water and freeze into ice, the process would exactly reverse,with the temperature again constant during each phase transition.


Chapter 1 | Temperature and Heat 29




Figure 1.15 Temperature versus heat. The system is constructed so that no vapor evaporates while icewarms to become liquid water, and so that, when vaporization occurs, the vapor remains in the system. Thelong stretches of constant temperatures at 0 °C and 100 °C reflect the large amounts of heat needed to
cause melting and vaporization, respectively.


Where does the heat added during melting or boiling go, considering that the temperature does not change until the transitionis complete? Energy is required to melt a solid, because the attractive forces between the molecules in the solid must bebroken apart, so that in the liquid, the molecules can move around at comparable kinetic energies; thus, there is no rise intemperature. Energy is needed to vaporize a liquid for similar reasons. Conversely, work is done by attractive forces whenmolecules are brought together during freezing and condensation. That energy must be transferred out of the system, usuallyin the form of heat, to allow the molecules to stay together (Figure 1.18). Thus, condensation occurs in association withcold objects—the glass in Figure 1.16, for example.


Figure 1.16 Condensation forms on this glass of iced teabecause the temperature of the nearby air is reduced. The aircannot hold as much water as it did at room temperature, sowater condenses. Energy is released when the water condenses,speeding the melting of the ice in the glass. (credit: JennyDowning)


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The energy released when a liquid freezes is used by orange growers when the temperature approaches 0 °C . Growers
spray water on the trees so that the water freezes and heat is released to the growing oranges. This prevents the temperatureinside the orange from dropping below freezing, which would damage the fruit (Figure 1.17).


Figure 1.17 The ice on these trees released large amounts ofenergy when it froze, helping to prevent the temperature of thetrees from dropping below 0 °C . Water is intentionally sprayed
on orchards to help prevent hard frosts. (credit: HermannHammer)


The energy involved in a phase change depends on the number of bonds or force pairs and their strength. The number ofbonds is proportional to the number of molecules and thus to the mass of the sample. The energy per unit mass required tochange a substance from the solid phase to the liquid phase, or released when the substance changes from liquid to solid, isknown as the heat of fusion. The energy per unit mass required to change a substance from the liquid phase to the vaporphase is known as the heat of vaporization. The strength of the forces depends on the type of molecules. The heat Qabsorbed or released in a phase change in a sample of mass m is given by


(1.7)Q = mLf⎛⎝melting/freezing⎞⎠


(1.8)Q = mLv⎛⎝vaporization/condensation⎞⎠


where the latent heat of fusion Lf and latent heat of vaporization Lv are material constants that are determined
experimentally. (Latent heats are also called latent heat coefficients and heats of transformation.) These constants are“latent,” or hidden, because in phase changes, energy enters or leaves a system without causing a temperature change in thesystem, so in effect, the energy is hidden.


Chapter 1 | Temperature and Heat 31




Figure 1.18 (a) Energy is required to partially overcome the attractive forces (modeled as springs) between molecules in asolid to form a liquid. That same energy must be removed from the liquid for freezing to take place. (b) Molecules becomeseparated by large distances when going from liquid to vapor, requiring significant energy to completely overcome molecularattraction. The same energy must be removed from the vapor for condensation to take place.


Table 1.4 lists representative values of Lf and Lv in kJ/kg, together with melting and boiling points. Note that in general,
Lv > Lf . The table shows that the amounts of energy involved in phase changes can easily be comparable to or greater
than those involved in temperature changes, as Figure 1.15 and the accompanying discussion also showed.


Lf Lv


Substance Melting Point
(°C)


kJ/kg kcal/kg Boiling Point(°C) kJ/kg kcal/kg
Helium[2] −272.2 (0.95 K) 5.23 1.25 −268.9(4.2 K) 20.9 4.99
Hydrogen −259.3(13.9 K) 58.6 14.0 −252.9(20.2 K) 452 108
Nitrogen −210.0(63.2 K) 25.5 6.09 −195.8(77.4 K) 201 48.0
Oxygen −218.8(54.4 K) 13.8 3.30 −183.0(90.2 K) 213 50.9
Ethanol –114 104 24.9 78.3 854 204
Ammonia –75 332 79.3 –33.4 1370 327
Mercury –38.9 11.8 2.82 357 272 65.0
Water 0.00 334 79.8 100.0 2256[3] 539[4]
Sulfur 119 38.1 9.10 444.6 326 77.9
Lead 327 24.5 5.85 1750 871 208
Antimony 631 165 39.4 1440 561 134
Aluminum 660 380 90 2450 11400 2720
Silver 961 88.3 21.1 2193 2336 558


Table 1.4 Heats of Fusion and Vaporization[1] [1]Values quoted at the normal melting and boilingtemperatures at standard atmospheric pressure ( 1 atm ). [2]Helium has no solid phase at atmospheric pressure.
The melting point given is at a pressure of 2.5 MPa. [3]At 37.0 °C (body temperature), the heat of vaporization
Lv for water is 2430 kJ/kg or 580 kcal/kg. [4]At 37.0 °C (body temperature), the heat of vaporization, Lv for
water is 2430 kJ/kg or 580 kcal/kg.


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Lf Lv


Gold 1063 64.5 15.4 2660 1578 377
Copper 1083 134 32.0 2595 5069 1211
Uranium 1133 84 20 3900 1900 454
Tungsten 3410 184 44 5900 4810 1150


Table 1.4 Heats of Fusion and Vaporization[1] [1]Values quoted at the normal melting and boilingtemperatures at standard atmospheric pressure ( 1 atm ). [2]Helium has no solid phase at atmospheric pressure.
The melting point given is at a pressure of 2.5 MPa. [3]At 37.0 °C (body temperature), the heat of vaporization
Lv for water is 2430 kJ/kg or 580 kcal/kg. [4]At 37.0 °C (body temperature), the heat of vaporization, Lv for
water is 2430 kJ/kg or 580 kcal/kg.
Phase changes can have a strong stabilizing effect on temperatures that are not near the melting and boiling points, sinceevaporation and condensation occur even at temperatures below the boiling point. For example, air temperatures in humidclimates rarely go above approximately 38.0 °C because most heat transfer goes into evaporating water into the air.
Similarly, temperatures in humid weather rarely fall below the dew point—the temperature where condensation occursgiven the concentration of water vapor in the air—because so much heat is released when water vapor condenses.
More energy is required to evaporate water below the boiling point than at the boiling point, because the kinetic energy ofwater molecules at temperatures below 100 °C is less than that at 100 °C , so less energy is available from random thermal
motions. For example, at body temperature, evaporation of sweat from the skin requires a heat input of 2428 kJ/kg, whichis about 10% higher than the latent heat of vaporization at 100 °C . This heat comes from the skin, and this evaporative
cooling effect of sweating helps reduce the body temperature in hot weather. However, high humidity inhibits evaporation,so that body temperature might rise, while unevaporated sweat might be left on your brow.
Example 1.9


Calculating Final Temperature from Phase Change
Three ice cubes are used to chill a soda at 20 °C with mass msoda = 0.25 kg . The ice is at 0 °C and each ice
cube has a mass of 6.0 g. Assume that the soda is kept in a foam container so that heat loss can be ignored andthat the soda has the same specific heat as water. Find the final temperature when all ice has melted.
Strategy
The ice cubes are at the melting temperature of 0 °C. Heat is transferred from the soda to the ice for melting.
Melting yields water at 0 °C, so more heat is transferred from the soda to this water until the water plus soda
system reaches thermal equilibrium.
The heat transferred to the ice is


Qice = miceLf + mice cW

⎝Tf − 0 °C



⎠.


The heat given off by the soda is
Qsoda = msoda cW



⎝Tf − 20 °C



⎠.


Since no heat is lost, Qice = −Qsoda, as in Example 1.7, so that
miceLf + mice cW



⎝Tf − 0 °C



⎠ = −msoda cW



⎝Tf − 20 °C



⎠.


Solve for the unknown quantity Tf :
Tf =


msoda cw (20 °C) − miceLf
(msoda + mice)cw


.


Chapter 1 | Temperature and Heat 33




1.5


Solution
First we identify the known quantities. The mass of ice is mice = 3 × 6.0 g = 0.018 kg and the mass of soda is
msoda = 0.25 kg. Then we calculate the final temperature:


Tf =
20,930 J − 6012 J


1122 J/°C
= 13 °C.


Significance
This example illustrates the large energies involved during a phase change. The mass of ice is about 7% of themass of the soda but leads to a noticeable change in the temperature of the soda. Although we assumed that theice was at the freezing temperature, this is unrealistic for ice straight out of a freezer: The typical temperature is
−6 °C . However, this correction makes no significant change from the result we found. Can you explain why?


Like solid-liquid and and liquid-vapor transitions, direct solid-vapor transitions or sublimations involve heat. The energytransferred is given by the equation Q = mLs , where Ls is the heat of sublimation, analogous to Lf and Lv . The heat
of sublimation at a given temperature is equal to the heat of fusion plus the heat of vaporization at that temperature.
We can now calculate any number of effects related to temperature and phase change. In each case, it is necessary to identifywhich temperature and phase changes are taking place. Keep in mind that heat transfer and work can cause both temperatureand phase changes.


Problem-Solving Strategy: The Effects of Heat Transfer
1. Examine the situation to determine that there is a change in the temperature or phase. Is there heat transfer intoor out of the system? When it is not obvious whether a phase change occurs or not, you may wish to first solvethe problem as if there were no phase changes, and examine the temperature change obtained. If it is sufficientto take you past a boiling or melting point, you should then go back and do the problem in steps—temperaturechange, phase change, subsequent temperature change, and so on.
2. Identify and list all objects that change temperature or phase.
3. Identify exactly what needs to be determined in the problem (identify the unknowns). A written list is useful.
4. Make a list of what is given or what can be inferred from the problem as stated (identify the knowns). If thereis a temperature change, the transferred heat depends on the specific heat of the substance (Heat Transfer,Specific Heat, and Calorimetry), and if there is a phase change, the transferred heat depends on the latentheat of the substance (Table 1.4).
5. Solve the appropriate equation for the quantity to be determined (the unknown).
6. Substitute the knowns along with their units into the appropriate equation and obtain numerical solutionscomplete with units. You may need to do this in steps if there is more than one state to the process, such as atemperature change followed by a phase change. However, in a calorimetry problem, each step corresponds toa term in the single equation Qhot + Qcold = 0 .
7. Check the answer to see if it is reasonable. Does it make sense? As an example, be certain that any temperaturechange does not also cause a phase change that you have not taken into account.


Check Your Understanding Why does snow often remain even when daytime temperatures are higherthan the freezing temperature?


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1.6 | Mechanisms of Heat Transfer
Learning Objectives


By the end of this section, you will be able to:
• Explain some phenomena that involve conductive, convective, and radiative heat transfer
• Solve problems on the relationships between heat transfer, time, and rate of heat transfer
• Solve problems using the formulas for conduction and radiation


Just as interesting as the effects of heat transfer on a system are the methods by which it occurs. Whenever there is atemperature difference, heat transfer occurs. It may occur rapidly, as through a cooking pan, or slowly, as through the wallsof a picnic ice chest. So many processes involve heat transfer that it is hard to imagine a situation where no heat transferoccurs. Yet every heat transfer takes place by only three methods:
1. Conduction is heat transfer through stationary matter by physical contact. (The matter is stationary on amacroscopic scale—we know that thermal motion of the atoms and molecules occurs at any temperature aboveabsolute zero.) Heat transferred from the burner of a stove through the bottom of a pan to food in the pan istransferred by conduction.
2. Convection is the heat transfer by the macroscopic movement of a fluid. This type of transfer takes place in aforced-air furnace and in weather systems, for example.
3. Heat transfer by radiation occurs when microwaves, infrared radiation, visible light, or another form ofelectromagnetic radiation is emitted or absorbed. An obvious example is the warming of Earth by the Sun. A lessobvious example is thermal radiation from the human body.


In the illustration at the beginning of this chapter, the fire warms the snowshoers’ faces largely by radiation. Convectioncarries some heat to them, but most of the air flow from the fire is upward (creating the familiar shape of flames), carryingheat to the food being cooked and into the sky. The snowshoers wear clothes designed with low conductivity to prevent heatflow out of their bodies.
In this section, we examine these methods in some detail. Each method has unique and interesting characteristics, butall three have two things in common: They transfer heat solely because of a temperature difference, and the greater thetemperature difference, the faster the heat transfer (Figure 1.19).


Chapter 1 | Temperature and Heat 35




1.6


Figure 1.19 In a fireplace, heat transfer occurs by all three methods:conduction, convection, and radiation. Radiation is responsible for most ofthe heat transferred into the room. Heat transfer also occurs throughconduction into the room, but much slower. Heat transfer by convectionalso occurs through cold air entering the room around windows and hot airleaving the room by rising up the chimney.
Check Your Understanding Name an example from daily life (different from the text) for eachmechanism of heat transfer.


Conduction
As you walk barefoot across the living room carpet in a cold house and then step onto the kitchen tile floor, your feet feelcolder on the tile. This result is intriguing, since the carpet and tile floor are both at the same temperature. The differentsensation is explained by the different rates of heat transfer: The heat loss is faster for skin in contact with the tiles than withthe carpet, so the sensation of cold is more intense.
Some materials conduct thermal energy faster than others. Figure 1.20 shows a material that conducts heat slowly—it is agood thermal insulator, or poor heat conductor—used to reduce heat flow into and out of a house.


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Figure 1.20 Insulation is used to limit the conduction of heatfrom the inside to the outside (in winter) and from the outside tothe inside (in summer). (credit: Giles Douglas)


A molecular picture of heat conduction will help justify the equation that describes it. Figure 1.21 shows molecules intwo bodies at different temperatures, Th and Tc, for “hot” and “cold.” The average kinetic energy of a molecule in
the hot body is higher than in the colder body. If two molecules collide, energy transfers from the high-energy to thelow-energy molecule. In a metal, the picture would also include free valence electrons colliding with each other and withatoms, likewise transferring energy. The cumulative effect of all collisions is a net flux of heat from the hotter body tothe colder body. Thus, the rate of heat transfer increases with increasing temperature difference ΔT = Th − Tc. If the
temperatures are the same, the net heat transfer rate is zero. Because the number of collisions increases with increasing area,heat conduction is proportional to the cross-sectional area—a second factor in the equation.


Chapter 1 | Temperature and Heat 37




Figure 1.21 Molecules in two bodies at different temperatureshave different average kinetic energies. Collisions occurring atthe contact surface tend to transfer energy from high-temperature regions to low-temperature regions. In thisillustration, a molecule in the lower-temperature region (rightside) has low energy before collision, but its energy increasesafter colliding with a high-energy molecule at the contactsurface. In contrast, a molecule in the higher-temperature region(left side) has high energy before collision, but its energydecreases after colliding with a low-energy molecule at thecontact surface.


A third quantity that affects the conduction rate is the thickness of the material through which heat transfers. Figure 1.22shows a slab of material with a higher temperature on the left than on the right. Heat transfers from the left to the right bya series of molecular collisions. The greater the distance between hot and cold, the more time the material takes to transferthe same amount of heat.


Figure 1.22 Heat conduction occurs through any material, represented here by arectangular bar, whether window glass or walrus blubber.


All four of these quantities appear in a simple equation deduced from and confirmed by experiments. The rate ofconductive heat transfer through a slab of material, such as the one in Figure 1.22, is given by


(1.9)
P = dQ


dT
=


kA⎛⎝Th − Tc



d


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where P is the power or rate of heat transfer in watts or in kilocalories per second, A and d are its surface area and thickness,as shown in Figure 1.22, Th − Tc is the temperature difference across the slab, and k is the thermal conductivity of the
material. Table 1.5 gives representative values of thermal conductivity.
More generally, we can write


P = −kAdT
dx


,


where x is the coordinate in the direction of heat flow. Since in Figure 1.22, the power and area are constant, dT/dx isconstant, and the temperature decreases linearly from Th to Tc.
Substance Thermal Conductivity k (W/m· °C)
Diamond 2000
Silver 420
Copper 390
Gold 318
Aluminum 220
Steel iron 80
Steel (stainless) 14
Ice 2.2
Glass (average) 0.84
Concrete brick 0.84
Water 0.6
Fatty tissue (without blood) 0.2
Asbestos 0.16
Plasterboard 0.16
Wood 0.08–0.16
Snow (dry) 0.10
Cork 0.042
Glass wool 0.042
Wool 0.04
Down feathers 0.025
Air 0.023
Polystyrene foam 0.010


Table 1.5 Thermal Conductivities of Common Substances Values aregiven for temperatures near 0 °C .


Example 1.10
Calculating Heat Transfer through Conduction
A polystyrene foam icebox has a total area of 0.950 m2 and walls with an average thickness of 2.50 cm. The
box contains ice, water, and canned beverages at 0 °C. The inside of the box is kept cold by melting ice. How
much ice melts in one day if the icebox is kept in the trunk of a car at 35.0 ºC ?


Chapter 1 | Temperature and Heat 39




Strategy
This question involves both heat for a phase change (melting of ice) and the transfer of heat by conduction. Tofind the amount of ice melted, we must find the net heat transferred. This value can be obtained by calculating therate of heat transfer by conduction and multiplying by time.
Solution
First we identify the knowns.
k = 0.010W/m · °C for polystyrene foam; A = 0.950 m2; d = 2.50 cm = 0.0250 m; ; Tc = 0 °C;
Th = 35.0 °C ; t = 1 day = 24 hours - 84,400 s.
Then we identify the unknowns. We need to solve for the mass of the ice, m. We also need to solve for the netheat transferred to melt the ice, Q. The rate of heat transfer by conduction is given by


P = dQ
dT


=
kA⎛⎝Th − Tc





d
.


The heat used to melt the ice is Q = mLf .We insert the known values:
P =


(0.010W/m · °C)⎛⎝0.950 m
2⎞
⎠(35.0 °C − 0 °C)


0.0250 m
= 13.3W.


Multiplying the rate of heat transfer by the time ( 1 day = 86,400 s ), we obtain
Q = Pt = (13.3W)(86.400 s) = 1.15 × 106 J.


We set this equal to the heat transferred to melt the ice, Q = mLf, and solve for the mass m:
m = Q


Lf
= 1.15 × 10


6 J
334 × 103 J/kg


= 3.44 kg.


Significance
The result of 3.44 kg, or about 7.6 lb, seems about right, based on experience. You might expect to use about a 4kg (7–10 lb) bag of ice per day. A little extra ice is required if you add any warm food or beverages.
Table 1.5 shows that polystyrene foam is a very poor conductor and thus a good insulator. Other good insulatorsinclude fiberglass, wool, and goosedown feathers. Like polystyrene foam, these all contain many small pocketsof air, taking advantage of air’s poor thermal conductivity.


In developing insulation, the smaller the conductivity k and the larger the thickness d, the better. Thus, the ratio d/k, calledthe R factor, is large for a good insulator. The rate of conductive heat transfer is inversely proportional to R. R factors aremost commonly quoted for household insulation, refrigerators, and the like. Unfortunately, in the United States, R is still
in non-metric units of ft2 · °F · h/Btu , although the unit usually goes unstated [1 British thermal unit (Btu) is the amount
of energy needed to change the temperature of 1.0 lb of water by 1.0 °F , which is 1055.1 J]. A couple of representative
values are an R factor of 11 for 3.5-inch-thick fiberglass batts (pieces) of insulation and an R factor of 19 for 6.5-inch-thick fiberglass batts (Figure 1.23). In the US, walls are usually insulated with 3.5-inch batts, whereas ceilings are usuallyinsulated with 6.5-inch batts. In cold climates, thicker batts may be used.


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Figure 1.23 The fiberglass batt is used for insulation of wallsand ceilings to prevent heat transfer between the inside of thebuilding and the outside environment. (credit: Tracey Nicholls)


Note that in Table 1.5, most of the best thermal conductors—silver, copper, gold, and aluminum—are also the bestelectrical conductors, because they contain many free electrons that can transport thermal energy. (Diamond, an electricalinsulator, conducts heat by atomic vibrations.) Cooking utensils are typically made from good conductors, but the handlesof those used on the stove are made from good insulators (bad conductors).
Example 1.11


Two Conductors End to End
A steel rod and an aluminum rod, each of diameter 1.00 cm and length 25.0 cm, are welded end to end. One endof the steel rod is placed in a large tank of boiling water at 100 °C , while the far end of the aluminum rod is
placed in a large tank of water at 20 °C . The rods are insulated so that no heat escapes from their surfaces. What
is the temperature at the joint, and what is the rate of heat conduction through this composite rod?
Strategy
The heat that enters the steel rod from the boiling water has no place to go but through the steel rod, then throughthe aluminum rod, to the cold water. Therefore, we can equate the rate of conduction through the steel to the rateof conduction through the aluminum.
We repeat the calculation with a second method, in which we use the thermal resistance R of the rod, since itsimply adds when two rods are joined end to end. (We will use a similar method in the chapter on direct-currentcircuits.)
Solution1. Identify the knowns and convert them to SI units.The length of each rod is LA1 = Lsteel = 0.25 m, the cross-sectional area of each rod is


AA1 = Asteel = 7.85 × 10
−5 m2 , the thermal conductivity of aluminum is kA1 = 220W/m · °C , the


thermal conductivity of steel is ksteel = 80W/m · °C , the temperature at the hot end is T = 100 °C , and
the temperature at the cold end is T = 20 °C .


Chapter 1 | Temperature and Heat 41




2. Calculate the heat-conduction rate through the steel rod and the heat-conduction rate through thealuminum rod in terms of the unknown temperature T at the joint:
Psteel =


ksteel AsteelΔTsteel
Lsteel


=
(80W/m · °C)⎛⎝7.85 × 10


−5 m2⎞⎠(100 °C − T)


0.25 m
= (0.0251W/°C)(100 °C − T);


PA1 =
kAl AA1ΔTAl


LA1


=
(220W/m · °C)⎛⎝7.85 × 10


−5 m2⎞⎠(T − 20 °C)


0.25 m
= (0.0691W/°C)(T − 20 °C).


3. Set the two rates equal and solve for the unknown temperature:
(0.0691W/°C)(T − 20 °C) = (0.0251W/°C)(100 °C − T)


T = 41.3 °C.
4. Calculate either rate:


Psteel = (0.0251W/°C)(100 °C − 41.3 °C) = 1.47W.5. If desired, check your answer by calculating the other rate.
Solution1. Recall that R = L/k . Now P = AΔT /R, or ΔT = PR/A.


2. We know that ΔTsteel + ΔTAl = 100 °C − 20 °C = 80 °C . We also know that Psteel = PAl, and we
denote that rate of heat flow by P. Combine the equations:


PRsteel
A


+
PRAl
A


= 80 °C.


Thus, we can simply add R factors. Now, P = 80 °C
A⎛⎝Rsteel + RAl




.


3. Find the Rs from the known quantities:
R steel = 3.13 × 10


−3m2 · °C/W


and
RAl = 1.14 × 10


−3m2 · °C/W.


4. Substitute these values in to find P = 1.47W as before.
5. Determine ΔT for the aluminum rod (or for the steel rod) and use it to find T at the joint.


ΔTAl =
PRAl
A


=
(1.47W)⎛⎝1.14 × 10


−3 m2 · °C/W⎞⎠


7.85 × 10−5 m2
= 21.3 °C,


so T = 20 °C + 21.3 °C = 41.3 °C , as in Solution 1 .
6. If desired, check by determining ΔT for the other rod.


Significance
In practice, adding R values is common, as in calculating the R value of an insulated wall. In the analogoussituation in electronics, the resistance corresponds to AR in this problem and is additive even when the areas are


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1.7


unequal, as is common in electronics. Our equation for heat conduction can be used only when the areas are equal;otherwise, we would have a problem in three-dimensional heat flow, which is beyond our scope.
Check Your Understanding How does the rate of heat transfer by conduction change when all spatialdimensions are doubled?


Conduction is caused by the random motion of atoms and molecules. As such, it is an ineffective mechanism for heattransport over macroscopic distances and short times. For example, the temperature on Earth would be unbearably coldduring the night and extremely hot during the day if heat transport in the atmosphere were only through conduction. Also,car engines would overheat unless there was a more efficient way to remove excess heat from the pistons. The next modulediscusses the important heat-transfer mechanism in such situations.
Convection
In convection, thermal energy is carried by the large-scale flow of matter. It can be divided into two types. In forcedconvection, the flow is driven by fans, pumps, and the like. A simple example is a fan that blows air past you in hotsurroundings and cools you by replacing the air heated by your body with cooler air. A more complicated example is thecooling system of a typical car, in which a pump moves coolant through the radiator and engine to cool the engine and a fanblows air to cool the radiator.
In free or natural convection, the flow is driven by buoyant forces: hot fluid rises and cold fluid sinks because densitydecreases as temperature increases. The house in Figure 1.24 is kept warm by natural convection, as is the pot of wateron the stove in Figure 1.25. Ocean currents and large-scale atmospheric circulation, which result from the buoyancy ofwarm air and water, transfer hot air from the tropics toward the poles and cold air from the poles toward the tropics. (Earth’srotation interacts with those flows, causing the observed eastward flow of air in the temperate zones.)


Figure 1.24 Air heated by a so-called gravity furnace expands andrises, forming a convective loop that transfers energy to other parts ofthe room. As the air is cooled at the ceiling and outside walls, itcontracts, eventually becoming denser than room air and sinking to thefloor. A properly designed heating system using natural convection,like this one, can heat a home quite efficiently.


Chapter 1 | Temperature and Heat 43




Figure 1.25 Natural convection plays an important role inheat transfer inside this pot of water. Once conducted to theinside, heat transfer to other parts of the pot is mostly byconvection. The hotter water expands, decreases in density, andrises to transfer heat to other regions of the water, while colderwater sinks to the bottom. This process keeps repeating.
Natural convection like that of Figure 1.24 and Figure 1.25, but acting on rock in Earth’s mantle, drives platetectonics (https://openstaxcollege.org/l/21platetecton) that are the motions that have shaped Earth’ssurface.


Convection is usually more complicated than conduction. Beyond noting that the convection rate is often approximatelyproportional to the temperature difference, we will not do any quantitative work comparable to the formula for conduction.However, we can describe convection qualitatively and relate convection rates to heat and time. However, air is a poorconductor. Therefore, convection dominates heat transfer by air, and the amount of available space for airflow determineswhether air transfers heat rapidly or slowly. There is little heat transfer in a space filled with air with a small amount of othermaterial that prevents flow. The space between the inside and outside walls of a typical American house, for example, isabout 9 cm (3.5 in.)—large enough for convection to work effectively. The addition of wall insulation prevents airflow, soheat loss (or gain) is decreased. On the other hand, the gap between the two panes of a double-paned window is about 1 cm,which largely prevents convection and takes advantage of air’s low conductivity reduce heat loss. Fur, cloth, and fiberglassalso take advantage of the low conductivity of air by trapping it in spaces too small to support convection (Figure 1.26).


Figure 1.26 Fur is filled with air, breaking it up into manysmall pockets. Convection is very slow here, because the loopsare so small. The low conductivity of air makes fur a very goodlightweight insulator.


Some interesting phenomena happen when convection is accompanied by a phase change. The combination allows us to


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cool off by sweating even if the temperature of the surrounding air exceeds body temperature. Heat from the skin is requiredfor sweat to evaporate from the skin, but without air flow, the air becomes saturated and evaporation stops. Air flow causedby convection replaces the saturated air by dry air and evaporation continues.
Example 1.12


Calculating the Flow of Mass during Convection
The average person produces heat at the rate of about 120 W when at rest. At what rate must water evaporatefrom the body to get rid of all this energy? (For simplicity, we assume this evaporation occurs when a person issitting in the shade and surrounding temperatures are the same as skin temperature, eliminating heat transfer byother methods.)
Strategy
Energy is needed for this phase change (Q = mLv ). Thus, the energy loss per unit time is


Q
t =


mLV
t = 120W = 120 J/s.


We divide both sides of the equation by Lv to find that the mass evaporated per unit time is
m
t =


120 J/s
Lv


.


Solution
Insert the value of the latent heat from Table 1.4, Lv = 2430 kJ/kg = 2430 J/g . This yields


m
t =


120 J/s
2430 J/g


= 0.0494 g/s = 2.96 g/min.


Significance
Evaporating about 3 g/min seems reasonable. This would be about 180 g (about 7 oz.) per hour. If the air is verydry, the sweat may evaporate without even being noticed. A significant amount of evaporation also takes place inthe lungs and breathing passages.


Another important example of the combination of phase change and convection occurs when water evaporates from theoceans. Heat is removed from the ocean when water evaporates. If the water vapor condenses in liquid droplets as cloudsform, possibly far from the ocean, heat is released in the atmosphere. Thus, there is an overall transfer of heat from the oceanto the atmosphere. This process is the driving power behind thunderheads, those great cumulus clouds that rise as muchas 20.0 km into the stratosphere (Figure 1.27). Water vapor carried in by convection condenses, releasing tremendousamounts of energy. This energy causes the air to expand and rise to colder altitudes. More condensation occurs in theseregions, which in turn drives the cloud even higher. This mechanism is an example of positive feedback, since the processreinforces and accelerates itself. It sometimes produces violent storms, with lightning and hail. The same mechanism driveshurricanes.
This time-lapse video (https://openstaxcollege.org/l/21convthuncurr) shows convection currents in athunderstorm, including “rolling” motion similar to that of boiling water.


Chapter 1 | Temperature and Heat 45




1.8


Figure 1.27 Cumulus clouds are caused by water vapor thatrises because of convection. The rise of clouds is driven by apositive feedback mechanism. (credit: “Amada44”/WikimediaCommons)
Check Your Understanding Explain why using a fan in the summer feels refreshing.


Radiation
You can feel the heat transfer from the Sun. The space between Earth and the Sun is largely empty, so the Sun warmsus without any possibility of heat transfer by convection or conduction. Similarly, you can sometimes tell that the ovenis hot without touching its door or looking inside—it may just warm you as you walk by. In these examples, heat istransferred by radiation (Figure 1.28). That is, the hot body emits electromagnetic waves that are absorbed by the skin.No medium is required for electromagnetic waves to propagate. Different names are used for electromagnetic waves ofdifferent wavelengths: radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays, and gammarays.


Figure 1.28 Most of the heat transfer from this fire to theobservers occurs through infrared radiation. The visible light,although dramatic, transfers relatively little thermal energy.Convection transfers energy away from the observers as hot airrises, while conduction is negligibly slow here. Skin is verysensitive to infrared radiation, so you can sense the presence of afire without looking at it directly. (credit: Daniel O’Neil)


The energy of electromagnetic radiation varies over a wide range, depending on the wavelength: A shorter wavelength (orhigher frequency) corresponds to a higher energy. Because more heat is radiated at higher temperatures, higher temperaturesproduce more intensity at every wavelength but especially at shorter wavelengths. In visible light, wavelength determinescolor—red has the longest wavelength and violet the shortest—so a temperature change is accompanied by a color change.For example, an electric heating element on a stove glows from red to orange, while the higher-temperature steel in a


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blast furnace glows from yellow to white. Infrared radiation is the predominant form radiated by objects cooler than theelectric element and the steel. The radiated energy as a function of wavelength depends on its intensity, which is representedin Figure 1.29 by the height of the distribution. (Electromagnetic Waves explains more about the electromagneticspectrum, and Photons and Matter Waves (http://cnx.org/content/m58757/latest/) discusses why the decrease inwavelength corresponds to an increase in energy.)


Figure 1.29 (a) A graph of the spectrum of electromagnetic waves emitted from an idealradiator at three different temperatures. The intensity or rate of radiation emission increasesdramatically with temperature, and the spectrum shifts down in wavelength toward the visibleand ultraviolet parts of the spectrum. The shaded portion denotes the visible part of the spectrum.It is apparent that the shift toward the ultraviolet with temperature makes the visible appearanceshift from red to white to blue as temperature increases. (b) Note the variations in colorcorresponding to variations in flame temperature.


The rate of heat transfer by radiation also depends on the object’s color. Black is the most effective, and white is the leasteffective. On a clear summer day, black asphalt in a parking lot is hotter than adjacent gray sidewalk, because black absorbsbetter than gray (Figure 1.30). The reverse is also true—black radiates better than gray. Thus, on a clear summer night,the asphalt is colder than the gray sidewalk, because black radiates the energy more rapidly than gray. A perfectly blackobject would be an ideal radiator and an ideal absorber, as it would capture all the radiation that falls on it. In contrast,a perfectly white object or a perfect mirror would reflect all radiation, and a perfectly transparent object would transmit itall (Figure 1.31). Such objects would not emit any radiation. Mathematically, the color is represented by the emissivitye. A “blackbody” radiator would have an e = 1 , whereas a perfect reflector or transmitter would have e = 0 . For real
examples, tungsten light bulb filaments have an e of about 0.5, and carbon black (a material used in printer toner) has anemissivity of about 0.95.


Chapter 1 | Temperature and Heat 47




Figure 1.30 The darker pavement is hotter than the lighter pavement (much more of the ice on the right hasmelted), although both have been in the sunlight for the same time. The thermal conductivities of the pavements arethe same.


Figure 1.31 A black object is a good absorber and a good radiator, whereas a white, clear, or silver object is apoor absorber and a poor radiator.


To see that, consider a silver object and a black object that can exchange heat by radiation and are in thermal equilibrium.We know from experience that they will stay in equilibrium (the result of a principle that will be discussed at length inSecond Law of Thermodynamics). For the black object’s temperature to stay constant, it must emit as much radiationas it absorbs, so it must be as good at radiating as absorbing. Similar considerations show that the silver object must radiateas little as it absorbs. Thus, one property, emissivity, controls both radiation and absorption.
Finally, the radiated heat is proportional to the object’s surface area, since every part of the surface radiates. If you knockapart the coals of a fire, the radiation increases noticeably due to an increase in radiating surface area.
The rate of heat transfer by emitted radiation is described by the Stefan-Boltzmann law of radiation:


P = σAeT 4,


where σ = 5.67 × 10−8 J/s · m2 · K4 is the Stefan-Boltzmann constant, a combination of fundamental constants of nature;
A is the surface area of the object; and T is its temperature in kelvins.
The proportionality to the fourth power of the absolute temperature is a remarkably strong temperature dependence. Itallows the detection of even small temperature variations. Images called thermographs can be used medically to detectregions of abnormally high temperature in the body, perhaps indicative of disease. Similar techniques can be used to detectheat leaks in homes (Figure 1.32), optimize performance of blast furnaces, improve comfort levels in work environments,and even remotely map Earth’s temperature profile.


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Figure 1.32 A thermograph of part of a building shows temperature variations,indicating where heat transfer to the outside is most severe. Windows are a majorregion of heat transfer to the outside of homes. (credit: US Army)


The Stefan-Boltzmann equation needs only slight refinement to deal with a simple case of an object’s absorption of radiationfrom its surroundings. Assuming that an object with a temperature T1 is surrounded by an environment with uniform
temperature T2 , the net rate of heat transfer by radiation is


(1.10)Pnet = σeA⎛⎝T2 4 − T1 4⎞⎠,


where e is the emissivity of the object alone. In other words, it does not matter whether the surroundings are white, gray,or black: The balance of radiation into and out of the object depends on how well it emits and absorbs radiation. When
T2 > T1, the quantity Pnet is positive, that is, the net heat transfer is from hot to cold.
Before doing an example, we have a complication to discuss: different emissivities at different wavelengths. If the fractionof incident radiation an object reflects is the same at all visible wavelengths, the object is gray; if the fraction depends onthe wavelength, the object has some other color. For instance, a red or reddish object reflects red light more strongly thanother visible wavelengths. Because it absorbs less red, it radiates less red when hot. Differential reflection and absorption ofwavelengths outside the visible range have no effect on what we see, but they may have physically important effects. Skin isa very good absorber and emitter of infrared radiation, having an emissivity of 0.97 in the infrared spectrum. Thus, in spiteof the obvious variations in skin color, we are all nearly black in the infrared. This high infrared emissivity is why we canso easily feel radiation on our skin. It is also the basis for the effectiveness of night-vision scopes used by law enforcementand the military to detect human beings.
Example 1.13


Calculating the Net Heat Transfer of a Person
What is the rate of heat transfer by radiation of an unclothed person standing in a dark room whose ambient
temperature is 22.0 °C ? The person has a normal skin temperature of 33.0 °C and a surface area of 1.50 m2.
The emissivity of skin is 0.97 in the infrared, the part of the spectrum where the radiation takes place.
Strategy
We can solve this by using the equation for the rate of radiative heat transfer.


Chapter 1 | Temperature and Heat 49




Solution
Insert the temperature values T2 = 295 K and T1 = 306 K , so that


Q
t = σeA



⎝T2


4 − T1
4⎞


= ⎛⎝5.67 × 10
−8 J/s · m2 · K4⎞⎠(0.97)



⎝1.50 m


2⎞


⎣(295 K)


4 − (306 K)4⎤⎦


= −99 J/s = −99W.


Significance
This value is a significant rate of heat transfer to the environment (note the minus sign), considering that a personat rest may produce energy at the rate of 125 W and that conduction and convection are also transferring energy tothe environment. Indeed, we would probably expect this person to feel cold. Clothing significantly reduces heattransfer to the environment by all mechanisms, because clothing slows down both conduction and convection,and has a lower emissivity (especially if it is light-colored) than skin.


The average temperature of Earth is the subject of much current discussion. Earth is in radiative contact with both the Sunand dark space, so we cannot use the equation for an environment at a uniform temperature. Earth receives almost all itsenergy from radiation of the Sun and reflects some of it back into outer space. Conversely, dark space is very cold, about 3K, so that Earth radiates energy into the dark sky. The rate of heat transfer from soil and grasses can be so rapid that frostmay occur on clear summer evenings, even in warm latitudes.
The average temperature of Earth is determined by its energy balance. To a first approximation, it is the temperature atwhich Earth radiates heat to space as fast as it receives energy from the Sun.
An important parameter in calculating the temperature of Earth is its emissivity (e). On average, it is about 0.65, butcalculation of this value is complicated by the great day-to-day variation in the highly reflective cloud coverage. Becauseclouds have lower emissivity than either oceans or land masses, they reflect some of the radiation back to the surface, greatlyreducing heat transfer into dark space, just as they greatly reduce heat transfer into the atmosphere during the day. Thereis negative feedback (in which a change produces an effect that opposes that change) between clouds and heat transfer;higher temperatures evaporate more water to form more clouds, which reflect more radiation back into space, reducing thetemperature.
The often-mentioned greenhouse effect is directly related to the variation of Earth’s emissivity with wavelength (Figure1.33). The greenhouse effect is a natural phenomenon responsible for providing temperatures suitable for life on Earth andfor making Venus unsuitable for human life. Most of the infrared radiation emitted from Earth is absorbed by carbon dioxide( CO2 ) and water (H2O ) in the atmosphere and then re-radiated into outer space or back to Earth. Re-radiation back to
Earth maintains its surface temperature about 40 °C higher than it would be if there were no atmosphere. (The glass walls
and roof of a greenhouse increase the temperature inside by blocking convective heat losses, not radiative losses.)


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Figure 1.33 The greenhouse effect is the name given to the increase of Earth’stemperature due to absorption of radiation in the atmosphere. The atmosphere istransparent to incoming visible radiation and most of the Sun’s infrared. The Earthabsorbs that energy and re-emits it. Since Earth’s temperature is much lower than theSun’s, it re-emits the energy at much longer wavelengths, in the infrared. Theatmosphere absorbs much of that infrared radiation and radiates about half of theenergy back down, keeping Earth warmer than it would otherwise be. The amount oftrapping depends on concentrations of trace gases such as carbon dioxide, and anincrease in the concentration of these gases increases Earth’s surface temperature.


The greenhouse effect is central to the discussion of global warming due to emission of carbon dioxide and methane (andother greenhouse gases) into Earth’s atmosphere from industry, transportation, and farming. Changes in global climate couldlead to more intense storms, precipitation changes (affecting agriculture), reduction in rain forest biodiversity, and rising sealevels.
You can explore a simulation of the greenhouse effect (https://openstaxcollege.org/l/21simgreeneff)that takes the point of view that the atmosphere scatters (redirects) infrared radiation rather than absorbing it andreradiating it. You may want to run the simulation first with no greenhouse gases in the atmosphere and then lookat how adding greenhouse gases affects the infrared radiation from the Earth and the Earth’s temperature.


Problem-Solving Strategy: Effects of Heat Transfer
1. Examine the situation to determine what type of heat transfer is involved.
2. Identify the type(s) of heat transfer—conduction, convection, or radiation.
3. Identify exactly what needs to be determined in the problem (identify the unknowns). A written list is useful.
4. Make a list of what is given or what can be inferred from the problem as stated (identify the knowns).
5. Solve the appropriate equation for the quantity to be determined (the unknown).
6. For conduction, use the equation P = kAΔT


d
. Table 1.5 lists thermal conductivities. For convection,


determine the amount of matter moved and the equation Q = mcΔT , along with Q = mLf or Q = mLV if a


Chapter 1 | Temperature and Heat 51




1.9


substance changes phase. For radiation, the equation Pnet = σeA⎛⎝T2 4 − T1 4⎞⎠ gives the net heat transfer rate.
7. Substitute the knowns along with their units into the appropriate equation and obtain numerical solutionscomplete with units.
8. Check the answer to see if it is reasonable. Does it make sense?


Check Your Understanding How much greater is the rate of heat radiation when a body is at thetemperature 40 °C than when it is at the temperature 20 °C ?


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absolute temperature scale
absolute zero
calorie (cal)
calorimeter
calorimetry
Celsius scale
coefficient of linear expansion


coefficient of volume expansion
conduction
convection
critical point
critical pressure
critical temperature
degree Celsius
degree Fahrenheit
emissivity
Fahrenheit scale
greenhouse effect
heat
heat of fusion
heat of sublimation
heat of vaporization
heat transfer
Kelvin scale (K)
kilocalorie (kcal)
latent heat coefficient
mechanical equivalent of heat
net rate of heat transfer by radiation
phase diagram


CHAPTER 1 REVIEW
KEY TERMS


scale, such as Kelvin, with a zero point that is absolute zero
temperature at which the average kinetic energy of molecules is zero


energy needed to change the temperature of 1.00 g of water by 1.00 °C
container that prevents heat transfer in or out
study of heat transfer inside a container impervious to heat
temperature scale in which the freezing point of water is 0 °C and the boiling point of water is 100 °C


( α ) material property that gives the change in length, per unit length, per 1-°C
change in temperature; a constant used in the calculation of linear expansion; the coefficient of linear expansiondepends to some degree on the temperature of the material


( β ) similar to α but gives the change in volume, per unit volume, per 1-°C
change in temperature


heat transfer through stationary matter by physical contact
heat transfer by the macroscopic movement of fluid
for a given substance, the combination of temperature and pressure above which the liquid and gas phasesare indistinguishable


pressure at the critical point
temperature at the critical point


( °C ) unit on the Celsius temperature scale
( °F ) unit on the Fahrenheit temperature scale


measure of how well an object radiates
temperature scale in which the freezing point of water is 32 °F and the boiling point of water is


212 °F


warming of the earth that is due to gases such as carbon dioxide and methane that absorb infraredradiation from Earth’s surface and reradiate it in all directions, thus sending some of it back toward Earth
energy transferred solely due to a temperature difference


energy per unit mass required to change a substance from the solid phase to the liquid phase, or releasedwhen the substance changes from liquid to solid
energy per unit mass required to change a substance from the solid phase to the vapor phase
energy per unit mass required to change a substance from the liquid phase to the vapor phase


movement of energy from one place or material to another as a result of a difference in temperature
temperature scale in which 0 K is the lowest possible temperature, representing absolute zero
energy needed to change the temperature of 1.00 kg of water between 14.5 °C and 15.5 °C


general term for the heats of fusion, vaporization, and sublimation
work needed to produce the same effects as heat transfer


Pnet = σeA

⎝T2


4 − T1
4⎞


graph of pressure vs. temperature of a particular substance, showing at which pressures andtemperatures the phases of the substance occur


Chapter 1 | Temperature and Heat 53




radiation
rate of conductive heat transfer
specific heat
Stefan-Boltzmann law of radiation


sublimation
temperature
thermal conductivity
thermal equilibrium
thermal expansion
thermal stress
triple point
vapor
vapor pressure
zeroth law of thermodynamics


energy transferred by electromagnetic waves directly as a result of a temperature difference
rate of heat transfer from one material to another


amount of heat necessary to change the temperature of 1.00 kg of a substance by 1.00 °C ; also called
“specific heat capacity”


P = σAeT 4, where σ = 5.67 × 10−8 J/s · m2 · K4 is the Stefan-Boltzmann
constant, A is the surface area of the object, T is the absolute temperature, and e is the emissivity


phase change from solid to gas
quantity measured by a thermometer, which reflects the mechanical energy of molecules in a system


property of a material describing its ability to conduct heat
condition in which heat no longer flows between two objects that are in contact; the two objectshave the same temperature
change in size or volume of an object with change in temperature


stress caused by thermal expansion or contraction
pressure and temperature at which a substance exists in equilibrium as a solid, liquid, and gas


gas at a temperature below the boiling temperature
pressure at which a gas coexists with its solid or liquid phase


law that states that if two objects are in thermal equilibrium, and a third object is inthermal equilibrium with one of those objects, it is also in thermal equilibrium with the other object


KEY EQUATIONS
Linear thermal expansion ΔL = αLΔT
Thermal expansion in two dimensions ΔA = 2αAΔT
Thermal expansion in three dimensions ΔV = βVΔT
Heat transfer Q = mcΔT
Transfer of heat in a calorimeter Qcold + Qhot = 0
Heat due to phase change (melting and freezing) Q = mLf
Heat due to phase change (evaporation and condensation) Q = mLv
Rate of conductive heat transfer P = kA⎛⎝Th − Tc⎞⎠


d


Net rate of heat transfer by radiation Pnet = σeA⎛⎝T2 4 − T1 4⎞⎠
SUMMARY
1.1 Temperature and Thermal Equilibrium


• Temperature is operationally defined as the quantity measured by a thermometer. It is proportional to the averagekinetic energy of atoms and molecules in a system.
• Thermal equilibrium occurs when two bodies are in contact with each other and can freely exchange energy.Systems are in thermal equilibrium when they have the same temperature.
• The zeroth law of thermodynamics states that when two systems, A and B, are in thermal equilibrium with eachother, and B is in thermal equilibrium with a third system C, then A is also in thermal equilibrium with C.


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1.2 Thermometers and Temperature Scales
• Three types of thermometers are alcohol, liquid crystal, and infrared radiation (pyrometer).
• The three main temperature scales are Celsius, Fahrenheit, and Kelvin. Temperatures can be converted from onescale to another using temperature conversion equations.
• The three phases of water (ice, liquid water, and water vapor) can coexist at a single pressure and temperature knownas the triple point.


1.3 Thermal Expansion
• Thermal expansion is the increase of the size (length, area, or volume) of a body due to a change in temperature,usually a rise. Thermal contraction is the decrease in size due to a change in temperature, usually a fall intemperature.
• Thermal stress is created when thermal expansion or contraction is constrained.


1.4 Heat Transfer, Specific Heat, and Calorimetry
• Heat and work are the two distinct methods of energy transfer.
• Heat transfer to an object when its temperature changes is often approximated well by Q = mcΔT , where m is the
object’s mass and c is the specific heat of the substance.


1.5 Phase Changes
• Most substances have three distinct phases (under ordinary conditions on Earth), and they depend on temperatureand pressure.
• Two phases coexist (i.e., they are in thermal equilibrium) at a set of pressures and temperatures.
• Phase changes occur at fixed temperatures for a given substance at a given pressure, and these temperatures arecalled boiling, freezing (or melting), and sublimation points.


1.6 Mechanisms of Heat Transfer
• Heat is transferred by three different methods: conduction, convection, and radiation.
• Heat conduction is the transfer of heat between two objects in direct contact with each other.
• The rate of heat transfer P (energy per unit time) is proportional to the temperature difference Th − Tc and the
contact area A and inversely proportional to the distance d between the objects.


• Convection is heat transfer by the macroscopic movement of mass. Convection can be natural or forced, andgenerally transfers thermal energy faster than conduction. Convection that occurs along with a phase change cantransfer energy from cold regions to warm ones.
• Radiation is heat transfer through the emission or absorption of electromagnetic waves.
• The rate of radiative heat transfer is proportional to the emissivity e. For a perfect blackbody, e = 1 , whereas a
perfectly white, clear, or reflective body has e = 0 , with real objects having values of e between 1 and 0.


• The rate of heat transfer depends on the surface area and the fourth power of the absolute temperature:
P = σeAT 4,


where σ = 5.67 × 10−8 J/s · m2 · K4 is the Stefan-Boltzmann constant and e is the emissivity of the body. The net
rate of heat transfer from an object by radiation is


Qnet
t = σeA



⎝T2


4 − T1
4⎞
⎠,


where T1 is the temperature of the object surrounded by an environment with uniform temperature T2 and e is the


Chapter 1 | Temperature and Heat 55




emissivity of the object.


CONCEPTUAL QUESTIONS
1.1 Temperature and Thermal Equilibrium
1. What does it mean to say that two systems are inthermal equilibrium?
2. Give an example in which A has some kind of non-thermal equilibrium relationship with B, and B has the samerelationship with C, but A does not have that relationshipwith C.


1.2 Thermometers and Temperature Scales
3. If a thermometer is allowed to come to equilibrium withthe air, and a glass of water is not in equilibrium with theair, what will happen to the thermometer reading when it isplaced in the water?
4. Give an example of a physical property that varieswith temperature and describe how it is used to measuretemperature.


1.3 Thermal Expansion
5. Pouring cold water into hot glass or ceramic cookwarecan easily break it. What causes the breaking? Explainwhy Pyrex®, a glass with a small coefficient of linearexpansion, is less susceptible.
6. One method of getting a tight fit, say of a metal peg ina hole in a metal block, is to manufacture the peg slightlylarger than the hole. The peg is then inserted when at adifferent temperature than the block. Should the block behotter or colder than the peg during insertion? Explain youranswer.
7. Does it really help to run hot water over a tight metal lidon a glass jar before trying to open it? Explain your answer.
8. When a cold alcohol thermometer is placed in a hotliquid, the column of alcohol goes down slightly beforegoing up. Explain why.
9. Calculate the length of a 1-meter rod of a material withthermal expansion coefficient α when the temperature is
raised from 300 K to 600 K. Taking your answer as the newinitial length, find the length after the rod is cooled backdown to 300 K. Is your answer 1 meter? Should it be? Howcan you account for the result you got?
10. Noting the large stresses that can be caused by thermalexpansion, an amateur weapon inventor decides to use it to


make a new kind of gun. He plans to jam a bullet againstan aluminum rod inside a closed invar tube. When he heatsthe tube, the rod will expand more than the tube and a verystrong force will build up. Then, by a method yet to bedetermined, he will open the tube in a split second and letthe force of the rod launch the bullet at very high speed.What is he overlooking?


1.4 Heat Transfer, Specific Heat, and
Calorimetry
11. How is heat transfer related to temperature?
12. Describe a situation in which heat transfer occurs.
13. When heat transfers into a system, is the energy storedas heat? Explain briefly.
14. The brakes in a car increase in temperature by ΔT
when bringing the car to rest from a speed v. How muchgreater would ΔT be if the car initially had twice the
speed? You may assume the car stops fast enough that noheat transfers out of the brakes.


1.5 Phase Changes
15. A pressure cooker contains water and steam inequilibrium at a pressure greater than atmospheric pressure.How does this greater pressure increase cooking speed?
16. As shown below, which is the phase diagram forcarbon dioxide, what is the vapor pressure of solid carbondioxide (dry ice) at −78.5 °C? (Note that the axes in the
figure are nonlinear and the graph is not to scale.)


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17. Can carbon dioxide be liquefied at room temperature( 20 °C )? If so, how? If not, why not? (See the phase
diagram in the preceding problem.)
18. What is the distinction between gas and vapor?
19. Heat transfer can cause temperature and phasechanges. What else can cause these changes?
20. How does the latent heat of fusion of water helpslow the decrease of air temperatures, perhaps preventingtemperatures from falling significantly below 0 °C, in the
vicinity of large bodies of water?
21. What is the temperature of ice right after it is formedby freezing water?
22. If you place 0 °C ice into 0 °C water in an insulated
container, what will the net result be? Will there be less iceand more liquid water, or more ice and less liquid water, orwill the amounts stay the same?
23. What effect does condensation on a glass of ice waterhave on the rate at which the ice melts? Will thecondensation speed up the melting process or slow itdown?
24. In Miami, Florida, which has a very humid climateand numerous bodies of water nearby, it is unusual fortemperatures to rise above about 38 °C ( 100 °F ). In the
desert climate of Phoenix, Arizona, however, temperaturesrise above that almost every day in July and August.Explain how the evaporation of water helps limit hightemperatures in humid climates.


25. In winter, it is often warmer in San Francisco thanin Sacramento, 150 km inland. In summer, it is nearlyalways hotter in Sacramento. Explain how the bodies ofwater surrounding San Francisco moderate its extremetemperatures.
26. Freeze-dried foods have been dehydrated in a vacuum.During the process, the food freezes and must be heated tofacilitate dehydration. Explain both how the vacuum speedsup dehydration and why the food freezes as a result.
27. In a physics classroom demonstration, an instructorinflates a balloon by mouth and then cools it in liquidnitrogen. When cold, the shrunken balloon has a smallamount of light blue liquid in it, as well as some snow-likecrystals. As it warms up, the liquid boils, and part of thecrystals sublime, with some crystals lingering for a whileand then producing a liquid. Identify the blue liquid and thetwo solids in the cold balloon. Justify your identificationsusing data from Table 1.4.


1.6 Mechanisms of Heat Transfer
28. What are the main methods of heat transfer from thehot core of Earth to its surface? From Earth’s surface toouter space?
29. When our bodies get too warm, they respond bysweating and increasing blood circulation to the surface totransfer thermal energy away from the core. What effectwill those processes have on a person in a 40.0-°C hot
tub?
30. Shown below is a cut-away drawing of a thermosbottle (also known as a Dewar flask), which is a devicedesigned specifically to slow down all forms of heattransfer. Explain the functions of the various parts, such asthe vacuum, the silvering of the walls, the thin-walled longglass neck, the rubber support, the air layer, and the stopper.


Chapter 1 | Temperature and Heat 57




31. Some electric stoves have a flat ceramic surface withheating elements hidden beneath. A pot placed over aheating element will be heated, while the surface only a fewcentimeters away is safe to touch. Why is ceramic, with aconductivity less than that of a metal but greater than thatof a good insulator, an ideal choice for the stove top?
32. Loose-fitting white clothing covering most of thebody, shown below, is ideal for desert dwellers, both inthe hot Sun and during cold evenings. Explain how suchclothing is advantageous during both day and night.


33. One way to make a fireplace more energy-efficient isto have room air circulate around the outside of the fire boxand back into the room. Detail the methods of heat transferinvolved.
34. On cold, clear nights horses will sleep under the coverof large trees. How does this help them keep warm?
35. When watching a circus during the day in a large,dark-colored tent, you sense significant heat transfer fromthe tent. Explain why this occurs.
36. Satellites designed to observe the radiation from cold(3 K) dark space have sensors that are shaded from theSun, Earth, and the Moon and are cooled to very lowtemperatures. Why must the sensors be at lowtemperature?
37. Why are thermometers that are used in weatherstations shielded from the sunshine? What does athermometer measure if it is shielded from the sunshine?What does it measure if it is not?
38. Putting a lid on a boiling pot greatly reduces the heattransfer necessary to keep it boiling. Explain why.
39. Your house will be empty for a while in cold weather,and you want to save energy and money. Should you turnthe thermostat down to the lowest level that will protect thehouse from damage such as freezing pipes, or leave it atthe normal temperature? (If you don’t like coming back to acold house, imagine that a timer controls the heating systemso the house will be warm when you get back.) Explainyour answer.
40. You pour coffee into an unlidded cup, intending todrink it 5 minutes later. You can add cream when you pourthe cup or right before you drink it. (The cream is at thesame temperature either way. Assume that the cream andcoffee come into thermal equilibrium with each other veryquickly.) Which way will give you hotter coffee? Whatfeature of this question is different from the previous one?
41. Broiling is a method of cooking by radiation, whichproduces somewhat different results from cooking byconduction or convection. A gas flame or electric heatingelement produces a very high temperature close to the foodand above it. Why is radiation the dominant heat-transfermethod in this situation?
42. On a cold winter morning, why does the metal of abike feel colder than the wood of a porch?


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PROBLEMS
1.2 Thermometers and Temperature Scales
43. While traveling outside the United States, you feelsick. A companion gets you a thermometer, which saysyour temperature is 39. What scale is that on? What is yourFahrenheit temperature? Should you seek medical help?
44. What are the following temperatures on the Kelvinscale?
(a) 68.0 °F, an indoor temperature sometimes
recommended for energy conservation in winter
(b) 134 °F, one of the highest atmospheric temperatures
ever recorded on Earth (Death Valley, California, 1913)
(c) 9890 °F, the temperature of the surface of the Sun
45. (a) Suppose a cold front blows into your locale anddrops the temperature by 40.0 Fahrenheit degrees. Howmany degrees Celsius does the temperature decrease whenit decreases by 40.0 °F ? (b) Show that any change in
temperature in Fahrenheit degrees is nine-fifths the changein Celsius degrees
46. An Associated Press article on climate change said,“Some of the ice shelf’s disappearance was probably duringtimes when the planet was 36 degrees Fahrenheit (2 degreesCelsius) to 37 degrees Fahrenheit (3 degrees Celsius)warmer than it is today.” What mistake did the reportermake?
47. (a) At what temperature do the Fahrenheit and Celsiusscales have the same numerical value? (b) At whattemperature do the Fahrenheit and Kelvin scales have thesame numerical value?
48. A person taking a reading of the temperature in afreezer in Celsius makes two mistakes: first omitting thenegative sign and then thinking the temperature isFahrenheit. That is, the person reads – x °C as x °F .
Oddly enough, the result is the correct Fahrenheittemperature. What is the original Celsius reading? Roundyour answer to three significant figures.


1.3 Thermal Expansion
49. The height of the Washington Monument is measuredto be 170.00 m on a day when the temperature is 35.0 °C.
What will its height be on a day when the temperaturefalls to −10.0 °C ? Although the monument is made of
limestone, assume that its coefficient of thermal expansionis the same as that of marble. Give your answer to fivesignificant figures.


50. How much taller does the Eiffel Tower become atthe end of a day when the temperature has increased by
15 °C? Its original height is 321 m and you can assume it
is made of steel.
51. What is the change in length of a 3.00-cm-longcolumn of mercury if its temperature changes from
37.0 °C to 40.0 °C , assuming the mercury is constrained
to a cylinder but unconstrained in length? Your answerwill show why thermometers contain bulbs at the bottominstead of simple columns of liquid.
52. How large an expansion gap should be left betweensteel railroad rails if they may reach a maximumtemperature 35.0 °C greater than when they were laid?
Their original length is 10.0 m.
53. You are looking to buy a small piece of land in HongKong. The price is “only” $60,000 per square meter. Theland title says the dimensions are 20 m × 30 m . By how
much would the total price change if you measured theparcel with a steel tape measure on a day when thetemperature was 20 °C above the temperature that the tape
measure was designed for? The dimensions of the land donot change.
54. Global warming will produce rising sea levels partlydue to melting ice caps and partly due to the expansion ofwater as average ocean temperatures rise. To get some ideaof the size of this effect, calculate the change in length ofa column of water 1.00 km high for a temperature increaseof 1.00 °C . Assume the column is not free to expand
sideways. As a model of the ocean, that is a reasonableapproximation, as only parts of the ocean very close tothe surface can expand sideways onto land, and only to alimited degree. As another approximation, neglect the factthat ocean warming is not uniform with depth.
55. (a) Suppose a meter stick made of steel and one madeof aluminum are the same length at 0 °C . What is their
difference in length at 22.0 °C ? (b) Repeat the calculation
for two 30.0-m-long surveyor’s tapes.
56. (a) If a 500-mL glass beaker is filled to the brim withethyl alcohol at a temperature of 5.00 °C , how much will
overflow when the alcohol’s temperature reaches the roomtemperature of 22.0 °C ? (b) How much less water would
overflow under the same conditions?
57. Most cars have a coolant reservoir to catch radiatorfluid that may overflow when the engine is hot. A radiatoris made of copper and is filled to its 16.0-L capacity whenat 10.0 °C . What volume of radiator fluid will overflow


Chapter 1 | Temperature and Heat 59




when the radiator and fluid reach a temperature of
95.0 °C, given that the fluid’s volume coefficient of
expansion is β = 400 × 10−6 /°C ? (Your answer will be a
conservative estimate, as most car radiators have operatingtemperatures greater than 95.0 °C ).
58. A physicist makes a cup of instant coffee and noticesthat, as the coffee cools, its level drops 3.00 mm in theglass cup. Show that this decrease cannot be due to thermalcontraction by calculating the decrease in level if the
350 cm3 of coffee is in a 7.00-cm-diameter cup and
decreases in temperature from 95.0 °C to 45.0 °C . (Most
of the drop in level is actually due to escaping bubbles ofair.)
59. The density of water at 0 °C is very nearly
1000 kg/m3 (it is actually 999.84 kg/m3 ), whereas the
density of ice at 0 °C is 917 kg/m3. Calculate the
pressure necessary to keep ice from expanding when itfreezes, neglecting the effect such a large pressure wouldhave on the freezing temperature. (This problem gives youonly an indication of how large the forces associated withfreezing water might be.)
60. Show that β = 3α, by calculating the infinitesimal
change in volume dV of a cube with sides of length L whenthe temperature changes by dT.


1.4 Heat Transfer, Specific Heat, and
Calorimetry
61. On a hot day, the temperature of an 80,000-Lswimming pool increases by 1.50 °C . What is the net
heat transfer during this heating? Ignore any complications,such as loss of water by evaporation.
62. To sterilize a 50.0-g glass baby bottle, we must raiseits temperature from 22.0 °C to 95.0 °C . How much heat
transfer is required?
63. The same heat transfer into identical masses ofdifferent substances produces different temperaturechanges. Calculate the final temperature when 1.00 kcalof heat transfers into 1.00 kg of the following, originallyat 20.0 °C : (a) water; (b) concrete; (c) steel; and (d)
mercury.
64. Rubbing your hands together warms them byconverting work into thermal energy. If a woman rubs her


hands back and forth for a total of 20 rubs, at a distanceof 7.50 cm per rub, and with an average frictional forceof 40.0 N, what is the temperature increase? The mass oftissues warmed is only 0.100 kg, mostly in the palms andfingers.
65. A 0.250-kg block of a pure material is heated from
20.0 °C to 65.0 °C by the addition of 4.35 kJ of energy.
Calculate its specific heat and identify the substance ofwhich it is most likely composed.
66. Suppose identical amounts of heat transfer intodifferent masses of copper and water, causing identicalchanges in temperature. What is the ratio of the mass ofcopper to water?
67. (a) The number of kilocalories in food is determinedby calorimetry techniques in which the food is burned andthe amount of heat transfer is measured. How manykilocalories per gram are there in a 5.00-g peanut if theenergy from burning it is transferred to 0.500 kg of waterheld in a 0.100-kg aluminum cup, causing a 54.9-°C
temperature increase? Assume the process takes place inan ideal calorimeter, in other words a perfectly insulatedcontainer. (b) Compare your answer to the followinglabeling information found on a package of dry roastedpeanuts: a serving of 33 g contains 200 calories. Commenton whether the values are consistent.
68. Following vigorous exercise, the body temperatureof an 80.0 kg person is 40.0 °C . At what rate in watts
must the person transfer thermal energy to reduce the bodytemperature to 37.0 °C in 30.0 min, assuming the body
continues to produce energy at the rate of 150 W?

⎝1 watt = 1 joule/second or 1W = 1 J/s⎞⎠


69. In a study of healthy young men[1], doing 20 push-ups in 1 minute burned an amount of energy per kg thatfor a 70.0-kg man corresponds to 8.06 calories (kcal). Howmuch would a 70.0-kg man’s temperature rise if he did notlose any heat during that time?
70. A 1.28-kg sample of water at 10.0 °C is in a
calorimeter. You drop a piece of steel with a mass of 0.385kg at 215 °C into it. After the sizzling subsides, what is
the final equilibrium temperature? (Make the reasonableassumptions that any steam produced condenses into liquidwater during the process of equilibration and that theevaporation and condensation don’t affect the outcome, aswe’ll see in the next section.)
71. Repeat the preceding problem, assuming the water


1. JW Vezina, “An examination of the differences between two methods of estimating energy expenditure in resistancetraining activities,” Journal of Strength and Conditioning Research, April 28, 2014, http://www.ncbi.nlm.nih.gov/pubmed/24402448


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is in a glass beaker with a mass of 0.200 kg, which inturn is in a calorimeter. The beaker is initially at the sametemperature as the water. Before doing the problem, shouldthe answer be higher or lower than the preceding answer?Comparing the mass and specific heat of the beaker tothose of the water, do you think the beaker will make muchdifference?


1.5 Phase Changes
72. How much heat transfer (in kilocalories) is required tothaw a 0.450-kg package of frozen vegetables originally at
0 °C if their heat of fusion is the same as that of water?
73. A bag containing 0 °C ice is much more effective in
absorbing energy than one containing the same amount of
0 °C water. (a) How much heat transfer is necessary to
raise the temperature of 0.800 kg of water from 0 °C to
30.0 °C ? (b) How much heat transfer is required to first
melt 0.800 kg of 0 °C ice and then raise its temperature?
(c) Explain how your answer supports the contention thatthe ice is more effective.
74. (a) How much heat transfer is required to raise thetemperature of a 0.750-kg aluminum pot containing 2.50 kgof water from 30.0 °C to the boiling point and then boil
away 0.750 kg of water? (b) How long does this take if therate of heat transfer is 500 W?
75. Condensation on a glass of ice water causes the iceto melt faster than it would otherwise. If 8.00 g of vaporcondense on a glass containing both water and 200 g of ice,how many grams of the ice will melt as a result? Assumeno other heat transfer occurs. Use Lv for water at 37 °C
as a better approximation than Lv for water at 100 °C .)
76. On a trip, you notice that a 3.50-kg bag of ice lastsan average of one day in your cooler. What is the averagepower in watts entering the ice if it starts at 0 °C and
completely melts to 0 °C water in exactly one day?
77. On a certain dry sunny day, a swimming pool’stemperature would rise by 1.50 °C if not for evaporation.
What fraction of the water must evaporate to carry awayprecisely enough energy to keep the temperature constant?
78. (a) How much heat transfer is necessary to raise thetemperature of a 0.200-kg piece of ice from −20.0 °C to
130.0 °C , including the energy needed for phase changes?
(b) How much time is required for each stage, assuming aconstant 20.0 kJ/s rate of heat transfer? (c) Make a graph oftemperature versus time for this process.
79. In 1986, an enormous iceberg broke away from the


Ross Ice Shelf in Antarctica. It was an approximatelyrectangular prism 160 km long, 40.0 km wide, and 250 mthick. (a) What is the mass of this iceberg, given that the
density of ice is 917 kg/m3 ? (b) How much heat transfer
(in joules) is needed to melt it? (c) How many years wouldit take sunlight alone to melt ice this thick, if the ice absorbs
an average of 100W/m2 , 12.00 h per day?
80. How many grams of coffee must evaporate from 350g of coffee in a 100-g glass cup to cool the coffee and thecup from 95.0 °C to 45.0 °C ? Assume the coffee has the
same thermal properties as water and that the average heatof vaporization is 2340 kJ/kg (560 kcal/g). Neglect heatlosses through processes other than evaporation, as well asthe change in mass of the coffee as it cools. Do the lattertwo assumptions cause your answer to be higher or lowerthan the true answer?
81. (a) It is difficult to extinguish a fire on a crude oiltanker, because each liter of crude oil releases
2.80 × 107 J of energy when burned. To illustrate this
difficulty, calculate the number of liters of water that mustbe expended to absorb the energy released by burning 1.00L of crude oil, if the water’s temperature rises from
20.0 °C to 100 °C , it boils, and the resulting steam’s
temperature rises to 300 °C at constant pressure. (b)
Discuss additional complications caused by the fact thatcrude oil is less dense than water.
82. The energy released from condensation inthunderstorms can be very large. Calculate the energyreleased into the atmosphere for a small storm of radius 1km, assuming that 1.0 cm of rain is precipitated uniformlyover this area.
83. To help prevent frost damage, 4.00 kg of water at
0 °C is sprayed onto a fruit tree. (a) How much heat
transfer occurs as the water freezes? (b) How much wouldthe temperature of the 200-kg tree decrease if this amountof heat transferred from the tree? Take the specific heatto be 3.35 kJ/kg · °C , and assume that no phase change
occurs in the tree.
84. A 0.250-kg aluminum bowl holding 0.800 kg of soup
at 25.0 °C is placed in a freezer. What is the final
temperature if 388 kJ of energy is transferred from the bowland soup, assuming the soup’s thermal properties are thesame as that of water?
85. A 0.0500-kg ice cube at −30.0 °C is placed in 0.400
kg of 35.0-°C water in a very well-insulated container.
What is the final temperature?


Chapter 1 | Temperature and Heat 61




86. If you pour 0.0100 kg of 20.0 °C water onto a
1.20-kg block of ice (which is initially at −15.0 °C ), what
is the final temperature? You may assume that the watercools so rapidly that effects of the surroundings arenegligible.
87. Indigenous people sometimes cook in watertightbaskets by placing hot rocks into water to bring it to a boil.What mass of 500-°C granite must be placed in 4.00 kg
of 15.0-°C water to bring its temperature to 100 °C , if
0.0250 kg of water escapes as vapor from the initial sizzle?You may neglect the effects of the surroundings.
88. What would the final temperature of the pan and waterbe in Example 1.7 if 0.260 kg of water were placed inthe pan and 0.0100 kg of the water evaporated immediately,leaving the remainder to come to a common temperaturewith the pan?


1.6 Mechanisms of Heat Transfer
89. (a) Calculate the rate of heat conduction through housewalls that are 13.0 cm thick and have an average thermalconductivity twice that of glass wool. Assume there are
no windows or doors. The walls’ surface area is 120 m2
and their inside surface is at 18.0 °C , while their outside
surface is at 5.00 °C . (b) How many 1-kW room heaters
would be needed to balance the heat transfer due toconduction?
90. The rate of heat conduction out of a window on awinter day is rapid enough to chill the air next to it. To seejust how rapidly the windows transfer heat by conduction,calculate the rate of conduction in watts through a
3.00-m2 window that is 0.634 cm thick (1/4 in.) if the
temperatures of the inner and outer surfaces are 5.00 °C
and −10.0 °C , respectively. (This rapid rate will not be
maintained—the inner surface will cool, even to the pointof frost formation.)
91. Calculate the rate of heat conduction out of the humanbody, assuming that the core internal temperature is
37.0 °C , the skin temperature is 34.0 °C , the thickness of
the fatty tissues between the core and the skin averages 1.00
cm, and the surface area is 1.40 m2 .
92. Suppose you stand with one foot on ceramic flooringand one foot on a wool carpet, making contact over an
area of 80.0 cm2 with each foot. Both the ceramic and the
carpet are 2.00 cm thick and are 10.0 °C on their bottom
sides. At what rate must heat transfer occur from each footto keep the top of the ceramic and carpet at 33.0 °C ?


93. A man consumes 3000 kcal of food in one day,converting most of it to thermal energy to maintain bodytemperature. If he loses half this energy by evaporatingwater (through breathing and sweating), how manykilograms of water evaporate?
94. A firewalker runs across a bed of hot coals withoutsustaining burns. Calculate the heat transferred byconduction into the sole of one foot of a firewalker giventhat the bottom of the foot is a 3.00-mm-thick callus witha conductivity at the low end of the range for wood and its
density is 300 kg/m3 . The area of contact is 25.0 cm2,
the temperature of the coals is 700 °C , and the time in
contact is 1.00 s. Ignore the evaporative cooling of sweat.
95. (a) What is the rate of heat conduction through the
3.00-cm-thick fur of a large animal having a 1.40-m2
surface area? Assume that the animal’s skin temperature is
32.0 °C , that the air temperature is −5.00 °C , and that
fur has the same thermal conductivity as air. (b) What foodintake will the animal need in one day to replace this heattransfer?
96. A walrus transfers energy by conduction through itsblubber at the rate of 150 W when immersed in −1.00 °C
water. The walrus’s internal core temperature is 37.0 °C ,
and it has a surface area of 2.00 m2 . What is the average
thickness of its blubber, which has the conductivity of fattytissues without blood?
97. Compare the rate of heat conduction through a
13.0-cm-thick wall that has an area of 10.0 m2 and a
thermal conductivity twice that of glass wool with the rateof heat conduction through a 0.750-cm-thick window that
has an area of 2.00 m2 , assuming the same temperature
difference across each.
98. Suppose a person is covered head to foot by woolclothing with average thickness of 2.00 cm and istransferring energy by conduction through the clothing atthe rate of 50.0 W. What is the temperature difference
across the clothing, given the surface area is 1.40 m2 ?
99. Some stove tops are smooth ceramic for easy cleaning.If the ceramic is 0.600 cm thick and heat conduction occursthrough the same area and at the same rate as computed inExample 1.11, what is the temperature difference acrossit? Ceramic has the same thermal conductivity as glass andbrick.
100. One easy way to reduce heating (and cooling) costsis to add extra insulation in the attic of a house. Suppose asingle-story cubical house already had 15 cm of fiberglassinsulation in the attic and in all the exterior surfaces. If


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you added an extra 8.0 cm of fiberglass to the attic, bywhat percentage would the heating cost of the house drop?Take the house to have dimensions 10 m by 15 m by 3.0m. Ignore air infiltration and heat loss through windowsand doors, and assume that the interior is uniformly at onetemperature and the exterior is uniformly at another.
101. Many decisions are made on the basis of the paybackperiod: the time it will take through savings to equal the


capital cost of an investment. Acceptable payback timesdepend upon the business or philosophy one has. (For someindustries, a payback period is as small as 2 years.) Supposeyou wish to install the extra insulation in the precedingproblem. If energy cost $1.00 per million joules and the
insulation was $4.00 per square meter, then calculate thesimple payback time. Take the average ΔT for the 120-day
heating season to be 15.0 °C.


ADDITIONAL PROBLEMS
102. In 1701, the Danish astronomer Ole Rømer proposeda temperature scale with two fixed points, freezing water at7.5 degrees, and boiling water at 60.0 degrees. What is theboiling point of oxygen, 90.2 K, on the Rømer scale?
103. What is the percent error of thinking the meltingpoint of tungsten is 3695 °C instead of the correct value of
3695 K?
104. An engineer wants to design a structure in which thedifference in length between a steel beam and an aluminumbeam remains at 0.500 m regardless of temperature, forordinary temperatures. What must the lengths of the beamsbe?
105. How much stress is created in a steel beam if itstemperature changes from –15 °C to 40 °C but it cannot
expand? For steel, the Young’s modulus
Y = 210 × 109 N/m2 from Stress, Strain, and
Elastic Modulus (http://cnx.org/content/m58342/latest/#fs-id1163713086230) . (Ignore the change inarea resulting from the expansion.)
106. A brass rod ⎛⎝Y = 90 × 109 N/m2⎞⎠, with a diameter
of 0.800 cm and a length of 1.20 m when the temperatureis 25 °C , is fixed at both ends. At what temperature is the
force in it at 36,000 N?
107. A mercury thermometer still in use for meteorology
has a bulb with a volume of 0.780 cm3 and a tube for the
mercury to expand into of inside diameter 0.130 mm. (a)Neglecting the thermal expansion of the glass, what is thespacing between marks 1 °C apart? (b) If the thermometer
is made of ordinary glass (not a good idea), what is thespacing?
108. Even when shut down after a period of normal use,a large commercial nuclear reactor transfers thermal energyat the rate of 150 MW by the radioactive decay of fissionproducts. This heat transfer causes a rapid increase intemperature if the cooling system fails



⎝1 watt = 1 joule/second or 1W = 1 J/s and
1MW = 1megawatt ⎞⎠. (a) Calculate the rate of
temperature increase in degrees Celsius per second (°C/s)
if the mass of the reactor core is 1.60 × 105 kg and it
has an average specific heat of 0.3349 kJ/kg · °C . (b) How
long would it take to obtain a temperature increase of
2000 °C , which could cause some metals holding the
radioactive materials to melt? (The initial rate oftemperature increase would be greater than that calculatedhere because the heat transfer is concentrated in a smallermass. Later, however, the temperature increase would slowdown because the 500,000-kg steel containment vesselwould also begin to heat up.)
109. You leave a pastry in the refrigerator on a plate andask your roommate to take it out before you get homeso you can eat it at room temperature, the way you likeit. Instead, your roommate plays video games for hours.When you return, you notice that the pastry is still cold, butthe game console has become hot. Annoyed, and knowingthat the pastry will not be good if it is microwaved, youwarm up the pastry by unplugging the console and puttingit in a clean trash bag (which acts as a perfect calorimeter)with the pastry on the plate. After a while, you find thatthe equilibrium temperature is a nice, warm 38.3 °C . You
know that the game console has a mass of 2.1 kg.Approximate it as having a uniform initial temperature of
45 °C . The pastry has a mass of 0.16 kg and a specific heat
of 3.0 k J/(kg · ºC), and is at a uniform initial temperature
of 4.0 °C . The plate is at the same temperature and has
a mass of 0.24 kg and a specific heat of 0.90 J/(kg · ºC) .
What is the specific heat of the console?
110. Two solid spheres, A and B, made of the samematerial, are at temperatures of 0 °C and 100 °C ,
respectively. The spheres are placed in thermal contact inan ideal calorimeter, and they reach an equilibriumtemperature of 20 °C . Which is the bigger sphere? What is
the ratio of their diameters?


Chapter 1 | Temperature and Heat 63




111. In some countries, liquid nitrogen is used on dairytrucks instead of mechanical refrigerators. A 3.00-hourdelivery trip requires 200 L of liquid nitrogen, which has
a density of 808 kg/m3. (a) Calculate the heat transfer
necessary to evaporate this amount of liquid nitrogen andraise its temperature to 3.00 °C . (Use cP and assume it
is constant over the temperature range.) This value is theamount of cooling the liquid nitrogen supplies. (b) What isthis heat transfer rate in kilowatt-hours? (c) Compare theamount of cooling obtained from melting an identical massof 0-°C ice with that from evaporating the liquid nitrogen.
112. Some gun fanciers make their own bullets, whichinvolves melting lead and casting it into lead slugs. Howmuch heat transfer is needed to raise the temperature andmelt 0.500 kg of lead, starting from 25.0 °C ?
113. A 0.800-kg iron cylinder at a temperature of
1.00 × 103 °C is dropped into an insulated chest of 1.00
kg of ice at its melting point. What is the final temperature,and how much ice has melted?
114. Repeat the preceding problem with 2.00 kg of iceinstead of 1.00 kg.
115. Repeat the preceding problem with 0.500 kg of ice,assuming that the ice is initially in a copper container ofmass 1.50 kg in equilibrium with the ice.
116. A 30.0-g ice cube at its melting point is dropped intoan aluminum calorimeter of mass 100.0 g in equilibriumat 24.0 °C with 300.0 g of an unknown liquid. The final
temperature is 4.0 °C . What is the heat capacity of the
liquid?
117. (a) Calculate the rate of heat conduction through
a double-paned window that has a 1.50-m2 area and is
made of two panes of 0.800-cm-thick glass separated bya 1.00-cm air gap. The inside surface temperature is
15.0 °C, while that on the outside is −10.0 °C. (Hint:
There are identical temperature drops across the two glasspanes. First find these and then the temperature drop acrossthe air gap. This problem ignores the increased heat transferin the air gap due to convection.) (b) Calculate the rateof heat conduction through a 1.60-cm-thick window of thesame area and with the same temperatures. Compare youranswer with that for part (a).
118. (a) An exterior wall of a house is 3 m tall and 10m wide. It consists of a layer of drywall with an R factorof 0.56, a layer 3.5 inches thick filled with fiberglass batts,and a layer of insulated siding with an R factor of 2.6. Thewall is built so well that there are no leaks of air through it.When the inside of the wall is at 22 °C and the outside is at


−2 °C , what is the rate of heat flow through the wall? (b)
More realistically, the 3.5-inch space also contains 2-by-4studs—wooden boards 1.5 inches by 3.5 inches orientedso that 3.5-inch dimension extends from the drywall to thesiding. They are “on 16-inch centers,” that is, the centers ofthe studs are 16 inches apart. What is the heat current in thissituation? Don’t worry about one stud more or less.
119. For the human body, what is the rate of heat transferby conduction through the body’s tissue with the followingconditions: the tissue thickness is 3.00 cm, the difference
in temperature is 2.00 °C , and the skin area is 1.50 m2 .
How does this compare with the average heat transfer rateto the body resulting from an energy intake of about 2400kcal per day? (No exercise is included.)
120. You have a Dewar flask (a laboratory vacuum flask)that has an open top and straight sides, as shown below. Youfill it with water and put it into the freezer. It is effectivelya perfect insulator, blocking all heat transfer, except on thetop. After a time, ice forms on the surface of the water. Theliquid water and the bottom surface of the ice, in contactwith the liquid water, are at 0 °C . The top surface of the ice
is at the same temperature as the air in the freezer, −18 °C.
Set the rate of heat flow through the ice equal to the rateof loss of heat of fusion as the water freezes. When the icelayer is 0.700 cm thick, find the rate in m/s at which the iceis thickening.


121. An infrared heater for a sauna has a surface area
of 0.050 m2 and an emissivity of 0.84. What temperature
must it run at if the required power is 360 W? Neglect thetemperature of the environment.
122. (a) Determine the power of radiation from the Sunby noting that the intensity of the radiation at the distance
of Earth is 1370W/m2 . Hint: That intensity will be found
everywhere on a spherical surface with radius equal to thatof Earth’s orbit. (b) Assuming that the Sun’s temperature is


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5780 K and that its emissivity is 1, find its radius.


CHALLENGE PROBLEMS
123. A pendulum is made of a rod of length L andnegligible mass, but capable of thermal expansion, anda weight of negligible size. (a) Show that when thetemperature increases by dT, the period of the pendulumincreases by a fraction αLdT /2 . (b) A clock controlled
by a brass pendulum keeps time correctly at 10 °C . If the
room temperature is 30 °C , does the clock run faster or
slower? What is its error in seconds per day?
124. At temperatures of a few hundred kelvins the specificheat capacity of copper approximately follows the
empirical formula c = α + βT + δT−2, where
α = 349 J/kg · K, β = 0.107 J/kg · K2, and
δ = 4.58 × 105 J · kg · K. How much heat is needed to
raise the temperature of a 2.00-kg piece of copper from
20 °C to 250 °C ?
125. In a calorimeter of negligible heat capacity, 200 g ofsteam at 150 °C and 100 g of ice at −40 °C are mixed.
The pressure is maintained at 1 atm. What is the finaltemperature, and how much steam, ice, and water arepresent?
126. An astronaut performing an extra-vehicular activity(space walk) shaded from the Sun is wearing a spacesuitthat can be approximated as perfectly white (e = 0) except
for a 5 cm × 8 cm patch in the form of the astronaut’s
national flag. The patch has emissivity 0.300. The spacesuitunder the patch is 0.500 cm thick, with a thermalconductivity k = 0.0600W/m °C , and its inner surface is
at a temperature of 20.0 °C . What is the temperature of the
patch, and what is the rate of heat loss through it? Assumethe patch is so thin that its outer surface is at the sametemperature as the outer surface of the spacesuit under it.Also assume the temperature of outer space is 0 K. You willget an equation that is very hard to solve in closed form,so you can solve it numerically with a graphing calculator,with software, or even by trial and error with a calculator.
127. The goal in this problem is to find the growth ofan ice layer as a function of time. Call the thickness ofthe ice layer L. (a) Derive an equation for dL/dt in termsof L , the temperature T above the ice, and the propertiesof ice (which you can leave in symbolic form instead ofsubstituting the numbers). (b) Solve this differentialequation assuming that at t = 0 , you have L = 0. If you
have studied differential equations, you will know atechnique for solving equations of this type: manipulate the


equation to get dL/dt multiplied by a (very simple) functionof L on one side, and integrate both sides with respect totime. Alternatively, you may be able to use your knowledgeof the derivatives of various functions to guess the solution,which has a simple dependence on t. (c) Will the watereventually freeze to the bottom of the flask?
128. As the very first rudiment of climatology, estimatethe temperature of Earth. Assume it is a perfect sphere andits temperature is uniform. Ignore the greenhouse effect.Thermal radiation from the Sun has an intensity (the “solar
constant” S) of about 1370W/m2 at the radius of Earth’s
orbit. (a) Assuming the Sun’s rays are parallel, what areamust S be multiplied by to get the total radiation interceptedby Earth? It will be easiest to answer in terms of Earth’sradius, R. (b) Assume that Earth reflects about 30% ofthe solar energy it intercepts. In other words, Earth hasan albedo with a value of A = 0.3 . In terms of S, A,
and R, what is the rate at which Earth absorbs energyfrom the Sun? (c) Find the temperature at which Earthradiates energy at the same rate. Assume that at the infraredwavelengths where it radiates, the emissivity e is 1. Doesyour result show that the greenhouse effect is important?(d) How does your answer depend on the the area of Earth?
129. Let’s stop ignoring the greenhouse effect andincorporate it into the previous problem in a very roughway. Assume the atmosphere is a single layer, a sphericalshell around Earth, with an emissivity e = 0.77 (chosen
simply to give the right answer) at infrared wavelengthsemitted by Earth and by the atmosphere. However, theatmosphere is transparent to the Sun’s radiation (that is,assume the radiation is at visible wavelengths with noinfrared), so the Sun’s radiation reaches the surface. Thegreenhouse effect comes from the difference between theatmosphere’s transmission of visible light and its ratherstrong absorption of infrared. Note that the atmosphere’sradius is not significantly different from Earth’s, but sincethe atmosphere is a layer above Earth, it emits radiationboth upward and downward, so it has twice Earth’s area.There are three radiative energy transfers in this problem:solar radiation absorbed by Earth’s surface; infraredradiation from the surface, which is absorbed by theatmosphere according to its emissivity; and infraredradiation from the atmosphere, half of which is absorbedby Earth and half of which goes out into space. Applythe method of the previous problem to get an equation forEarth’s surface and one for the atmosphere, and solve themfor the two unknown temperatures, surface and atmosphere.


Chapter 1 | Temperature and Heat 65




a. In terms of Earth’s radius, the constant σ , and
the unknown temperature Ts of the surface, what
is the power of the infrared radiation from thesurface?b. What is the power of Earth’s radiation absorbedby the atmosphere?c. In terms of the unknown temperature Te of the
atmosphere, what is the power radiated from theatmosphere?d. Write an equation that says the power of theradiation the atmosphere absorbs from Earth equalsthe power of the radiation it emits.e. Half of the power radiated by the atmospherehits Earth. Write an equation that says that thepower Earth absorbs from the atmosphere and theSun equals the power that it emits.f. Solve your two equations for the unknowntemperature of Earth.For steps that make this model less crude, see forexample the lectures(https://openstaxcollege.org/l/21paulgormlec) by Paul O’Gorman.


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2 | THE KINETIC THEORY OFGASES


Figure 2.1 A volcanic eruption releases tons of gas and dust into the atmosphere. Most of the gas is water vapor, but severalother gases are common, including greenhouse gases such as carbon dioxide and acidic pollutants such as sulfur dioxide.However, the emission of volcanic gas is not all bad: Many geologists believe that in the earliest stages of Earth’s formation,volcanic emissions formed the early atmosphere. (credit: modification of work by “Boaworm”/Wikimedia Commons)


Chapter Outline
2.1 Molecular Model of an Ideal Gas
2.2 Pressure, Temperature, and RMS Speed
2.3 Heat Capacity and Equipartition of Energy
2.4 Distribution of Molecular Speeds


Introduction
Gases are literally all around us—the air that we breathe is a mixture of gases. Other gases include those that make breadsand cakes soft, those that make drinks fizzy, and those that burn to heat many homes. Engines and refrigerators depend onthe behaviors of gases, as we will see in later chapters.
As we discussed in the preceding chapter, the study of heat and temperature is part of an area of physics known as
thermodynamics, in which we require a system to be macroscopic, that is, to consist of a huge number (such as 1023 ) of
molecules. We begin by considering some macroscopic properties of gases: volume, pressure, and temperature. The simplemodel of a hypothetical “ideal gas” describes these properties of a gas very accurately under many conditions. We movefrom the ideal gas model to a more widely applicable approximation, called the Van der Waals model.
To understand gases even better, we must also look at them on the microscopic scale of molecules. In gases, the moleculesinteract weakly, so the microscopic behavior of gases is relatively simple, and they serve as a good introduction to systemsof many molecules. The molecular model of gases is called the kinetic theory of gases and is one of the classic examples ofa molecular model that explains everyday behavior.


Chapter 2 | The Kinetic Theory of Gases 67




2.1 | Molecular Model of an Ideal Gas
Learning Objectives


By the end of this section, you will be able to:
• Apply the ideal gas law to situations involving the pressure, volume, temperature, and thenumber of molecules of a gas
• Use the unit of moles in relation to numbers of molecules, and molecular and macroscopicmasses
• Explain the ideal gas law in terms of moles rather than numbers of molecules
• Apply the van der Waals gas law to situations where the ideal gas law is inadequate


In this section, we explore the thermal behavior of gases. Our word “gas” comes from the Flemish word meaning “chaos,”first used for vapors by the seventeenth-century chemist J. B. van Helmont. The term was more appropriate than he knew,because gases consist of molecules moving and colliding with each other at random. This randomness makes the connectionbetween the microscopic and macroscopic domains simpler for gases than for liquids or solids.
How do gases differ from solids and liquids? Under ordinary conditions, such as those of the air around us, the differenceis that the molecules of gases are much farther apart than those of solids and liquids. Because the typical distances betweenmolecules are large compared to the size of a molecule, as illustrated in Figure 2.2, the forces between them are considerednegligible, except when they come into contact with each other during collisions. Also, at temperatures well above theboiling temperature, the motion of molecules is fast, and the gases expand rapidly to occupy all of the accessible volume.In contrast, in liquids and solids, molecules are closer together, and the behavior of molecules in liquids and solids is highlyconstrained by the molecules’ interactions with one another. The macroscopic properties of such substances depend stronglyon the forces between the molecules, and since many molecules are interacting, the resulting “many-body problems” can beextremely complicated (see Condensed Matter Physics (http://cnx.org/content/m58591/latest/) ).


Figure 2.2 Atoms and molecules in a gas are typically widelyseparated. Because the forces between them are quite weak atthese distances, the properties of a gas depend more on thenumber of atoms per unit volume and on temperature than onthe type of atom.
The Gas Laws
In the previous chapter, we saw one consequence of the large intermolecular spacing in gases: Gases are easily compressed.Table 1.2 shows that gases have larger coefficients of volume expansion than either solids or liquids. These largecoefficients mean that gases expand and contract very rapidly with temperature changes. We also saw (in the section onthermal expansion) that most gases expand at the same rate or have the same coefficient of volume expansion, β . This
raises a question: Why do all gases act in nearly the same way, when all the various liquids and solids have widely varyingexpansion rates?
To study how the pressure, temperature, and volume of a gas relate to one another, consider what happens when you pumpair into a deflated car tire. The tire’s volume first increases in direct proportion to the amount of air injected, without muchincrease in the tire pressure. Once the tire has expanded to nearly its full size, the tire’s walls limit its volume expansion.If we continue to pump air into the tire, the pressure increases. When the car is driven and the tires flex, their temperatureincreases, and therefore the pressure increases even further (Figure 2.3).


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Figure 2.3 (a) When air is pumped into a deflated tire, its volume first increases without much increase in pressure. (b) Whenthe tire is filled to a certain point, the tire walls resist further expansion, and the pressure increases with more air. (c) Once thetire is inflated, its pressure increases with temperature.


Figure 2.4 shows data from the experiments of Robert Boyle (1627–1691), illustrating what is now called Boyle’s law:At constant temperature and number of molecules, the absolute pressure of a gas and its volume are inversely proportional.(Recall from Fluid Mechanics (http://cnx.org/content/m58624/latest/) that the absolute pressure is the true pressureand the gauge pressure is the absolute pressure minus the ambient pressure, typically atmospheric pressure.) The graph inFigure 2.4 displays this relationship as an inverse proportionality of volume to pressure.


Figure 2.4 Robert Boyle and his assistant found that volume and pressure areinversely proportional. Here their data are plotted as V versus 1/p; the linearity of thegraph shows the inverse proportionality. The number shown as the volume is actuallythe height in inches of air in a cylindrical glass tube. The actual volume was thatheight multiplied by the cross-sectional area of the tube, which Boyle did not publish.The data are from Boyle’s book A Defence of the Doctrine Touching the Spring andWeight of the Air…, p. 60.[1]


Figure 2.5 shows experimental data illustrating what is called Charles’s law, after Jacques Charles (1746–1823). Charles’slaw states that at constant pressure and number of molecules, the volume of a gas is proportional to its absolute temperature.


1. http://bvpb.mcu.es/en/consulta/registro.cmd?id=406806


Chapter 2 | The Kinetic Theory of Gases 69




Figure 2.5 Experimental data showing that at constant pressure, volume isapproximately proportional to temperature. The best-fit line passes approximatelythrough the origin.[2]


Similar is Amonton’s or Gay-Lussac’s law, which states that at constant volume and number of molecules, the pressure isproportional to the temperature. That law is the basis of the constant-volume gas thermometer, discussed in the previouschapter. (The histories of these laws and the appropriate credit for them are more complicated than can be discussed here.)
It is known experimentally that for gases at low density (such that their molecules occupy a negligible fraction of thetotal volume) and at temperatures well above the boiling point, these proportionalities hold to a good approximation. Notsurprisingly, with the other quantities held constant, either pressure or volume is proportional to the number of molecules.More surprisingly, when the proportionalities are combined into a single equation, the constant of proportionality isindependent of the composition of the gas. The resulting equation for all gases applies in the limit of low density and hightemperature; it’s the same for oxygen as for helium or uranium hexafluoride. A gas at that limit is called an ideal gas; itobeys the ideal gas law, which is also called the equation of state of an ideal gas.


Ideal Gas Law
The ideal gas law states that


(2.1)pV = NkBT ,
where p is the absolute pressure of a gas, V is the volume it occupies, N is the number of molecules in the gas, and T isits absolute temperature.


The constant kB is called the Boltzmann constant in honor of the Austrian physicist Ludwig Boltzmann (1844–1906) and
has the value


kB = 1.38 × 10
−23 J/K.


The ideal gas law describes the behavior of any real gas when its density is low enough or its temperature high enough thatit is far from liquefaction. This encompasses many practical situations. In the next section, we’ll see why it’s independentof the type of gas.
In many situations, the ideal gas law is applied to a sample of gas with a constant number of molecules; for instance, thegas may be in a sealed container. If N is constant, then solving for N shows that pV /T is constant. We can write that fact ina convenient form:
2. http://chemed.chem.purdue.edu/genchem/history/charles.html


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(2.2)p1V1
T1


=
p2V2
T2


,


where the subscripts 1 and 2 refer to any two states of the gas at different times. Again, the temperature must be expressedin kelvin and the pressure must be absolute pressure, which is the sum of gauge pressure and atmospheric pressure.
Example 2.1


Calculating Pressure Changes Due to Temperature Changes
Suppose your bicycle tire is fully inflated, with an absolute pressure of 7.00 × 105 Pa (a gauge pressure of just
under 90.0 lb/in.2 ) at a temperature of 18.0 °C. What is the pressure after its temperature has risen to 35.0 °C
on a hot day? Assume there are no appreciable leaks or changes in volume.
Strategy
The pressure in the tire is changing only because of changes in temperature. We know the initial pressure
p0 = 7.00 × 10


5 Pa, the initial temperature T0 = 18.0 °C, and the final temperature Tf = 35.0 °C. We
must find the final pressure pf. Since the number of molecules is constant, we can use the equation


pf Vf
Tf


=
p0V0
T0


.


Since the volume is constant, Vf and V0 are the same and they divide out. Therefore,
pf
Tf


=
p0
T0


.


We can then rearrange this to solve for pf :
pf = p0


Tf
T0


,


where the temperature must be in kelvin.
Solution1. Convert temperatures from degrees Celsius to kelvin


T0 = (18.0 + 273)K = 291 K,


Tf = (35.0 + 273)K = 308 K.2. Substitute the known values into the equation,
pf = p0


Tf
T0


= 7.00 × 105 Pa ⎛⎝
308 K
291 K



⎠ = 7.41 × 10


5 Pa.


Significance
The final temperature is about 6% greater than the original temperature, so the final pressure is about 6% greater
as well. Note that absolute pressure (see Fluid Mechanics (http://cnx.org/content/m58624/latest/) ) andabsolute temperature (see Temperature and Heat) must be used in the ideal gas law.


Chapter 2 | The Kinetic Theory of Gases 71




Example 2.2
Calculating the Number of Molecules in a Cubic Meter of Gas
How many molecules are in a typical object, such as gas in a tire or water in a glass? This calculation can giveus an idea of how large N typically is. Let’s calculate the number of molecules in the air that a typical healthyyoung adult inhales in one breath, with a volume of 500 mL, at standard temperature and pressure (STP), whichis defined as 0 ºC and atmospheric pressure. (Our young adult is apparently outside in winter.)
Strategy
Because pressure, volume, and temperature are all specified, we can use the ideal gas law, pV = NkBT , to find
N.
Solution1. Identify the knowns.


T = 0 °C = 273 K, p = 1.01 × 105 Pa, V = 500 mL = 5 × 10−4 m3, kB = 1.38 × 10
−23 J/K


2. Substitute the known values into the equation and solve for N.
N =


pV
kBT


= (1.01 × 10
5 Pa) (5 × 10−4 m3)


(1.38 × 10−23 J/K) (273 K)
= 1.34 × 1022 molecules


Significance
N is huge, even in small volumes. For example, 1 cm3 of a gas at STP contains 2.68 × 1019 molecules. Once
again, note that our result for N is the same for all types of gases, including mixtures.
As we observed in the chapter on fluid mechanics, pascals are N/m2 , so Pa ·m3 = N ·m = J. Thus, our result
for N is dimensionless, a pure number that could be obtained by counting (in principle) rather than measuring.As it is the number of molecules, we put “molecules” after the number, keeping in mind that it is an aid tocommunication rather than a unit.


Moles and Avogadro’s Number
It is often convenient to measure the amount of substance with a unit on a more human scale than molecules. The SI unit forthis purpose was developed by the Italian scientist Amedeo Avogadro (1776–1856). (He worked from the hypothesis thatequal volumes of gas at equal pressure and temperature contain equal numbers of molecules, independent of the type of gas.As mentioned above, this hypothesis has been confirmed when the ideal gas approximation applies.) A mole (abbreviatedmol) is defined as the amount of any substance that contains as many molecules as there are atoms in exactly 12 grams(0.012 kg) of carbon-12. (Technically, we should say “formula units,” not “molecules,” but this distinction is irrelevant forour purposes.) The number of molecules in one mole is called Avogadro’s number (NA), and the value of Avogadro’s
number is now known to be


NA = 6.02 × 10
23 mol−1.


We can now write N = NAn , where n represents the number of moles of a substance.
Avogadro’s number relates the mass of an amount of substance in grams to the number of protons and neutrons in an atomor molecule (12 for a carbon-12 atom), which roughly determine its mass. It’s natural to define a unit of mass such that themass of an atom is approximately equal to its number of neutrons and protons. The unit of that kind accepted for use withthe SI is the unified atomic mass unit (u), also called the dalton. Specifically, a carbon-12 atom has a mass of exactly 12u, so that its molar mass M in grams per mole is numerically equal to the mass of one carbon-12 atom in u. That equalityholds for any substance. In other words, NA is not only the conversion from numbers of molecules to moles, but it is also
the conversion from u to grams: 6.02 × 1023 u = 1 g. See Figure 2.6.


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2.1


2.2


Figure 2.6 How big is a mole? On a macroscopic level, Avogadro’s number of table tennis ballswould cover Earth to a depth of about 40 km.


Now letting ms stand for the mass of a sample of a substance, we have ms = nM. Letting m stand for the mass of a
molecule, we have M = NAm.


Check Your Understanding The recommended daily amount of vitamin B3 or niacin, C6NH5O2,
for women who are not pregnant or nursing, is 14 mg. Find the number of molecules of niacin in that amount.


Check Your Understanding The density of air in a classroom (p = 1.00 atm and T = 20 °C) is
1.28 kg/m3 . At what pressure is the density 0.600 kg/m3 if the temperature is kept constant?


The Ideal Gas Law Restated using Moles
A very common expression of the ideal gas law uses the number of moles in a sample, n, rather than the number ofmolecules, N. We start from the ideal gas law,


pV = NkBT ,


and multiply and divide the right-hand side of the equation by Avogadro’s number NA. This gives us
pV = N


NA
NA kBT .


Note that n = N/NA is the number of moles. We define the universal gas constant as R = NA kB, and obtain the ideal
gas law in terms of moles.


Ideal Gas Law (in terms of moles)
In terms of number of moles n, the ideal gas law is written as


(2.3)pV = nRT .
In SI units,


R = NA kB =

⎝6.02 × 10


23 mol−1⎞⎠

⎝1.38 × 10


−23 J
K

⎠ = 8.31


J
mol · K


.


In other units,
R = 1.99 cal


mol · K
= 0.0821L · atm


mol · K
.


You can use whichever value of R is most convenient for a particular problem.


Chapter 2 | The Kinetic Theory of Gases 73




2.3


Example 2.3
Density of Air at STP and in a Hot Air Balloon
Calculate the density of dry air (a) under standard conditions and (b) in a hot air balloon at a temperature of
120 ºC . Dry air is approximately 78%N2, 21%O2, and 1%Ar .
Strategy and Solutiona. We are asked to find the density, or mass per cubic meter. We can begin by finding the molar mass. If wehave a hundred molecules, of which 78 are nitrogen, 21 are oxygen, and 1 is argon, the average molecular


mass is 78mN2 + 21mO2 + mAr
100


, or the mass of each constituent multiplied by its percentage. The
same applies to the molar mass, which therefore is


M = 0.78MN2 + 0.21MO2 + 0.01MAr = 29.0 g/mol.


Now we can find the number of moles per cubic meter. We use the ideal gas law in terms of moles,
pV = nRT , with p = 1.00 atm , T = 273 K , V = 1m3 , and R = 8.31 J/mol · K . The most
convenient choice for R in this case is R = 8.31 J/mol · K because the known quantities are in SI units:


n =
pV
RT


= (1.01 × 10
5 Pa) (1 m3)


(8.31 J/mol · K) (273 K)
= 44.5 mol.


Then, the mass ms of that air is
ms = nM = (44.5 mol)(29.0 g/mol) = 1290 g = 1.29 kg.


Finally the density of air at STP is
ρ = ms


V
=


1.29 kg


1 m3
= 1.29 kg/m3.


b. The air pressure inside the balloon is still 1 atm because the bottom of the balloon is open to theatmosphere. The calculation is the same except that we use a temperature of 120 ºC , which is 393 K.
We can repeat the calculation in (a), or simply observe that the density is proportional to the number ofmoles, which is inversely proportional to the temperature. Then using the subscripts 1 for air at STP and2 for the hot air, we have


ρ2 =
T1
T2


ρ1 =
273 K
393 K


(1.29 kg/m3) = 0.896 kg/m3.


Significance
Using the methods of Archimedes’ Principle and Buoyancy (http://cnx.org/content/m58356/latest/)
, we can find that the net force on 2200 m3 of air at 120 ºC is
Fb − Fg = ρatmosphereVg − ρhot airVg = 8.49 × 10


3 N, or enough to lift about 867 kg. The mass density and
molar density of air at STP, found above, are often useful numbers. From the molar density, we can easilydetermine another useful number, the volume of a mole of any ideal gas at STP, which is 22.4 L.


Check Your Understanding Liquids and solids have densities on the order of 1000 times greater thangases. Explain how this implies that the distances between molecules in gases are on the order of 10 timesgreater than the size of their molecules.
The ideal gas law is closely related to energy: The units on both sides of the equation are joules. The right-hand side ofthe ideal gas law equation is NkBT . This term is roughly the total translational kinetic energy (which, when discussing
gases, refers to the energy of translation of a molecule, not that of vibration of its atoms or rotation) of N molecules at


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an absolute temperature T, as we will see formally in the next section. The left-hand side of the ideal gas law equation ispV. As mentioned in the example on the number of molecules in an ideal gas, pressure multiplied by volume has units ofenergy. The energy of a gas can be changed when the gas does work as it increases in volume, something we explored inthe preceding chapter, and the amount of work is related to the pressure. This is the process that occurs in gasoline or steamengines and turbines, as we’ll see in the next chapter.
Problem-Solving Strategy: The Ideal Gas Law
Step 1. Examine the situation to determine that an ideal gas is involved. Most gases are nearly ideal unless they areclose to the boiling point or at pressures far above atmospheric pressure.
Step 2. Make a list of what quantities are given or can be inferred from the problem as stated (identify the knownquantities).
Step 3. Identify exactly what needs to be determined in the problem (identify the unknown quantities). A written listis useful.
Step 4. Determine whether the number of molecules or the number of moles is known or asked for to decide whether touse the ideal gas law as pV = NkBT , where N is the number of molecules, or pV = nRT , where n is the number
of moles.
Step 5. Convert known values into proper SI units (K for temperature, Pa for pressure, m3 for volume, molecules for
N, and moles for n). If the units of the knowns are consistent with one of the non-SI values of R, you can leave them inthose units. Be sure to use absolute temperature and absolute pressure.
Step 6. Solve the ideal gas law for the quantity to be determined (the unknown quantity). You may need to take a ratioof final states to initial states to eliminate the unknown quantities that are kept fixed.
Step 7. Substitute the known quantities, along with their units, into the appropriate equation and obtain numericalsolutions complete with units.
Step 8. Check the answer to see if it is reasonable: Does it make sense?


The Van der Waals Equation of State
We have repeatedly noted that the ideal gas law is an approximation. How can it be improved upon? The van der Waalsequation of state (named after the Dutch physicist Johannes van der Waals, 1837−1923) improves it by taking into accounttwo factors. First, the attractive forces between molecules, which are stronger at higher density and reduce the pressure,are taken into account by adding to the pressure a term equal to the square of the molar density multiplied by a positivecoefficient a. Second, the volume of the molecules is represented by a positive constant b, which can be thought of as thevolume of a mole of molecules. This is subtracted from the total volume to give the remaining volume that the moleculescan move in. The constants a and b are determined experimentally for each gas. The resulting equation is


(2.4)⎡

⎢p + a⎛⎝


n
V



2⎤



⎥(V − nb) = nRT .


In the limit of low density (small n), the a and b terms are negligible, and we have the ideal gas law, as we should for lowdensity. On the other hand, if V − nb is small, meaning that the molecules are very close together, the pressure must be
higher to give the same nRT, as we would expect in the situation of a highly compressed gas. However, the increase in
pressure is less than that argument would suggest, because at high density the (n/V)2 term is significant. Since it’s positive,
it causes a lower pressure to give the same nRT.
The van der Waals equation of state works well for most gases under a wide variety of conditions. As we’ll see in the nextmodule, it even predicts the gas-liquid transition.
pV Diagrams
We can examine aspects of the behavior of a substance by plotting a pV diagram, which is a graph of pressure versus


Chapter 2 | The Kinetic Theory of Gases 75




volume. When the substance behaves like an ideal gas, the ideal gas law pV = nRT describes the relationship between its
pressure and volume. On a pV diagram, it’s common to plot an isotherm, which is a curve showing p as a function of V withthe number of molecules and the temperature fixed. Then, for an ideal gas, pV = constant. For example, the volume of
the gas decreases as the pressure increases. The resulting graph is a hyperbola.
However, if we assume the van der Waals equation of state, the isotherms become more interesting, as shown in Figure 2.7.At high temperatures, the curves are approximately hyperbolas, representing approximately ideal behavior at various fixedtemperatures. At lower temperatures, the curves look less and less like hyperbolas—that is, the gas is not behaving ideally.There is a critical temperature Tc at which the curve has a point with zero slope. Below that temperature, the curves do
not decrease monotonically; instead, they each have a “hump,” meaning that for a certain range of volume, increasing thevolume increases the pressure.


Figure 2.7 pV diagram for a Van der Waals gas at various temperatures. The red curvesare calculated at temperatures above the critical temperature and the blue curves attemperatures below it. The blue curves have an oscillation in which volume (V) increaseswith increasing temperature (T), an impossible situation, so they must be corrected as inFigure 2.8. (credit: “Eman”/Wikimedia Commons)


Such behavior would be completely unphysical. Instead, the curves are understood as describing a liquid-gas phasetransition. The oscillating part of the curve is replaced by a horizontal line, showing that as the volume increases at constanttemperature, the pressure stays constant. That behavior corresponds to boiling and condensation; when a substance is at itsboiling temperature for a particular pressure, it can increase in volume as some of the liquid turns to gas, or decrease assome of the gas turns to liquid, without any change in temperature or pressure.
Figure 2.8 shows similar isotherms that are more realistic than those based on the van der Waals equation. The steep partsof the curves to the left of the transition region show the liquid phase, which is almost incompressible—a slight decrease involume requires a large increase in pressure. The flat parts show the liquid-gas transition; the blue regions that they definerepresent combinations of pressure and volume where liquid and gas can coexist.


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Figure 2.8 pV diagrams. (a) Each curve (isotherm) represents the relationship between p and V at a fixed temperature; theupper curves are at higher temperatures. The lower curves are not hyperbolas because the gas is no longer an ideal gas. (b) Anexpanded portion of the pV diagram for low temperatures, where the phase can change from a gas to a liquid. The term “vapor”refers to the gas phase when it exists at a temperature below the boiling temperature.


The isotherms above Tc do not go through the liquid-gas transition. Therefore, liquid cannot exist above that temperature,
which is the critical temperature (described in the chapter on temperature and heat). At sufficiently low pressure above thattemperature, the gas has the density of a liquid but will not condense; the gas is said to be supercritical. At higher pressure,it is solid. Carbon dioxide, for example, has no liquid phase at a temperature above 31.0 ºC . The critical pressure is the
maximum pressure at which the liquid can exist. The point on the pV diagram at the critical pressure and temperature isthe critical point (which you learned about in the chapter on temperature and heat). Table 2.1 lists representative criticaltemperatures and pressures.


Substance Critical temperature Critical pressure
K °C Pa atm


Water 647.4 374.3 22.12 × 106 219.0
Sulfur dioxide 430.7 157.6 7.88 × 106 78.0
Ammonia 405.5 132.4 11.28 × 106 111.7
Carbon dioxide 304.2 31.1 7.39 × 106 73.2
Oxygen 154.8 –118.4 5.08 × 106 50.3
Nitrogen 126.2 –146.9 3.39 × 106 33.6
Hydrogen 33.3 –239.9 1.30 × 106 12.9
Helium 5.3 –267.9 0.229 × 106 2.27


Table 2.1 Critical Temperatures and Pressures for VariousSubstances


Chapter 2 | The Kinetic Theory of Gases 77




2.2 | Pressure, Temperature, and RMS Speed
Learning Objectives


By the end of this section, you will be able to:
• Explain the relations between microscopic and macroscopic quantities in a gas
• Solve problems involving mixtures of gases
• Solve problems involving the distance and time between a gas molecule’s collisions


We have examined pressure and temperature based on their macroscopic definitions. Pressure is the force divided by thearea on which the force is exerted, and temperature is measured with a thermometer. We can gain a better understanding ofpressure and temperature from the kinetic theory of gases, the theory that relates the macroscopic properties of gases to themotion of the molecules they consist of. First, we make two assumptions about molecules in an ideal gas.
1. There is a very large number N of molecules, all identical and each having mass m.
2. The molecules obey Newton’s laws and are in continuous motion, which is random and isotropic, that is, the samein all directions.


To derive the ideal gas law and the connection between microscopic quantities such as the energy of a typical molecule andmacroscopic quantities such as temperature, we analyze a sample of an ideal gas in a rigid container, about which we maketwo further assumptions:
3. The molecules are much smaller than the average distance between them, so their total volume is much less thanthat of their container (which has volume V). In other words, we take the Van der Waals constant b, the volume of amole of gas molecules, to be negligible compared to the volume of a mole of gas in the container.
4. The molecules make perfectly elastic collisions with the walls of the container and with each other. Other forceson them, including gravity and the attractions represented by the Van der Waals constant a, are negligible (as isnecessary for the assumption of isotropy).


The collisions between molecules do not appear in the derivation of the ideal gas law. They do not disturb the derivationeither, since collisions between molecules moving with random velocities give new random velocities. Furthermore, if thevelocities of gas molecules in a container are initially not random and isotropic, molecular collisions are what make themrandom and isotropic.
We make still further assumptions that simplify the calculations but do not affect the result. First, we let the container be arectangular box. Second, we begin by considering monatomic gases, those whose molecules consist of single atoms, suchas helium. Then, we can assume that the atoms have no energy except their translational kinetic energy; for instance, theyhave neither rotational nor vibrational energy. (Later, we discuss the validity of this assumption for real monatomic gasesand dispense with it to consider diatomic and polyatomic gases.)
Figure 2.9 shows a collision of a gas molecule with the wall of a container, so that it exerts a force on the wall (byNewton’s third law). These collisions are the source of pressure in a gas. As the number of molecules increases, the numberof collisions, and thus the pressure, increases. Similarly, if the average velocity of the molecules is higher, the gas pressureis higher.


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Figure 2.9 When a molecule collides with a rigid wall, thecomponent of its momentum perpendicular to the wall isreversed. A force is thus exerted on the wall, creating pressure.


In a sample of gas in a container, the randomness of the molecular motion causes the number of collisions of moleculeswith any part of the wall in a given time to fluctuate. However, because a huge number of molecules collide with the wallin a short time, the number of collisions on the scales of time and space we measure fluctuates by only a tiny, usuallyunobservable fraction from the average. We can compare this situation to that of a casino, where the outcomes of the betsare random and the casino’s takings fluctuate by the minute and the hour. However, over long times such as a year, thecasino’s takings are very close to the averages expected from the odds. A tank of gas has enormously more molecules thana casino has bettors in a year, and the molecules make enormously more collisions in a second than a casino has bets.
A calculation of the average force exerted by molecules on the walls of the box leads us to the ideal gas law and to theconnection between temperature and molecular kinetic energy. (In fact, we will take two averages: one over time to getthe average force exerted by one molecule with a given velocity, and then another average over molecules with differentvelocities.) This approach was developed by Daniel Bernoulli (1700–1782), who is best known in physics for his work onfluid flow (hydrodynamics). Remarkably, Bernoulli did this work before Dalton established the view of matter as consistingof atoms.
Figure 2.10 shows a container full of gas and an expanded view of an elastic collision of a gas molecule with a wall ofthe container, broken down into components. We have assumed that a molecule is small compared with the separation ofmolecules in the gas, and that its interaction with other molecules can be ignored. Under these conditions, the ideal gas lawis experimentally valid. Because we have also assumed the wall is rigid and the particles are points, the collision is elastic(by conservation of energy—there’s nowhere for a particle’s kinetic energy to go). Therefore, the molecule’s kinetic energyremains constant, and hence, its speed and the magnitude of its momentum remain constant as well. This assumption is notalways valid, but the results in the rest of this module are also obtained in models that let the molecules exchange energyand momentum with the wall.


Chapter 2 | The Kinetic Theory of Gases 79




Figure 2.10 Gas in a box exerts an outward pressure on itswalls. A molecule colliding with a rigid wall has its velocity andmomentum in the x-direction reversed. This direction isperpendicular to the wall. The components of its velocitymomentum in the y- and z-directions are not changed, whichmeans there is no force parallel to the wall.


If the molecule’s velocity changes in the x-direction, its momentum changes from −mvx to +mvx. Thus, its change in
momentum is Δmv = + mvx − (−mvx) = 2mvx. According to the impulse-momentum theorem given in the chapter on
linear momentum and collisions, the force exerted on the ith molecule, where i labels the molecules from 1 to N, is given by


Fi =
Δpi
Δt


=
2mvix
Δt


.


(In this equation alone, p represents momentum, not pressure.) There is no force between the wall and the molecule exceptwhile the molecule is touching the wall. During the short time of the collision, the force between the molecule and wallis relatively large, but that is not the force we are looking for. We are looking for the average force, so we take Δt to be
the average time between collisions of the given molecule with this wall, which is the time in which we expect to find onecollision. Let l represent the length of the box in the x-direction. Then Δt is the time the molecule would take to go across
the box and back, a distance 2l, at a speed of vx. Thus Δt = 2l/vx, and the expression for the force becomes


Fi =
2mvix
2l/vix


=
mvix


2


l
.


This force is due to one molecule. To find the total force on the wall, F, we need to add the contributions of all N molecules:
F = ∑


i = 1


N


Fi = ∑
i = 1


N mvix
2


l
= m


l

i = 1


N


vix
2 .


We now use the definition of the average, which we denote with a bar, to find the force:
F = Nm


l





⎜1
N

i = 1


N


vix
2



⎟ = Nmvx


2


l
.


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We want the force in terms of the speed v, rather than the x-component of the velocity. Note that the total velocity squaredis the sum of the squares of its components, so that
v2


= vx
2


+ vy
2


+ vz
2

.


With the assumption of isotropy, the three averages on the right side are equal, so
v2


= 3vix
2

.


Substituting this into the expression for F gives
F = Nmv


2


3l
.


The pressure is F/A, so we obtain
p = F


A
= Nmv


2


3Al
= Nmv


2


3V
,


where we used V = Al for the volume. This gives the important result


(2.5)pV = 1
3
Nmv2



.


Combining this equation with pV = NkBT gives
1
3
Nmv2



= NkBT .


We can get the average kinetic energy of a molecule, 1
2
mv2


– , from the left-hand side of the equation by dividing out N and
multiplying by 3/2.


Average Kinetic Energy per Molecule
The average kinetic energy of a molecule is directly proportional to its absolute temperature:


(2.6)K– = 1
2
mv2



= 3


2
kBT .


The equation K– = 3
2
kBT is the average kinetic energy per molecule. Note in particular that nothing in this equation


depends on the molecular mass (or any other property) of the gas, the pressure, or anything but the temperature. If samplesof helium and xenon gas, with very different molecular masses, are at the same temperature, the molecules have the sameaverage kinetic energy.
The internal energy of a thermodynamic system is the sum of the mechanical energies of all of the molecules in it. We cannow give an equation for the internal energy of a monatomic ideal gas. In such a gas, the molecules’ only energy is their
translational kinetic energy. Therefore, denoting the internal energy by Eint, we simply have Eint = NK–, or


(2.7)Eint = 32 NkBT .


Often we would like to use this equation in terms of moles:
Eint =


3
2
nRT .


Chapter 2 | The Kinetic Theory of Gases 81




We can solve K– = 1
2
mv2



= 3


2
kBT for a typical speed of a molecule in an ideal gas in terms of temperature to determine


what is known as the root-mean-square (rms) speed of a molecule.
RMS Speed of a Molecule
The root-mean-square (rms) speed of a molecule, or the square root of the average of the square of the speed v2– , is


(2.8)
vrms = v


2


=
3kBT
m .


The rms speed is not the average or the most likely speed of molecules, as we will see in Distribution of MolecularSpeeds, but it provides an easily calculated estimate of the molecules’ speed that is related to their kinetic energy. Againwe can write this equation in terms of the gas constant R and the molar mass M in kg/mol:


(2.9)vrms = 3 RTM .


We digress for a moment to answer a question that may have occurred to you: When we apply the model to atoms instead oftheoretical point particles, does rotational kinetic energy change our results? To answer this question, we have to appeal toquantum mechanics. In quantum mechanics, rotational kinetic energy cannot take on just any value; it’s limited to a discreteset of values, and the smallest value is inversely proportional to the rotational inertia. The rotational inertia of an atom is
tiny because almost all of its mass is in the nucleus, which typically has a radius less than 10−14 m . Thus the minimum
rotational energy of an atom is much more than 1


2
kBT for any attainable temperature, and the energy available is not


enough to make an atom rotate. We will return to this point when discussing diatomic and polyatomic gases in the nextsection.
Example 2.4


Calculating Kinetic Energy and Speed of a Gas Molecule
(a) What is the average kinetic energy of a gas molecule at 20.0 ºC (room temperature)? (b) Find the rms speed
of a nitrogen molecule ⎛⎝N2⎞⎠ at this temperature.
Strategy
(a) The known in the equation for the average kinetic energy is the temperature:


K


= 1
2
mv2



= 3


2
kBT .


Before substituting values into this equation, we must convert the given temperature into kelvin:
T = (20.0 + 273) K = 293 K. We can find the rms speed of a nitrogen molecule by using the equation


vrms = v
2


=
3kBT
m ,


but we must first find the mass of a nitrogen molecule. Obtaining the molar mass of nitrogen N2 from the
periodic table, we find


m = M
NA


=
2 (14.0067) × 10−3 kg/mol)


6.02 × 1023 mol-1
= 4.65 × 10−26 kg.


Solutiona. The temperature alone is sufficient for us to find the average translational kinetic energy. Substituting thetemperature into the translational kinetic energy equation gives


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K


= 3
2
kBT =


3
2
(1.38 × 10−23 J/K)(293 K) = 6.07 × 10−21 J.


b. Substituting this mass and the value for kB into the equation for vrms yields
vrms =


3kBT
m =


3(1.38 × 10−23 J/K)(293 K)


4.65 × 10−26 kg
= 511 m/s.


Significance
Note that the average kinetic energy of the molecule is independent of the type of molecule. The averagetranslational kinetic energy depends only on absolute temperature. The kinetic energy is very small compared tomacroscopic energies, so that we do not feel when an air molecule is hitting our skin. On the other hand, it ismuch greater than the typical difference in gravitational potential energy when a molecule moves from, say, thetop to the bottom of a room, so our neglect of gravitation is justified in typical real-world situations. The rmsspeed of the nitrogen molecule is surprisingly large. These large molecular velocities do not yield macroscopicmovement of air, since the molecules move in all directions with equal likelihood. The mean free path (thedistance a molecule moves on average between collisions, discussed a bit later in this section) of molecules in airis very small, so the molecules move rapidly but do not get very far in a second. The high value for rms speedis reflected in the speed of sound, which is about 340 m/s at room temperature. The higher the rms speed ofair molecules, the faster sound vibrations can be transferred through the air. The speed of sound increases withtemperature and is greater in gases with small molecular masses, such as helium (see Figure 2.11).


Figure 2.11 (a) In an ordinary gas, so many molecules move so fast that they collidebillions of times every second. (b) Individual molecules do not move very far in asmall amount of time, but disturbances like sound waves are transmitted at speedsrelated to the molecular speeds.


Example 2.5
Calculating Temperature: Escape Velocity of Helium Atoms
To escape Earth’s gravity, an object near the top of the atmosphere (at an altitude of 100 km) must travel awayfrom Earth at 11.1 km/s. This speed is called the escape velocity. At what temperature would helium atoms havean rms speed equal to the escape velocity?
Strategy
Identify the knowns and unknowns and determine which equations to use to solve the problem.
Solution1. Identify the knowns: v is the escape velocity, 11.1 km/s.


Chapter 2 | The Kinetic Theory of Gases 83




2. Identify the unknowns: We need to solve for temperature, T. We also need to solve for the mass m of thehelium atom.
3. Determine which equations are needed.


◦ To get the mass m of the helium atom, we can use information from the periodic table:
m = M


NA
.


◦ To solve for temperature T, we can rearrange
1
2
mv2



= 3


2
kBT


to yield
T = mv


2


3kB
.


4. Substitute the known values into the equations and solve for the unknowns,
m = M


NA
=


4.0026 × 10−3 kg/mol


6.02 × 1023 mol
= 6.65 × 10−27 kg


and
T =


(6.65 × 10−27 kg⎞⎠

⎝11.1 × 10


3 m/s⎞⎠
2


3 (1.38 × 10−23 J/K)
= 1.98 × 104 K.


Significance
This temperature is much higher than atmospheric temperature, which is approximately 250 K
(−25 °C or − 10 °F⎞⎠ at high elevation. Very few helium atoms are left in the atmosphere, but many were present
when the atmosphere was formed, and more are always being created by radioactive decay (see the chapter onnuclear physics). The reason for the loss of helium atoms is that a small number of helium atoms have speedshigher than Earth’s escape velocity even at normal temperatures. The speed of a helium atom changes from onecollision to the next, so that at any instant, there is a small but nonzero chance that the atom’s speed is greater thanthe escape velocity. The chance is high enough that over the lifetime of Earth, almost all the helium atoms thathave been in the atmosphere have reached escape velocity at high altitudes and escaped from Earth’s gravitationalpull. Heavier molecules, such as oxygen, nitrogen, and water, have smaller rms speeds, and so it is much lesslikely that any of them will have speeds greater than the escape velocity. In fact, the likelihood is so small thatbillions of years are required to lose significant amounts of heavier molecules from the atmosphere. Figure 2.12shows the effect of a lack of an atmosphere on the Moon. Because the gravitational pull of the Moon is muchweaker, it has lost almost its entire atmosphere. The atmospheres of Earth and other bodies are compared in thischapter’s exercises.


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2.4


Figure 2.12 This photograph of Apollo 17 CommanderEugene Cernan driving the lunar rover on the Moon in 1972looks as though it was taken at night with a large spotlight. Infact, the light is coming from the Sun. Because the accelerationdue to gravity on the Moon is so low (about 1/6 that of Earth),the Moon’s escape velocity is much smaller. As a result, gasmolecules escape very easily from the Moon, leaving it withvirtually no atmosphere. Even during the daytime, the sky isblack because there is no gas to scatter sunlight. (credit:Harrison H. Schmitt/NASA)


Check Your Understanding If you consider a very small object, such as a grain of pollen, in a gas, thenthe number of molecules striking its surface would also be relatively small. Would you expect the grain ofpollen to experience any fluctuations in pressure due to statistical fluctuations in the number of gas moleculesstriking it in a given amount of time?


Vapor Pressure, Partial Pressure, and Dalton’s Law
The pressure a gas would create if it occupied the total volume available is called the gas’s partial pressure. If two or moregases are mixed, they will come to thermal equilibrium as a result of collisions between molecules; the process is analogousto heat conduction as described in the chapter on temperature and heat. As we have seen from kinetic theory, when thegases have the same temperature, their molecules have the same average kinetic energy. Thus, each gas obeys the idealgas law separately and exerts the same pressure on the walls of a container that it would if it were alone. Therefore, in amixture of gases, the total pressure is the sum of partial pressures of the component gases, assuming ideal gas behavior andno chemical reactions between the components. This law is known as Dalton’s law of partial pressures, after the Englishscientist John Dalton (1766–1844) who proposed it. Dalton’s law is consistent with the fact that pressures add according toPascal’s principle.
In a mixture of ideal gases in thermal equilibrium, the number of molecules of each gas is proportional to its partial pressure.This result follows from applying the ideal gas law to each in the form p/n = RT /V . Because the right-hand side is the
same for any gas at a given temperature in a container of a given volume, the left-hand side is the same as well.


• Partial pressure is the pressure a gas would create if it existed alone.
• Dalton’s law states that the total pressure is the sum of the partial pressures of all of the gases present.
• For any two gases (labeled 1 and 2) in equilibrium in a container, p1n1 = p2n2 .


Chapter 2 | The Kinetic Theory of Gases 85




An important application of partial pressure is that, in chemistry, it functions as the concentration of a gas in determining therate of a reaction. Here, we mention only that the partial pressure of oxygen in a person’s lungs is crucial to life and health.Breathing air that has a partial pressure of oxygen below 0.16 atm can impair coordination and judgment, particularly inpeople not acclimated to a high elevation. Lower partial pressures of O2 have more serious effects; partial pressures below
0.06 atm can be quickly fatal, and permanent damage is likely even if the person is rescued. However, the sensation ofneeding to breathe, as when holding one’s breath, is caused much more by high concentrations of carbon dioxide in theblood than by low concentrations of oxygen. Thus, if a small room or closet is filled with air having a low concentration ofoxygen, perhaps because a leaking cylinder of some compressed gas is stored there, a person will not feel any “choking”sensation and may go into convulsions or lose consciousness without noticing anything wrong. Safety engineers giveconsiderable attention to this danger.
Another important application of partial pressure is vapor pressure, which is the partial pressure of a vapor at which it isin equilibrium with the liquid (or solid, in the case of sublimation) phase of the same substance. At any temperature, thepartial pressure of the water in the air cannot exceed the vapor pressure of the water at that temperature, because wheneverthe partial pressure reaches the vapor pressure, water condenses out of the air. Dew is an example of this condensation. Thetemperature at which condensation occurs for a sample of air is called the dew point. It is easily measured by slowly coolinga metal ball; the dew point is the temperature at which condensation first appears on the ball.
The vapor pressures of water at some temperatures of interest for meteorology are given in Table 2.2.


T (°C) Vapor Pressure (Pa)
0 610.5
3 757.9
5 872.3
8 1073
10 1228
13 1497
15 1705
18 2063
20 2338
23 2809
25 3167
30 4243
35 5623
40 7376


Table 2.2 Vapor Pressure of Water atVarious Temperatures
The relative humidity (R.H.) at a temperature T is defined by


R.H. =
Partial pressure of water vapor at T


Vapor pressure of water at T
× 100%.


A relative humidity of 100% means that the partial pressure of water is equal to the vapor pressure; in other words, the air
is saturated with water.
Example 2.6


Calculating Relative Humidity
What is the relative humidity when the air temperature is 25 ºC and the dew point is 15 ºC ?


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Strategy
We simply look up the vapor pressure at the given temperature and that at the dew point and find the ratio.
Solution


R.H. =
Partial pressure of water vapor at 15 °C
Partial pressure of water vapor at 25 °C


× 100% = 1705 Pa
3167 Pa


× 100% = 53.8%.


Significance
R.H. is important to our comfort. The value of 53.8% is within the range of 40% to 60% recommended for
comfort indoors.
As noted in the chapter on temperature and heat, the temperature seldom falls below the dew point, because whenit reaches the dew point or frost point, water condenses and releases a relatively large amount of latent heat ofvaporization.


Mean Free Path and Mean Free Time
We now consider collisions explicitly. The usual first step (which is all we’ll take) is to calculate the mean free path, λ,
the average distance a molecule travels between collisions with other molecules, and the mean free time τ , the average
time between the collisions of a molecule. If we assume all the molecules are spheres with a radius r, then a molecule willcollide with another if their centers are within a distance 2r of each other. For a given particle, we say that the area of a
circle with that radius, 4πr2 , is the “cross-section” for collisions. As the particle moves, it traces a cylinder with that cross-
sectional area. The mean free path is the length λ such that the expected number of other molecules in a cylinder of length
λ and cross-section 4πr2 is 1. If we temporarily ignore the motion of the molecules other than the one we’re looking at,
the expected number is the number density of molecules, N/V, times the volume, and the volume is 4πr2 λ , so we have
(N/V)4πr2 λ = 1, or


λ = V
4πr2N


.


Taking the motion of all the molecules into account makes the calculation much harder, but the only change is a factor of
2. The result is


(2.10)λ = V
4 2πr2N


.


In an ideal gas, we can substitute V /N = kBT /p to obtain


(2.11)
λ =


kBT


4 2πr2 p
.


The mean free time τ is simply the mean free path divided by a typical speed, and the usual choice is the rms speed. Then


(2.12)
τ =


kBT


4 2πr2 pvrms
.


Chapter 2 | The Kinetic Theory of Gases 87




2.5


Example 2.7
Calculating Mean Free Time
Find the mean free time for argon atoms (M = 39.9 g/mol⎞⎠ at a temperature of 0 °C and a pressure of 1.00 atm.
Take the radius of an argon atom to be 1.70 × 10−10 m.
Solution1. Identify the knowns and convert into SI units. We know the molar mass is 0.0399 kg/mol, the temperature


is 273 K, the pressure is 1.01 × 105 Pa, and the radius is 1.70 × 10−10 m.
2. Find the rms speed: vrms = 3RTM = 413 ms .
3. Substitute into the equation for the mean free time:


τ =
kBT


4 2πr2 pvrms
= (1.38 × 10


−23 J/K) (273 K)


4 2π(1.70 × 10−10 m)2(1.01 × 105 Pa)(413 m/s)
= 1.76 × 10−10 s.


Significance
We can hardly compare this result with our intuition about gas molecules, but it gives us a picture of moleculescolliding with extremely high frequency.


Check Your Understanding Which has a longer mean free path, liquid water or water vapor in the air?


2.3 | Heat Capacity and Equipartition of Energy
Learning Objectives


By the end of this section, you will be able to:
• Solve problems involving heat transfer to and from ideal monatomic gases whose volumes areheld constant
• Solve similar problems for non-monatomic ideal gases based on the number of degrees offreedom of a molecule
• Estimate the heat capacities of metals using a model based on degrees of freedom


In the chapter on temperature and heat, we defined the specific heat capacity with the equation Q = mcΔT , or
c = (1/m)Q/ΔT . However, the properties of an ideal gas depend directly on the number of moles in a sample, so here we
define specific heat capacity in terms of the number of moles, not the mass. Furthermore, when talking about solids andliquids, we ignored any changes in volume and pressure with changes in temperature—a good approximation for solids andliquids, but for gases, we have to make some condition on volume or pressure changes. Here, we focus on the heat capacitywith the volume held constant. We can calculate it for an ideal gas.
Heat Capacity of an Ideal Monatomic Gas at Constant Volume
We define the molar heat capacity at constant volume CV as


CV =
1
n


Q
ΔT


, with V held constant.


This is often expressed in the form


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2.6


(2.13)Q = nCVΔT .


If the volume does not change, there is no overall displacement, so no work is done, and the only change in internalenergy is due to the heat flow ΔEint = Q. (This statement is discussed further in the next chapter.) We use the equation
Eint = 3nRT /2 to write ΔEint = 3nRΔT /2 and substitute ΔE for Q to find Q = 3nRΔT /2 , which gives the following
simple result for an ideal monatomic gas:


CV =
3
2
R.


It is independent of temperature, which justifies our use of finite differences instead of a derivative. This formula agreeswell with experimental results.
In the next chapter we discuss the molar specific heat at constant pressure C p, which is always greater than CV.
Example 2.8


Calculating Temperature
A sample of 0.125 kg of xenon is contained in a rigid metal cylinder, big enough that the xenon can be modeledas an ideal gas, at a temperature of 20.0 °C . The cylinder is moved outside on a hot summer day. As the xenon
comes into equilibrium by reaching the temperature of its surroundings, 180 J of heat are conducted to it throughthe cylinder walls. What is the equilibrium temperature? Ignore the expansion of the metal cylinder.
Solution1. Identify the knowns: We know the initial temperature T1 is 20.0 °C , the heat Q is 180 J, and the mass


m of the xenon is 0.125 kg.
2. Identify the unknown. We need the final temperature, so we’ll need ΔT .
3. Determine which equations are needed. Because xenon gas is monatomic, we can use Q = 3nRΔT /2.


Then we need the number of moles, n = m/M.
4. Substitute the known values into the equations and solve for the unknowns.The molar mass of xenon is 131.3 g, so we obtain


n =
125 g


131.3 g/mol
= 0.952 mol,


ΔT = 2Q
3nR


= 2(180 J)
3(0.952 mol)(8.31 J/mol · °C)


= 15.2 °C.


Therefore, the final temperature is 35.2 °C . The problem could equally well be solved in kelvin; as a
kelvin is the same size as a degree Celsius of temperature change, you would get ΔT = 15.2 K.


Significance
The heating of an ideal or almost ideal gas at constant volume is important in car engines and many other practicalsystems.


Check Your Understanding Suppose 2 moles of helium gas at 200 K are mixed with 2 moles of kryptongas at 400 K in a calorimeter. What is the final temperature?


We would like to generalize our results to ideal gases with more than one atom per molecule. In such systems, the moleculescan have other forms of energy beside translational kinetic energy, such as rotational kinetic energy and vibrational kineticand potential energies. We will see that a simple rule lets us determine the average energies present in these forms and solve


Chapter 2 | The Kinetic Theory of Gases 89




problems in much the same way as we have for monatomic gases.
Degrees of Freedom
In the previous section, we found that 1


2
mv2



= 3


2
kBT and v2– = 3vx2– , from which it follows that 12mvx2



= 1


2
kBT . The


same equation holds for vy2– and for vz2– . Thus, we can look at our energy of 32kBT as the sum of contributions of 12kBT
from each of the three dimensions of translational motion. Shifting to the gas as a whole, we see that the 3 in the formula
CV =


3
2
R also reflects those three dimensions. We define a degree of freedom as an independent possible motion of a


molecule, such as each of the three dimensions of translation. Then, letting d represent the number of degrees of freedom,
the molar heat capacity at constant volume of a monatomic ideal gas is CV = d2R, where d = 3 .
The branch of physics called statistical mechanics tells us, and experiment confirms, that CV of any ideal gas is given
by this equation, regardless of the number of degrees of freedom. This fact follows from a more general result, theequipartition theorem, which holds in classical (non-quantum) thermodynamics for systems in thermal equilibrium undertechnical conditions that are beyond our scope. Here, we mention only that in a system, the energy is shared among thedegrees of freedom by collisions.


Equipartition Theorem
The energy of a thermodynamic system in equilibrium is partitioned equally among its degrees of freedom.Accordingly, the molar heat capacity of an ideal gas is proportional to its number of degrees of freedom, d:


(2.14)CV = d2R.


This result is due to the Scottish physicist James Clerk Maxwell (1831−1871), whose name will appear several more timesin this book.
For example, consider a diatomic ideal gas (a good model for nitrogen, N2, and oxygen, O2). Such a gas has more
degrees of freedom than a monatomic gas. In addition to the three degrees of freedom for translation, it has two degreesof freedom for rotation perpendicular to its axis. Furthermore, the molecule can vibrate along its axis. This motion is oftenmodeled by imagining a spring connecting the two atoms, and we know from simple harmonic motion that such motion hasboth kinetic and potential energy. Each of these forms of energy corresponds to a degree of freedom, giving two more.
We might expect that for a diatomic gas, we should use 7 as the number of degrees of freedom; classically, if the moleculesof a gas had only translational kinetic energy, collisions between molecules would soon make them rotate and vibrate.However, as explained in the previous module, quantum mechanics controls which degrees of freedom are active. Theresult is shown in Figure 2.13. Both rotational and vibrational energies are limited to discrete values. For temperaturesbelow about 60 K, the energies of hydrogen molecules are too low for a collision to bring the rotational state or vibrationalstate of a molecule from the lowest energy to the second lowest, so the only form of energy is translational kinetic energy,and d = 3 or CV = 3R/2 as in a monatomic gas. Above that temperature, the two rotational degrees of freedom begin
to contribute, that is, some molecules are excited to the rotational state with the second-lowest energy. (This temperatureis much lower than that where rotations of monatomic gases contribute, because diatomic molecules have much higherrotational inertias and hence much lower rotational energies.) From about room temperature (a bit less than 300 K) to about600 K, the rotational degrees of freedom are fully active, but the vibrational ones are not, and d = 5 . Then, finally, above
about 3000 K, the vibrational degrees of freedom are fully active, and d = 7 as the classical theory predicted.


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Figure 2.13 The molar heat capacity of hydrogen as a function of temperature (on a logarithmicscale). The three “steps” or “plateaus” show different numbers of degrees of freedom that thetypical energies of molecules must achieve to activate. Translational kinetic energy corresponds tothree degrees of freedom, rotational to another two, and vibrational to yet another two.


Polyatomic molecules typically have one additional rotational degree of freedom at room temperature, since they havecomparable moments of inertia around any axis. Thus, at room temperature, they have d = 6, and at high temperature,
d = 8. We usually assume that gases have the theoretical room-temperature values of d.
As shown in Table 2.3, the results agree well with experiments for many monatomic and diatomic gases, but the agreementfor triatomic gases is only fair. The differences arise from interactions that we have ignored between and within molecules.


Gas CV /R at 25 °C and 1 atm
Ar 1.50
He 1.50
Ne 1.50
CO 2.50
H2 2.47
N2 2.50
O2 2.53
F2 2.8
CO2 3.48
H2S 3.13
N2O 3.66


Table 2.3 CV /R for Various Monatomic,
Diatomic, and Triatomic Gases


What about internal energy for diatomic and polyatomic gases? For such gases, CV is a function of temperature (Figure
2.13), so we do not have the kind of simple result we have for monatomic ideal gases.


Chapter 2 | The Kinetic Theory of Gases 91




Molar Heat Capacity of Solid Elements
The idea of equipartition leads to an estimate of the molar heat capacity of solid elements at ordinary temperatures. We canmodel the atoms of a solid as attached to neighboring atoms by springs (Figure 2.14).


Figure 2.14 In a simple model of a solid element, each atomis attached to others by six springs, two for each possiblemotion: x, y, and z. Each of the three motions corresponds to twodegrees of freedom, one for kinetic energy and one for potentialenergy. Thus d = 6.


Analogously to the discussion of vibration in the previous module, each atom has six degrees of freedom: one kinetic andone potential for each of the x-, y-, and z-directions. Accordingly, the molar specific heat of a metal should be 3R. Thisresult, known as the Law of Dulong and Petit, works fairly well experimentally at room temperature. (For every element,it fails at low temperatures for quantum-mechanical reasons. Since quantum effects are particularly important for low-massparticles, the Law of Dulong and Petit already fails at room temperature for some light elements, such as beryllium andcarbon. It also fails for some heavier elements for various reasons beyond what we can cover.)
Problem-Solving Strategy: Heat Capacity and Equipartition
The strategy for solving these problems is the same as the one in Phase Changes for the effects of heat transfer.The only new feature is that you should determine whether the case just presented—ideal gases at constantvolume—applies to the problem. (For solid elements, looking up the specific heat capacity is generally better thanestimating it from the Law of Dulong and Petit.) In the case of an ideal gas, determine the number d of degrees offreedom from the number of atoms in the gas molecule and use it to calculate CV (or use CV to solve for d).


Example 2.9
Calculating Temperature: Calorimetry with an Ideal Gas
A 300-g piece of solid gallium (a metal used in semiconductor devices) at its melting point of only 30.0 °C is
in contact with 12.0 moles of air (assumed diatomic) at 95.0 °C in an insulated container. When the air reaches
equilibrium with the gallium, 202 g of the gallium have melted. Based on those data, what is the heat of fusion ofgallium? Assume the volume of the air does not change and there are no other heat transfers.


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Strategy
We’ll use the equation Qhot + Qcold = 0. As some of the gallium doesn’t melt, we know the final temperature
is still the melting point. Then the only Qhot is the heat lost as the air cools, Qhot = nairCVΔT , where
CV = 5R/2. The only Qcold is the latent heat of fusion of the gallium, Qcold = mGaLf. It is positive because
heat flows into the gallium.
Solution1. Set up the equation:


nairCVΔT + mGa Lf = 0.


2. Substitute the known values and solve:
(12.0 mol)⎛⎝


5
2



⎝8.31


J
mol · °C



⎠(30.0 °C − 95.0 °C) + (0.202 kg)Lf = 0.


We solve to find that the heat of fusion of gallium is 80.2 kJ/kg.


2.4 | Distribution of Molecular Speeds
Learning Objectives


By the end of this section, you will be able to:
• Describe the distribution of molecular speeds in an ideal gas
• Find the average and most probable molecular speeds in an ideal gas


Particles in an ideal gas all travel at relatively high speeds, but they do not travel at the same speed. The rms speed isone kind of average, but many particles move faster and many move slower. The actual distribution of speeds has severalinteresting implications for other areas of physics, as we will see in later chapters.
The Maxwell-Boltzmann Distribution
The motion of molecules in a gas is random in magnitude and direction for individual molecules, but a gas of manymolecules has a predictable distribution of molecular speeds. This predictable distribution of molecular speeds is known asthe Maxwell-Boltzmann distribution, after its originators, who calculated it based on kinetic theory, and it has since beenconfirmed experimentally (Figure 2.15).
To understand this figure, we must define a distribution function of molecular speeds, since with a finite number ofmolecules, the probability that a molecule will have exactly a given speed is 0.


Chapter 2 | The Kinetic Theory of Gases 93




Figure 2.15 The Maxwell-Boltzmann distribution of molecular speeds in anideal gas. The most likely speed vp is less than the rms speed vrms . Although
very high speeds are possible, only a tiny fraction of the molecules have speedsthat are an order of magnitude greater than vrms.


We define the distribution function f (v) by saying that the expected number N(v1, v2) of particles with speeds between
v1 and v2 is given by


N(v1, v2) = N∫
v1


v2
f (v)dv.


[Since N is dimensionless, the unit of f(v) is seconds per meter.] We can write this equation conveniently in differentialform:
dN = N f (v)dv.


In this form, we can understand the equation as saying that the number of molecules with speeds between v and v + dv is
the total number of molecules in the sample times f(v) times dv. That is, the probability that a molecule’s speed is betweenv and v + dv is f(v)dv.
We can now quote Maxwell’s result, although the proof is beyond our scope.


Maxwell-Boltzmann Distribution of Speeds
The distribution function for speeds of particles in an ideal gas at temperature T is


(2.15)
f (v) = 4π





m
2kBT



3/2


v2 e
−mv2 /2kBT.


The factors before the v2 are a normalization constant; they make sure that N(0, ∞) = N by making sure that


0



f (v)dv = 1. Let’s focus on the dependence on v. The factor of v2 means that f (0) = 0 and for small v, the curve


looks like a parabola. The factor of e−m0 v2 /2kBT means that lim
v → ∞


f (v) = 0 and the graph has an exponential tail, which
indicates that a few molecules may move at several times the rms speed. The interaction of these factors gives the functionthe single-peaked shape shown in the figure.


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Example 2.10
Calculating the Ratio of Numbers of Molecules Near Given Speeds
In a sample of nitrogen (N2, with a molar mass of 28.0 g/mol) at a temperature of 273 °C , find the ratio of the
number of molecules with a speed very close to 300 m/s to the number with a speed very close to 100 m/s.
Strategy
Since we’re looking at a small range, we can approximate the number of molecules near 100 m/s as
dN100 = f (100 m/s)dv. Then the ratio we want is


dN300
dN100


=
f (300 m/s)dv
f (100 m/s)dv


=
f (300 m/s)
f (100 m/s)


.


All we have to do is take the ratio of the two f values.
Solution1. Identify the knowns and convert to SI units if necessary.


T = 300 K, kB = 1.38 × 10
−23 J/K


M = 0.0280 kg/mol som = 4.65 × 10−26 kg


2. Substitute the values and solve.
f (300 m/s)
f (100 m/s)


=


4
π



m
2kBT



3/2


(300 m/s)2 exp[−m(300 m/s)2 /2kBT]


4
π



m
2kBT



3/2


(100 m/s)2 exp[−m(100 m/s)2 /2kBT]


=
(300 m/s)2 exp[−(4.65 × 10−26 kg)(300 m/s)2 /2(1.38 × 10−23 J/K)(300 K)]


(100 m/s)2 exp[−(4.65 × 10−26 kg)(100 m/s)2 /2(1.38 × 10−23 J/K)(300 K)]


= 32 exp



⎢−


(4.65 × 10−26 kg)[(300 m/s)2 − (100 ms)2]


2(1.38 × 10−23 J/K)(300 K)







= 5.74


Figure 2.16 shows that the curve is shifted to higher speeds at higher temperatures, with a broader range of speeds.


Figure 2.16 The Maxwell-Boltzmann distribution is shifted tohigher speeds and broadened at higher temperatures.


Chapter 2 | The Kinetic Theory of Gases 95




With only a relatively small number of molecules, the distribution of speeds fluctuates around the Maxwell-Boltzmann distribution. However, you can view this simulation (https://openstaxcollege.org/l/21maxboltzdisim) to see the essential features that more massive molecules move slower and have a narrowerdistribution. Use the set-up “2 Gases, Random Speeds”. Note the display at the bottom comparing histograms ofthe speed distributions with the theoretical curves.
We can use a probability distribution to calculate average values by multiplying the distribution function by the quantityto be averaged and integrating the product over all possible speeds. (This is analogous to calculating averages of discretedistributions, where you multiply each value by the number of times it occurs, add the results, and divide by the numberof values. The integral is analogous to the first two steps, and the normalization is analogous to dividing by the number ofvalues.) Thus the average velocity is


(2.16)
v̄ = ⌠
⌡0




v f (v)dv = 8π
kBT
m =


8
π
RT
M


.


Similarly,
vrms = v


2


= ∫
0



v2 f (v)dv =


3kBT
m =


3RT
M


as in Pressure, Temperature, and RMS Speed. The most probable speed, also called the peak speed v p, is the
speed at the peak of the velocity distribution. (In statistics it would be called the mode.) It is less than the rms speed vrms.
The most probable speed can be calculated by the more familiar method of setting the derivative of the distribution function,with respect to v, equal to 0. The result is


(2.17)
v p =


2kBT
m =


2RT
M


,


which is less than vrms. In fact, the rms speed is greater than both the most probable speed and the average speed.
The peak speed provides a sometimes more convenient way to write the Maxwell-Boltzmann distribution function:


(2.18)
f (v) = 4v


2


πv p
3
e
−v2 /v p


2


In the factor e−mv2 /2kBT , it is easy to recognize the translational kinetic energy. Thus, that expression is equal to
e
−K/kBT. The distribution f(v) can be transformed into a kinetic energy distribution by requiring that f (K)dK = f (v)dv.


Boltzmann showed that the resulting formula is much more generally applicable if we replace the kinetic energy oftranslation with the total mechanical energy E. Boltzmann’s result is
f (E) = 2π(kBT)


−3/2 Ee
−E/kBT = 2


π(kBT)
3/2


E


e
E/kBT


.


The first part of this equation, with the negative exponential, is the usual way to write it. We give the second part only to
remark that eE/kBT in the denominator is ubiquitous in quantum as well as classical statistical mechanics.


Problem-Solving Strategy: Speed Distribution
Step 1. Examine the situation to determine that it relates to the distribution of molecular speeds.


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Step 2. Make a list of what quantities are given or can be inferred from the problem as stated (identify the knownquantities).
Step 3. Identify exactly what needs to be determined in the problem (identify the unknown quantities). A written listis useful.
Step 4. Convert known values into proper SI units (K for temperature, Pa for pressure, m3 for volume, molecules for
N, and moles for n). In many cases, though, using R and the molar mass will be more convenient than using kB and
the molecular mass.
Step 5. Determine whether you need the distribution function for velocity or the one for energy, and whether you areusing a formula for one of the characteristic speeds (average, most probably, or rms), finding a ratio of values of thedistribution function, or approximating an integral.
Step 6. Solve the appropriate equation for the ideal gas law for the quantity to be determined (the unknown quantity).Note that if you are taking a ratio of values of the distribution function, the normalization factors divide out. Or ifapproximating an integral, use the method asked for in the problem.
Step 7. Substitute the known quantities, along with their units, into the appropriate equation and obtain numericalsolutions complete with units.


We can now gain a qualitative understanding of a puzzle about the composition of Earth’s atmosphere. Hydrogen is by farthe most common element in the universe, and helium is by far the second-most common. Moreover, helium is constantlyproduced on Earth by radioactive decay. Why are those elements so rare in our atmosphere? The answer is that gasmolecules that reach speeds above Earth’s escape velocity, about 11 km/s, can escape from the atmosphere into space.Because of the lower mass of hydrogen and helium molecules, they move at higher speeds than other gas molecules, such asnitrogen and oxygen. Only a few exceed escape velocity, but far fewer heavier molecules do. Thus, over the billions of yearsthat Earth has existed, far more hydrogen and helium molecules have escaped from the atmosphere than other molecules,and hardly any of either is now present.
We can also now take another look at evaporative cooling, which we discussed in the chapter on temperature and heat.Liquids, like gases, have a distribution of molecular energies. The highest-energy molecules are those that can escape fromthe intermolecular attractions of the liquid. Thus, when some liquid evaporates, the molecules left behind have a loweraverage energy, and the liquid has a lower temperature.


Chapter 2 | The Kinetic Theory of Gases 97




Avogadro’s number
Boltzmann constant


critical temperature
Dalton’s law of partial pressures
degree of freedom
equipartition theorem
ideal gas
ideal gas law
internal energy
kinetic theory of gases
Maxwell-Boltzmann distribution
mean free path
mean free time
mole
most probable speed
partial pressure
peak speed
pV diagram
root-mean-square (rms) speed
supercritical
universal gas constant
van der Waals equation of state
vapor pressure


CHAPTER 2 REVIEW
KEY TERMS


NA, the number of molecules in one mole of a substance; NA = 6.02 × 1023 particles/mole
kB, a physical constant that relates energy to temperature and appears in the ideal gas law;


kB = 1.38 × 10
−23 J/K


Tc at which the isotherm has a point with zero slope
physical law that states that the total pressure of a gas is the sum of partialpressures of the component gases


independent kind of motion possessing energy, such as the kinetic energy of motion in one of thethree orthogonal spatial directions
theorem that the energy of a classical thermodynamic system is shared equally among itsdegrees of freedom


gas at the limit of low density and high temperature
physical law that relates the pressure and volume of a gas, far from liquefaction, to the number of gasmolecules or number of moles of gas and the temperature of the gas
sum of the mechanical energies of all of the molecules in it


theory that derives the macroscopic properties of gases from the motion of the molecules theyconsist of
function that can be integrated to give the probability of finding ideal gas moleculeswith speeds in the range between the limits of integration


average distance between collisions of a particle
average time between collisions of a particle


quantity of a substance whose mass (in grams) is equal to its molecular mass
speed near which the speeds of most molecules are found, the peak of the speed distributionfunction


pressure a gas would create if it occupied the total volume of space available
same as “most probable speed”
graph of pressure vs. volume


square root of the average of the square (of a quantity)
condition of a fluid being at such a high temperature and pressure that the liquid phase cannot exist


R, the constant that appears in the ideal gas law expressed in terms of moles, given by
R = NA kB


equation, typically approximate, which relates the pressure and volume of a gas tothe number of gas molecules or number of moles of gas and the temperature of the gas
partial pressure of a vapor at which it is in equilibrium with the liquid (or solid, in the case ofsublimation) phase of the same substance


KEY EQUATIONS
Ideal gas law in terms of molecules pV = NkBT


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Ideal gas law ratios if the amount of gas is constant p1V1
T1


=
p2V2
T2


Ideal gas law in terms of moles pV = nRT
Van der Waals equation ⎡



⎢p + a⎛⎝


n
V



2⎤



⎥(V−nb) = nRT


Pressure, volume, and molecular speed pV = 1
3
Nmv2




Root-mean-square speed
vrms = 3RTM


=
3kBT
m


Mean free path
λ = V


4 2πr2N
=


kBT


4 2πr2 p


Mean free time
τ =


kBT


4 2πr2 pvrms


The following two equations apply only to a monatomic ideal gas:
Average kinetic energy of a molecule K– = 3


2
kBT


Internal energy Eint = 32NkBT .
Heat in terms of molar heat capacity at constant volume Q = nCVΔT
Molar heat capacity at constant volume for an ideal gas with ddegrees of freedom CV = d2R
Maxwell–Boltzmann speed distribution


f (v) = 4π



m
2kBT



3/2


v2 e
−mv2 /2kBT


Average velocity of a molecule
v̄ = 8π


kBT
m =


8
π
RT
M


Peak velocity of a molecule
v p =


2kBT
m =


2RT
M


SUMMARY
2.1 Molecular Model of an Ideal Gas


• The ideal gas law relates the pressure and volume of a gas to the number of gas molecules and the temperature ofthe gas.
• A mole of any substance has a number of molecules equal to the number of atoms in a 12-g sample of carbon-12.The number of molecules in a mole is called Avogadro’s number NA,


NA = 6.02 × 10
23 mol−1.


• A mole of any substance has a mass in grams numerically equal to its molecular mass in unified mass units, whichcan be determined from the periodic table of elements. The ideal gas law can also be written and solved in terms ofthe number of moles of gas:
pV = nRT ,


Chapter 2 | The Kinetic Theory of Gases 99




where n is the number of moles and R is the universal gas constant,
R = 8.31 J/mol · K.


• The ideal gas law is generally valid at temperatures well above the boiling temperature.
• The van der Waals equation of state for gases is valid closer to the boiling point than the ideal gas law.
• Above the critical temperature and pressure for a given substance, the liquid phase does not exist, and the sample is“supercritical.”


2.2 Pressure, Temperature, and RMS Speed
• Kinetic theory is the atomic description of gases as well as liquids and solids. It models the properties of matter interms of continuous random motion of molecules.
• The ideal gas law can be expressed in terms of the mass of the gas’s molecules and v2– , the average of the
molecular speed squared, instead of the temperature.


• The temperature of gases is proportional to the average translational kinetic energy of molecules. Hence, the typicalspeed of gas molecules vrms is proportional to the square root of the temperature and inversely proportional to the
square root of the molecular mass.


• In a mixture of gases, each gas exerts a pressure equal to the total pressure times the fraction of the mixture that thegas makes up.
• The mean free path (the average distance between collisions) and the mean free time of gas molecules areproportional to the temperature and inversely proportional to the molar density and the molecules’ cross-sectionalarea.


2.3 Heat Capacity and Equipartition of Energy
• Every degree of freedom of an ideal gas contributes 1


2
kBT per atom or molecule to its changes in internal energy.


• Every degree of freedom contributes 1
2
R to its molar heat capacity at constant volume CV.


• Degrees of freedom do not contribute if the temperature is too low to excite the minimum energy of the degree offreedom as given by quantum mechanics. Therefore, at ordinary temperatures, d = 3 for monatomic gases, d = 5
for diatomic gases, and d ≈ 6 for polyatomic gases.


2.4 Distribution of Molecular Speeds
• The motion of individual molecules in a gas is random in magnitude and direction. However, a gas of manymolecules has a predictable distribution of molecular speeds, known as the Maxwell-Boltzmann distribution.
• The average and most probable velocities of molecules having the Maxwell-Boltzmann speed distribution, as wellas the rms velocity, can be calculated from the temperature and molecular mass.


CONCEPTUAL QUESTIONS
2.1 Molecular Model of an Ideal Gas
1. Two H2 molecules can react with one O2 molecule
to produce two H2O molecules. How many moles of
hydrogen molecules are needed to react with one mole ofoxygen molecules?
2. Under what circumstances would you expect a gas tobehave significantly differently than predicted by the idealgas law?


3. A constant-volume gas thermometer contains a fixedamount of gas. What property of the gas is measured toindicate its temperature?
4. Inflate a balloon at room temperature. Leave theinflated balloon in the refrigerator overnight. What happensto the balloon, and why?
5. In the last chapter, free convection was explained as theresult of buoyant forces on hot fluids. Explain the upwardmotion of air in flames based on the ideal gas law.


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2.2 Pressure, Temperature, and RMS Speed
6. How is momentum related to the pressure exerted bya gas? Explain on the molecular level, considering thebehavior of molecules.
7. If one kind of molecule has double the radius of anotherand eight times the mass, how do their mean free pathsunder the same conditions compare? How do their meanfree times compare?
8. What is the average velocity of the air molecules in theroom where you are right now?
9. Why do the atmospheres of Jupiter, Saturn, Uranus, andNeptune, which are much more massive and farther fromthe Sun than Earth is, contain large amounts of hydrogenand helium?
10. Statistical mechanics says that in a gas maintained at aconstant temperature through thermal contact with a biggersystem (a “reservoir”) at that temperature, the fluctuationsin internal energy are typically a fraction 1/ N of the
internal energy. As a fraction of the total internal energy ofa mole of gas, how big are the fluctuations in the internalenergy? Are we justified in ignoring them?
11. Which is more dangerous, a closet where tanks ofnitrogen are stored, or one where tanks of carbon dioxideare stored?


2.3 Heat Capacity and Equipartition of Energy
12. Experimentally it appears that many polyatomic


molecules’ vibrational degrees of freedom can contributeto some extent to their energy at room temperature. Wouldyou expect that fact to increase or decrease their heatcapacity from the value R? Explain.
13. One might think that the internal energy of diatomicgases is given by Eint = 5RT /2. Do diatomic gases near
room temperature have more or less internal energy thanthat? Hint: Their internal energy includes the total energyadded in raising the temperature from the boiling point(very low) to room temperature.
14. You mix 5 moles of H2 at 300 K with 5 moles of
He at 360 K in a perfectly insulated calorimeter. Is the finaltemperature higher or lower than 330 K?


2.4 Distribution of Molecular Speeds
15. One cylinder contains helium gas and another containskrypton gas at the same temperature. Mark each of thesestatements true, false, or impossible to determine from thegiven information. (a) The rms speeds of atoms in thetwo gases are the same. (b) The average kinetic energiesof atoms in the two gases are the same. (c) The internalenergies of 1 mole of gas in each cylinder are the same. (d)The pressures in the two cylinders are the same.
16. Repeat the previous question if one gas is still heliumbut the other is changed to fluorine, F2 .
17. An ideal gas is at a temperature of 300 K. To doublethe average speed of its molecules, what does thetemperature need to be changed to?


PROBLEMS
2.1 Molecular Model of an Ideal Gas
18. The gauge pressure in your car tires is
2.50 × 105 N/m2 at a temperature of 35.0 °C when you
drive it onto a ship in Los Angeles to be sent to Alaska.What is their gauge pressure on a night in Alaska whentheir temperature has dropped to −40.0 °C ? Assume the
tires have not gained or lost any air.
19. Suppose a gas-filled incandescent light bulb ismanufactured so that the gas inside the bulb is atatmospheric pressure when the bulb has a temperature of
20.0 °C . (a) Find the gauge pressure inside such a bulb
when it is hot, assuming its average temperature is 60.0 °C
(an approximation) and neglecting any change in volumedue to thermal expansion or gas leaks. (b) The actual finalpressure for the light bulb will be less than calculated in


part (a) because the glass bulb will expand. Is this effectsignificant?
20. People buying food in sealed bags at high elevationsoften notice that the bags are puffed up because the airinside has expanded. A bag of pretzels was packed at apressure of 1.00 atm and a temperature of 22.0 °C. When
opened at a summer picnic in Santa Fe, New Mexico, at atemperature of 32.0 °C, the volume of the air in the bag is
1.38 times its original volume. What is the pressure of theair?
21. How many moles are there in (a) 0.0500 g of N2
gas (M = 28.0 g/mol⎞⎠? (b) 10.0 g of CO2 gas
(M = 44.0 g/mol)? (c) How many molecules are present
in each case?


Chapter 2 | The Kinetic Theory of Gases 101




22. A cubic container of volume 2.00 L holds 0.500 molof nitrogen gas at a temperature of 25.0 °C. What is the
net force due to the nitrogen on one wall of the container?Compare that force to the sample’s weight.
23. Calculate the number of moles in the 2.00-L volumeof air in the lungs of the average person. Note that the airis at 37.0 °C (body temperature) and that the total volume
in the lungs is several times the amount inhaled in a typicalbreath as given in Example 2.2.
24. An airplane passenger has 100 cm3 of air in his
stomach just before the plane takes off from a sea-levelairport. What volume will the air have at cruising altitude if
cabin pressure drops to 7.50 × 104 N/m2 ?
25. A company advertises that it delivers helium at a
gauge pressure of 1.72 × 107 Pa in a cylinder of volume
43.8 L. How many balloons can be inflated to a volume of4.00 L with that amount of helium? Assume the pressure
inside the balloons is 1.01 × 105 Pa and the temperature
in the cylinder and the balloons is 25.0 °C .
26. According to http://hyperphysics.phy-astr.gsu.edu/hbase/solar/venusenv.html, the atmosphere of Venus isapproximately 96.5% CO2 and 3.5% N2 by volume. On
the surface, where the temperature is about 750 K andthe pressure is about 90 atm, what is the density of theatmosphere?
27. An expensive vacuum system can achieve a pressure
as low as 1.00 × 10−7 N/m2 at 20.0 °C. How many
molecules are there in a cubic centimeter at this pressureand temperature?
28. The number density N/V of gas molecules at a certainlocation in the space above our planet is about
1.00 × 1011 m−3, and the pressure is
2.75 × 10−10 N/m2 in this space. What is the temperature
there?
29. A bicycle tire contains 2.00 L of gas at an absolute
pressure of 7.00 × 105 N/m2 and a temperature of
18.0 °C . What will its pressure be if you let out an amount
of air that has a volume of 100 cm3 at atmospheric
pressure? Assume tire temperature and volume remainconstant.
30. In a common demonstration, a bottle is heated andstoppered with a hard-boiled egg that’s a little bigger thanthe bottle’s neck. When the bottle is cooled, the pressuredifference between inside and outside forces the egg into


the bottle. Suppose the bottle has a volume of 0.500 L andthe temperature inside it is raised to 80.0 °C while the
pressure remains constant at 1.00 atm because the bottleis open. (a) How many moles of air are inside? (b) Nowthe egg is put in place, sealing the bottle. What is thegauge pressure inside after the air cools back to the ambienttemperature of 25 °C but before the egg is forced into the
bottle?
31. A high-pressure gas cylinder contains 50.0 L of toxic
gas at a pressure of 1.40 × 107 N/m2 and a temperature
of 25.0 °C . The cylinder is cooled to dry ice temperature
(−78.5 °C) to reduce the leak rate and pressure so that
it can be safely repaired. (a) What is the final pressure inthe tank, assuming a negligible amount of gas leaks whilebeing cooled and that there is no phase change? (b) Whatis the final pressure if one-tenth of the gas escapes? (c) Towhat temperature must the tank be cooled to reduce thepressure to 1.00 atm (assuming the gas does not changephase and that there is no leakage during cooling)? (d)Does cooling the tank as in part (c) appear to be a practicalsolution?
32. Find the number of moles in 2.00 L of gas at 35.0 °C
and under 7.41 × 107 N/m2 of pressure.
33. Calculate the depth to which Avogadro’s number oftable tennis balls would cover Earth. Each ball has adiameter of 3.75 cm. Assume the space between balls addsan extra 25.0% to their volume and assume they are not
crushed by their own weight.
34. (a) What is the gauge pressure in a 25.0 °C car tire
containing 3.60 mol of gas in a 30.0-L volume? (b) Whatwill its gauge pressure be if you add 1.00 L of gas originallyat atmospheric pressure and 25.0 °C ? Assume the
temperature remains at 25.0 °C and the volume remains
constant.


2.2 Pressure, Temperature, and RMS Speed
In the problems in this section, assume all gases are ideal.
35. A person hits a tennis ball with a mass of 0.058 kgagainst a wall. The average component of the ball’s velocityperpendicular to the wall is 11 m/s, and the ball hits thewall every 2.1 s on average, rebounding with the oppositeperpendicular velocity component. (a) What is the averageforce exerted on the wall? (b) If the part of the wall the
person hits has an area of 3.0 m2, what is the average
pressure on that area?
36. A person is in a closed room (a racquetball court)


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with V = 453 m3 hitting a ball (m = 42.0 g⎞⎠ around at
random without any pauses. The average kinetic energyof the ball is 2.30 J. (a) What is the average value of
vx
2? Does it matter which direction you take to be x?


(b) Applying the methods of this chapter, find the averagepressure on the walls? (c) Aside from the presence of onlyone “molecule” in this problem, what is the mainassumption in Pressure, Temperature, and RMSSpeed that does not apply here?
37. Five bicyclists are riding at the following speeds: 5.4m/s, 5.7 m/s, 5.8 m/s, 6.0 m/s, and 6.5 m/s. (a) What is theiraverage speed? (b) What is their rms speed?
38. Some incandescent light bulbs are filled with argongas. What is vrms for argon atoms near the filament,
assuming their temperature is 2500 K?
39. Typical molecular speeds (vrms) are large, even at
low temperatures. What is vrms for helium atoms at 5.00
K, less than one degree above helium’s liquefactiontemperature?
40. What is the average kinetic energy in joules ofhydrogen atoms on the 5500 °C surface of the Sun? (b)
What is the average kinetic energy of helium atoms ina region of the solar corona where the temperature is
6.00 × 105 K ?
41. What is the ratio of the average translational kineticenergy of a nitrogen molecule at a temperature of 300K to the gravitational potential energy of a nitrogen-molecule−Earth system at the ceiling of a 3-m-tall roomwith respect to the same system with the molecule at thefloor?
42. What is the total translational kinetic energy of the air
molecules in a room of volume 23 m3 if the pressure is
9.5 × 104 Pa (the room is at fairly high elevation) and the
temperature is 21 °C ? Is any item of data unnecessary for
the solution?
43. The product of the pressure and volume of a sample ofhydrogen gas at 0.00 °C is 80.0 J. (a) How many moles of
hydrogen are present? (b) What is the average translationalkinetic energy of the hydrogen molecules? (c) What is thevalue of the product of pressure and volume at 200 °C?
44. What is the gauge pressure inside a tank of
4.86 × 104 mol of compressed nitrogen with a volume of
6.56 m3 if the rms speed is 514 m/s?


45. If the rms speed of oxygen molecules inside a
refrigerator of volume 22.0 ft.3 is 465 m/s, what is the
partial pressure of the oxygen? There are 5.71 moles ofoxygen in the refrigerator, and the molar mass of oxygen is32.0 g/mol.
46. The escape velocity of any object from Earth is 11.1km/s. At what temperature would oxygen molecules (molarmass is equal to 32.0 g/mol) have root-mean-squarevelocity vrms equal to Earth’s escape velocity of 11.1 km/
s?
47. The escape velocity from the Moon is much smallerthan that from the Earth, only 2.38 km/s. At whattemperature would hydrogen molecules (molar mass isequal to 2.016 g/mol) have a root-mean-square velocity
vrms equal to the Moon’s escape velocity?
48. Nuclear fusion, the energy source of the Sun,hydrogen bombs, and fusion reactors, occurs much morereadily when the average kinetic energy of the atoms ishigh—that is, at high temperatures. Suppose you want theatoms in your fusion experiment to have average kinetic
energies of 6.40 × 10−14 J . What temperature is needed?
49. Suppose that the typical speed (vrms) of carbon
dioxide molecules (molar mass is 44.0 g/mol) in a flame isfound to be 1350 m/s. What temperature does this indicate?
50. (a) Hydrogen molecules (molar mass is equal to 2.016g/mol) have vrms equal to 193 m/s. What is the
temperature? (b) Much of the gas near the Sun is atomichydrogen (H rather than H2). Its temperature would have
to be 1.5 × 107 K for the rms speed vrms to equal the
escape velocity from the Sun. What is that velocity?
51. There are two important isotopes of uranium, 235U
and 238U ; these isotopes are nearly identical chemically
but have different atomic masses. Only 235U is very
useful in nuclear reactors. Separating the isotopes is calleduranium enrichment (and is often in the news as of thiswriting, because of concerns that some countries areenriching uranium with the goal of making nuclearweapons.) One of the techniques for enrichment, gasdiffusion, is based on the different molecular speeds ofuranium hexafluoride gas, UF6 . (a) The molar masses of
235U and 238UF6 are 349.0 g/mol and 352.0 g/mol,
respectively. What is the ratio of their typical speeds vrms ?
(b) At what temperature would their typical speeds differ by1.00 m/s? (c) Do your answers in this problem imply thatthis technique may be difficult?


Chapter 2 | The Kinetic Theory of Gases 103




52. The partial pressure of carbon dioxide in the lungs isabout 470 Pa when the total pressure in the lungs is 1.0 atm.What percentage of the air molecules in the lungs is carbondioxide? Compare your result to the percentage of carbondioxide in the atmosphere, about 0.033%.
53. Dry air consists of approximately
78% nitrogen, 21% oxygen, and 1% argon by mole, with
trace amounts of other gases. A tank of compressed dry airhas a volume of 1.76 cubic feet at a gauge pressure of 2200pounds per square inch and a temperature of 293 K. Howmuch oxygen does it contain in moles?
54. (a) Using data from the previous problem, find themass of nitrogen, oxygen, and argon in 1 mol of dry air. Themolar mass of N2 is 28.0 g/mol, that of O2 is 32.0 g/mol,
and that of argon is 39.9 g/mol. (b) Dry air is mixed withpentane (C5H12, molar mass 72.2 g/mol), an important
constituent of gasoline, in an air-fuel ratio of 15:1 by mass(roughly typical for car engines). Find the partial pressureof pentane in this mixture at an overall pressure of 1.00atm.
55. (a) Given that air is 21% oxygen, find the minimum
atmospheric pressure that gives a relatively safe partialpressure of oxygen of 0.16 atm. (b) What is the minimumpressure that gives a partial pressure of oxygen above thequickly fatal level of 0.06 atm? (c) The air pressure at thesummit of Mount Everest (8848 m) is 0.334 atm. Why havea few people climbed it without oxygen, while some whohave tried, even though they had trained at high elevation,had to turn back?
56. (a) If the partial pressure of water vapor is 8.05 torr,what is the dew point? (760 torr = 1 atm = 101, 325 Pa⎞⎠
(b) On a warm day when the air temperature is 35 °C and
the dew point is 25 °C , what are the partial pressure of the
water in the air and the relative humidity?


2.3 Heat Capacity and Equipartition of Energy
57. To give a helium atom nonzero angular momentumrequires about 21.2 eV of energy (that is, 21.2 eV is thedifference between the energies of the lowest-energy orground state and the lowest-energy state with angularmomentum). The electron-volt or eV is defined as
1.60 × 10−19 J. Find the temperature T where this
amount of energy equals kBT /2. Does this explain why
we can ignore the rotational energy of helium for mostpurposes? (The results for other monatomic gases, and fordiatomic gases rotating around the axis connecting the twoatoms, have comparable orders of magnitude.)
58. (a) How much heat must be added to raise the


temperature of 1.5 mol of air from 25.0 °C to 33.0 °C at
constant volume? Assume air is completely diatomic. (b)Repeat the problem for the same number of moles of xenon,Xe.
59. A sealed, rigid container of 0.560 mol of an unknownideal gas at a temperature of 30.0 °C is cooled to
−40.0 °C . In the process, 980 J of heat are removed from
the gas. Is the gas monatomic, diatomic, or polyatomic?
60. A sample of neon gas (Ne, molar mass
M = 20.2 g/mol) at a temperature of 13.0 °C is put into
a steel container of mass 47.2 g that’s at a temperature of
−40.0 °C . The final temperature is−28.0 °C . (No heat is
exchanged with the surroundings, and you can neglect anychange in the volume of the container.) What is the mass ofthe sample of neon?
61. A steel container of mass 135 g contains 24.0 g ofammonia, NH3 , which has a molar mass of 17.0 g/mol.
The container and gas are in equilibrium at 12.0 °C . How
much heat has to be removed to reach a temperature of
−20.0 °C ? Ignore the change in volume of the steel.
62. A sealed room has a volume of 24 m3 . It’s filled with
air, which may be assumed to be diatomic, at a temperature
of 24 °C and a pressure of 9.83 × 104 Pa. A 1.00-kg
block of ice at its melting point is placed in the room.Assume the walls of the room transfer no heat. What is theequilibrium temperature?
63. Heliox, a mixture of helium and oxygen, is sometimesgiven to hospital patients who have trouble breathing,because the low mass of helium makes it easier to breathethan air. Suppose helium at 25 °C is mixed with oxygen
at 35 °C to make a mixture that is 70% helium by mole.
What is the final temperature? Ignore any heat flow to orfrom the surroundings, and assume the final volume is thesum of the initial volumes.
64. Professional divers sometimes use heliox, consistingof 79% helium and 21% oxygen by mole. Suppose a
perfectly rigid scuba tank with a volume of 11 L contains
heliox at an absolute pressure of 2.1 × 107 Pa at a
temperature of 31 °C . (a) How many moles of helium and
how many moles of oxygen are in the tank? (b) The divergoes down to a point where the sea temperature is 27 °C
while using a negligible amount of the mixture. As the gasin the tank reaches this new temperature, how much heat isremoved from it?
65. In car racing, one advantage of mixing liquid nitrousoxide (N2O) with air is that the boiling of the “nitrous”


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absorbs latent heat of vaporization and thus cools the airand ultimately the fuel-air mixture, allowing more fuel-airmixture to go into each cylinder. As a very rough lookat this process, suppose 1.0 mol of nitrous oxide gas atits boiling point, −88 °C , is mixed with 4.0 mol of air
(assumed diatomic) at 30 °C . What is the final
temperature of the mixture? Use the measured heat capacityof N2O at 25 °C , which is 30.4 J/mol °C . (The primary
advantage of nitrous oxide is that it consists of 1/3 oxygen,which is more than air contains, so it supplies more oxygento burn the fuel. Another advantage is that itsdecomposition into nitrogen and oxygen releases energy inthe cylinder.)


2.4 Distribution of Molecular Speeds
66. In a sample of hydrogen sulfide (M = 34.1 g/mol) at
a temperature of 3.00 × 102 K, estimate the ratio of the
number of molecules that have speeds very close to vrms
to the number that have speeds very close to 2vrms.
67. Using the approximation

v1


v1 + Δv


f (v)dv ≈ f (v1)Δv for small Δv , estimate the
fraction of nitrogen molecules at a temperature of
3.00 × 102 K that have speeds between 290 m/s and 291
m/s.
68. Using the method of the preceding problem, estimatethe fraction of nitric oxide (NO) molecules at a temperature
of 250 K that have energies between 3.45 × 10−21 J and
3.50 × 10−21 J .
69. By counting squares in the following figure, estimatethe fraction of argon atoms at T = 300 K that have speeds
between 600 m/s and 800 m/s. The curve is correctlynormalized. The value of a square is its length as measuredon the x-axis times its height as measured on the y-axis,


with the units given on those axes.


70. Using a numerical integration method such asSimpson’s rule, find the fraction of molecules in a sampleof oxygen gas at a temperature of 250 K that have speedsbetween 100 m/s and 150 m/s. The molar mass of oxygen

⎝O2



⎠ is 32.0 g/mol. A precision to two significant digits is


enough.
71. Find (a) the most probable speed, (b) the averagespeed, and (c) the rms speed for nitrogen molecules at 295K.
72. Repeat the preceding problem for nitrogen moleculesat 2950 K.
73. At what temperature is the average speed of carbondioxide molecules (M = 44.0 g/mol) 510 m/s?
74. The most probable speed for molecules of a gas at 296K is 263 m/s. What is the molar mass of the gas? (Youmight like to figure out what the gas is likely to be.)
75. a) At what temperature do oxygen molecules have thesame average speed as helium atoms (M = 4.00 g/mol)
have at 300 K? b) What is the answer to the same questionabout most probable speeds? c) What is the answer to thesame question about rms speeds?


ADDITIONAL PROBLEMS
76. In the deep space between galaxies, the density ofmolecules (which are mostly single atoms) can be as low
as 106 atoms/m3, and the temperature is a frigid 2.7
K. What is the pressure? (b) What volume (in m3 ) is
occupied by 1 mol of gas? (c) If this volume is a cube, whatis the length of its sides in kilometers?
77. (a) Find the density in SI units of air at a pressureof 1.00 atm and a temperature of 20 °C , assuming that


air is 78%N2, 21% O2, and 1%Ar , (b) Find the density
of the atmosphere on Venus, assuming that it’s
96% CO2 and 4%N2 , with a temperature of 737 K and a
pressure of 92.0 atm.
78. The air inside a hot-air balloon has a temperatureof 370 K and a pressure of 101.3 kPa, the same as thatof the air outside. Using the composition of air as
78%N2, 21%O2, and 1%Ar , find the density of the air


Chapter 2 | The Kinetic Theory of Gases 105




inside the balloon.
79. When an air bubble rises from the bottom to the topof a freshwater lake, its volume increases by 80% . If the
temperatures at the bottom and the top of the lake are 4.0and 10 °C , respectively, how deep is the lake?
80. (a) Use the ideal gas equation to estimate thetemperature at which 1.00 kg of steam (molar mass
M = 18.0 g/mol ) at a pressure of 1.50 × 106 Pa
occupies a volume of 0.220 m3 . (b) The van der Waals
constants for water are a = 0.5537 Pa ·m6 /mol2 and
b = 3.049 × 10−5 m3 /mol . Use the Van der Waals
equation of state to estimate the temperature under the sameconditions. (c) The actual temperature is 779 K. Whichestimate is better?
81. One process for decaffeinating coffee uses carbondioxide (M = 44.0 g/mol) at a molar density of about
14,600 mol/m3 and a temperature of about 60 °C . (a) Is
CO2 a solid, liquid, gas, or supercritical fluid under thoseconditions? (b) The van der Waals constants for carbon
dioxide are a = 0.3658 Pa ·m6 /mol2 and
b = 4.286 × 10−5 m3 /mol. Using the van der Waals
equation, estimate the pressure of CO2 at that temperature
and density.
82. On a winter day when the air temperature is 0 °C,
the relative humidity is 50% . Outside air comes inside
and is heated to a room temperature of 20 °C . What is
the relative humidity of the air inside the room. (Does thisproblem show why inside air is so dry in winter?)
83. On a warm day when the air temperature is 30 °C , a
metal can is slowly cooled by adding bits of ice to liquidwater in it. Condensation first appears when the can reaches
15 °C . What is the relative humidity of the air?
84. (a) People often think of humid air as “heavy.”Compare the densities of air with 0% relative humidity
and 100% relative humidity when both are at 1 atm and
30 °C . Assume that the dry air is an ideal gas composed
of molecules with a molar mass of 29.0 g/mol and themoist air is the same gas mixed with water vapor. (b) Asdiscussed in the chapter on the applications of Newton’slaws, the air resistance felt by projectiles such as baseballs
and golf balls is approximately FD = CρAv2 /2 , where ρ
is the mass density of the air, A is the cross-sectional area ofthe projectile, and C is the projectile’s drag coefficient. Fora fixed air pressure, describe qualitatively how the range ofa projectile changes with the relative humidity. (c) When


a thunderstorm is coming, usually the humidity is highand the air pressure is low. Do those conditions give anadvantage or disadvantage to home-run hitters?
85. The mean free path for helium at a certain temperature
and pressure is 2.10 × 10−7 m. The radius of a helium
atom can be taken as 1.10 × 10−11 m . What is the
measure of the density of helium under those conditions (a)in molecules per cubic meter and (b) in moles per cubicmeter?
86. The mean free path for methane at a temperature
of 269 K and a pressure of 1.11 × 105 Pa is
4.81 × 10−8 m. Find the effective radius r of the methane
molecule.
87. In the chapter on fluid mechanics, Bernoulli’s equationfor the flow of incompressible fluids was explained interms of changes affecting a small volume dV of fluid.Such volumes are a fundamental idea in the study of theflow of compressible fluids such as gases as well. For theequations of hydrodynamics to apply, the mean free pathmust be much less than the linear size of such a volume,
a ≈ dV 1/3. For air in the stratosphere at a temperature of
220 K and a pressure of 5.8 kPa, how big should a be forit to be 100 times the mean free path? Take the effective
radius of air molecules to be 1.88 × 10−11 m, which is
roughly correct for N2 .
88. Find the total number of collisions between moleculesin 1.00 s in 1.00 L of nitrogen gas at standard temperature
and pressure ( 0 °C , 1.00 atm). Use 1.88 × 10−10 m as
the effective radius of a nitrogen molecule. (The number ofcollisions per second is the reciprocal of the collision time.)Keep in mind that each collision involves two molecules,so if one molecule collides once in a certain period of time,the collision of the molecule it hit cannot be counted.
89. (a) Estimate the specific heat capacity of sodium fromthe Law of Dulong and Petit. The molar mass of sodium is23.0 g/mol. (b) What is the percent error of your estimatefrom the known value, 1230 J/kg · °C ?
90. A sealed, perfectly insulated container contains 0.630mol of air at 20.0 °C and an iron stirring bar of mass 40.0
g. The stirring bar is magnetically driven to a kinetic energyof 50.0 J and allowed to slow down by air resistance. Whatis the equilibrium temperature?
91. Find the ratio f (vp)/ f (vrms) for hydrogen gas
(M = 2.02 g/mol) at a temperature of 77.0 K.


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92. Unreasonable results. (a) Find the temperature of0.360 kg of water, modeled as an ideal gas, at a pressure of
1.01 × 105 Pa if it has a volume of 0.615 m3 . (b) What
is unreasonable about this answer? How could you get abetter answer?


93. Unreasonable results. (a) Find the average speed ofhydrogen sulfide, H2S , molecules at a temperature of 250
K. Its molar mass is 31.4 g/mol (b) The result isn’t veryunreasonable, but why is it less reliable than those for, say,neon or nitrogen?


CHALLENGE PROBLEMS
94. An airtight dispenser for drinking water is
25 cm × 10 cm in horizontal dimensions and 20 cm tall.
It has a tap of negligible volume that opens at the level ofthe bottom of the dispenser. Initially, it contains water to alevel 3.0 cm from the top and air at the ambient pressure,1.00 atm, from there to the top. When the tap is opened,water will flow out until the gauge pressure at the bottomof the dispenser, and thus at the opening of the tap, is 0.What volume of water flows out? Assume the temperatureis constant, the dispenser is perfectly rigid, and the water
has a constant density of 1000 kg/m3 .
95. Eight bumper cars, each with a mass of 322 kg, arerunning in a room 21.0 m long and 13.0 m wide. They haveno drivers, so they just bounce around on their own. Therms speed of the cars is 2.50 m/s. Repeating the argumentsof Pressure, Temperature, and RMS Speed, find theaverage force per unit length (analogous to pressure) thatthe cars exert on the walls.
96. Verify that v p = 2kBTm .


97. Verify the normalization equation ∫
0



f (v)dv = 1.


In doing the integral, first make the substitution
u = m


2kBT
v = vv p. This “scaling” transformation gives


you all features of the answer except for the integral, whichis a dimensionless numerical factor. You’ll need theformula


0



x2 e−x


2
dx = π


4


to find the numerical factor and verify the normalization.


98. Verify that v̄ = 8π kBTm . Make the same scaling
transformation as in the preceding problem.


99. Verify that vrms = v2– = 3kBTm .


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3 | THE FIRST LAW OFTHERMODYNAMICS


Figure 3.1 A weak cold front of air pushes all the smog in northeastern China into a giant smog blanket over the Yellow Sea,as captured by NASA’s Terra satellite in 2012. To understand changes in weather and climate, such as the event shown here, youneed a thorough knowledge of thermodynamics. (credit: modification of work by NASA)


Chapter Outline
3.1 Thermodynamic Systems
3.2 Work, Heat, and Internal Energy
3.3 First Law of Thermodynamics
3.4 Thermodynamic Processes
3.5 Heat Capacities of an Ideal Gas
3.6 Adiabatic Processes for an Ideal Gas


Introduction
Heat is energy in transit, and it can be used to do work. It can also be converted into any other form of energy. A carengine, for example, burns gasoline. Heat is produced when the burned fuel is chemically transformed into mostly CO2
and H2O, which are gases at the combustion temperature. These gases exert a force on a piston through a displacement,
doing work and converting the piston’s kinetic energy into a variety of other forms—into the car’s kinetic energy; intoelectrical energy to run the spark plugs, radio, and lights; and back into stored energy in the car’s battery.
Energy is conserved in all processes, including those associated with thermodynamic systems. The roles of heat transfer andinternal energy change vary from process to process and affect how work is done by the system in that process. We will seethat the first law of thermodynamics puts a limit on the amount of work that can be delivered by the system when the amountof internal energy change or heat transfer is constrained. Understanding the laws that govern thermodynamic processes andthe relationship between the system and its surroundings is therefore paramount in gaining scientific knowledge of energyand energy consumption.


Chapter 3 | The First Law of Thermodynamics 109




3.1 | Thermodynamic Systems
Learning Objectives


By the end of this section, you will be able to:
• Define a thermodynamic system, its boundary, and its surroundings
• Explain the roles of all the components involved in thermodynamics
• Define thermal equilibrium and thermodynamic temperature
• Link an equation of state to a system


A thermodynamic system includes anything whose thermodynamic properties are of interest. It is embedded in itssurroundings or environment; it can exchange heat with, and do work on, its environment through a boundary, whichis the imagined wall that separates the system and the environment (Figure 3.2). In reality, the immediate surroundingsof the system are interacting with it directly and therefore have a much stronger influence on its behavior and properties.For example, if we are studying a car engine, the burning gasoline inside the cylinder of the engine is the thermodynamicsystem; the piston, exhaust system, radiator, and air outside form the surroundings of the system. The boundary then consistsof the inner surfaces of the cylinder and piston.


Figure 3.2 (a) A system, which can include any relevant process or value, is self-contained in an area.The surroundings may also have relevant information; however, the surroundings are important to studyonly if the situation is an open system. (b) The burning gasoline in the cylinder of a car engine is anexample of a thermodynamic system.


Normally, a system must have some interactions with its surroundings. A system is called an isolated or closed system if itis completely separated from its environment—for example, a gas that is surrounded by immovable and thermally insulatingwalls. In reality, a closed system does not exist unless the entire universe is treated as the system, or it is used as a modelfor an actual system that has minimal interactions with its environment. Most systems are known as an open system, whichcan exchange energy and/or matter with its surroundings (Figure 3.3).


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Figure 3.3 (a) This boiling tea kettle is an open thermodynamic system. It transfersheat and matter (steam) to its surroundings. (b) A pressure cooker is a goodapproximation to a closed system. A little steam escapes through the top valve to preventexplosion. (credit a: modification of work by Gina Hamilton)


When we examine a thermodynamic system, we ignore the difference in behavior from place to place inside the system fora given moment. In other words, we concentrate on the macroscopic properties of the system, which are the averages ofthe microscopic properties of all the molecules or entities in the system. Any thermodynamic system is therefore treated asa continuum that has the same behavior everywhere inside. We assume the system is in equilibrium. You could have, forexample, a temperature gradient across the system. However, when we discuss a thermodynamic system in this chapter, westudy those that have uniform properties throughout the system.
Before we can carry out any study on a thermodynamic system, we need a fundamental characterization of the system.When we studied a mechanical system, we focused on the forces and torques on the system, and their balances dictatedthe mechanical equilibrium of the system. In a similar way, we should examine the heat transfer between a thermodynamicsystem and its environment or between the different parts of the system, and its balance should dictate the thermalequilibrium of the system. Intuitively, such a balance is reached if the temperature becomes the same for different objectsor parts of the system in thermal contact, and the net heat transfer over time becomes zero.
Thus, when we say two objects (a thermodynamic system and its environment, for example) are in thermal equilibrium, wemean that they are at the same temperature, as we discussed in Temperature and Heat. Let us consider three objects attemperatures T1, T2, and T3, respectively. How do we know whether they are in thermal equilibrium? The governing
principle here is the zeroth law of thermodynamics, as described in Temperature and Heat on temperature and heat:
If object 1 is in thermal equilibrium with objects 2 and 3, respectively, then objects 2 and 3 must also be in thermalequilibrium.
Mathematically, we can simply write the zeroth law of thermodynamics as


(3.1)If T1 = T2 and T1 = T3, then T2 = T3.
This is the most fundamental way of defining temperature: Two objects must be at the same temperature thermodynamicallyif the net heat transfer between them is zero when they are put in thermal contact and have reached a thermal equilibrium.
The zeroth law of thermodynamics is equally applicable to the different parts of a closed system and requires that thetemperature everywhere inside the system be the same if the system has reached a thermal equilibrium. To simplify ourdiscussion, we assume the system is uniform with only one type of material—for example, water in a tank. The measurableproperties of the system at least include its volume, pressure, and temperature. The range of specific relevant variablesdepends upon the system. For example, for a stretched rubber band, the relevant variables would be length, tension, andtemperature. The relationship between these three basic properties of the system is called the equation of state of the systemand is written symbolically for a closed system as


Chapter 3 | The First Law of Thermodynamics 111




(3.2)f (p, V , T) = 0,


where V, p, and T are the volume, pressure, and temperature of the system at a given condition.
In principle, this equation of state exists for any thermodynamic system but is not always readily available. The forms of
f (p, V , T) = 0 for many materials have been determined either experimentally or theoretically. In the preceding chapter,
we saw an example of an equation of state for an ideal gas, f (p, V , T) = pV − nRT = 0.
We have so far introduced several physical properties that are relevant to the thermodynamics of a thermodynamic system,such as its volume, pressure, and temperature. We can separate these quantities into two generic categories. The quantityassociated with an amount of matter is an extensive variable, such as the volume and the number of moles. The otherproperties of a system are intensive variables, such as the pressure and temperature. An extensive variable doubles its valueif the amount of matter in the system doubles, provided all the intensive variables remain the same. For example, the volumeor total energy of the system doubles if we double the amount of matter in the system while holding the temperature andpressure of the system unchanged.
3.2 | Work, Heat, and Internal Energy


Learning Objectives
By the end of this section, you will be able to:
• Describe the work done by a system, heat transfer between objects, and internal energychange of a system
• Calculate the work, heat transfer, and internal energy change in a simple process


We discussed the concepts of work and energy earlier in mechanics. Examples and related issues of heat transfer betweendifferent objects have also been discussed in the preceding chapters. Here, we want to expand these concepts to athermodynamic system and its environment. Specifically, we elaborated on the concepts of heat and heat transfer in theprevious two chapters. Here, we want to understand how work is done by or to a thermodynamic system; how heat istransferred between a system and its environment; and how the total energy of the system changes under the influence ofthe work done and heat transfer.
Work Done by a System
A force created from any source can do work by moving an object through a displacement. Then how does a thermodynamicsystem do work? Figure 3.4 shows a gas confined to a cylinder that has a movable piston at one end. If the gas expandsagainst the piston, it exerts a force through a distance and does work on the piston. If the piston compresses the gas asit is moved inward, work is also done—in this case, on the gas. The work associated with such volume changes can bedetermined as follows: Let the gas pressure on the piston face be p. Then the force on the piston due to the gas is pA, whereA is the area of the face. When the piston is pushed outward an infinitesimal distance dx, the magnitude of the work doneby the gas is


dW = F dx = pA dx.


Since the change in volume of the gas is dV = A dx, this becomes
(3.3)dW = pdV .


For a finite change in volume from V1 to V2, we can integrate this equation from V1 to V2 to find the net work:


(3.4)
W = ∫


V1


V2
pdV .


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Figure 3.4 The work done by a confined gas in moving apiston a distance dx is given by dW = Fdx = pdV .


This integral is only meaningful for a quasi-static process, which means a process that takes place in infinitesimally smallsteps, keeping the system at thermal equilibrium. (We examine this idea in more detail later in this chapter.) Only then doesa well-defined mathematical relationship (the equation of state) exist between the pressure and volume. This relationshipcan be plotted on a pV diagram of pressure versus volume, where the curve is the change of state. We can approximatesuch a process as one that occurs slowly, through a series of equilibrium states. The integral is interpreted graphically as thearea under the pV curve (the shaded area of Figure 3.5). Work done by the gas is positive for expansion and negative forcompression.


Figure 3.5 When a gas expands slowly from V1 to V2, the
work done by the system is represented by the shaded area underthe pV curve.


Consider the two processes involving an ideal gas that are represented by paths AC and ABC in Figure 3.6. The firstprocess is an isothermal expansion, with the volume of the gas changing its volume from V1 to V2 . This isothermal process
is represented by the curve between points A and C. The gas is kept at a constant temperature T by keeping it in thermalequilibrium with a heat reservoir at that temperature. From Equation 3.4 and the ideal gas law,


W = ∫
V1


V2
pdV = ⌠


⌡V1


V2⎛

nRT
V

⎠ dV .


Chapter 3 | The First Law of Thermodynamics 113




Figure 3.6 The paths ABC, AC, and ADC represent threedifferent quasi-static transitions between the equilibrium states Aand C.


The expansion is isothermal, so T remains constant over the entire process. Since n and R are also constant, the only variablein the integrand is V, so the work done by an ideal gas in an isothermal process is
W = nRT⌠


⌡V1


V2
dV
V


= nRTln
V2
V1


.


Notice that if V2 > V1 (expansion), W is positive, as expected.
The straight lines from A to B and then from B to C represent a different process. Here, a gas at a pressure p1 first expands
isobarically (constant pressure) and quasi-statically from V1 to V2 , after which it cools quasi-statically at the constant
volume V2 until its pressure drops to p2 . From A to B, the pressure is constant at p, so the work over this part of the path
is


W = ∫
V1


V2
pdV = p1∫


V1


V2
dV = p1(V2 − V1).


From B to C, there is no change in volume and therefore no work is done. The net work over the path ABC is then
W = p1(V2 − V1) + 0 = p1(V2 − V1).


A comparison of the expressions for the work done by the gas in the two processes of Figure 3.6 shows that they are quitedifferent. This illustrates a very important property of thermodynamic work: It is path dependent. We cannot determine thework done by a system as it goes from one equilibrium state to another unless we know its thermodynamic path. Differentvalues of the work are associated with different paths.
Example 3.1


Isothermal Expansion of a van der Waals Gas
Studies of a van der Waals gas require an adjustment to the ideal gas law that takes into consideration that gasmolecules have a definite volume (see The Kinetic Theory of Gases). One mole of a van der Waals gas hasan equation of state



⎝p +


a
V 2

⎠(V − b) = RT ,


where a and b are two parameters for a specific gas. Suppose the gas expands isothermally and quasi-staticallyfrom volume V1 to volume V2. How much work is done by the gas during the expansion?


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3.1


Strategy
Because the equation of state is given, we can use Equation 3.4 to express the pressure in terms of V and T.Furthermore, temperature T is a constant under the isothermal condition, so V becomes the only changing variableunder the integral.
Solution
To evaluate this integral, we must express p as a function of V. From the given equation of state, the gas pressureis


p = RT
V − b


− a
V 2


.


Because T is constant under the isothermal condition, the work done by 1 mol of a van der Waals gas in expandingfrom a volume V1 to a volume V2 is thus


W = ⌠

V1


V2



RT
V − b


− a
V 2

⎠ = |RTln(V − b) + aV |V1


V2


= RTln


V2 − b
V1 − b

⎠+ a


1
V2


− 1
V1



⎠.


Significance
By taking into account the volume of molecules, the expression for work is much more complex. If, however,we set a = 0 and b = 0, we see that the expression for work matches exactly the work done by an isothermal
process for one mole of an ideal gas.


Check Your Understanding How much work is done by the gas, as given in Figure 3.6, when itexpands quasi-statically along the path ADC?


Internal Energy
The internal energy Eint of a thermodynamic system is, by definition, the sum of the mechanical energies of all the
molecules or entities in the system. If the kinetic and potential energies of molecule i are Ki and Ui, respectively, then
the internal energy of the system is the average of the total mechanical energy of all the entities:


(3.5)Eint = ∑
i


(K

i + U



i),


where the summation is over all the molecules of the system, and the bars over K and U indicate average values. The kineticenergy Ki of an individual molecule includes contributions due to its rotation and vibration, as well as its translational
energy mi vi2/2, where vi is the molecule’s speed measured relative to the center of mass of the system. The potential
energy Ui is associated only with the interactions between molecule i and the other molecules of the system. In fact, neither
the system’s location nor its motion is of any consequence as far as the internal energy is concerned. The internal energy ofthe system is not affected by moving it from the basement to the roof of a 100-story building or by placing it on a movingtrain.
In an ideal monatomic gas, each molecule is a single atom. Consequently, there is no rotational or vibrational kinetic energy
and Ki = mi vi2/2 . Furthermore, there are no interatomic interactions (collisions notwithstanding), so Ui = constant ,
which we set to zero. The internal energy is therefore due to translational kinetic energy only and


Chapter 3 | The First Law of Thermodynamics 115




Eint = ∑
i
K

i = ∑


i


1
2
mi vi


2.


From the discussion in the preceding chapter, we know that the average kinetic energy of a molecule in an ideal monatomicgas is
1
2
mi vi


2


= 3
2
kBT ,


where T is the Kelvin temperature of the gas. Consequently, the average mechanical energy per molecule of an idealmonatomic gas is also 3kBT /2, that is,
Ki + Ui = Ki



= 3


2
kBT .


The internal energy is just the number of molecules multiplied by the average mechanical energy per molecule. Thus for nmoles of an ideal monatomic gas,


(3.6)Eint = nNA ⎛⎝32kBT⎞⎠ = 32nRT .


Notice that the internal energy of a given quantity of an ideal monatomic gas depends on just the temperature and iscompletely independent of the pressure and volume of the gas. For other systems, the internal energy cannot be expressedso simply. However, an increase in internal energy can often be associated with an increase in temperature.
We know from the zeroth law of thermodynamics that when two systems are placed in thermal contact, they eventuallyreach thermal equilibrium, at which point they are at the same temperature. As an example, suppose we mix two monatomicideal gases. Now, the energy per molecule of an ideal monatomic gas is proportional to its temperature. Thus, when the twogases are mixed, the molecules of the hotter gas must lose energy and the molecules of the colder gas must gain energy.This continues until thermal equilibrium is reached, at which point, the temperature, and therefore the average translationalkinetic energy per molecule, is the same for both gases. The approach to equilibrium for real systems is somewhat morecomplicated than for an ideal monatomic gas. Nevertheless, we can still say that energy is exchanged between the systemsuntil their temperatures are the same.
3.3 | First Law of Thermodynamics


Learning Objectives
By the end of this section, you will be able to:
• State the first law of thermodynamics and explain how it is applied
• Explain how heat transfer, work done, and internal energy change are related in anythermodynamic process


Now that we have seen how to calculate internal energy, heat, and work done for a thermodynamic system undergoingchange during some process, we can see how these quantities interact to affect the amount of change that can occur. Thisinteraction is given by the first law of thermodynamics. British scientist and novelist C. P. Snow (1905–1980) is creditedwith a joke about the four laws of thermodynamics. His humorous statement of the first law of thermodynamics is stated“you can’t win,” or in other words, you cannot get more energy out of a system than you put into it. We will see in thischapter how internal energy, heat, and work all play a role in the first law of thermodynamics.
Suppose Q represents the heat exchanged between a system and the environment, and W is the work done by or on thesystem. The first law states that the change in internal energy of that system is given by Q −W . Since added heat increases
the internal energy of a system, Q is positive when it is added to the system and negative when it is removed from thesystem.
When a gas expands, it does work and its internal energy decreases. Thus, W is positive when work is done by the systemand negative when work is done on the system. This sign convention is summarized in Table 3.1. The first law of


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thermodynamics is stated as follows:
First Law of Thermodynamics
Associated with every equilibrium state of a system is its internal energy Eint. The change in Eint for any transition
between two equilibrium states is


(3.7)ΔEint = Q −W
where Q and W represent, respectively, the heat exchanged by the system and the work done by or on the system.


Thermodynamic Sign Conventions for Heat and Work
Process Convention
Heat added to system Q > 0
Heat removed from system Q < 0
Work done by system W > 0
Work done on system W < 0


Table 3.1
The first law is a statement of energy conservation. It tells us that a system can exchange energy with its surroundingsby the transmission of heat and by the performance of work. The net energy exchanged is then equal to the change in thetotal mechanical energy of the molecules of the system (i.e., the system’s internal energy). Thus, if a system is isolated, itsinternal energy must remain constant.
Although Q and W both depend on the thermodynamic path taken between two equilibrium states, their difference Q −W
does not. Figure 3.7 shows the pV diagram of a system that is making the transition from A to B repeatedly along differentthermodynamic paths. Along path 1, the system absorbs heat Q1 and does work W1; along path 2, it absorbs heat Q2
and does work W2, and so on. The values of Qi and Wi may vary from path to path, but we have


Q1 −W1 = Q2 −W2 = ⋯ = Qi −Wi = ⋯,


or
ΔEint1 = ΔEint2 = ⋯ = ΔEinti = ⋯.


That is, the change in the internal energy of the system between A and B is path independent. In the chapter on potentialenergy and the conservation of energy, we encountered another path-independent quantity: the change in potential energybetween two arbitrary points in space. This change represents the negative of the work done by a conservative forcebetween the two points. The potential energy is a function of spatial coordinates, whereas the internal energy is a functionof thermodynamic variables. For example, we might write Eint(T , p) for the internal energy. Functions such as internal
energy and potential energy are known as state functions because their values depend solely on the state of the system.


Chapter 3 | The First Law of Thermodynamics 117




Figure 3.7 Different thermodynamic paths taken by a systemin going from state A to state B. For all transitions, the change inthe internal energy of the system ΔEint = Q −W is the same.


Often the first law is used in its differential form, which is
(3.8)dEint = dQ − dW.


Here dEint is an infinitesimal change in internal energy when an infinitesimal amount of heat dQ is exchanged with the
system and an infinitesimal amount of work dW is done by (positive in sign) or on (negative in sign) the system.
Example 3.2


Changes of State and the First Law
During a thermodynamic process, a system moves from state A to state B, it is supplied with 400 J of heat anddoes 100 J of work. (a) For this transition, what is the system’s change in internal energy? (b) If the system thenmoves from state B back to state A, what is its change in internal energy? (c) If in moving from A to B along adifferent path, W′AB = 400 J of work is done on the system, how much heat does it absorb?
Strategy
The first law of thermodynamics relates the internal energy change, work done by the system, and the heattransferred to the system in a simple equation. The internal energy is a function of state and is therefore fixed atany given point regardless of how the system reaches the state.
Solutiona. From the first law, the change in the system’s internal energy is


ΔEintAB = QAB −WAB = 400 J − 100 J = 300 J.


b. Consider a closed path that passes through the states A and B. Internal energy is a state function, so ΔEint
is zero for a closed path. Thus


ΔEint = ΔEintAB + ΔEintBA = 0,


and
ΔEintAB = −ΔEintBA.


This yields
ΔEintBA = −300 J.


c. The change in internal energy is the same for any path, so


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ΔEintAB = ΔE′intAB = Q′AB – W′AB;


300 J = Q′AB – (−400 J),


and the heat exchanged is
Q′AB = −100 J.


The negative sign indicates that the system loses heat in this transition.
Significance
When a closed cycle is considered for the first law of thermodynamics, the change in internal energy around thewhole path is equal to zero. If friction were to play a role in this example, less work would result from this heatadded. Example 3.3 takes into consideration what happens if friction plays a role.


Notice that in Example 3.2, we did not assume that the transitions were quasi-static. This is because the first law is notsubject to such a restriction. It describes transitions between equilibrium states but is not concerned with the intermediatestates. The system does not have to pass through only equilibrium states. For example, if a gas in a steel container at awell-defined temperature and pressure is made to explode by means of a spark, some of the gas may condense, different gasmolecules may combine to form new compounds, and there may be all sorts of turbulence in the container—but eventually,the system will settle down to a new equilibrium state. This system is clearly not in equilibrium during its transition;however, its behavior is still governed by the first law because the process starts and ends with the system in equilibriumstates.
Example 3.3


Polishing a Fitting
A machinist polishes a 0.50-kg copper fitting with a piece of emery cloth for 2.0 min. He moves the cloth acrossthe fitting at a constant speed of 1.0 m/s by applying a force of 20 N, tangent to the surface of the fitting. (a) Whatis the total work done on the fitting by the machinist? (b) What is the increase in the internal energy of the fitting?Assume that the change in the internal energy of the cloth is negligible and that no heat is exchanged between thefitting and its environment. (c) What is the increase in the temperature of the fitting?
Strategy
The machinist’s force over a distance that can be calculated from the speed and time given is the work done onthe system. The work, in turn, increases the internal energy of the system. This energy can be interpreted as theheat that raises the temperature of the system via its heat capacity. Be careful with the sign of each quantity.
Solutiona. The power created by a force on an object or the rate at which the machinist does frictional work on the


fitting is F→ · v→ = −Fv . Thus, in an elapsed time Δt (2.0 min), the work done on the fitting is
W = −FvΔt = −(20 N)(0.1 m/s)(1.2 × 102 s)


= −2.4 × 103 J.


b. By assumption, no heat is exchanged between the fitting and its environment, so the first law gives forthe change in the internal energy of the fitting:
ΔEint = −W = 2.4 × 10


3 J.


c. Since ΔEint is path independent, the effect of the 2.4 × 103 J of work is the same as if it were supplied
at atmospheric pressure by a transfer of heat. Thus,


2.4 × 103 J = mcΔT = (0.50 kg)(3.9 × 102 J/kg · °C)ΔT ,


and the increase in the temperature of the fitting is


Chapter 3 | The First Law of Thermodynamics 119




3.2


ΔT = 12 °C,


where we have used the value for the specific heat of copper, c = 3.9 × 102 J/kg · °C .
Significance
If heat were released, the change in internal energy would be less and cause less of a temperature change thanwhat was calculated in the problem.


Check Your Understanding The quantities below represent four different transitions between the sameinitial and final state. Fill in the blanks.
Q (J) W (J) ΔEint(J)
–80 –120
90


40
–40


Table 3.2


Example 3.4
An Ideal Gas Making Transitions between Two States
Consider the quasi-static expansions of an ideal gas between the equilibrium states A and C of Figure 3.6. If515 J of heat are added to the gas as it traverses the path ABC, how much heat is required for the transition
along ADC? Assume that p1 = 2.10 × 105 N/m2 , p2 = 1.05 × 105 N/m2 , V1 = 2.25 × 10−3 m3 , and
V2 = 4.50 × 10


−3 m3.


Strategy
The difference in work done between process ABC and process ADC is the area enclosed by ABCD. Because thechange of the internal energy (a function of state) is the same for both processes, the difference in work is thusthe same as the difference in heat transferred to the system.
Solution
For path ABC, the heat added is QABC = 515 J and the work done by the gas is the area under the path on the
pV diagram, which is


WABC = p1(V2 − V1) = 473 J.


Along ADC, the work done by the gas is again the area under the path:
WADC = p2(V2 − V1) = 236 J.


Then using the strategy we just described, we have
QADC − QABC = WADC −WABC,


which leads to
QADC = QABC +WADC −WABC = (515 + 236 − 473) J = 278 J.


Significance
The work calculations in this problem are made simple since no work is done along AD and BC and along AB and


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3.3


DC; the pressure is constant over the volume change, so the work done is simply pΔV . An isothermal line could
also have been used, as we have derived the work for an isothermal process as W = nRTlnV2


V1
.


Example 3.5
Isothermal Expansion of an Ideal Gas
Heat is added to 1 mol of an ideal monatomic gas confined to a cylinder with a movable piston at one end. Thegas expands quasi-statically at a constant temperature of 300 K until its volume increases from V to 3V. (a) Whatis the change in internal energy of the gas? (b) How much work does the gas do? (c) How much heat is added tothe gas?
Strategy
(a) Because the system is an ideal gas, the internal energy only changes when the temperature changes. (b) Theheat added to the system is therefore purely used to do work that has been calculated in Work, Heat, andInternal Energy. (c) Lastly, the first law of thermodynamics can be used to calculate the heat added to the gas.
Solutiona. We saw in the preceding section that the internal energy of an ideal monatomic gas is a function only oftemperature. Since ΔT = 0 , for this process, ΔEint = 0.


b. The quasi-static isothermal expansion of an ideal gas was considered in the preceding section and wasfound to be
W = nRTln


V2
V1


= nRTln3V
V


= (1.00 mol)(8.314 J/K ·mol)(300 K)(ln3) = 2.74 × 103 J.


c. With the results of parts (a) and (b), we can use the first law to determine the heat added:
ΔEint = Q −W = 0,


which leads to
Q = W = 2.74 × 103 J.


Significance
An isothermal process has no change in the internal energy. Based on that, the first law of thermodynamicsreduces to Q = W .


Check Your Understanding Why was it necessary to state that the process of Example 3.5 is quasi-static?


Example 3.6
Vaporizing Water
When 1.00 g of water at 100 °C changes from the liquid to the gas phase at atmospheric pressure, its change in
volume is 1.67 × 10−3 m3 . (a) How much heat must be added to vaporize the water? (b) How much work is
done by the water against the atmosphere in its expansion? (c) What is the change in the internal energy of thewater?


Chapter 3 | The First Law of Thermodynamics 121




3.4


Strategy
We can first figure out how much heat is needed from the latent heat of vaporization of the water. From thevolume change, we can calculate the work done from W = pΔV because the pressure is constant. Then, the first
law of thermodynamics provides us with the change in the internal energy.
Solutiona. With Lv representing the latent heat of vaporization, the heat required to vaporize the water is


Q = mLv = (1.00 g)(2.26 × 10
3 J/g) = 2.26 × 103 J.


b. Since the pressure on the system is constant at 1.00 atm = 1.01 × 105 N/m2 , the work done by the
water as it is vaporized is


W = pΔV = (1.01 × 105 N/m2)(1.67 × 10−3 m3) = 169 J.


c. From the first law, the thermal energy of the water during its vaporization changes by
ΔEint = Q −W = 2.26 × 10


3 J − 169 J = 2.09 × 103 J.


Significance
We note that in part (c), we see a change in internal energy, yet there is no change in temperature. Ideal gases thatare not undergoing phase changes have the internal energy proportional to temperature. Internal energy in generalis the sum of all energy in the system.


Check Your Understanding When 1.00 g of ammonia boils at atmospheric pressure and −33.0 °C, its
volume changes from 1.47 to 1130 cm3 . Its heat of vaporization at this pressure is 1.37 × 106 J/kg. What is
the change in the internal energy of the ammonia when it vaporizes?


View this site (https://openstaxcollege.org/l/211stlawthermo) to learn about how the first law ofthermodynamics. First, pump some heavy species molecules into the chamber. Then, play around by doing work(pushing the wall to the right where the person is located) to see how the internal energy changes (as seen bytemperature). Then, look at how heat added changes the internal energy. Finally, you can set a parameter constantsuch as temperature and see what happens when you do work to keep the temperature constant (Note: You mightsee a change in these variables initially if you are moving around quickly in the simulation, but ultimately, thisvalue will return to its equilibrium value).


3.4 | Thermodynamic Processes
Learning Objectives


By the end of this section, you will be able to:
• Define a thermodynamic process
• Distinguish between quasi-static and non-quasi-static processes
• Calculate physical quantities, such as the heat transferred, work done, and internal energychange for isothermal, adiabatic, and cyclical thermodynamic processes


In solving mechanics problems, we isolate the body under consideration, analyze the external forces acting on it, and thenuse Newton’s laws to predict its behavior. In thermodynamics, we take a similar approach. We start by identifying the partof the universe we wish to study; it is also known as our system. (We defined a system at the beginning of this chapter asanything whose properties are of interest to us; it can be a single atom or the entire Earth.) Once our system is selected,we determine how the environment, or surroundings, interact with the system. Finally, with the interaction understood, westudy the thermal behavior of the system with the help of the laws of thermodynamics.


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The thermal behavior of a system is described in terms of thermodynamic variables. For an ideal gas, these variables arepressure, volume, temperature, and the number of molecules or moles of the gas. Different types of systems are generallycharacterized by different sets of variables. For example, the thermodynamic variables for a stretched rubber band aretension, length, temperature, and mass.
The state of a system can change as a result of its interaction with the environment. The change in a system can be fast orslow and large or small. The manner in which a state of a system can change from an initial state to a final state is calleda thermodynamic process. For analytical purposes in thermodynamics, it is helpful to divide up processes as either quasi-static or non-quasi-static, as we now explain.
Quasi-static and Non-quasi-static Processes
A quasi-static process refers to an idealized or imagined process where the change in state is made infinitesimally slowly sothat at each instant, the system can be assumed to be at a thermodynamic equilibrium with itself and with the environment.For instance, imagine heating 1 kg of water from a temperature 20 °C to 21 °C at a constant pressure of 1 atmosphere. To
heat the water very slowly, we may imagine placing the container with water in a large bath that can be slowly heated suchthat the temperature of the bath can rise infinitesimally slowly from 20 °C to 21 °C . If we put 1 kg of water at 20 °C
directly into a bath at 21 °C , the temperature of the water will rise rapidly to 21 °C in a non-quasi-static way.
Quasi-static processes are done slowly enough that the system remains at thermodynamic equilibrium at each instant,despite the fact that the system changes over time. The thermodynamic equilibrium of the system is necessary for the systemto have well-defined values of macroscopic properties such as the temperature and the pressure of the system at each instantof the process. Therefore, quasi-static processes can be shown as well-defined paths in state space of the system.
Since quasi-static processes cannot be completely realized for any finite change of the system, all processes in natureare non-quasi-static. Examples of quasi-static and non-quasi-static processes are shown in Figure 3.8. Despite the factthat all finite changes must occur essentially non-quasi-statically at some stage of the change, we can imagine performinginfinitely many quasi-static process corresponding to every quasi-static process. Since quasi-static processes can beanalyzed analytically, we mostly study quasi-static processes in this book. We have already seen that in a quasi-static processthe work by a gas is given by pdV.


Figure 3.8 Quasi-static and non-quasi-static processesbetween states A and B of a gas. In a quasi-static process, thepath of the process between A and B can be drawn in a statediagram since all the states that the system goes through areknown. In a non-quasi-static process, the states between A and Bare not known, and hence no path can be drawn. It may followthe dashed line as shown in the figure or take a very differentpath.
Isothermal Processes
An isothermal process is a change in the state of the system at a constant temperature. This process is accomplished bykeeping the system in thermal equilibrium with a large heat bath during the process. Recall that a heat bath is an idealized“infinitely” large system whose temperature does not change. In practice, the temperature of a finite bath is controlled byeither adding or removing a finite amount of energy as the case may be.
As an illustration of an isothermal process, consider a cylinder of gas with a movable piston immersed in a large water tank


Chapter 3 | The First Law of Thermodynamics 123




whose temperature is maintained constant. Since the piston is freely movable, the pressure inside Pin is balanced by the
pressure outside Pout by some weights on the piston, as in Figure 3.9.


Figure 3.9 Expanding a system at a constant temperature. Removing weights onthe piston leads to an imbalance of forces on the piston, which causes the piston tomove up. As the piston moves up, the temperature is lowered momentarily, whichcauses heat to flow from the heat bath to the system. The energy to move the pistoneventually comes from the heat bath.


As weights on the piston are removed, an imbalance of forces on the piston develops. The net nonzero force on the pistonwould cause the piston to accelerate, resulting in an increase in volume. The expansion of the gas cools the gas to a lowertemperature, which makes it possible for the heat to enter from the heat bath into the system until the temperature of thegas is reset to the temperature of the heat bath. If weights are removed in infinitesimal steps, the pressure in the systemdecreases infinitesimally slowly. This way, an isothermal process can be conducted quasi-statically. An isothermal line ona (p, V) diagram is represented by a curved line from starting point A to finishing point B, as seen in Figure 3.10. For an
ideal gas, an isothermal process is hyperbolic, since for an ideal gas at constant temperature, p ∝ 1


V
.


Figure 3.10 An isothermal expansion from a state labeled Ato another state labeled B on a pV diagram. The curve representsthe relation between pressure and volume in an ideal gas atconstant temperature.


An isothermal process studied in this chapter is quasi-statically performed, since to be isothermal throughout the changeof volume, you must be able to state the temperature of the system at each step, which is possible only if the system isin thermal equilibrium continuously. The system must go out of equilibrium for the state to change, but for quasi-staticprocesses, we imagine that the process is conducted in infinitesimal steps such that these departures from equilibrium canbe made as brief and as small as we like.
Other quasi-static processes of interest for gases are isobaric and isochoric processes. An isobaric process is a processwhere the pressure of the system does not change, whereas an isochoric process is a process where the volume of thesystem does not change.
Adiabatic Processes
In an adiabatic process, the system is insulated from its environment so that although the state of the system changes,no heat is allowed to enter or leave the system, as seen in Figure 3.11. An adiabatic process can be conducted eitherquasi-statically or non-quasi-statically. When a system expands adiabatically, it must do work against the outside world, and


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therefore its energy goes down, which is reflected in the lowering of the temperature of the system. An adiabatic expansionleads to a lowering of temperature, and an adiabatic compression leads to an increase of temperature. We discuss adiabaticexpansion again in Adiabatic Processes for an ideal Gas.


Figure 3.11 An insulated piston with a hot, compressed gas isreleased. The piston moves up, the volume expands, and thepressure and temperature decrease. The internal energy goes intowork. If the expansion occurs within a time frame in whichnegligible heat can enter the system, then the process is calledadiabatic. Ideally, during an adiabatic process no heat enters orexits the system.
Cyclic Processes
We say that a system goes through a cyclic process if the state of the system at the end is same as the state at the beginning.Therefore, state properties such as temperature, pressure, volume, and internal energy of the system do not change over acomplete cycle:


ΔEint = 0.


When the first law of thermodynamics is applied to a cyclic process, we obtain a simple relation between heat into thesystem and the work done by the system over the cycle:
Q = W ⎛⎝cyclic process⎞⎠.


Thermodynamic processes are also distinguished by whether or not they are reversible. A reversible process is one that canbe made to retrace its path by differential changes in the environment. Such a process must therefore also be quasi-static.Note, however, that a quasi-static process is not necessarily reversible, since there may be dissipative forces involved. Forexample, if friction occurred between the piston and the walls of the cylinder containing the gas, the energy lost to frictionwould prevent us from reproducing the original states of the system.
We considered several thermodynamic processes:


1. An isothermal process, during which the system’s temperature remains constant
2. An adiabatic process, during which no heat is transferred to or from the system
3. An isobaric process, during which the system’s pressure does not change
4. An isochoric process, during which the system’s volume does not change


Many other processes also occur that do not fit into any of these four categories.
View this site (https://openstaxcollege.org/l/21idegaspvdiag) to set up your own process in a pV diagram.See if you can calculate the values predicted by the simulation for heat, work, and change in internal energy.


Chapter 3 | The First Law of Thermodynamics 125




3.5 | Heat Capacities of an Ideal Gas
Learning Objectives


By the end of this section, you will be able to:
• Define heat capacity of an ideal gas for a specific process
• Calculate the specific heat of an ideal gas for either an isobaric or isochoric process
• Explain the difference between the heat capacities of an ideal gas and a real gas
• Estimate the change in specific heat of a gas over temperature ranges


We learned about specific heat and molar heat capacity in Temperature and Heat; however, we have not considered aprocess in which heat is added. We do that in this section. First, we examine a process where the system has a constantvolume, then contrast it with a system at constant pressure and show how their specific heats are related.
Let’s start with looking at Figure 3.12, which shows two vessels A and B, each containing 1 mol of the same type of idealgas at a temperature T and a volume V. The only difference between the two vessels is that the piston at the top of A is fixed,whereas the one at the top of B is free to move against a constant external pressure p. We now consider what happens whenthe temperature of the gas in each vessel is slowly increased to T + dT with the addition of heat.


Figure 3.12 Two vessels are identical except that the piston atthe top of A is fixed, whereas that atop B is free to move againsta constant external pressure p.


Since the piston of vessel A is fixed, the volume of the enclosed gas does not change. Consequently, the gas does no work,and we have from the first law
dEint = dQ − dW = dQ.


We represent the fact that the heat is exchanged at constant volume by writing
dQ = CV dT ,


where CV is the molar heat capacity at constant volume of the gas. In addition, since dEint = dQ for this particular
process,


(3.9)dEint = CV dT .
We obtained this equation assuming the volume of the gas was fixed. However, internal energy is a state function thatdepends on only the temperature of an ideal gas. Therefore, dEint = CV dT gives the change in internal energy of an ideal
gas for any process involving a temperature change dT.
When the gas in vessel B is heated, it expands against the movable piston and does work dW = pdV . In this case, the heat
is added at constant pressure, and we write


dQ = C pdT ,


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where C p is the molar heat capacity at constant pressure of the gas. Furthermore, since the ideal gas expands against a
constant pressure,


d(pV) = d(RT)


becomes
pdV = RdT .


Finally, inserting the expressions for dQ and pdV into the first law, we obtain
dEint = dQ − pdV = (C p − R)dT .


We have found dEint for both an isochoric and an isobaric process. Because the internal energy of an ideal gas depends
only on the temperature, dEint must be the same for both processes. Thus,


CV dT = (C p − R)dT ,


and


(3.10)C p = CV + R.


The derivation of Equation 3.10 was based only on the ideal gas law. Consequently, this relationship is approximatelyvalid for all dilute gases, whether monatomic like He, diatomic like O2, or polyatomic like CO2 or NH3 .
In the preceding chapter, we found the molar heat capacity of an ideal gas under constant volume to be


CV =
d
2
R,


where d is the number of degrees of freedom of a molecule in the system. Table 3.3 shows the molar heat capacities ofsome dilute ideal gases at room temperature. The heat capacities of real gases are somewhat higher than those predicted bythe expressions of CV and C p given in Equation 3.10. This indicates that vibrational motion in polyatomic molecules
is significant, even at room temperature. Nevertheless, the difference in the molar heat capacities, C p − CV, is very close
to R, even for the polyatomic gases.


Molar Heat Capacities of Dilute Ideal Gases at Room Temperature
Type of Molecule Gas C p


(J/mol K)
CV


(J/mol K)
C p−CV


(J/mol K)
Monatomic Ideal 5


2
R = 20.79 3


2
R = 12.47 R = 8.31


Diatomic Ideal 7
2
R = 29.10 5


2
R = 20.79 R = 8.31


Polyatomic Ideal 4R = 33.26 3R = 24.94 R = 8.31
Table 3.3


Chapter 3 | The First Law of Thermodynamics 127




3.6 | Adiabatic Processes for an Ideal Gas
Learning Objectives


By the end of this section, you will be able to:
• Define adiabatic expansion of an ideal gas
• Demonstrate the qualitative difference between adiabatic and isothermal expansions


When an ideal gas is compressed adiabatically (Q = 0), work is done on it and its temperature increases; in an adiabatic
expansion, the gas does work and its temperature drops. Adiabatic compressions actually occur in the cylinders of a car,where the compressions of the gas-air mixture take place so quickly that there is no time for the mixture to exchange heatwith its environment. Nevertheless, because work is done on the mixture during the compression, its temperature does risesignificantly. In fact, the temperature increases can be so large that the mixture can explode without the addition of a spark.Such explosions, since they are not timed, make a car run poorly—it usually “knocks.” Because ignition temperature riseswith the octane of gasoline, one way to overcome this problem is to use a higher-octane gasoline.
Another interesting adiabatic process is the free expansion of a gas. Figure 3.13 shows a gas confined by a membrane toone side of a two-compartment, thermally insulated container. When the membrane is punctured, gas rushes into the emptyside of the container, thereby expanding freely. Because the gas expands “against a vacuum” (p = 0) , it does no work, and
because the vessel is thermally insulated, the expansion is adiabatic. With Q = 0 and W = 0 in the first law, ΔEint = 0,
so Eint i = Eint f for the free expansion.


Figure 3.13 The gas in the left chamber expands freely into the right chamber when the membrane is punctured.


If the gas is ideal, the internal energy depends only on the temperature. Therefore, when an ideal gas expands freely, itstemperature does not change.
A quasi-static, adiabatic expansion of an ideal gas is represented in Figure 3.14, which shows an insulated cylinder thatcontains 1 mol of an ideal gas. The gas is made to expand quasi-statically by removing one grain of sand at a time fromthe top of the piston. When the gas expands by dV, the change in its temperature is dT. The work done by the gas in theexpansion is dW = pdV; dQ = 0 because the cylinder is insulated; and the change in the internal energy of the gas is,
from Equation 3.9, dEint = CV dT . Therefore, from the first law,


CV dT = 0 − pdV = −pdV ,


so
dT = −


pdV
CV


.


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Figure 3.14 When sand is removed from the piston one grainat a time, the gas expands adiabatically and quasi-statically inthe insulated vessel.


Also, for 1 mol of an ideal gas,
d(pV) = d(RT),


so
pdV + Vdp = RdT


and
dT =


pdV + Vdp
R


.


We now have two equations for dT. Upon equating them, we find that
CV Vdp + (CV + R)pdV = 0.


Now, we divide this equation by pV and use C p = CV + R . We are then left with
CV


dp
p + C p


dV
V


= 0,


which becomes
dp
p + γ


dV
V


= 0,


where we define γ as the ratio of the molar heat capacities:


(3.11)
γ =


C p
CV


.


Chapter 3 | The First Law of Thermodynamics 129




Thus,


dp
p + γ




dV
V


= 0


and
ln p + γln V = constant.


Finally, using ln(Ax) = xlnA and ln AB = ln A + ln B , we can write this in the form


(3.12)pV γ = constant.


This equation is the condition that must be obeyed by an ideal gas in a quasi-static adiabatic process. For example, if anideal gas makes a quasi-static adiabatic transition from a state with pressure and volume p1 and V1 to a state with p2
and V2, then it must be true that p1V1γ = p2V2γ.
The adiabatic condition of Equation 3.12 can be written in terms of other pairs of thermodynamic variables by combiningit with the ideal gas law. In doing this, we find that


(3.13)p1 − γ T γ = constant
and


(3.14)TV γ − 1 = constant.
A reversible adiabatic expansion of an ideal gas is represented on the pV diagram of Figure 3.15. The slope of the curveat any point is


dp
dV


= d
dV


constant


V γ

⎠ = −γ


p
V
.


Figure 3.15 Quasi-static adiabatic and isothermal expansionsof an ideal gas.


The dashed curve shown on this pV diagram represents an isothermal expansion where T (and therefore pV) is constant. Theslope of this curve is useful when we consider the second law of thermodynamics in the next chapter. This slope is
dp
dV


= d
dV


nRT
V


= −
p
V
.


Because γ > 1, the isothermal curve is not as steep as that for the adiabatic expansion.


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Example 3.7
Compression of an Ideal Gas in an Automobile Engine
Gasoline vapor is injected into the cylinder of an automobile engine when the piston is in its expanded position.
The temperature, pressure, and volume of the resulting gas-air mixture are 20 °C , 1.00 × 105 N/m2 , and
240 cm3 , respectively. The mixture is then compressed adiabatically to a volume of 40 cm3 . Note that in
the actual operation of an automobile engine, the compression is not quasi-static, although we are making thatassumption here. (a) What are the pressure and temperature of the mixture after the compression? (b) How muchwork is done by the mixture during the compression?
Strategy
Because we are modeling the process as a quasi-static adiabatic compression of an ideal gas, we have
pV γ = constant and pV = nRT . The work needed can then be evaluated with W = ∫


V1


V2
pdV .


Solutiona. For an adiabatic compression we have
p2 = p1




V1
V2





γ


,


so after the compression, the pressure of the mixture is
p2 = (1.00 × 10


5 N/m2)


240 × 10−6 m3


40 × 10−6 m3



1.40


= 1.23 × 106 N/m2 .


From the ideal gas law, the temperature of the mixture after the compression is
T2 =


p2V2
p1V1



⎠T1


= (1.23 × 10
6 N/m2)(40 × 10−6 m3)


(1.00 × 105 N/m2)(240 × 10−6 m3)
· 293 K


= 600 K = 328 °C.


b. The work done by the mixture during the compression is
W = ∫


V1


V2
pdV .


With the adiabatic condition of Equation 3.12, we may write p as K/V γ, where K = p1V1γ = p2V2γ.
The work is therefore


Chapter 3 | The First Law of Thermodynamics 131




W = ⌠
⌡V1


V2
K
V γ


dV


= K
1 − γ





⎜ 1
V2


γ − 1
− 1
V1


γ − 1







= 1
1 − γ





⎜ p2V2


γ


V2
γ − 1



p1V1


γ


V1
γ − 1







= 1
1 − γ


(p2V2 − p1V1)


= 1
1 − 1.40


[(1.23 × 106 N/m2)(40 × 10−6 m3)


−(1.00 × 105 N/m2)(240 × 10−6m3)]
= −63 J.


Significance
The negative sign on the work done indicates that the piston does work on the gas-air mixture. The engine wouldnot work if the gas-air mixture did work on the piston.


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adiabatic process
boundary
closed system
cyclic process
environment
equation of state
equilibrium
extensive variable
first law of thermodynamics
intensive variable
internal energy
isobaric process
isochoric process
isothermal process
molar heat capacity at constant pressure
molar heat capacity at constant volume
open system
quasi-static process
reversible process
surroundings
thermodynamic process
thermodynamic system


CHAPTER 3 REVIEW
KEY TERMS


process during which no heat is transferred to or from the system
imagined walls that separate the system and its surroundings


system that is mechanically and thermally isolated from its environment
process in which the state of the system at the end is same as the state at the beginning


outside of the system being studied
describes properties of matter under given physical conditions


thermal balance established between two objects or parts within a system
variable that is proportional to the amount of matter in the system


the change in internal energy for any transition between two equilibrium states is
ΔEint = Q −W


variable that is independent of the amount of matter in the system
average of the total mechanical energy of all the molecules or entities in the system
process during which the system’s pressure does not change
process during which the system’s volume does not change
process during which the system’s temperature remains constant


quantifies the ratio of the amount of heat added removed to thetemperature while measuring at constant pressure
quantifies the ratio of the amount of heat added removed to thetemperature while measuring at constant volume


system that can exchange energy and/or matter with its surroundings
evolution of a system that goes so slowly that the system involved is always in thermodynamicequilibrium


process that can be reverted to restore both the system and its environment back to their originalstates together
environment that interacts with an open system


manner in which a state of a system can change from initial state to final state
object and focus of thermodynamic study


KEY EQUATIONS
Equation of state for a closed system f (p, V , T) = 0
Net work for a finite change in volume


W = ∫
V1


V2
pdV


Internal energy of a system (average total energy)
Eint = ∑


i
(K
¯
i + U


¯
i),


Internal energy of a monatomic ideal gas Eint = nNA ⎛⎝32kBT⎞⎠ = 32nRT
First law of thermodynamics ΔEint = Q −W


Chapter 3 | The First Law of Thermodynamics 133




Molar heat capacity at constant pressure C p = CV + R
Ratio of molar heat capacities γ = C p /CV
Condition for an ideal gas in a quasi-static adiabatic process pV γ = constant


SUMMARY
3.1 Thermodynamic Systems


• A thermodynamic system, its boundary, and its surroundings must be defined with all the roles of the componentsfully explained before we can analyze a situation.
• Thermal equilibrium is reached with two objects if a third object is in thermal equilibrium with the other twoseparately.
• A general equation of state for a closed system has the form f (p, V , T) = 0, with an ideal gas as an illustrative
example.


3.2 Work, Heat, and Internal Energy
• Positive (negative) work is done by a thermodynamic system when it expands (contracts) under an external pressure.
• Heat is the energy transferred between two objects (or two parts of a system) because of a temperature difference.
• Internal energy of a thermodynamic system is its total mechanical energy.


3.3 First Law of Thermodynamics
• The internal energy of a thermodynamic system is a function of state and thus is unique for every equilibrium stateof the system.
• The increase in the internal energy of the thermodynamic system is given by the heat added to the system less thework done by the system in any thermodynamics process.


3.4 Thermodynamic Processes
• The thermal behavior of a system is described in terms of thermodynamic variables. For an ideal gas, these variablesare pressure, volume, temperature, and number of molecules or moles of the gas.
• For systems in thermodynamic equilibrium, the thermodynamic variables are related by an equation of state.
• A heat reservoir is so large that when it exchanges heat with other systems, its temperature does not change.
• A quasi-static process takes place so slowly that the system involved is always in thermodynamic equilibrium.
• A reversible process is one that can be made to retrace its path and both the temperature and pressure are uniformthroughout the system.
• There are several types of thermodynamic processes, including (a) isothermal, where the system’s temperature isconstant; (b) adiabatic, where no heat is exchanged by the system; (c) isobaric, where the system’s pressure isconstant; and (d) isochoric, where the system’s volume is constant.
• As a consequence of the first law of thermodymanics, here is a summary of the thermodymaic processes: (a)isothermal: ΔEint = 0, Q = W; (b) adiabatic: Q = 0, ΔEint = −W; (c) isobaric: ΔEint = Q −W; and (d)
isochoric: W = 0, ΔEint = Q.


3.5 Heat Capacities of an Ideal Gas
• For an ideal gas, the molar capacity at constant pressure C p is given by C p = CV + R = dR/2 + R , where d is
the number of degrees of freedom of each molecule/entity in the system.


• A real gas has a specific heat close to but a little bit higher than that of the corresponding ideal gas with


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C p ≃ CV + R.


3.6 Adiabatic Processes for an Ideal Gas
• A quasi-static adiabatic expansion of an ideal gas produces a steeper pV curve than that of the correspondingisotherm.
• A realistic expansion can be adiabatic but rarely quasi-static.


CONCEPTUAL QUESTIONS
3.1 Thermodynamic Systems
1. Consider these scenarios and state whether work isdone by the system on the environment (SE) or by theenvironment on the system (ES): (a) opening a carbonatedbeverage; (b) filling a flat tire; (c) a sealed empty gas canexpands on a hot day, bowing out the walls.


3.2 Work, Heat, and Internal Energy
2. Is it possible to determine whether a change in internalenergy is caused by heat transferred, by work performed, orby a combination of the two?
3. When a liquid is vaporized, its change in internal energyis not equal to the heat added. Why?
4. Why does a bicycle pump feel warm as you inflate yourtire?
5. Is it possible for the temperature of a system to remainconstant when heat flows into or out of it? If so, giveexamples.


3.3 First Law of Thermodynamics
6. What does the first law of thermodynamics tell us aboutthe energy of the universe?
7. Does adding heat to a system always increase itsinternal energy?
8. A great deal of effort, time, and money has been spentin the quest for a so-called perpetual-motion machine,which is defined as a hypothetical machine that operatesor produces useful work indefinitely and/or a hypotheticalmachine that produces more work or energy than itconsumes. Explain, in terms of the first law ofthermodynamics, why or why not such a machine is likelyto be constructed.


3.4 Thermodynamic Processes
9. When a gas expands isothermally, it does work. What is


the source of energy needed to do this work?
10. If the pressure and volume of a system are given, is thetemperature always uniquely determined?
11. It is unlikely that a process can be isothermal unlessit is a very slow process. Explain why. Is the same true forisobaric and isochoric processes? Explain your answer.


3.5 Heat Capacities of an Ideal Gas
12. How can an object transfer heat if the object does notpossess a discrete quantity of heat?
13. Most materials expand when heated. One notableexception is water between 0 °C and 4 °C, which actually
decreases in volume with the increase in temperature.Which is greater for water in this temperature region, C p
or CV ?
14. Why are there two specific heats for gases C p and
CV , yet only one given for solid?


3.6 Adiabatic Processes for an Ideal Gas
15. Is it possible for γ to be smaller than unity?
16. Would you expect γ to be larger for a gas or a solid?
Explain.
17. There is no change in the internal energy of an idealgas undergoing an isothermal process since the internalenergy depends only on the temperature. Is it thereforecorrect to say that an isothermal process is the same as anadiabatic process for an ideal gas? Explain your answer.
18. Does a gas do any work when it expandsadiabatically? If so, what is the source of the energy neededto do this work?


Chapter 3 | The First Law of Thermodynamics 135




PROBLEMS
3.1 Thermodynamic Systems
19. A gas follows pV = bp + cT on an isothermal curve,
where p is the pressure, V is the volume, b is a constant,and c is a function of temperature. Show that a temperaturescale under an isochoric process can be established withthis gas and is identical to that of an ideal gas.
20. A mole of gas has isobaric expansion coefficient
dV /dT = R/p and isochoric pressure-temperature
coefficient dp/dT = p/T . Find the equation of state of the
gas.
21. Find the equation of state of a solid that has an isobaricexpansion coefficient dV /dT = 2cT − bp and an
isothermal pressure-volume coefficient dV /dp = −bT .


3.2 Work, Heat, and Internal Energy
22. A gas at a pressure of 2.00 atm undergoes a quasi-static isobaric expansion from 3.00 to 5.00 L. How muchwork is done by the gas?
23. It takes 500 J of work to compress quasi-statically0.50 mol of an ideal gas to one-fifth its original volume.Calculate the temperature of the gas, assuming it remainsconstant during the compression.
24. It is found that, when a dilute gas expands quasi-statically from 0.50 to 4.0 L, it does 250 J of work.Assuming that the gas temperature remains constant at 300K, how many moles of gas are present?
25. In a quasi-static isobaric expansion, 500 J of work aredone by the gas. If the gas pressure is 0.80 atm, what isthe fractional increase in the volume of the gas, assuming itwas originally at 20.0 L?
26. When a gas undergoes a quasi-static isobaric changein volume from 10.0 to 2.0 L, 15 J of work from an externalsource are required. What is the pressure of the gas?
27. An ideal gas expands quasi-statically and isothermallyfrom a state with pressure p and volume V to a state withvolume 4V. Show that the work done by the gas in theexpansion is pV(ln 4).
28. As shown below, calculate the work done by the gas inthe quasi-static processes represented by the paths (a) AB;(b) ADB; (c) ACB; and (d) ADCB.


29. (a) Calculate the work done by the gas along theclosed path shown below. The curved section between Rand S is semicircular. (b) If the process is carried out in theopposite direction, what is the work done by the gas?


30. An ideal gas expands quasi-statically to three timesits original volume. Which process requires more workfrom the gas, an isothermal process or an isobaric one?Determine the ratio of the work done in these processes.
31. A dilute gas at a pressure of 2.0 atm and a volumeof 4.0 L is taken through the following quasi-static steps:(a) an isobaric expansion to a volume of 10.0 L, (b) anisochoric change to a pressure of 0.50 atm, (c) an isobariccompression to a volume of 4.0 L, and (d) an isochoricchange to a pressure of 2.0 atm. Show these steps on a pVdiagram and determine from your graph the net work doneby the gas.
32. What is the average mechanical energy of the atoms ofan ideal monatomic gas at 300 K?
33. What is the internal energy of 6.00 mol of an idealmonatomic gas at 200 °C ?
34. Calculate the internal energy of 15 mg of helium at atemperature of 0 °C.
35. Two monatomic ideal gases A and B are at the sametemperature. If 1.0 g of gas A has the same internal energyas 0.10 g of gas B, what are (a) the ratio of the number ofmoles of each gas and (b) the ration of the atomic massesof the two gases?


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36. The van der Waals coefficients for oxygen are
a = 0.138 J ·m3 /mol2 and b = 3.18 × 10−5 m3 /mol .
Use these values to draw a van der Waals isotherm ofoxygen at 100 K. On the same graph, draw isotherms of onemole of an ideal gas.
37. Find the work done in the quasi-static processes shownbelow. The states are given as (p, V) values for the pointsin the pV plane: 1 (3 atm, 4 L), 2 (3 atm, 6 L), 3 (5 atm, 4L), 4 (2 atm, 6 L), 5 (4 atm, 2 L), 6 (5 atm, 5 L), and 7 (2atm, 5 L).


3.3 First Law of Thermodynamics
38. When a dilute gas expands quasi-statically from 0.50to 4.0 L, it does 250 J of work. Assuming that the gastemperature remains constant at 300 K, (a) what is thechange in the internal energy of the gas? (b) How muchheat is absorbed by the gas in this process?
39. In a quasi-static isobaric expansion, 500 J of workare done by the gas. The gas pressure is 0.80 atm and itwas originally at 20.0 L. If the internal energy of the gasincreased by 80 J in the expansion, how much heat does thegas absorb?
40. An ideal gas expands quasi-statically and isothermally


from a state with pressure p and volume V to a state withvolume 4V. How much heat is added to the expanding gas?
41. As shown below, if the heat absorbed by the gas alongAB is 400 J, determine the quantities of heat absorbedalong (a) ADB; (b) ACB; and (c) ADCB.


42. During the isobaric expansion from A to B representedbelow, 130 J of heat are removed from the gas. What is thechange in its internal energy?


43. (a) What is the change in internal energy for theprocess represented by the closed path shown below? (b)How much heat is exchanged? (c) If the path is traversed inthe opposite direction, how much heat is exchanged?


44. When a gas expands along path AC shown below,it does 400 J of work and absorbs either 200 or 400 J ofheat. (a) Suppose you are told that along path ABC, the gasabsorbs either 200 or 400 J of heat. Which of these valuesis correct? (b) Give the correct answer from part (a), howmuch work is done by the gas along ABC? (c) Along CD,the internal energy of the gas decreases by 50 J. How muchheat is exchanged by the gas along this path?


Chapter 3 | The First Law of Thermodynamics 137




45. When a gas expands along AB (see below), it does 500J of work and absorbs 250 J of heat. When the gas expandsalong AC, it does 700 J of work and absorbs 300 J of heat.(a) How much heat does the gas exchange along BC? (b)When the gas makes the transmission from C to A alongCDA, 800 J of work are done on it from C to D. How muchheat does it exchange along CDA?


46. A dilute gas is stored in the left chamber of a containerwhose walls are perfectly insulating (see below), and theright chamber is evacuated. When the partition is removed,the gas expands and fills the entire container. Calculate thework done by the gas. Does the internal energy of the gaschange in this process?


47. Ideal gases A and B are stored in the left and rightchambers of an insulated container, as shown below. Thepartition is removed and the gases mix. Is any work donein this process? If the temperatures of A and B are initiallyequal, what happens to their common temperature afterthey are mixed?


48. An ideal monatomic gas at a pressure of
2.0 × 105 N/m2 and a temperature of 300 K undergoes
a quasi-static isobaric expansion from
2.0 × 103 to 4.0 × 103 cm3. (a) What is the work done
by the gas? (b) What is the temperature of the gas after theexpansion? (c) How many moles of gas are there? (d) Whatis the change in internal energy of the gas? (e) How muchheat is added to the gas?
49. Consider the process for steam in a cylinder shownbelow. Suppose the change in the internal energy in thisprocess is 30 kJ. Find the heat entering the system.


50. The state of 30 moles of steam in a cylinder is changedin a cyclic manner from a-b-c-a, where the pressure andvolume of the states are: a (30 atm, 20 L), b (50 atm, 20L), and c (50 atm, 45 L). Assume each change takes placealong the line connecting the initial and final states in thepV plane. (a) Display the cycle in the pV plane. (b) Findthe net work done by the steam in one cycle. (c) Find thenet amount of heat flow in the steam over the course of onecycle.
51. A monatomic ideal gas undergoes a quasi-staticprocess that is described by the function
p(V) = p1 + 3(V − V1) , where the starting state is

⎝p1, V1



⎠ and the final state ⎛⎝p2, V2⎞⎠ . Assume the system


consists of n moles of the gas in a container that canexchange heat with the environment and whose volume canchange freely. (a) Evaluate the work done by the gas duringthe change in the state. (b) Find the change in internalenergy of the gas. (c) Find the heat input to the gas duringthe change. (d) What are initial and final temperatures?


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52. A metallic container of fixed volume of
2.5 × 10−3 m3 immersed in a large tank of temperature
27 °C contains two compartments separated by a freely
movable wall. Initially, the wall is kept in place by a stopperso that there are 0.02 mol of the nitrogen gas on one sideand 0.03 mol of the oxygen gas on the other side, eachoccupying half the volume. When the stopper is removed,the wall moves and comes to a final position. Themovement of the wall is controlled so that the wall movesin infinitesimal quasi-static steps. (a) Find the finalvolumes of the two sides assuming the ideal gas behaviorfor the two gases. (b) How much work does each gas do onthe other? (c) What is the change in the internal energy ofeach gas? (d) Find the amount of heat that enters or leaveseach gas.
53. A gas in a cylindrical closed container is adiabaticallyand quasi-statically expanded from a state A (3 MPa, 2L) to a state B with volume of 6 L along the path
1.8 pV = constant. (a) Plot the path in the pV plane. (b)
Find the amount of work done by the gas and the change inthe internal energy of the gas during the process.


3.4 Thermodynamic Processes
54. Two moles of a monatomic ideal gas at (5 MPa, 5 L) isexpanded isothermally until the volume is doubled (step 1).Then it is cooled isochorically until the pressure is 1 MPa(step 2). The temperature drops in this process. The gas isnow compressed isothermally until its volume is back to 5L, but its pressure is now 2 MPa (step 3). Finally, the gas isheated isochorically to return to the initial state (step 4). (a)Draw the four processes in the pV plane. (b) Find the totalwork done by the gas.
55. Consider a transformation from point A to B in a two-step process. First, the pressure is lowered from 3 MPa atpoint A to a pressure of 1 MPa, while keeping the volumeat 2 L by cooling the system. The state reached is labeledC. Then the system is heated at a constant pressure toreach a volume of 6 L in the state B. (a) Find the amountof work done on the ACB path. (b) Find the amount ofheat exchanged by the system when it goes from A to Bon the ACB path. (c) Compare the change in the internalenergy when the AB process occurs adiabatically with theAB change through the two-step process on the ACB path.
56. Consider a cylinder with a movable piston containingn moles of an ideal gas. The entire apparatus is immersedin a constant temperature bath of temperature T kelvin. Thepiston is then pushed slowly so that the pressure of thegas changes quasi-statically from p1 to p2 at constant
temperature T. Find the work done by the gas in terms of n,R, T, p1, and p2.


57. An ideal gas expands isothermally along AB and does700 J of work (see below). (a) How much heat does the gasexchange along AB? (b) The gas then expands adiabaticallyalong BC and does 400 J of work. When the gas returns toA along CA, it exhausts 100 J of heat to its surroundings.How much work is done on the gas along this path?


58. Consider the processes shown below. In the processesAB and BC, 3600 J and 2400 J of heat are added to thesystem, respectively. (a) Find the work done in each ofthe processes AB, BC, AD, and DC. (b) Find the internalenergy change in processes AB and BC. (c) Find theinternal energy difference between states C and A. (d) Findthe total heat added in the ADC process. (e) From theinformation give, can you find the heat added in processAD? Why or why not?


59. Two moles of helium gas are placed in a cylindricalcontainer with a piston. The gas is at room temperature
25 °C and under a pressure of 3.0 × 105 Pa. When the
pressure from the outside is decreased while keeping thetemperature the same as the room temperature, the volumeof the gas doubles. (a) Find the work the external agentdoes on the gas in the process. (b) Find the heat exchangedby the gas and indicate whether the gas takes in or gives upheat. Assume ideal gas behavior.
60. An amount of n moles of a monatomic ideal gas ina conducting container with a movable piston is placed ina large thermal heat bath at temperature T1 and the gas
is allowed to come to equilibrium. After the equilibriumis reached, the pressure on the piston is lowered so that


Chapter 3 | The First Law of Thermodynamics 139




the gas expands at constant temperature. The process iscontinued quasi-statically until the final pressure is 4/3 ofthe initial pressure p1. (a) Find the change in the internal
energy of the gas. (b) Find the work done by the gas. (c)Find the heat exchanged by the gas, and indicate, whetherthe gas takes in or gives up heat.


3.5 Heat Capacities of an Ideal Gas
61. The temperature of an ideal monatomic gas rises by8.0 K. What is the change in the internal energy of 1 mol ofthe gas at constant volume?
62. For a temperature increase of 10 °C at constant
volume, what is the heat absorbed by (a) 3.0 mol of a dilutemonatomic gas; (b) 0.50 mol of a dilute diatomic gas; and(c) 15 mol of a dilute polyatomic gas?
63. If the gases of the preceding problem are initially at300 K, what are their internal energies after they absorb theheat?
64. Consider 0.40 mol of dilute carbon dioxide at apressure of 0.50 atm and a volume of 50 L. What is theinternal energy of the gas?
65. When 400 J of heat are slowly added to 10 mol of anideal monatomic gas, its temperature rises by 10 °C . What
is the work done on the gas?
66. One mole of a dilute diatomic gas occupying a volumeof 10.00 L expands against a constant pressure of 2.000 atmwhen it is slowly heated. If the temperature of the gas risesby 10.00 K and 400.0 J of heat are added in the process,what is its final volume?


3.6 Adiabatic Processes for an Ideal Gas
67. A monatomic ideal gas undergoes a quasi-staticadiabatic expansion in which its volume is doubled. How isthe pressure of the gas changed?
68. An ideal gas has a pressure of 0.50 atm and a volumeof 10 L. It is compressed adiabatically and quasi-staticallyuntil its pressure is 3.0 atm and its volume is 2.8 L. Is thegas monatomic, diatomic, or polyatomic?
69. Pressure and volume measurements of a dilute gasundergoing a quasi-static adiabatic expansion are shownbelow. Plot ln p vs. V and determine γ for this gas from
your graph.


P (atm) V (L)
20.0 1.0
17.0 1.1
14.0 1.3
11.0 1.5
8.0 2.0
5.0 2.6
2.0 5.2
1.0 8.4


70. An ideal monatomic gas at 300 K expandsadiabatically and reversibly to twice its volume. What is itsfinal temperature?
71. An ideal diatomic gas at 80 K is slowly compressedadiabatically and reversibly to twice its volume. What is itsfinal temperature?
72. An ideal diatomic gas at 80 K is slowly compressedadiabatically to one-third its original volume. What is itsfinal temperature?
73. Compare the charge in internal energy of an idealgas for a quasi-static adiabatic expansion with that for aquasi-static isothermal expansion. What happens to thetemperature of an ideal gas in an adiabatic expansion?
74. The temperature of n moles of an ideal gas changesfrom T1 to T2 in a quasi-static adiabatic transition. Show
that the work done by the gas is given by
W = nR


γ − 1
(T1 − T2).


75. A dilute gas expands quasi-statically to three timesits initial volume. Is the final gas pressure greater for anisothermal or an adiabatic expansion? Does your answerdepend on whether the gas is monatomic, diatomic, orpolyatomic?
76. (a) An ideal gas expands adiabatically from a volume
of 2.0 × 10−3 m3 to 2.5 × 10−3 m3 . If the initial
pressure and temperature were 5.0 × 105 Pa and 300 K,
respectively, what are the final pressure and temperatureof the gas? Use γ = 5/3 for the gas. (b) In an isothermal
process, an ideal gas expands from a volume of
2.0 × 10−3 m3 to 2.5 × 10−3 m3 . If the initial pressure
and temperature were 5.0 × 105 Pa and 300 K,
respectively, what are the final pressure and temperature of


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the gas?
77. On an adiabatic process of an ideal gas pressure,volume and temperature change such that pV γ is constant
with γ = 5/3 for monatomic gas such as helium and
γ = 7/5 for diatomic gas such as hydrogen at room
temperature. Use numerical values to plot two isotherms of1 mol of helium gas using ideal gas law and two adiabaticprocesses mediating between them. Use


T1 = 500 K, V1 = 1 L, and T2 = 300 K for your plot.
78. Two moles of a monatomic ideal gas such as heliumis compressed adiabatically and reversibly from a state (3atm, 5 L) to a state with pressure 4 atm. (a) Find the volumeand temperature of the final state. (b) Find the temperatureof the initial state of the gas. (c) Find the work done by thegas in the process. (d) Find the change in internal energy ofthe gas in the process.


ADDITIONAL PROBLEMS
79. Consider the process shown below. During steps ABand BC, 3600 J and 2400 J of heat, respectively, are addedto the system. (a) Find the work done in each of theprocesses AB, BC, AD, and DC. (b) Find the internal energychange in processes AB and BC. (c) Find the internal energydifference between states C and A. (d) Find the total heatadded in the ADC process. (e) From the information given,can you find the heat added in process AD? Why or whynot?


80. A car tire contains 0.0380 m3 of air at a pressure of
2.20 × 105 Pa (about 32 psi). How much more internal
energy does this gas have than the same volume has at zerogauge pressure (which is equivalent to normal atmosphericpressure)?
81. A helium-filled toy balloon has a gauge pressure of0.200 atm and a volume of 10.0 L. How much greater is theinternal energy of the helium in the balloon than it wouldbe at zero gauge pressure?
82. Steam to drive an old-fashioned steam locomotive is
supplied at a constant gauge pressure of 1.75 × 106N/m2
(about 250 psi) to a piston with a 0.200-m radius. (a) Bycalculating pΔV , find the work done by the steam when
the piston moves 0.800 m. Note that this is the net workoutput, since gauge pressure is used. (b) Now find theamount of work by calculating the force exerted times thedistance traveled. Is the answer the same as in part (a)?


83. A hand-driven tire pump has a piston with a 2.50-cmdiameter and a maximum stroke of 30.0 cm. (a) How muchwork do you do in one stroke if the average gauge pressure
is 2.4 × 105 N/m2 (about 35 psi)? (b) What average force
do you exert on the piston, neglecting friction andgravitational force?
84. Calculate the net work output of a heat enginefollowing path ABCDA as shown below.


85. What is the net work output of a heat engine thatfollows path ABDA in the preceding problem with a straightline from B to D? Why is the work output less than for pathABCDA?
86. Five moles of a monatomic ideal gas in a cylinder at
27 °C is expanded isothermally from a volume of 5 L to
10 L. (a) What is the change in internal energy? (b) Howmuch work was done on the gas in the process? (c) Howmuch heat was transferred to the gas?
87. Four moles of a monatomic ideal gas in a cylinderat 27 °C is expanded at constant pressure equal to 1 atm
until its volume doubles. (a) What is the change in internalenergy? (b) How much work was done by the gas in theprocess? (c) How much heat was transferred to the gas?
88. Helium gas is cooled from 20 °C to 10 °C by
expanding from 40 atm to 1 atm. If there is 1.4 mol of


Chapter 3 | The First Law of Thermodynamics 141




helium, (a) What is the final volume of helium? (b) What isthe change in internal energy?
89. In an adiabatic process, oxygen gas in a containeris compressed along a path that can be described by thefollowing pressure in atm as a function of volume V, with
V0 = 1L : p = (3.0 atm)(V /V0 )−1.2 . The initial and final
volumes during the process were 2 L and 1.5 L,respectively. Find the amount of work done on the gas.
90. A cylinder containing three moles of a monatomicideal gas is heated at a constant pressure of 2 atm. Thetemperature of the gas changes from 300 K to 350 K as aresult of the expansion. Find work done (a) on the gas; and(b) by the gas.
91. A cylinder containing three moles of nitrogen gas isheated at a constant pressure of 2 atm. The temperature ofthe gas changes from 300 K to 350 K as a result of theexpansion. Find work done (a) on the gas, and (b) by the


gas by using van der Waals equation of state instead of idealgas law.
92. Two moles of a monatomic ideal gas such as oxygenis compressed adiabatically and reversibly from a state (3atm, 5 L) to a state with a pressure of 4 atm. (a) Findthe volume and temperature of the final state. (b) Find thetemperature of the initial state. (c) Find work done by thegas in the process. (d) Find the change in internal energy inthe process. Assume CV = 5R and C p = CV + R for the
diatomic ideal gas in the conditions given.
93. An insulated vessel contains 1.5 moles of argon at 2atm. The gas initially occupies a volume of 5 L. As a resultof the adiabatic expansion the pressure of the gas is reducedto 1 atm. (a) Find the volume and temperature of the finalstate. (b) Find the temperature of the gas in the initial state.(c) Find the work done by the gas in the process. (d) Findthe change in the internal energy of the gas in the process.


CHALLENGE PROBLEMS
94. One mole of an ideal monatomic gas occupies a
volume of 1.0 × 10−2 m3 at a pressure of
2.0 × 105 N/m2. (a) What is the temperature of the gas?
(b) The gas undergoes a quasi-static adiabatic compression
until its volume is decreased to 5.0 × 10−3 m3. What is
the new gas temperature? (c) How much work is done onthe gas during the compression? (d) What is the change inthe internal energy of the gas?
95. One mole of an ideal gas is initially in a chamber of
volume 1.0 × 10−2 m3 and at a temperature of 27 °C .
(a) How much heat is absorbed by the gas when it slowlyexpands isothermally to twice its initial volume? (b)Suppose the gas is slowly transformed to the same finalstate by first decreasing the pressure at constant volume andthen expanding it isobarically. What is the heat transferredfor this case? (c) Calculate the heat transferred when thegas is transformed quasi-statically to the same final stateby expanding it isobarically, then decreasing its pressure atconstant volume.
96. A bullet of mass 10 g is traveling horizontally at 200m/s when it strikes and embeds in a pendulum bob of mass2.0 kg. (a) How much mechanical energy is dissipated inthe collision? (b) Assuming that Cv for the bob plus bullet
is 3R, calculate the temperature increase of the system dueto the collision. Take the molecular mass of the system tobe 200 g/mol.
97. The insulated cylinder shown below is closed at both


ends and contains an insulating piston that is free to moveon frictionless bearings. The piston divides the chamberinto two compartments containing gases A and B.Originally, each compartment has a volume of
5.0 × 10−2 m3 and contains a monatomic ideal gas at a
temperature of 0 °C and a pressure of 1.0 atm. (a) How
many moles of gas are in each compartment? (b) Heat Q isslowly added to A so that it expands and B is compresseduntil the pressure of both gases is 3.0 atm. Use the factthat the compression of B is adiabatic to determine the finalvolume of both gases. (c) What are their final temperatures?(d) What is the value of Q?


98. In a diesel engine, the fuel is ignited without a sparkplug. Instead, air in a cylinder is compressed adiabaticallyto a temperature above the ignition temperature of the fuel;at the point of maximum compression, the fuel is injectedinto the cylinder. Suppose that air at 20 °C is taken into the
cylinder at a volume V1 and then compressed adiabatically
and quasi-statically to a temperature of 600 °C and a


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volume V2. If γ = 1.4, what is the ratio V1/V2? (Note:
In an operating diesel engine, the compression is not quasi-


static.)


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4 | THE SECOND LAW OFTHERMODYNAMICS


Figure 4.1 A xenon ion engine from the Jet Propulsion Laboratory shows the faint blue glow of charged atoms emitted fromthe engine. The ion propulsion engine is the first nonchemical propulsion to be used as the primary means of propelling aspacecraft.


Chapter Outline
4.1 Reversible and Irreversible Processes
4.2 Heat Engines
4.3 Refrigerators and Heat Pumps
4.4 Statements of the Second Law of Thermodynamics
4.5 The Carnot Cycle
4.6 Entropy
4.7 Entropy on a Microscopic Scale


Introduction
According to the first law of thermodynamics, the only processes that can occur are those that conserve energy. But thiscannot be the only restriction imposed by nature, because many seemingly possible thermodynamic processes that wouldconserve energy do not occur. For example, when two bodies are in thermal contact, heat never flows from the colder bodyto the warmer one, even though this is not forbidden by the first law. So some other thermodynamic principles must becontrolling the behavior of physical systems.
One such principle is the second law of thermodynamics, which limits the use of energy within a source. Energy cannotarbitrarily pass from one object to another, just as we cannot transfer heat from a cold object to a hot one without doingany work. We cannot unmix cream from coffee without a chemical process that changes the physical characteristics of thesystem or its environment. We cannot use internal energy stored in the air to propel a car, or use the energy of the ocean torun a ship, without disturbing something around that object.
In the chapter covering the first law of thermodynamics, we started our discussion with a joke by C. P. Snow stating that


Chapter 4 | The Second Law of Thermodynamics 145




the first law means “you can’t win.” He paraphrased the second law as “you can’t break even, except on a very cold day.”Unless you are at zero kelvin, you cannot convert 100% of thermal energy into work. We start by discussing spontaneousprocesses and explain why some processes require work to occur even if energy would have been conserved.
4.1 | Reversible and Irreversible Processes


Learning Objectives
By the end of this section, you will be able to:
• Define reversible and irreversible processes
• State the second law of thermodynamics via an irreversible process


Consider an ideal gas that is held in half of a thermally insulated container by a wall in the middle of the container. Theother half of the container is under vacuum with no molecules inside. Now, if we remove the wall in the middle quickly, thegas expands and fills up the entire container immediately, as shown in Figure 4.2.


Figure 4.2 A gas expanding from half of a container to the entire container (a) before and (b) after the wall in the middle isremoved.


Because half of the container is under vacuum before the gas expands there, we do not expect any work to be done by thesystem—that is, W = 0—because no force from the vacuum is exerted on the gas during the expansion. If the container
is thermally insulated from the rest of the environment, we do not expect any heat transfer to the system either, so Q = 0 .
Then the first law of thermodynamics leads to the change of the internal energy of the system,


ΔEint = Q −W = 0.


For an ideal gas, if the internal energy doesn’t change, then the temperature stays the same. Thus, the equation of state ofthe ideal gas gives us the final pressure of the gas, p = nRT /V = p0 /2, where p0 is the pressure of the gas before the
expansion. The volume is doubled and the pressure is halved, but nothing else seems to have changed during the expansion.
All of this discussion is based on what we have learned so far and makes sense. Here is what puzzles us: Can all themolecules go backward to the original half of the container in some future time? Our intuition tells us that this is going to bevery unlikely, even though nothing we have learned so far prevents such an event from happening, regardless of how smallthe probability is. What we are really asking is whether the expansion into the vacuum half of the container is reversible.
A reversible process is a process in which the system and environment can be restored to exactly the same initial states thatthey were in before the process occurred, if we go backward along the path of the process. The necessary condition for areversible process is therefore the quasi-static requirement. Note that it is quite easy to restore a system to its original state;the hard part is to have its environment restored to its original state at the same time. For example, in the example of an idealgas expanding into vacuum to twice its original volume, we can easily push it back with a piston and restore its temperatureand pressure by removing some heat from the gas. The problem is that we cannot do it without changing something in itssurroundings, such as dumping some heat there.
A reversible process is truly an ideal process that rarely happens. We can make certain processes close to reversible andtherefore use the consequences of the corresponding reversible processes as a starting point or reference. In reality, almost


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all processes are irreversible, and some properties of the environment are altered when the properties of the system arerestored. The expansion of an ideal gas, as we have just outlined, is irreversible because the process is not even quasi-static,that is, not in an equilibrium state at any moment of the expansion.
From the microscopic point of view, a particle described by Newton’s second law can go backward if we flip the direction
of time. But this is not the case, in practical terms, in a macroscopic system with more than 1023 particles or molecules,
where numerous collisions between these molecules tend to erase any trace of memory of the initial trajectory of each ofthe particles. For example, we can actually estimate the chance for all the particles in the expanded gas to go back to theoriginal half of the container, but the current age of the universe is still not long enough for it to happen even once.
An irreversible process is what we encounter in reality almost all the time. The system and its environment cannot berestored to their original states at the same time. Because this is what happens in nature, it is also called a natural process.The sign of an irreversible process comes from the finite gradient between the states occurring in the actual process. Forexample, when heat flows from one object to another, there is a finite temperature difference (gradient) between the twoobjects. More importantly, at any given moment of the process, the system most likely is not at equilibrium or in a well-defined state. This phenomenon is called irreversibility.
Let us see another example of irreversibility in thermal processes. Consider two objects in thermal contact: one attemperature T1 and the other at temperature T2 > T1 , as shown in Figure 4.3.


Figure 4.3 Spontaneous heat flow from an object at highertemperature T2 to another at lower temperature T1.


We know from common personal experience that heat flows from a hotter object to a colder one. For example, when wehold a few pieces of ice in our hands, we feel cold because heat has left our hands into the ice. The opposite is true whenwe hold one end of a metal rod while keeping the other end over a fire. Based on all of the experiments that have been doneon spontaneous heat transfer, the following statement summarizes the governing principle:
Second Law of Thermodynamics (Clausius statement)
Heat never flows spontaneously from a colder object to a hotter object.


This statement turns out to be one of several different ways of stating the second law of thermodynamics. The form of thisstatement is credited to German physicist Rudolf Clausius (1822−1888) and is referred to as the Clausius statement of thesecond law of thermodynamics. The word “spontaneously” here means no other effort has been made by a third party, orone that is neither the hotter nor colder object. We will introduce some other major statements of the second law and showthat they imply each other. In fact, all the different statements of the second law of thermodynamics can be shown to beequivalent, and all lead to the irreversibility of spontaneous heat flow between macroscopic objects of a very large numberof molecules or particles.
Both isothermal and adiabatic processes sketched on a pV graph (discussed in The First Law of Thermodynamics) arereversible in principle because the system is always at an equilibrium state at any point of the processes and can go forwardor backward along the given curves. Other idealized processes can be represented by pV curves; Table 4.1 summarizes themost common reversible processes.


Process Constant Quantity and Resulting Fact
Isobaric Constant pressure W = pΔV
Isochoric Constant volume W = 0


Table 4.1 Summary of Simple Thermodynamic Processes


Chapter 4 | The Second Law of Thermodynamics 147




Process Constant Quantity and Resulting Fact
Isothermal Constant temperature ΔT = 0
Adiabatic No heat transfer Q = 0


Table 4.1 Summary of Simple Thermodynamic Processes
4.2 | Heat Engines


Learning Objectives
By the end of this section, you will be able to:
• Describe the function and components of a heat engine
• Explain the efficiency of an engine
• Calculate the efficiency of an engine for a given cycle of an ideal gas


A heat engine is a device used to extract heat from a source and then convert it into mechanical work that is used forall sorts of applications. For example, a steam engine on an old-style train can produce the work needed for driving thetrain. Several questions emerge from the construction and application of heat engines. For example, what is the maximumpercentage of the heat extracted that can be used to do work? This turns out to be a question that can only be answeredthrough the second law of thermodynamics.
The second law of thermodynamics can be formally stated in several ways. One statement presented so far is about thedirection of spontaneous heat flow, known as the Clausius statement. A couple of other statements are based on heat engines.Whenever we consider heat engines and associated devices such as refrigerators and heat pumps, we do not use the normalsign convention for heat and work. For convenience, we assume that the symbols Qh, Qc, and W represent only the
amounts of heat transferred and work delivered, regardless what the givers or receivers are. Whether heat is entering orleaving a system and work is done to or by a system are indicated by proper signs in front of the symbols and by thedirections of arrows in diagrams.
It turns out that we need more than one heat source/sink to construct a heat engine. We will come back to this point laterin the chapter, when we compare different statements of the second law of thermodynamics. For the moment, we assumethat a heat engine is constructed between a heat source (high-temperature reservoir or hot reservoir) and a heat sink (low-temperature reservoir or cold reservoir), represented schematically in Figure 4.4. The engine absorbs heat Qh from a heat
source ( hot reservoir) of Kelvin temperature Th, uses some of that energy to produce useful work W, and then discards
the remaining energy as heat Qc into a heat sink ( cold reservoir) of Kelvin temperature Tc. Power plants and internal
combustion engines are examples of heat engines. Power plants use steam produced at high temperature to drive electricgenerators, while exhausting heat to the atmosphere or a nearby body of water in the role of the heat sink. In an internalcombustion engine, a hot gas-air mixture is used to push a piston, and heat is exhausted to the nearby atmosphere in a similarmanner.


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Figure 4.4 A schematic representation of a heat engine.Energy flows from the hot reservoir to the cold reservoir whiledoing work.


Actual heat engines have many different designs. Examples include internal combustion engines, such as those used in mostcars today, and external combustion engines, such as the steam engines used in old steam-engine trains. Figure 4.5 showsa photo of a nuclear power plant in operation. The atmosphere around the reactors acts as the cold reservoir, and the heatgenerated from the nuclear reaction provides the heat from the hot reservoir.


Figure 4.5 The heat exhausted from a nuclear power plantgoes to the cooling towers, where it is released into theatmosphere.


Heat engines operate by carrying a working substance through a cycle. In a steam power plant, the working substance iswater, which starts as a liquid, becomes vaporized, is then used to drive a turbine, and is finally condensed back into theliquid state. As is the case for all working substances in cyclic processes, once the water returns to its initial state, it repeatsthe same sequence.
For now, we assume that the cycles of heat engines are reversible, so there is no energy loss to friction or other irreversibleeffects. Suppose that the engine of Figure 4.4 goes through one complete cycle and that Qh, Qc, and W represent the
heats exchanged and the work done for that cycle. Since the initial and final states of the system are the same, ΔEint = 0
for the cycle. We therefore have from the first law of thermodynamics,


W = Q − ΔEint = (Qh − Qc) − 0,


so that


(4.1)W = Qh − Qc.


The most important measure of a heat engine is its efficiency (e), which is simply “what we get out” divided by “what weput in” during each cycle, as defined by e = Wout /Qin.


Chapter 4 | The Second Law of Thermodynamics 149




With a heat engine working between two heat reservoirs, we get out W and put in Qh, so the efficiency of the engine is


(4.2)e = W
Qh


= 1 − Qc
Qh


.


Here, we used Equation 4.1, W = Qh − Qc, in the final step of this expression for the efficiency.
Example 4.1


A Lawn Mower
A lawn mower is rated to have an efficiency of 25.0% and an average power of 3.00 kW. What are (a) the
average work and (b) the minimum heat discharge into the air by the lawn mower in one minute of use?
Strategy
From the average power—that is, the rate of work production—we can figure out the work done in the givenelapsed time. Then, from the efficiency given, we can figure out the minimum heat discharge Qc = Qh(1 − e)
with Qh = Qc +W.
Solutiona. The average work delivered by the lawn mower is


W = PΔt = 3.00 × 103 × 60 × 1.00 J = 180 kJ.
b. The minimum heat discharged into the air is given by


Qc = Qh(1 − e) = (Qc +W)(1 − e),


which leads to
Qc = W(1/e − 1) = 180 × (1/0.25 − 1) kJ = 540 kJ.


Significance
As the efficiency rises, the minimum heat discharged falls. This helps our environment and atmosphere by nothaving as much waste heat expelled.


4.3 | Refrigerators and Heat Pumps
Learning Objectives


By the end of this section, you will be able to:
• Describe a refrigerator and a heat pump and list their differences
• Calculate the performance coefficients of simple refrigerators and heat pumps


The cycles we used to describe the engine in the preceding section are all reversible, so each sequence of steps can just aseasily be performed in the opposite direction. In this case, the engine is known as a refrigerator or a heat pump, dependingon what is the focus: the heat removed from the cold reservoir or the heat dumped to the hot reservoir. Either a refrigeratoror a heat pump is an engine running in reverse. For a refrigerator, the focus is on removing heat from a specific area. Fora heat pump, the focus is on dumping heat to a specific area.
We first consider a refrigerator (Figure 4.6). The purpose of this engine is to remove heat from the cold reservoir, whichis the space inside the refrigerator for an actual household refrigerator or the space inside a building for an air-conditioningunit.


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Figure 4.6 A schematic representation of a refrigerator (or aheat pump). The arrow next to work (W) indicates work beingput into the system.


A refrigerator (or heat pump) absorbs heat Qc from the cold reservoir at Kelvin temperature Tc and discards heat Qh
to the hot reservoir at Kelvin temperature Th, while work W is done on the engine’s working substance, as shown by
the arrow pointing toward the system in the figure. A household refrigerator removes heat from the food within it whileexhausting heat to the surrounding air. The required work, for which we pay in our electricity bill, is performed by the motorthat moves a coolant through the coils. A schematic sketch of a household refrigerator is given in Figure 4.7.


Chapter 4 | The Second Law of Thermodynamics 151




Figure 4.7 A schematic diagram of a household refrigerator. A coolant with aboiling temperature below the freezing point of water is sent through the cycle(clockwise in this diagram). The coolant extracts heat from the refrigerator at theevaporator, causing coolant to vaporize. It is then compressed and sent through thecondenser, where it exhausts heat to the outside.


The effectiveness or coefficient of performance KR of a refrigerator is measured by the heat removed from the cold
reservoir divided by the work done by the working substance cycle by cycle:


(4.3)KR = QcW = QcQh − Qc.


Note that we have used the condition of energy conservation, W = Qh − Qc, in the final step of this expression.
The effectiveness or coefficient of performance KP of a heat pump is measured by the heat dumped to the hot reservoir
divided by the work done to the engine on the working substance cycle by cycle:


(4.4)
KP =


Qh
W


=
Qh


Qh − Qc
.


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Once again, we use the energy conservation condition W = Qh − Qc to obtain the final step of this expression.


4.4 | Statements of the Second Law of Thermodynamics
Learning Objectives


By the end of this section, you will be able to:
• Contrast the second law of thermodynamics statements according to Kelvin and Clausiusformulations
• Interpret the second of thermodynamics via irreversibility


Earlier in this chapter, we introduced the Clausius statement of the second law of thermodynamics, which is based onthe irreversibility of spontaneous heat flow. As we remarked then, the second law of thermodynamics can be stated inseveral different ways, and all of them can be shown to imply the others. In terms of heat engines, the second law ofthermodynamics may be stated as follows:
Second Law of Thermodynamics (Kelvin statement)
It is impossible to convert the heat from a single source into work without any other effect.


This is known as the Kelvin statement of the second law of thermodynamics. This statement describes an unattainable“ perfect engine,” as represented schematically in Figure 4.8(a). Note that “without any other effect” is a very strongrestriction. For example, an engine can absorb heat and turn it all into work, but not if it completes a cycle. Withoutcompleting a cycle, the substance in the engine is not in its original state and therefore an “other effect” has occurred.Another example is a chamber of gas that can absorb heat from a heat reservoir and do work isothermally against a pistonas it expands. However, if the gas were returned to its initial state (that is, made to complete a cycle), it would have to becompressed and heat would have to be extracted from it.
The Kelvin statement is a manifestation of a well-known engineering problem. Despite advancing technology, we are notable to build a heat engine that is 100% efficient. The first law does not exclude the possibility of constructing a perfect
engine, but the second law forbids it.


Figure 4.8 (a) A “perfect heat engine” converts all input heat into work. (b) A “perfectrefrigerator” transports heat from a cold reservoir to a hot reservoir without work input. Neitherof these devices is achievable in reality.


We can show that the Kelvin statement is equivalent to the Clausius statement if we view the two objects in the Clausiusstatement as a cold reservoir and a hot reservoir. Thus, the Clausius statement becomes: It is impossible to construct arefrigerator that transfers heat from a cold reservoir to a hot reservoir without aid from an external source. The Clausiusstatement is related to the everyday observation that heat never flows spontaneously from a cold object to a hot object. Heattransfer in the direction of increasing temperature always requires some energy input. A “ perfect refrigerator,” shown inFigure 4.8(b), which works without such external aid, is impossible to construct.


Chapter 4 | The Second Law of Thermodynamics 153




To prove the equivalence of the Kelvin and Clausius statements, we show that if one statement is false, it necessarily followsthat the other statement is also false. Let us first assume that the Clausius statement is false, so that the perfect refrigeratorof Figure 4.8(b) does exist. The refrigerator removes heat Q from a cold reservoir at a temperature Tc and transfers all
of it to a hot reservoir at a temperature Th. Now consider a real heat engine working in the same temperature range. It
extracts heat Q + ΔQ from the hot reservoir, does work W, and discards heat Q to the cold reservoir. From the first law,
these quantities are related by W = (Q + ΔQ) − Q = ΔQ .
Suppose these two devices are combined as shown in Figure 4.9. The net heat removed from the hot reservoir is ΔQ ,
no net heat transfer occurs to or from the cold reservoir, and work W is done on some external body. Since W = ΔQ , the
combination of a perfect refrigerator and a real heat engine is itself a perfect heat engine, thereby contradicting the Kelvinstatement. Thus, if the Clausius statement is false, the Kelvin statement must also be false.


Figure 4.9 Combining a perfect refrigerator and a real heatengine yields a perfect heat engine because W = ΔQ.


Using the second law of thermodynamics, we now prove two important properties of heat engines operating between twoheat reservoirs. The first property is that any reversible engine operating between two reservoirs has a greater efficiencythan any irreversible engine operating between the same two reservoirs.
The second property to be demonstrated is that all reversible engines operating between the same two reservoirs have thesame efficiency. To show this, we start with the two engines D and E of Figure 4.10(a), which are operating betweentwo common heat reservoirs at temperatures Th and Tc. First, we assume that D is a reversible engine and that E is a
hypothetical irreversible engine that has a higher efficiency than D. If both engines perform the same amount of work W
per cycle, it follows from Equation 4.2 that Qh > Qh′ . It then follows from the first law that Qc > Qc′ .


Figure 4.10 (a) Two uncoupled engines D and E working between the same reservoirs. (b) The coupled engines, with Dworking in reverse.


Suppose the cycle of D is reversed so that it operates as a refrigerator, and the two engines are coupled such that the workoutput of E is used to drive D, as shown in Figure 4.10(b). Since Qh > Qh′ and Qc > Qc′ , the net result of each cycle is


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4.1


4.2


equivalent to a spontaneous transfer of heat from the cold reservoir to the hot reservoir, a process the second law does notallow. The original assumption must therefore be wrong, and it is impossible to construct an irreversible engine such that Eis more efficient than the reversible engine D.
Now it is quite easy to demonstrate that the efficiencies of all reversible engines operating between the same reservoirs areequal. Suppose that D and E are both reversible engines. If they are coupled as shown in Figure 4.10(b), the efficiencyof E cannot be greater than the efficiency of D, or the second law would be violated. If both engines are then reversed, thesame reasoning implies that the efficiency of D cannot be greater than the efficiency of E. Combining these results leads tothe conclusion that all reversible engines working between the same two reservoirs have the same efficiency.


Check Your Understanding What is the efficiency of a perfect heat engine? What is the coefficient ofperformance of a perfect refrigerator?


Check Your Understanding Show that Qh − Qh′ = Qc − Qc′ for the hypothetical engine of Figure
4.10(b).


4.5 | The Carnot Cycle
Learning Objectives


• Describe the Carnot cycle with the roles of all four processes involved
• Outline the Carnot principle and its implications
• Demonstrate the equivalence of the Carnot principle and the second law of thermodynamics


In the early 1820s, Sadi Carnot (1786−1832), a French engineer, became interested in improving the efficiencies of practicalheat engines. In 1824, his studies led him to propose a hypothetical working cycle with the highest possible efficiencybetween the same two reservoirs, known now as the Carnot cycle. An engine operating in this cycle is called a Carnotengine. The Carnot cycle is of special importance for a variety of reasons. At a practical level, this cycle represents areversible model for the steam power plant and the refrigerator or heat pump. Yet, it is also very important theoretically,for it plays a major role in the development of another important statement of the second law of thermodynamics. Finally,because only two reservoirs are involved in its operation, it can be used along with the second law of thermodynamics todefine an absolute temperature scale that is truly independent of any substance used for temperature measurement.
With an ideal gas as the working substance, the steps of the Carnot cycle, as represented by Figure 4.11, are as follows.


1. Isothermal expansion. The gas is placed in thermal contact with a heat reservoir at a temperature Th. The gas
absorbs heat Qh from the heat reservoir and is allowed to expand isothermally, doing work W1. Because the
internal energy Eint of an ideal gas is a function of the temperature only, the change of the internal energy is zero,
that is, ΔEint = 0 during this isothermal expansion. With the first law of thermodynamics, ΔEint = Q −W, we
find that the heat absorbed by the gas is


Qh = W1 = nRTh ln
VN
VM


.


Chapter 4 | The Second Law of Thermodynamics 155




Figure 4.11 The four processes of the Carnot cycle. The working substance is assumed to be anideal gas whose thermodynamic path MNOP is represented in Figure 4.12.


Figure 4.12 The total work done by the gas in the Carnot cycleis shown and given by the area enclosed by the loop MNOPM.


2. Adiabatic expansion. The gas is thermally isolated and allowed to expand further, doing work W2. Because this
expansion is adiabatic, the temperature of the gas falls—in this case, from Th to Tc. From pV γ = constant and
the equation of state for an ideal gas, pV = nRT , we have


TV
γ − 1


= constant,


so that
ThVN


γ − 1
= TcVO


γ − 1
.


3. Isothermal compression. The gas is placed in thermal contact with a cold reservoir at temperature Tc and
compressed isothermally. During this process, work W3 is done on the gas and it gives up heat Qc to the cold
reservoir. The reasoning used in step 1 now yields


Qc = nRTc ln
VO
VP


,


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where Qc is the heat dumped to the cold reservoir by the gas.
4. Adiabatic compression. The gas is thermally isolated and returned to its initial state by compression. In this process,work W4 is done on the gas. Because the compression is adiabatic, the temperature of the gas rises—from


Tc to Th in this particular case. The reasoning of step 2 now gives
TcVP


γ − 1
= ThVM


γ − 1
.


The total work done by the gas in the Carnot cycle is given by
W = W1 +W2 −W3 −W4.


This work is equal to the area enclosed by the loop shown in the pV diagram of Figure 4.12. Because the initial and finalstates of the system are the same, the change of the internal energy of the gas in the cycle must be zero, that is, ΔEint = 0 .
The first law of thermodynamics then gives


W = Q − ΔEint = (Qh − Qc) − 0,


and
W = Qh − Qc.


To find the efficiency of this engine, we first divide Qc by Qh :
Qc
Qh


= Tc
Th


lnVO/VP
lnVN/VM


.


When the adiabatic constant from step 2 is divided by that of step 4, we find
VO
VP


=
VN
VM


.


Substituting this into the equation for Qc/Qh, we obtain
Qc
Qh


= Tc
Th


.


Finally, with Equation 4.2, we find that the efficiency of this ideal gas Carnot engine is given by


(4.5)e = 1 − Tc
Th


.


An engine does not necessarily have to follow a Carnot engine cycle. All engines, however, have the same net effect,namely the absorption of heat from a hot reservoir, the production of work, and the discarding of heat to a cold reservoir.This leads us to ask: Do all reversible cycles operating between the same two reservoirs have the same efficiency? Theanswer to this question comes from the second law of thermodynamics discussed earlier: All reversible engine cyclesproduce exactly the same efficiency. Also, as you might expect, all real engines operating between two reservoirs are lessefficient than reversible engines operating between the same two reservoirs. This too is a consequence of the second law ofthermodynamics shown earlier.
The cycle of an ideal gas Carnot refrigerator is represented by the pV diagram of Figure 4.13. It is a Carnot engineoperating in reverse. The refrigerator extracts heat Qc from a cold-temperature reservoir at Tc when the ideal gas
expands isothermally. The gas is then compressed adiabatically until its temperature reaches Th, after which an isothermal
compression of the gas results in heat Qh being discarded to a high-temperature reservoir at Th. Finally, the cycle is
completed by an adiabatic expansion of the gas, causing its temperature to drop to Tc.


Chapter 4 | The Second Law of Thermodynamics 157




Figure 4.13 The work done on the gas in one cycle of theCarnot refrigerator is shown and given by the area enclosed bythe loop MPONM.


The work done on the ideal gas is equal to the area enclosed by the path of the pV diagram. From the first law, this work isgiven by
W = Qh − Qc.


An analysis just like the analysis done for the Carnot engine gives
Qc
Tc


=
Qh
Th


.


When combined with Equation 4.3, this yields


(4.6)KR = TcTh − Tc


for the coefficient of performance of the ideal-gas Carnot refrigerator. Similarly, we can work out the coefficient ofperformance for a Carnot heat pump as


(4.7)
KP =


Qh
Qh − Qc


=
Th


Th − Tc
.


We have just found equations representing the efficiency of a Carnot engine and the coefficient of performance of a Carnotrefrigerator or a Carnot heat pump, assuming an ideal gas for the working substance in both devices. However, theseequations are more general than their derivations imply. We will soon show that they are both valid no matter what theworking substance is.
Carnot summarized his study of the Carnot engine and Carnot cycle into what is now known as Carnot’s principle:


Carnot’s Principle
No engine working between two reservoirs at constant temperatures can have a greater efficiency than a reversibleengine.


This principle can be viewed as another statement of the second law of thermodynamics and can be shown to be equivalentto the Kelvin statement and the Clausius statement.


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Example 4.2
The Carnot Engine
A Carnot engine has an efficiency of 0.60 and the temperature of its cold reservoir is 300 K. (a) What is thetemperature of the hot reservoir? (b) If the engine does 300 J of work per cycle, how much heat is removed fromthe high-temperature reservoir per cycle? (c) How much heat is exhausted to the low-temperature reservoir percycle?
Strategy
From the temperature dependence of the thermal efficiency of the Carnot engine, we can find the temperatureof the hot reservoir. Then, from the definition of the efficiency, we can find the heat removed when the workdone by the engine is given. Finally, energy conservation will lead to how much heat must be dumped to the coldreservoir.
Solutiona. From e = 1 − Tc /Th we have


0.60 = 1 − 300 K
Th


,


so that the temperature of the hot reservoir is
Th =


300 K
1 − 0.60


= 750 K.


b. By definition, the efficiency of the engine is e = W/Q , so that the heat removed from the high-
temperature reservoir per cycle is


Qh =
W
e =


300 J
0.60


= 500 J.


c. From the first law, the heat exhausted to the low-temperature reservoir per cycle by the engine is
Qc = Qh −W = 500 J − 300 J = 200 J.


Significance
A Carnot engine has the maximum possible efficiency of converting heat into work between two reservoirs, butthis does not necessarily mean it is 100% efficient. As the difference in temperatures of the hot and cold reservoir
increases, the efficiency of a Carnot engine increases.


Example 4.3
A Carnot Heat Pump
Imagine a Carnot heat pump operates between an outside temperature of 0 °C and an inside temperature of
20.0 °C . What is the work needed if the heat delivered to the inside of the house is 30.0 kJ?
Strategy
Because the heat pump is assumed to be a Carnot pump, its performance coefficient is given by
KP = Qh /W = Th /(Th − Tc). Thus, we can find the work W from the heat delivered Qh.
Solution
The work needed is obtained from


W = Qh /KP = Qh(Th − Tc)/Th = 30 kJ × (293 K − 273 K)/293 K = 2 kJ.


Significance
We note that this work depends not only on the heat delivered to the house but also on the temperaturesoutside and inside. The dependence on the temperature outside makes them impractical to use in areas where the


Chapter 4 | The Second Law of Thermodynamics 159




4.3


4.4


temperature is much colder outside than room temperature.
In terms of energy costs, the heat pump is a very economical means for heating buildings (Figure 4.14). Contrast thismethod with turning electrical energy directly into heat with resistive heating elements. In this case, one unit of electricalenergy furnishes at most only one unit of heat. Unfortunately, heat pumps have problems that do limit their usefulness. Theyare quite expensive to purchase compared to resistive heating elements, and, as the performance coefficient for a Carnotheat pump shows, they become less effective as the outside temperature decreases. In fact, below about –10 °C , the heat
they furnish is less than the energy used to operate them.


Figure 4.14 A photograph of a heat pump (large box) locatedoutside a house. This heat pump is located in a warm climatearea, like the southern United States, since it would be far tooinefficient located in the northern half of the United States.(credit: modification of work by Peter Stevens)
Check Your Understanding A Carnot engine operates between reservoirs at 400 °C and 30 °C . (a)


What is the efficiency of the engine? (b) If the engine does 5.0 J of work per cycle, how much heat per cycledoes it absorb from the high-temperature reservoir? (c) How much heat per cycle does it exhaust to the cold-temperature reservoir? (d) What temperatures at the cold reservoir would give the minimum and maximumefficiency?


Check Your Understanding A Carnot refrigerator operates between two heat reservoirs whosetemperatures are 0 °C and 25 °C . (a) What is the coefficient of performance of the refrigerator? (b) If 200 J of
work are done on the working substance per cycle, how much heat per cycle is extracted from the coldreservoir? (c) How much heat per cycle is discarded to the hot reservoir?


4.6 | Entropy
Learning Objectives


By the end of this section you will be able to:
• Describe the meaning of entropy
• Calculate the change of entropy for some simple processes


The second law of thermodynamics is best expressed in terms of a change in the thermodynamic variable known as entropy,which is represented by the symbol S. Entropy, like internal energy, is a state function. This means that when a systemmakes a transition from one state into another, the change in entropy ΔS is independent of path and depends only on the


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thermodynamic variables of the two states.
We first consider ΔS for a system undergoing a reversible process at a constant temperature. In this case, the change in
entropy of the system is given by


(4.8)ΔS = Q
T
,


where Q is the heat exchanged by the system kept at a temperature T (in kelvin). If the system absorbs heat—that is, with
Q > 0—the entropy of the system increases. As an example, suppose a gas is kept at a constant temperature of 300 K
while it absorbs 10 J of heat in a reversible process. Then from Equation 4.8, the entropy change of the gas is


ΔS = 10 J
300 K


= 0.033 J/K.


Similarly, if the gas loses 5.0 J of heat; that is, Q = −5.0 J , at temperature T = 200 K , we have the entropy change of the
system given by


ΔS = −5.0 J
200 K


= −0.025 J/K.


Example 4.4
Entropy Change of Melting Ice
Heat is slowly added to a 50-g chunk of ice at 0 °C until it completely melts into water at the same temperature.
What is the entropy change of the ice?
Strategy
Because the process is slow, we can approximate it as a reversible process. The temperature is a constant, and wecan therefore use Equation 4.8 in the calculation.
Solution
The ice is melted by the addition of heat:


Q = mLf = 50 g × 335 J/g = 16.8 kJ.


In this reversible process, the temperature of the ice-water mixture is fixed at 0 °C or 273 K. Now from
ΔS = Q/T , the entropy change of the ice is


ΔS = 16.8 kJ
273 K


= 61.5 J/K


when it melts to water at 0 °C .
Significance
During a phase change, the temperature is constant, allowing us to use Equation 4.8 to solve this problem. Thesame equation could also be used if we changed from a liquid to a gas phase, since the temperature does notchange during that process either.


The change in entropy of a system for an arbitrary, reversible transition for which the temperature is not necessarily constantis defined by modifying ΔS = Q/T . Imagine a system making a transition from state A to B in small, discrete steps. The
temperatures associated with these states are TA and TB, respectively. During each step of the transition, the system
exchanges heat ΔQi reversibly at a temperature Ti. This can be accomplished experimentally by placing the system in
thermal contact with a large number of heat reservoirs of varying temperatures Ti , as illustrated in Figure 4.15. The
change in entropy for each step is ΔSi = Qi /Ti. The net change in entropy of the system for the transition is


Chapter 4 | The Second Law of Thermodynamics 161




(4.9)
ΔS = SB − SA = ∑


i
ΔSi = ∑


i


ΔQi
Ti


.


We now take the limit as ΔQi → 0 , and the number of steps approaches infinity. Then, replacing the summation by an
integral, we obtain


(4.10)
ΔS = SB − SA =



⌡A


B
dQ
T


,


where the integral is taken between the initial state A and the final state B. This equation is valid only if the transition fromA to B is reversible.


Figure 4.15 The gas expands at constant pressure as its temperature is increased in small steps through theuse of a series of heat reservoirs.


As an example, let us determine the net entropy change of a reversible engine while it undergoes a single Carnot cycle. In the
adiabatic steps 2 and 4 of the cycle shown in Figure 4.11, no heat exchange takes place, so ΔS2 = ΔS4 = ∫ dQ/T = 0.
In step 1, the engine absorbs heat Qh at a temperature Th, so its entropy change is ΔS1 = Qh /Th. Similarly, in step 3,
ΔS3 = −Qc /Tc. The net entropy change of the engine in one cycle of operation is then


ΔSE = ΔS1 + ΔS2 + ΔS3 + ΔS4 =
Qh
Th


− Qc
Tc


.


However, we know that for a Carnot engine,
Qh
Th


= Qc
Tc


,


so
ΔSE = 0.


There is no net change in the entropy of the Carnot engine over a complete cycle. Although this result was obtained fora particular case, its validity can be shown to be far more general: There is no net change in the entropy of a systemundergoing any complete reversible cyclic process. Mathematically, we write this statement as


(4.11)∮ dS = ∮ dQT = 0


where ∮ represents the integral over a closed reversible path.


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We can use Equation 4.11 to show that the entropy change of a system undergoing a reversible process between two givenstates is path independent. An arbitrary, closed path for a reversible cycle that passes through the states A and B is shown
in Figure 4.16. From Equation 4.11, ∮ dS = 0 for this closed path. We may split this integral into two segments, one
along I, which leads from A to B, the other along II, which leads from B to A. Then



⎣∫A


B
dS

⎦I +



⎢∫


B


A
dS



⎥ II = 0.


Since the process is reversible,

⎣∫A


B
dS

⎦I =

⎣∫A


B
dS

⎦II.


Figure 4.16 The closed loop passing through states A and Brepresents a reversible cycle.


Hence, the entropy change in going from A to B is the same for paths I and II. Since paths I and II are arbitrary, reversiblepaths, the entropy change in a transition between two equilibrium states is the same for all the reversible processes joiningthese states. Entropy, like internal energy, is therefore a state function.
What happens if the process is irreversible? When the process is irreversible, we expect the entropy of a closed system, orthe system and its environment (the universe), to increase. Therefore we can rewrite this expression as


(4.12)ΔS ≥ 0,


where S is the total entropy of the closed system or the entire universe, and the equal sign is for a reversible process. Thefact is the entropy statement of the second law of thermodynamics:
Second Law of Thermodynamics (Entropy statement)
The entropy of a closed system and the entire universe never decreases.


We can show that this statement is consistent with the Kelvin statement, the Clausius statement, and the Carnot principle.
Example 4.5


Entropy Change of a System during an Isobaric Process
Determine the entropy change of an object of mass m and specific heat c that is cooled rapidly (and irreversibly)


Chapter 4 | The Second Law of Thermodynamics 163




at constant pressure from Th to Tc.
Strategy
The process is clearly stated as an irreversible process; therefore, we cannot simply calculate the entropy changefrom the actual process. However, because entropy of a system is a function of state, we can imagine a reversibleprocess that starts from the same initial state and ends at the given final state. Then, the entropy change of the
system is given by Equation 4.10, ΔS = ∫


A


B
dQ/T .


Solution
To replace this rapid cooling with a process that proceeds reversibly, we imagine that the hot object is put intothermal contact with successively cooler heat reservoirs whose temperatures range from Th to Tc. Throughout
the substitute transition, the object loses infinitesimal amounts of heat dQ, so we have


ΔS = ⌠
⌡Th


Tc
dQ
T


.


From the definition of heat capacity, an infinitesimal exchange dQ for the object is related to its temperaturechange dT by
dQ = mc dT .


Substituting this dQ into the expression for ΔS , we obtain the entropy change of the object as it is cooled at
constant pressure from Th to Tc :


ΔS = ⌠
⌡Th


Tc
mc dT
T


= mc lnTc
Th


.


Note that ΔS < 0 here because Tc < Th. In other words, the object has lost some entropy. But if we count
whatever is used to remove the heat from the object, we would still end up with ΔSuniverse > 0 because the
process is irreversible.
Significance
If the temperature changes during the heat flow, you must keep it inside the integral to solve for the change inentropy. If, however, the temperature is constant, you can simply calculate the entropy change as the heat flowdivided by the temperature.


Example 4.6
Stirling Engine
The steps of a reversible Stirling engine are as follows. For this problem, we will use 0.0010 mol of a monatomic
gas that starts at a temperature of 133 °C and a volume of 0.10 m3 , which will be called point A. Then it goes
through the following steps:


1. Step AB: isothermal expansion at 133 °C from 0.10 m3 to 0.20 m3
2. Step BC: isochoric cooling to 33 °C
3. Step CD: isothermal compression at 33 °C from 0.20 m3 to 0.10 m3
4. Step DA: isochoric heating back to 133 °C and 0.10 m3


(a) Draw the pV diagram for the Stirling engine with proper labels.
(b) Fill in the following table.


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Step W (J) Q (J) ΔS (J/K)
Step AB
Step BC
Step CD
Step DA
Complete cycle


(c) How does the efficiency of the Stirling engine compare to the Carnot engine working within the same twoheat reservoirs?
Strategy
Using the ideal gas law, calculate the pressure at each point so that they can be labeled on the pV diagram.
Isothermal work is calculated using W = nRT ln⎛⎝V2V1



⎠, and an isochoric process has no work done. The


heat flow is calculated from the first law of thermodynamics, Q = ΔEint −W where ΔEint = 32nRΔT for
monatomic gasses. Isothermal steps have a change in entropy of Q/T, whereas isochoric steps have
ΔS = 3


2
nR ln


T2
T1



⎠. The efficiency of a heat engine is calculated by using eStir = W/Qh.


Solutiona. The graph is shown below.


b. The completed table is shown below.
Step W (J) Q (J) ΔS (J/K)
Step AB Isotherm 2.3 2.3 0.0057
Step BC Isochoric 0 –1.2 0.0035
Step CD Isotherm –1.8 –1.8 –0.0059
Step DA Isochoric 0 1.2 –0.0035
Complete cycle 0.5 0.5 ~ 0


c. The efficiency of the Stirling heat engine is


Chapter 4 | The Second Law of Thermodynamics 165




eStir = W/Qh = (QAB + QCD)/(QAB + QDA) = 0.5/4.5 = 0.11.


If this were a Carnot engine operating between the same heat reservoirs, its efficiency would be
eCar = 1 −




Tc
Th



⎠ = 0.25.


Therefore, the Carnot engine would have a greater efficiency than the Stirling engine.
Significance
In the early days of steam engines, accidents would occur due to the high pressure of the steam in the boiler.Robert Stirling developed an engine in 1816 that did not use steam and therefore was safer. The Stirling enginewas commonly used in the nineteenth century, but developments in steam and internal combustion engines havemade it difficult to broaden the use of the Stirling engine.
The Stirling engine uses compressed air as the working substance, which passes back and forth between twochambers with a porous plug, called the regenerator, which is made of material that does not conduct heat as well.In two of the steps, pistons in the two chambers move in phase.


4.7 | Entropy on a Microscopic Scale
Learning Objectives


By the end of this section you will be able to:
• Interpret the meaning of entropy at a microscopic scale
• Calculate a change in entropy for an irreversible process of a system and contrast with thechange in entropy of the universe
• Explain the third law of thermodynamics


We have seen how entropy is related to heat exchange at a particular temperature. In this section, we consider entropy from astatistical viewpoint. Although the details of the argument are beyond the scope of this textbook, it turns out that entropy canbe related to how disordered or randomized a system is—the more it is disordered, the higher is its entropy. For example, anew deck of cards is very ordered, as the cards are arranged numerically by suit. In shuffling this new deck, we randomizethe arrangement of the cards and therefore increase its entropy (Figure 4.17). Thus, by picking one card off the top of thedeck, there would be no indication of what the next selected card will be.


Figure 4.17 The entropy of a new deck of cards goes up afterthe dealer shuffles them. (credit: “Rommel SK”/YouTube)


The second law of thermodynamics requires that the entropy of the universe increase in any irreversible process. Thus, interms of order, the second law may be stated as follows:
In any irreversible process, the universe becomes more disordered. For example, the irreversible free expansion of an idealgas, shown in Figure 4.2, results in a larger volume for the gas molecules to occupy. A larger volume means more possiblearrangements for the same number of atoms, so disorder is also increased. As a result, the entropy of the gas has gone


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up. The gas in this case is a closed system, and the process is irreversible. Changes in phase also illustrate the connectionbetween entropy and disorder.
Example 4.7


Entropy Change of the Universe
Suppose we place 50 g of ice at 0 °C in contact with a heat reservoir at 20 °C . Heat spontaneously flows from
the reservoir to the ice, which melts and eventually reaches a temperature of 20 °C . Find the change in entropy
of (a) the ice and (b) the universe.
Strategy
Because the entropy of a system is a function of its state, we can imagine two reversible processes for the ice:(1) ice is melted at 0 °C(TA); and (2) melted ice (water) is warmed up from 0 °C to 20 °C(TB) under constant
pressure. Then, we add the change in entropy of the reservoir when we calculate the change in entropy of theuniverse.
Solutiona. From Equation 4.10, the increase in entropy of the ice is


ΔSice = ΔS1 + ΔS2


=
mL f
TA


+ mc⌠
⌡A


B
dT
T


= ⎛⎝
50 × 335


273
+ 50 × 4.19 × ln293


273

⎠ J/K


= 76.3 J/K.
b. During this transition, the reservoir gives the ice an amount of heat equal to


Q = mL f + mc(TB − TA)


= 50 × (335 + 4.19 × 20) J


= 2.10 × 104 J.


This leads to a change (decrease) in entropy of the reservoir:
ΔSreservoir =


−Q
TB


= −71.7 J/K.


The increase in entropy of the universe is therefore
ΔSuniverse = 76.3 J/K − 71.7 J/K = 4.6 J/K > 0.


Significance
The entropy of the universe therefore is greater than zero since the ice gains more entropy than the reservoir loses.If we considered only the phase change of the ice into water and not the temperature increase, the entropy changeof the ice and reservoir would be the same, resulting in the universe gaining no entropy.


This process also results in a more disordered universe. The ice changes from a solid with molecules located at specific sitesto a liquid whose molecules are much freer to move. The molecular arrangement has therefore become more randomized.Although the change in average kinetic energy of the molecules of the heat reservoir is negligible, there is nevertheless asignificant decrease in the entropy of the reservoir because it has many more molecules than the melted ice cube. However,the reservoir’s decrease in entropy is still not as large as the increase in entropy of the ice. The increased disorder of the icemore than compensates for the increased order of the reservoir, and the entropy of the universe increases by 4.6 J/K.
You might suspect that the growth of different forms of life might be a net ordering process and therefore a violation of thesecond law. After all, a single cell gathers molecules and eventually becomes a highly structured organism, such as a humanbeing. However, this ordering process is more than compensated for by the disordering of the rest of the universe. The netresult is an increase in entropy and an increase in the disorder of the universe.


Chapter 4 | The Second Law of Thermodynamics 167




4.5 Check Your Understanding In Example 4.7, the spontaneous flow of heat from a hot object to a coldobject results in a net increase in entropy of the universe. Discuss how this result can be related to an increase indisorder of the system.
The second law of thermodynamics makes clear that the entropy of the universe never decreases during any thermodynamicprocess. For any other thermodynamic system, when the process is reversible, the change of the entropy is given by
ΔS = Q/T . But what happens if the temperature goes to zero, T → 0 ? It turns out this is not a question that can be
answered by the second law.
A fundamental issue still remains: Is it possible to cool a system all the way down to zero kelvin? We understand thatthe system must be at its lowest energy state because lowering temperature reduces the kinetic energy of the constituentsin the system. What happens to the entropy of a system at the absolute zero temperature? It turns out the absolute zerotemperature is not reachable—at least, not though a finite number of cooling steps. This is a statement of the third law ofthermodynamics, whose proof requires quantum mechanics that we do not present here. In actual experiments, physicists
have continuously pushed that limit downward, with the lowest temperature achieved at about 1 × 10−10 K in a low-
temperature lab at the Helsinki University of Technology in 2008.
Like the second law of thermodynamics, the third law of thermodynamics can be stated in different ways. One of thecommon statements of the third law of thermodynamics is: The absolute zero temperature cannot be reached through anyfinite number of cooling steps.
In other words, the temperature of any given physical system must be finite, that is, T > 0 . This produces a very interesting
question in physics: Do we know how a system would behave if it were at the absolute zero temperature?
The reason a system is unable to reach 0 K is fundamental and requires quantum mechanics to fully understand its origin.But we can certainly ask what happens to the entropy of a system when we try to cool it down to 0 K. Because the amountof heat that can be removed from the system becomes vanishingly small, we expect that the change in entropy of the systemalong an isotherm approaches zero, that is,


(4.13)lim
T → 0


(ΔS)T = 0.


This can be viewed as another statement of the third law, with all the isotherms becoming isentropic, or into a reversibleideal adiabat. We can put this expression in words: A system becomes perfectly ordered when its temperature approachesabsolute zero and its entropy approaches its absolute minimum.
The third law of thermodynamics puts another limit on what can be done when we look for energy resources. If there couldbe a reservoir at the absolute zero temperature, we could have engines with efficiency of 100% , which would, of course,
violate the second law of thermodynamics.
Example 4.8


Entropy Change of an Ideal Gas in Free Expansion
An ideal gas occupies a partitioned volume V1 inside a box whose walls are thermally insulating, as shown in
Figure 4.18(a). When the partition is removed, the gas expands and fills the entire volume V2 of the box, as
shown in part (b). What is the entropy change of the universe (the system plus its environment)?


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Figure 4.18 The adiabatic free expansion of an ideal gas from volume V1 to volume V2 .


Strategy
The adiabatic free expansion of an ideal gas is an irreversible process. There is no change in the internal energy(and hence temperature) of the gas in such an expansion because no work or heat transfer has happened. Thus, aconvenient reversible path connecting the same two equilibrium states is a slow, isothermal expansion from V1
to V2 . In this process, the gas could be expanding against a piston while in thermal contact with a heat reservoir,
as in step 1 of the Carnot cycle.
Solution
Since the temperature is constant, the entropy change is given by ΔS = Q/T , where


Q = W = ∫
V1


V2
pdV


because ΔEint = 0. Now, with the help of the ideal gas law, we have


Q = nRT⌠
⌡V1


V2
dV
V


= nRT ln
V2
V1


,


so the change in entropy of the gas is
ΔS = Q


T
= nR ln


V2
V1


.


Because V2 > V1 , ΔS is positive, and the entropy of the gas has gone up during the free expansion.
Significance
What about the environment? The walls of the container are thermally insulating, so no heat exchange takes placebetween the gas and its surroundings. The entropy of the environment is therefore constant during the expansion.The net entropy change of the universe is then simply the entropy change of the gas. Since this is positive, theentropy of the universe increases in the free expansion of the gas.


Example 4.9
Entropy Change during Heat Transfer
Heat flows from a steel object of mass 4.00 kg whose temperature is 400 K to an identical object at 300 K.Assuming that the objects are thermally isolated from the environment, what is the net entropy change of theuniverse after thermal equilibrium has been reached?


Chapter 4 | The Second Law of Thermodynamics 169




4.6


4.7


Strategy
Since the objects are identical, their common temperature at equilibrium is 350 K. To calculate the entropychanges associated with their transitions, we substitute the irreversible process of the heat transfer by two isobaric,reversible processes, one for each of the two objects. The entropy change for each object is then given by
ΔS = mc ln(TB /TA).


Solution
Using c = 450 J/kg · K , the specific heat of steel, we have for the hotter object


ΔSh =

⌡T1


T2
mc dT
T


= mc ln
T2
T1


= (4.00 kg)(450 J/kg · K)ln350 K
400 K


= −240 J/K.


Similarly, the entropy change of the cooler object is
ΔSc = (4.00 kg)(450 J/kg · K) ln350 K300 K


= 277 J/K.


The net entropy change of the two objects during the heat transfer is then
ΔSh + ΔSc = 37 J/K.


Significance
The objects are thermally isolated from the environment, so its entropy must remain constant. Thus, the entropyof the universe also increases by 37 J/K.


Check Your Understanding A quantity of heat Q is absorbed from a reservoir at a temperature Th by a
cooler reservoir at a temperature Tc. What is the entropy change of the hot reservoir, the cold reservoir, and the
universe?


Check Your Understanding A 50-g copper piece at a temperature of 20 °C is placed into a large
insulated vat of water at 100 °C . (a) What is the entropy change of the copper piece when it reaches thermal
equilibrium with the water? (b) What is the entropy change of the water? (c) What is the entropy change of theuniverse?


View this site (https://openstaxcollege.org/l/21reversereact) to learn about entropy and microstates. Startwith a large barrier in the middle and 1000 molecules in only the left chamber. What is the total entropy of thesystem? Now remove the barrier and let the molecules travel from the left to the right hand side? What is the totalentropy of the system now? Lastly, add heat and note what happens to the temperature. Did this increase entropyof the system?


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Carnot cycle
Carnot engine
Carnot principle


Clausius statement of the second law of thermodynamics
coefficient of performance
cold reservoir
disorder
efficiency (e)
entropy
entropy statement of the second law of thermodynamics
heat engine
heat pump
hot reservoir
irreversibility
irreversible process
isentropic
Kelvin statement of the second law of thermodynamics
perfect engine
perfect refrigerator (heat pump)
refrigerator
reversible process
third law of thermodynamics


CHAPTER 4 REVIEW
KEY TERMS


cycle that consists of two isotherms at the temperatures of two reservoirs and two adiabatic processesconnecting the isotherms
Carnot heat engine, refrigerator, or heat pump that operates on a Carnot cycle
principle governing the efficiency or performance of a heat device operating on a Carnot cycle: anyreversible heat device working between two reservoirs must have the same efficiency or performance coefficient,greater than that of an irreversible heat device operating between the same two reservoirs


heat never flows spontaneously from a colder objectto a hotter object
measure of effectiveness of a refrigerator or heat pump


sink of heat used by a heat engine
measure of order in a system; the greater the disorder is, the higher the entropy


output work from the engine over the input heat to the engine from the hot reservoir
state function of the system that changes when heat is transferred between the system and the environment


entropy of a closed system or the entire universenever decreases
device that converts heat into work
device that delivers heat to a hot reservoir
source of heat used by a heat engine
phenomenon associated with a natural process


process in which neither the system nor its environment can be restored to their original states atthe same time
reversible adiabatic process where the process is frictionless and no heat is transferred


it is impossible to convert the heat from a single sourceinto work without any other effect
engine that can convert heat into work with 100% efficiency


refrigerator (heat pump) that can remove (dump) heat without any input of work
device that removes heat from a cold reservoir


process in which both the system and the external environment theoretically can be returned to theiroriginal states
absolute zero temperature cannot be reached through any finite number of cooling steps


KEY EQUATIONS
Result of energy conservation W = Qh − Qc
Efficiency of a heat engine e = W


Qh
= 1 − Qc


Qh


Coefficient of performance of a refrigerator KR = QcW = QcQh − Qc


Chapter 4 | The Second Law of Thermodynamics 171




Coefficient of performance of a heat pump
KP =


Qh
W


=
Qh


Qh − Qc


Resulting efficiency of a Carnot cycle e = 1 − Tc
Th


Performance coefficient of a reversible refrigerator KR = TcTh − Tc
Performance coefficient of a reversible heat pump


KP =
Th


Th − Tc


Entropy of a system undergoing a reversible process at a constanttemperature ΔS = QT
Change of entropy of a system under a reversible process


ΔS = SB − SA = ∫
A


B
dQ/T


Entropy of a system undergoing any complete reversible cyclic process ∮ dS = ∮ dQT = 0
Change of entropy of a closed system under an irreversible process ΔS ≥ 0
Change in entropy of the system along an isotherm lim


T → 0
(ΔS)T = 0


SUMMARY
4.1 Reversible and Irreversible Processes


• A reversible process is one in which both the system and its environment can return to exactly the states they werein by following the reverse path.
• An irreversible process is one in which the system and its environment cannot return together to exactly the statesthat they were in.
• The irreversibility of any natural process results from the second law of thermodynamics.


4.2 Heat Engines
• The work done by a heat engine is the difference between the heat absorbed from the hot reservoir and the heatdischarged to the cold reservoir, that is, W = Qh − Qc.
• The ratio of the work done by the engine and the heat absorbed from the hot reservoir provides the efficiency of theengine, that is, e = W/Qh = 1 − Qc /Qh.


4.3 Refrigerators and Heat Pumps
• A refrigerator or a heat pump is a heat engine run in reverse.
• The focus of a refrigerator is on removing heat from the cold reservoir with a coefficient of performance KR.
• The focus of a heat pump is on dumping heat to the hot reservoir with a coefficient of performance KP.


4.4 Statements of the Second Law of Thermodynamics
• The Kelvin statement of the second law of thermodynamics: It is impossible to convert the heat from a single sourceinto work without any other effect.
• The Kelvin statement and Clausius statement of the second law of thermodynamics are equivalent.


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4.5 The Carnot Cycle
• The Carnot cycle is the most efficient engine for a reversible cycle designed between two reservoirs.
• The Carnot principle is another way of stating the second law of thermodynamics.


4.6 Entropy
• The change in entropy for a reversible process at constant temperature is equal to the heat divided by the
temperature. The entropy change of a system under a reversible process is given by ΔS = ∫


A


B
dQ/T .


• A system’s change in entropy between two states is independent of the reversible thermodynamic path taken by thesystem when it makes a transition between the states.
4.7 Entropy on a Microscopic Scale


• Entropy can be related to how disordered a system is—the more it is disordered, the higher is its entropy. In anyirreversible process, the universe becomes more disordered.
• According to the third law of thermodynamics, absolute zero temperature is unreachable.


CONCEPTUAL QUESTIONS
4.1 Reversible and Irreversible Processes
1. State an example of a process that occurs in nature thatis as close to reversible as it can be.


4.2 Heat Engines
2. Explain in practical terms why efficiency is defined as
W/Qh.


4.3 Refrigerators and Heat Pumps
3. If the refrigerator door is left open, what happens to thetemperature of the kitchen?
4. Is it possible for the efficiency of a reversible engineto be greater than 1.0? Is it possible for the coefficient ofperformance of a reversible refrigerator to be less than 1.0?


4.4 Statements of the Second Law of
Thermodynamics
5. In the text, we showed that if the Clausius statement isfalse, the Kelvin statement must also be false. Now showthe reverse, such that if the Kelvin statement is false, itfollows that the Clausius statement is false.
6. Why don’t we operate ocean liners by extracting heatfrom the ocean or operate airplanes by extracting heat fromthe atmosphere?
7. Discuss the practical advantages and disadvantages of


heat pumps and electric heating.
8. The energy output of a heat pump is greater than theenergy used to operate the pump. Why doesn’t thisstatement violate the first law of thermodynamics?
9. Speculate as to why nuclear power plants are lessefficient than fossil-fuel plants based on temperaturearguments.
10. An ideal gas goes from state (pi, Vi) to state
(pf, Vf) when it is allowed to expand freely. Is it possible
to represent the actual process on a pV diagram? Explain.


4.5 The Carnot Cycle
11. To increase the efficiency of a Carnot engine, shouldthe temperature of the hot reservoir be raised or lowered?What about the cold reservoir?
12. How could you design a Carnot engine with 100%
efficiency?
13. What type of processes occur in a Carnot cycle?


4.6 Entropy
14. Does the entropy increase for a Carnot engine for eachcycle?
15. Is it possible for a system to have an entropy changeif it neither absorbs nor emits heat during a reversible


Chapter 4 | The Second Law of Thermodynamics 173




transition? What happens if the process is irreversible?


4.7 Entropy on a Microscopic Scale
16. Are the entropy changes of the systems in thefollowing processes positive or negative? (a) water vaporthat condenses on a cold surface; (b) gas in a container that


leaks into the surrounding atmosphere; (c) an ice cube thatmelts in a glass of lukewarm water; (d) the lukewarm waterof part (c); (e) a real heat engine performing a cycle; (f)food cooled in a refrigerator.
17. Discuss the entropy changes in the systems ofQuestion 21.10 in terms of disorder.


PROBLEMS
4.1 Reversible and Irreversible Processes
18. A tank contains 111.0 g chlorine gas (Cl2), which
is at temperature 82.0 °C and absolute pressure
5.70 × 105 Pa. The temperature of the air outside the tank
is 20.0 °C . The molar mass of Cl2 is 70.9 g/mol. (a)
What is the volume of the tank? (b) What is the internalenergy of the gas? (c) What is the work done by the gasif the temperature and pressure inside the tank drop to
31.0 °C and 3.80 × 105 Pa , respectively, due to a leak?
19. A mole of ideal monatomic gas at 0 °C and 1.00 atm
is warmed up to expand isobarically to triple its volume.How much heat is transferred during the process?
20. A mole of an ideal gas at pressure 4.00 atm andtemperature 298 K expands isothermally to double itsvolume. What is the work done by the gas?
21. After a free expansion to quadruple its volume, a moleof ideal diatomic gas is compressed back to its originalvolume isobarically and then cooled down to its originaltemperature. What is the minimum heat removed from thegas in the final step to restoring its state?


4.2 Heat Engines
22. An engine is found to have an efficiency of 0.40. If itdoes 200 J of work per cycle, what are the correspondingquantities of heat absorbed and rejected?
23. In performing 100.0 J of work, an engine rejects 50.0J of heat. What is the efficiency of the engine?
24. An engine with an efficiency of 0.30 absorbs 500 Jof heat per cycle. (a) How much work does it perform percycle? (b) How much heat does it reject per cycle?
25. It is found that an engine rejects 100.0 J whileabsorbing 125.0 J each cycle of operation. (a) What isthe efficiency of the engine? (b) How much work does itperform per cycle?


26. The temperature of the cold reservoir of the engine is300 K. It has an efficiency of 0.30 and absorbs 500 J of heatper cycle. (a) How much work does it perform per cycle?(b) How much heat does it reject per cycle?
27. The Kelvin temperature of the hot reservoir of anengine is twice that of the cold reservoir, and work done bythe engine per cycle is 50 J. Calculate (a) the efficiency ofthe engine, (b) the heat absorbed per cycle, and (c) the heatrejected per cycle.
28. A coal power plant consumes 100,000 kg of coalper hour and produces 500 MW of power. If the heat ofcombustion of coal is 30 MJ/kg, what is the efficiency ofthe power plant?


4.3 Refrigerators and Heat Pumps
29. A refrigerator has a coefficient of performance of 3.0.(a) If it requires 200 J of work per cycle, how much heatper cycle does it remove the cold reservoir? (b) How muchheat per cycle is discarded to the hot reservoir?
30. During one cycle, a refrigerator removes 500 J from acold reservoir and rejects 800 J to its hot reservoir. (a) Whatis its coefficient of performance? (b) How much work percycle does it require to operate?
31. If a refrigerator discards 80 J of heat per cycle and itscoefficient of performance is 6.0, what are (a) the quantityoff heat it removes per cycle from a cold reservoir and (b)the amount of work per cycle required for its operation?
32. A refrigerator has a coefficient of performance of 3.0.(a) If it requires 200 J of work per cycle, how much heatper cycle does it remove the cold reservoir? (b) How muchheat per cycle is discarded to the hot reservoir?


4.5 The Carnot Cycle
33. The temperature of the cold and hot reservoirsbetween which a Carnot refrigerator operates are −73 °C
and 270 °C , respectively. Which is its coefficient of
performance?


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34. Suppose a Carnot refrigerator operates between
Tc and Th. Calculate the amount of work required to
extract 1.0 J of heat from the cold reservoir if (a)
Tc = 7 °C , Th = 27 °C ; (b) Tc = −73 °C ,
Th = 27 °C; (c) Tc = −173 °C , Th = 27 °C ; and (d)
Tc = −273 °C , Th = 27 °C .
35. A Carnot engine operates between reservoirs at 600and 300 K. If the engine absorbs 100 J per cycle at the hotreservoir, what is its work output per cycle?
36. A 500-W motor operates a Carnot refrigeratorbetween −5 °C and 30 °C . (a) What is the amount of heat
per second extracted from the inside of the refrigerator? (b)How much heat is exhausted to the outside air per second?
37. Sketch a Carnot cycle on a temperature-volumediagram.
38. A Carnot heat pump operates between 0 °C and
20 °C . How much heat is exhausted into the interior of a
house for every 1.0 J of work done by the pump?
39. An engine operating between heat reservoirs at 20 °C
and 200 °C extracts 1000 J per cycle from the hot
reservoir. (a) What is the maximum possible work thatengine can do per cycle? (b) For this maximum work, howmuch heat is exhausted to the cold reservoir per cycle?
40. Suppose a Carnot engine can be operated between tworeservoirs as either a heat engine or a refrigerator. How isthe coefficient of performance of the refrigerator related tothe efficiency of the heat engine?
41. A Carnot engine is used to measure the temperatureof a heat reservoir. The engine operates between the heatreservoir and a reservoir consisting of water at its triplepoint. (a) If 400 J per cycle are removed from the heatreservoir while 200 J per cycle are deposited in the triple-point reservoir, what is the temperature of the heatreservoir? (b) If 400 J per cycle are removed from thetriple-point reservoir while 200 J per cycle are depositedin the heat reservoir, what is the temperature of the heatreservoir?
42. What is the minimum work required of a refrigeratorif it is to extract 50 J per cycle from the inside of a freezerat −10 °C and exhaust heat to the air at 25 °C ?


4.6 Entropy
43. Two hundred joules of heat are removed from a heatreservoir at a temperature of 200 K. What is the entropychange of the reservoir?


44. In an isothermal reversible expansion at 27 °C , an
ideal gas does 20 J of work. What is the entropy change ofthe gas?
45. An ideal gas at 300 K is compressed isothermally toone-fifth its original volume. Determine the entropy changeper mole of the gas.
46. What is the entropy change of 10 g of steam at
100 °C when it condenses to water at the same
temperature?
47. A metal rod is used to conduct heat between tworeservoirs at temperatures Th and Tc, respectively. When
an amount of heat Q flows through the rod from the hot tothe cold reservoir, what is the net entropy change of the rod,the hot reservoir, the cold reservoir, and the universe?
48. For the Carnot cycle of Figure 4.12, what is theentropy change of the hot reservoir, the cold reservoir, andthe universe?
49. A 5.0-kg piece of lead at a temperature of 600 °C is
placed in a lake whose temperature is 15 °C . Determine
the entropy change of (a) the lead piece, (b) the lake, and(c) the universe.
50. One mole of an ideal gas doubles its volume in areversible isothermal expansion. (a) What is the change inentropy of the gas? (b) If 1500 J of heat are added in thisprocess, what is the temperature of the gas?
51. One mole of an ideal monatomic gas is confined to arigid container. When heat is added reversibly to the gas, itstemperature changes from T1 to T2. (a) How much heat is
added? (b) What is the change in entropy of the gas?
52. (a) A 5.0-kg rock at a temperature of 20 °C is
dropped into a shallow lake also at 20 °C from a height of
1.0 × 103 m . What is the resulting change in entropy of
the universe? (b) If the temperature of the rock is 100 °C
when it is dropped, what is the change of entropy of theuniverse? Assume that air friction is negligible (not a goodassumption) and that c = 860 J/kg · K is the specific heat
of the rock.


4.7 Entropy on a Microscopic Scale
53. A copper rod of cross-sectional area 5.0 cm2 and
length 5.0 m conducts heat from a heat reservoir at 373 Kto one at 273 K. What is the time rate of change of theuniverse’s entropy for this process?


Chapter 4 | The Second Law of Thermodynamics 175




54. Fifty grams of water at 20 °C is heated until it
becomes vapor at 100 °C . Calculate the change in entropy
of the water in this process.
55. Fifty grams of water at 0 °C are changed into vapor
at 100 °C . What is the change in entropy of the water in
this process?
56. In an isochoric process, heat is added to 10 mol ofmonoatomic ideal gas whose temperature increases from273 to 373 K. What is the entropy change of the gas?
57. Two hundred grams of water at 0 °C is brought into
contact with a heat reservoir at 80 °C . After thermal
equilibrium is reached, what is the temperature of thewater? Of the reservoir? How much heat has beentransferred in the process? What is the entropy change ofthe water? Of the reservoir? What is the entropy change ofthe universe?
58. Suppose that the temperature of the water in theprevious problem is raised by first bringing it to thermalequilibrium with a reservoir at a temperature of 40 °C
and then with a reservoir at 80 °C . Calculate the entropy
changes of (a) each reservoir, (b) of the water, and (c) of theuniverse.
59. Two hundred grams of water at 0 °C is brought into
contact into thermal equilibrium successively withreservoirs at 20 °C , 40 °C , 60 °C , and 80 °C . (a) What
is the entropy change of the water? (b) Of the reservoir? (c)What is the entropy change of the universe?
60. (a) Ten grams of H2O starts as ice at 0 °C . The ice
absorbs heat from the air (just above 0 °C ) until all of it
melts. Calculate the entropy change of the H2O , of the air,
and of the universe. (b) Suppose that the air in part (a) is at
20 °C rather than 0 °C and that the ice absorbs heat until
it becomes water at 20 °C . Calculate the entropy change
of the H2O , of the air, and of the universe. (c) Is either of
these processes reversible?
61. The Carnot cycle is represented by the temperature-entropy diagram shown below. (a) How much heat isabsorbed per cycle at the high-temperature reservoir? (b)How much heat is exhausted per cycle at the low-temperature reservoir? (c) How much work is done percycle by the engine? (d) What is the efficiency of theengine?


62. A Carnot engine operating between heat reservoirsat 500 and 300 K absorbs 1500 J per cycle at the high-temperature reservoir. (a) Represent the engine’s cycle ona temperature-entropy diagram. (b) How much work percycle is done by the engine?
63. A monoatomic ideal gas (n moles) goes through acyclic process shown below. Find the change in entropy ofthe gas in each step and the total entropy change over theentire cycle.


64. A Carnot engine has an efficiency of 0.60. When thetemperature of its cold reservoir changes, the efficiencydrops to 0.55. If initially Tc = 27 °C , determine (a) the
constant value of Th and (b) the final value of Tc .
65. A Carnot engine performs 100 J of work whilerejecting 200 J of heat each cycle. After the temperature ofthe hot reservoir only is adjusted, it is found that the enginenow does 130 J of work while discarding the same quantityof heat. (a) What are the initial and final efficiencies of theengine? (b) What is the fractional change in the temperatureof the hot reservoir?
66. A Carnot refrigerator exhausts heat to the air, which isat a temperature of 25 °C . How much power is used by the
refrigerator if it freezes 1.5 g of water per second? Assumethe water is at 0 °C .


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ADDITIONAL PROBLEMS
67. A 300-W heat pump operates between the ground,whose temperature is 0 °C , and the interior of a house at
22 °C . What is the maximum amount of heat per hour that
the heat pump can supply to the house?
68. An engineer must design a refrigerator that does 300J of work per cycle to extract 2100 J of heat per cyclefrom a freezer whose temperature is −10 °C . What is
the maximum air temperature for which this condition canbe met? Is this a reasonable condition to impose on thedesign?
69. A Carnot engine employs 1.5 mol of nitrogen gasas a working substance, which is considered as an idealdiatomic gas with γ = 7.5 at the working temperatures of
the engine. The Carnot cycle goes in the cycle ABCDAwith AB being an isothermal expansion. The volume at
points A and C of the cycle are 5.0 × 10−3 m3 and 0.15
L, respectively. The engine operates between two thermalbaths of temperature 500 K and 300 K. (a) Find the valuesof volume at B and D. (b) How much heat is absorbed bythe gas in the AB isothermal expansion? (c) How muchwork is done by the gas in the AB isothermal expansion? (d)How much heat is given up by the gas in the CD isothermalexpansion? (e) How much work is done by the gas in theCD isothermal compression? (f) How much work is doneby the gas in the BC adiabatic expansion? (g) How muchwork is done by the gas in the DA adiabatic compression?(h) Find the value of efficiency of the engine based onthe net work and heat input. Compare this value to theefficiency of a Carnot engine based on the temperatures ofthe two baths.
70. A 5.0-kg wood block starts with an initial speed of8.0 m/s and slides across the floor until friction stops it.Estimate the resulting change in entropy of the universe.Assume that everything stays at a room temperature of
20 °C .
71. A system consisting of 20.0 mol of a monoatomicideal gas is cooled at constant pressure from a volume of50.0 L to 10.0 L. The initial temperature was 300 K. Whatis the change in entropy of the gas?
72. A glass beaker of mass 400 g contains 500 g of waterat 27 °C . The beaker is heated reversibly so that the
temperature of the beaker and water rise gradually to
57 °C . Find the change in entropy of the beaker and water
together.
73. A Carnot engine operates between 550 °C and
20 °C baths and produces 300 kJ of energy in each cycle.


Find the change in entropy of the (a) hot bath and (b) coldbath, in each Carnot cycle?
74. An ideal gas at temperature T is stored in the left halfof an insulating container of volume V using a partition ofnegligible volume (see below). What is the entropy changeper mole of the gas in each of the following cases? (a) Thepartition is suddenly removed and the gas quickly fills theentire container. (b) A tiny hole is punctured in the partitionand after a long period, the gas reaches an equilibrium statesuch that there is no net flow through the hole. (c) Thepartition is moved very slowly and adiabatically all theway to the right wall so that the gas finally fills the entirecontainer.


75. A 0.50-kg piece of aluminum at 250 °C is dropped
into 1.0 kg of water at 20 °C . After equilibrium is reached,
what is the net entropy change of the system?
76. Suppose 20 g of ice at 0 °C is added to 300 g of
water at 60 °C . What is the total change in entropy of the
mixture after it reaches thermal equilibrium?
77. A heat engine operates between two temperatures suchthat the working substance of the engine absorbs 5000 Jof heat from the high-temperature bath and rejects 3000J to the low-temperature bath. The rest of the energy isconverted into mechanical energy of the turbine. Find (a)the amount of work produced by the engine and (b) theefficiency of the engine.
78. A thermal engine produces 4 MJ of electrical energywhile operating between two thermal baths of differenttemperatures. The working substance of the engine rejects5 MJ of heat to the cold temperature bath. What is theefficiency of the engine?
79. A coal power plant consumes 100,000 kg of coalper hour and produces 500 MW of power. If the heat ofcombustion of coal is 30 MJ/kg, what is the efficiency ofthe power plant?
80. A Carnot engine operates in a Carnot cycle between aheat source at 550 °C and a heat sink at 20 °C. Find the


Chapter 4 | The Second Law of Thermodynamics 177




efficiency of the Carnot engine.
81. A Carnot engine working between two heat baths oftemperatures 600 K and 273 K completes each cycle in 5sec. In each cycle, the engine absorbs 10 kJ of heat. Findthe power of the engine.


82. A Carnot cycle working between 100 °C and 30 °C
is used to drive a refrigerator between −10 °C and 30 °C.
How much energy must the Carnot engine produce persecond so that the refrigerator is able to discard 10 J ofenergy per second?


CHALLENGE PROBLEMS
83. (a) An infinitesimal amount of heat is added reversiblyto a system. By combining the first and second laws, showthat dU = TdS − dW . (b) When heat is added to an ideal
gas, its temperature and volume change from
T1 and V1 to T2 and V2 . Show that the entropy change of
n moles of the gas is given by
ΔS = nCv ln


T2
T1


+ nR ln
V2
V1


.


84. Using the result of the preceding problem, show that
for an ideal gas undergoing an adiabatic process, TV γ − 1
is constant.
85. With the help of the two preceding problems, showthat ΔS between states 1 and 2 of n moles an ideal gas is
given by
ΔS = nC p ln


T2
T1


− nR ln
p2
p1


.


86. A cylinder contains 500 g of helium at 120 atm and
20 °C . The valve is leaky, and all the gas slowly escapes
isothermally into the atmosphere. Use the results of thepreceding problem to determine the resulting change inentropy of the universe.
87. A diatomic ideal gas is brought from an initialequilibrium state at p1 = 0.50 atm and T1 = 300 K to a
final stage with p2 = 0.20 atm and T1 = 500 K. Use the
results of the previous problem to determine the entropychange per mole of the gas.
88. The gasoline internal combustion engine operates ina cycle consisting of six parts. Four of these parts involve,among other things, friction, heat exchange through finitetemperature differences, and accelerations of the piston; itis irreversible. Nevertheless, it is represented by the idealreversible Otto cycle, which is illustrated below. Theworking substance of the cycle is assumed to be air. The sixsteps of the Otto cycle are as follows:


i. Isobaric intake stroke (OA). A mixture ofgasoline and air is drawn into the combustionchamber at atmospheric pressure p0 as the piston
expands, increasing the volume of the cylinderfrom zero to VA .
ii. Adiabatic compression stroke (AB). Thetemperature of the mixture rises as the pistoncompresses it adiabatically from a volume
VA to VB .
iii. Ignition at constant volume (BC). The mixtureis ignited by a spark. The combustion happens sofast that there is essentially no motion of the piston.During this process, the added heat Q1 causes the
pressure to increase from pB to pC at the constant
volume VB( = VC) .
iv. Adiabatic expansion (CD). The heated mixtureof gasoline and air expands against the piston,increasing the volume from VC to VD . This is
called the power stroke, as it is the part of the cyclethat delivers most of the power to the crankshaft.v. Constant-volume exhaust (DA). When theexhaust valve opens, some of the combustionproducts escape. There is almost no movement ofthe piston during this part of the cycle, so thevolume remains constant at VA( = VD) . Most of
the available energy is lost here, as represented bythe heat exhaust Q2 .
vi. Isobaric compression (AO). The exhaust valveremains open, and the compression from VA to
zero drives out the remaining combustion products.


(a) Using (i) e = W/Q1 ; (ii) W = Q1 − Q2 ; and (iii)
Q1 = nCv(TC − TB) , Q2 = nCv(TD − TA) , show that
e = 1 −


TD − TA
TC − TB


.
(b) Use the fact that steps (ii) and (iv) are adiabatic to showthat
e = 1 − 1


r
γ − 1


,
where r = VA/VB . The quantity r is called the
compression ratio of the engine.


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(c) In practice, r is kept less than around 7. For largervalues, the gasoline-air mixture is compressed totemperatures so high that it explodes before the finelytimed spark is delivered. This preignition causes engineknock and loss of power. Show that for r = 6 and γ = 1.4
(the value for air), e = 0.51 , or an efficiency of 51%.
Because of the many irreversible processes, an actualinternal combustion engine has an efficiency much lessthan this ideal value. A typical efficiency for a tuned engineis about 25% to 30% .


89. An ideal diesel cycle is shown below. This cycleconsists of five strokes. In this case, only air is drawn intothe chamber during the intake stroke OA. The air is thencompressed adiabatically from state A to state B, raising itstemperature high enough so that when fuel is added duringthe power stroke BC, it ignites. After ignition ends at C,there is a further adiabatic power stroke CD. Finally, thereis an exhaust at constant volume as the pressure drops from
pD to pA , followed by a further exhaust when the piston
compresses the chamber volume to zero.
(a) Use W = Q1 − Q2 , Q1 = nC p(TC − TB) , and
Q2 = nCv(TD − TA) to show that
e = W


Q1
= 1 −


TD − TA
γ(TC − TB)


.
(b) Use the fact that A → B and C → D are adiabatic to
show that


e = 1 − 1γ




VC
VD





γ
− ⎛⎝


VB
VA





γ




VC
VD



⎠−


VB
VA





.


(c) Since there is no preignition (remember, the chamberdoes not contain any fuel during the compression), thecompression ratio can be larger than that for a gasolineengine. Typically, VA/VB = 15 and VD/VC = 5 . For these
values and γ = 1.4, show that ε = 0.56 , or an efficiency
of 56% . Diesel engines actually operate at an efficiency


of about 30% to 35% compared with 25% to 30% for
gasoline engines.


90. Consider an ideal gas Joule cycle, also called theBrayton cycle, shown below. Find the formula forefficiency of the engine using this cycle in terms of P1 ,
P2 , and γ .


91. Derive a formula for the coefficient of performanceof a refrigerator using an ideal gas as a working substanceoperating in the cycle shown below in terms of theproperties of the three states labeled 1, 2, and 3.


92. Two moles of nitrogen gas, with γ = 7/5 for ideal
diatomic gases, occupies a volume of 10−2m3 in an
insulated cylinder at temperature 300 K. The gas isadiabatically and reversibly compressed to a volume of 5L. The piston of the cylinder is locked in its place, and


Chapter 4 | The Second Law of Thermodynamics 179




the insulation around the cylinder is removed. The heat-conducting cylinder is then placed in a 300-K bath. Heatfrom the compressed gas leaves the gas, and thetemperature of the gas becomes 300 K again. The gas isthen slowly expanded at the fixed temperature 300 K until
the volume of the gas becomes 10−2m3 , thus making a
complete cycle for the gas. For the entire cycle, calculate(a) the work done by the gas, (b) the heat into or out of the


gas, (c) the change in the internal energy of the gas, and (d)the change in entropy of the gas.
93. A Carnot refrigerator, working between 0 °C and
30 °C is used to cool a bucket of water containing
10−2 m3 of water at 30 °C to 5 °C in 2 hours. Find the
total amount of work needed.


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5 | ELECTRIC CHARGESAND FIELDS


Figure 5.1 Electric charges exist all around us. They can cause objects to be repelled from each other or to be attracted to eachother. (credit: modification of work by Sean McGrath)


Chapter Outline
5.1 Electric Charge
5.2 Conductors, Insulators, and Charging by Induction
5.3 Coulomb's Law
5.4 Electric Field
5.5 Calculating Electric Fields of Charge Distributions
5.6 Electric Field Lines
5.7 Electric Dipoles


Introduction
Back when we were studying Newton’s laws, we identified several physical phenomena as forces. We did so based on theeffect they had on a physical object: Specifically, they caused the object to accelerate. Later, when we studied impulse andmomentum, we expanded this idea to identify a force as any physical phenomenon that changed the momentum of an object.In either case, the result is the same: We recognize a force by the effect that it has on an object.
In Gravitation (http://cnx.org/content/m58344/latest/) , we examined the force of gravity, which acts on all objectswith mass. In this chapter, we begin the study of the electric force, which acts on all objects with a property called charge.The electric force is much stronger than gravity (in most systems where both appear), but it can be a force of attraction or aforce of repulsion, which leads to very different effects on objects. The electric force helps keep atoms together, so it is of


Chapter 5 | Electric Charges and Fields 181




fundamental importance in matter. But it also governs most everyday interactions we deal with, from chemical interactionsto biological processes.
5.1 | Electric Charge


Learning Objectives
By the end of this section, you will be able to:
• Describe the concept of electric charge
• Explain qualitatively the force electric charge creates


You are certainly familiar with electronic devices that you activate with the click of a switch, from computers to cell phonesto television. And you have certainly seen electricity in a flash of lightning during a heavy thunderstorm. But you have alsomost likely experienced electrical effects in other ways, maybe without realizing that an electric force was involved. Let’stake a look at some of these activities and see what we can learn from them about electric charges and forces.
Discoveries
You have probably experienced the phenomenon of static electricity: When you first take clothes out of a dryer, many (notall) of them tend to stick together; for some fabrics, they can be very difficult to separate. Another example occurs if youtake a woolen sweater off quickly—you can feel (and hear) the static electricity pulling on your clothes, and perhaps evenyour hair. If you comb your hair on a dry day and then put the comb close to a thin stream of water coming out of a faucet,you will find that the water stream bends toward (is attracted to) the comb (Figure 5.2).


Figure 5.2 An electrically charged comb attracts a stream ofwater from a distance. Note that the water is not touching thecomb. (credit: Jane Whitney)


Suppose you bring the comb close to some small strips of paper; the strips of paper are attracted to the comb and even clingto it (Figure 5.3). In the kitchen, quickly pull a length of plastic cling wrap off the roll; it will tend to cling to most anynonmetallic material (such as plastic, glass, or food). If you rub a balloon on a wall for a few seconds, it will stick to thewall. Probably the most annoying effect of static electricity is getting shocked by a doorknob (or a friend) after shufflingyour feet on some types of carpeting.


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Figure 5.3 After being used to comb hair, this comb attractssmall strips of paper from a distance, without physical contact.Investigation of this behavior helped lead to the concept of theelectric force.


Many of these phenomena have been known for centuries. The ancient Greek philosopher Thales of Miletus (624–546 BCE)recorded that when amber (a hard, translucent, fossilized resin from extinct trees) was vigorously rubbed with a piece of fur,a force was created that caused the fur and the amber to be attracted to each other (Figure 5.4). Additionally, he found thatthe rubbed amber would not only attract the fur, and the fur attract the amber, but they both could affect other (nonmetallic)objects, even if not in contact with those objects (Figure 5.5).


Figure 5.4 Borneo amber is mined in Sabah, Malaysia, from shale-sandstone-mudstone veins.When a piece of amber is rubbed with a piece of fur, the amber gains more electrons, giving it anet negative charge. At the same time, the fur, having lost electrons, becomes positively charged.(credit: “Sebakoamber”/Wikimedia Commons)


Chapter 5 | Electric Charges and Fields 183




Figure 5.5 When materials are rubbed together, charges can be separated, particularly if one material has a greater affinity forelectrons than another. (a) Both the amber and cloth are originally neutral, with equal positive and negative charges. Only a tinyfraction of the charges are involved, and only a few of them are shown here. (b) When rubbed together, some negative charge istransferred to the amber, leaving the cloth with a net positive charge. (c) When separated, the amber and cloth now have netcharges, but the absolute value of the net positive and negative charges will be equal.


The English physicist William Gilbert (1544–1603) also studied this attractive force, using various substances. He workedwith amber, and, in addition, he experimented with rock crystal and various precious and semi-precious gemstones. He alsoexperimented with several metals. He found that the metals never exhibited this force, whereas the minerals did. Moreover,although an electrified amber rod would attract a piece of fur, it would repel another electrified amber rod; similarly, twoelectrified pieces of fur would repel each other.
This suggested there were two types of an electric property; this property eventually came to be called electric charge. Thedifference between the two types of electric charge is in the directions of the electric forces that each type of charge causes:These forces are repulsive when the same type of charge exists on two interacting objects and attractive when the chargesare of opposite types. The SI unit of electric charge is the coulomb (C), after the French physicist Charles Augustine deCoulomb (1736–1806).
The most peculiar aspect of this new force is that it does not require physical contact between the two objects in order tocause an acceleration. This is an example of a so-called “long-range” force. (Or, as Albert Einstein later phrased it, “actionat a distance.”) With the exception of gravity, all other forces we have discussed so far act only when the two interactingobjects actually touch.
The American physicist and statesman Benjamin Franklin found that he could concentrate charge in a “ Leyden jar,” whichwas essentially a glass jar with two sheets of metal foil, one inside and one outside, with the glass between them (Figure5.6). This created a large electric force between the two foil sheets.


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Figure 5.6 A Leyden jar (an early version of what is nowcalled a capacitor) allowed experimenters to store large amountsof electric charge. Benjamin Franklin used such a jar todemonstrate that lightning behaved exactly like the electricity hegot from the equipment in his laboratory.


Franklin pointed out that the observed behavior could be explained by supposing that one of the two types of chargeremained motionless, while the other type of charge flowed from one piece of foil to the other. He further suggested thatan excess of what he called this “electrical fluid” be called “positive electricity” and the deficiency of it be called “negativeelectricity.” His suggestion, with some minor modifications, is the model we use today. (With the experiments that he wasable to do, this was a pure guess; he had no way of actually determining the sign of the moving charge. Unfortunately, heguessed wrong; we now know that the charges that flow are the ones Franklin labeled negative, and the positive chargesremain largely motionless. Fortunately, as we’ll see, it makes no practical or theoretical difference which choice we make,as long as we stay consistent with our choice.)
Let’s list the specific observations that we have of this electric force:


• The force acts without physical contact between the two objects.
• The force can be either attractive or repulsive: If two interacting objects carry the same sign of charge, the force isrepulsive; if the charges are of opposite sign, the force is attractive. These interactions are referred to as electrostaticrepulsion and electrostatic attraction, respectively.
• Not all objects are affected by this force.
• The magnitude of the force decreases (rapidly) with increasing separation distance between the objects.


To be more precise, we find experimentally that the magnitude of the force decreases as the square of the distance betweenthe two interacting objects increases. Thus, for example, when the distance between two interacting objects is doubled, theforce between them decreases to one fourth what it was in the original system. We can also observe that the surroundings ofthe charged objects affect the magnitude of the force. However, we will explore this issue in a later chapter.
Properties of Electric Charge
In addition to the existence of two types of charge, several other properties of charge have been discovered.


Chapter 5 | Electric Charges and Fields 185




• Charge is quantized. This means that electric charge comes in discrete amounts, and there is a smallest possible
amount of charge that an object can have. In the SI system, this smallest amount is e ≡ 1.602 × 10−19 C . No free
particle can have less charge than this, and, therefore, the charge on any object—the charge on all objects—mustbe an integer multiple of this amount. All macroscopic, charged objects have charge because electrons have eitherbeen added or taken away from them, resulting in a net charge.


• The magnitude of the charge is independent of the type. Phrased another way, the smallest possible positive
charge (to four significant figures) is +1.602 × 10−19 C , and the smallest possible negative charge is
−1.602 × 10−19 C ; these values are exactly equal. This is simply how the laws of physics in our universe turned
out.


• Charge is conserved. Charge can neither be created nor destroyed; it can only be transferred from place to place,from one object to another. Frequently, we speak of two charges “canceling”; this is verbal shorthand. It means thatif two objects that have equal and opposite charges are physically close to each other, then the (oppositely directed)forces they apply on some other charged object cancel, for a net force of zero. It is important that you understandthat the charges on the objects by no means disappear, however. The net charge of the universe is constant.
• Charge is conserved in closed systems. In principle, if a negative charge disappeared from your lab bench andreappeared on the Moon, conservation of charge would still hold. However, this never happens. If the total chargeyou have in your local system on your lab bench is changing, there will be a measurable flow of charge into or outof the system. Again, charges can and do move around, and their effects can and do cancel, but the net charge inyour local environment (if closed) is conserved. The last two items are both referred to as the law of conservationof charge.


The Source of Charges: The Structure of the Atom
Once it became clear that all matter was composed of particles that came to be called atoms, it also quickly became clear thatthe constituents of the atom included both positively charged particles and negatively charged particles. The next questionwas, what are the physical properties of those electrically charged particles?
The negatively charged particle was the first one to be discovered. In 1897, the English physicist J. J. Thomson was studyingwhat was then known as cathode rays. Some years before, the English physicist William Crookes had shown that these“rays” were negatively charged, but his experiments were unable to tell any more than that. (The fact that they carried anegative electric charge was strong evidence that these were not rays at all, but particles.) Thomson prepared a pure beam ofthese particles and sent them through crossed electric and magnetic fields, and adjusted the various field strengths until thenet deflection of the beam was zero. With this experiment, he was able to determine the charge-to-mass ratio of the particle.This ratio showed that the mass of the particle was much smaller than that of any other previously known particle—1837times smaller, in fact. Eventually, this particle came to be called the electron.
Since the atom as a whole is electrically neutral, the next question was to determine how the positive and negative chargesare distributed within the atom. Thomson himself imagined that his electrons were embedded within a sort of positivelycharged paste, smeared out throughout the volume of the atom. However, in 1908, the New Zealand physicist ErnestRutherford showed that the positive charges of the atom existed within a tiny core—called a nucleus—that took up onlya very tiny fraction of the overall volume of the atom, but held over 99% of the mass. (See Linear Momentum andCollisions (http://cnx.org/content/m58317/latest/) .) In addition, he showed that the negatively charged electronsperpetually orbited about this nucleus, forming a sort of electrically charged cloud that surrounds the nucleus (Figure 5.7).Rutherford concluded that the nucleus was constructed of small, massive particles that he named protons.


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Figure 5.7 This simplified model of a hydrogen atom shows apositively charged nucleus (consisting, in the case of hydrogen, of asingle proton), surrounded by an electron “cloud.” The charge of theelectron cloud is equal (and opposite in sign) to the charge of thenucleus, but the electron does not have a definite location in space;hence, its representation here is as a cloud. Normal macroscopicamounts of matter contain immense numbers of atoms andmolecules, and, hence, even greater numbers of individual negativeand positive charges.


Since it was known that different atoms have different masses, and that ordinarily atoms are electrically neutral, it wasnatural to suppose that different atoms have different numbers of protons in their nucleus, with an equal number ofnegatively charged electrons orbiting about the positively charged nucleus, thus making the atoms overall electricallyneutral. However, it was soon discovered that although the lightest atom, hydrogen, did indeed have a single proton as itsnucleus, the next heaviest atom—helium—has twice the number of protons (two), but four times the mass of hydrogen.
This mystery was resolved in 1932 by the English physicist James Chadwick, with the discovery of the neutron. Theneutron is, essentially, an electrically neutral twin of the proton, with no electric charge, but (nearly) identical mass to theproton. The helium nucleus therefore has two neutrons along with its two protons. (Later experiments were to show thatalthough the neutron is electrically neutral overall, it does have an internal charge structure. Furthermore, although themasses of the neutron and the proton are nearly equal, they aren’t exactly equal: The neutron’s mass is very slightly largerthan the mass of the proton. That slight mass excess turned out to be of great importance. That, however, is a story that willhave to wait until our study of modern physics in Nuclear Physics (http://cnx.org/content/m58606/latest/) .)
Thus, in 1932, the picture of the atom was of a small, massive nucleus constructed of a combination of protons and neutrons,surrounded by a collection of electrons whose combined motion formed a sort of negatively charged “cloud” around thenucleus (Figure 5.8). In an electrically neutral atom, the total negative charge of the collection of electrons is equal to thetotal positive charge in the nucleus. The very low-mass electrons can be more or less easily removed or added to an atom,changing the net charge on the atom (though without changing its type). An atom that has had the charge altered in this wayis called an ion. Positive ions have had electrons removed, whereas negative ions have had excess electrons added. We alsouse this term to describe molecules that are not electrically neutral.


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Figure 5.8 The nucleus of a carbon atom is composed of sixprotons and six neutrons. As in hydrogen, the surrounding sixelectrons do not have definite locations and so can be considered tobe a sort of cloud surrounding the nucleus.


The story of the atom does not stop there, however. In the latter part of the twentieth century, many more subatomic particleswere discovered in the nucleus of the atom: pions, neutrinos, and quarks, among others. With the exception of the photon,none of these particles are directly relevant to the study of electromagnetism, so we defer further discussion of them untilthe chapter on particle physics (Particle Physics and Cosmology (http://cnx.org/content/m58767/latest/) ).
A Note on Terminology
As noted previously, electric charge is a property that an object can have. This is similar to how an object can have aproperty that we call mass, a property that we call density, a property that we call temperature, and so on. Technically, weshould always say something like, “Suppose we have a particle that carries a charge of 3 µC. ” However, it is very common
to say instead, “Suppose we have a 3-µC charge.” Similarly, we often say something like, “Six charges are located at the
vertices of a regular hexagon.” A charge is not a particle; rather, it is a property of a particle. Nevertheless, this terminologyis extremely common (and is frequently used in this book, as it is everywhere else). So, keep in the back of your mind whatwe really mean when we refer to a “charge.”
5.2 | Conductors, Insulators, and Charging by Induction


Learning Objectives
By the end of this section, you will be able to:
• Explain what a conductor is
• Explain what an insulator is
• List the differences and similarities between conductors and insulators
• Describe the process of charging by induction


In the preceding section, we said that scientists were able to create electric charge only on nonmetallic materials and neveron metals. To understand why this is the case, you have to understand more about the nature and structure of atoms. In thissection, we discuss how and why electric charges do—or do not—move through materials (Figure 5.9). A more completedescription is given in a later chapter.


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Figure 5.9 This power adapter uses metal wires and connectors to conduct electricity from thewall socket to a laptop computer. The conducting wires allow electrons to move freely through thecables, which are shielded by rubber and plastic. These materials act as insulators that don’t allowelectric charge to escape outward. (credit: modification of work by “Evan-Amos”/WikimediaCommons)
Conductors and Insulators
As discussed in the previous section, electrons surround the tiny nucleus in the form of a (comparatively) vast cloud ofnegative charge. However, this cloud does have a definite structure to it. Let’s consider an atom of the most commonly usedconductor, copper.
For reasons that will become clear in Atomic Structure (http://cnx.org/content/m58583/latest/) , there is anoutermost electron that is only loosely bound to the atom’s nucleus. It can be easily dislodged; it then moves to aneighboring atom. In a large mass of copper atoms (such as a copper wire or a sheet of copper), these vast numbersof outermost electrons (one per atom) wander from atom to atom, and are the electrons that do the moving whenelectricity flows. These wandering, or “free,” electrons are called conduction electrons, and copper is therefore an excellentconductor (of electric charge). All conducting elements have a similar arrangement of their electrons, with one or twoconduction electrons. This includes most metals.
Insulators, in contrast, are made from materials that lack conduction electrons; charge flows only with great difficulty, ifat all. Even if excess charge is added to an insulating material, it cannot move, remaining indefinitely in place. This is whyinsulating materials exhibit the electrical attraction and repulsion forces described earlier, whereas conductors do not; anyexcess charge placed on a conductor would instantly flow away (due to mutual repulsion from existing charges), leavingno excess charge around to create forces. Charge cannot flow along or through an insulator, so its electric forces remainfor long periods of time. (Charge will dissipate from an insulator, given enough time.) As it happens, amber, fur, and mostsemi-precious gems are insulators, as are materials like wood, glass, and plastic.
Charging by Induction
Let’s examine in more detail what happens in a conductor when an electrically charged object is brought close to it. Asmentioned, the conduction electrons in the conductor are able to move with nearly complete freedom. As a result, when acharged insulator (such as a positively charged glass rod) is brought close to the conductor, the (total) charge on the insulatorexerts an electric force on the conduction electrons. Since the rod is positively charged, the conduction electrons (whichthemselves are negatively charged) are attracted, flowing toward the insulator to the near side of the conductor (Figure5.10).
Now, the conductor is still overall electrically neutral; the conduction electrons have changed position, but they are still inthe conducting material. However, the conductor now has a charge distribution; the near end (the portion of the conductorclosest to the insulator) now has more negative charge than positive charge, and the reverse is true of the end farthest fromthe insulator. The relocation of negative charges to the near side of the conductor results in an overall positive charge in thepart of the conductor farthest from the insulator. We have thus created an electric charge distribution where one did not existbefore. This process is referred to as inducing polarization—in this case, polarizing the conductor. The resulting separationof positive and negative charge is called polarization, and a material, or even a molecule, that exhibits polarization is said tobe polarized. A similar situation occurs with a negatively charged insulator, but the resulting polarization is in the oppositedirection.


Chapter 5 | Electric Charges and Fields 189




Figure 5.10 Induced polarization. A positively charged glassrod is brought near the left side of the conducting sphere,attracting negative charge and leaving the other side of thesphere positively charged. Although the sphere is overall stillelectrically neutral, it now has a charge distribution, so it canexert an electric force on other nearby charges. Furthermore, thedistribution is such that it will be attracted to the glass rod.


The result is the formation of what is called an electric dipole, from a Latin phrase meaning “two ends.” The presence ofelectric charges on the insulator—and the electric forces they apply to the conduction electrons—creates, or “induces,” thedipole in the conductor.
Neutral objects can be attracted to any charged object. The pieces of straw attracted to polished amber are neutral, forexample. If you run a plastic comb through your hair, the charged comb can pick up neutral pieces of paper. Figure 5.11shows how the polarization of atoms and molecules in neutral objects results in their attraction to a charged object.


Figure 5.11 Both positive and negative objects attract a neutral object by polarizing its molecules. (a) A positive objectbrought near a neutral insulator polarizes its molecules. There is a slight shift in the distribution of the electrons orbiting themolecule, with unlike charges being brought nearer and like charges moved away. Since the electrostatic force decreases withdistance, there is a net attraction. (b) A negative object produces the opposite polarization, but again attracts the neutral object.(c) The same effect occurs for a conductor; since the unlike charges are closer, there is a net attraction.


When a charged rod is brought near a neutral substance, an insulator in this case, the distribution of charge in atoms andmolecules is shifted slightly. Opposite charge is attracted nearer the external charged rod, while like charge is repelled. Sincethe electrostatic force decreases with distance, the repulsion of like charges is weaker than the attraction of unlike charges,and so there is a net attraction. Thus, a positively charged glass rod attracts neutral pieces of paper, as will a negativelycharged rubber rod. Some molecules, like water, are polar molecules. Polar molecules have a natural or inherent separationof charge, although they are neutral overall. Polar molecules are particularly affected by other charged objects and showgreater polarization effects than molecules with naturally uniform charge distributions.
When the two ends of a dipole can be separated, this method of charging by induction may be used to create chargedobjects without transferring charge. In Figure 5.12, we see two neutral metal spheres in contact with one another butinsulated from the rest of the world. A positively charged rod is brought near one of them, attracting negative charge to thatside, leaving the other sphere positively charged.


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Figure 5.12 Charging by induction. (a) Two uncharged or neutral metal spheres are in contact with eachother but insulated from the rest of the world. (b) A positively charged glass rod is brought near the sphereon the left, attracting negative charge and leaving the other sphere positively charged. (c) The spheres areseparated before the rod is removed, thus separating negative and positive charges. (d) The spheres retainnet charges after the inducing rod is removed—without ever having been touched by a charged object.


Another method of charging by induction is shown in Figure 5.13. The neutral metal sphere is polarized when a chargedrod is brought near it. The sphere is then grounded, meaning that a conducting wire is run from the sphere to the ground.Since Earth is large and most of the ground is a good conductor, it can supply or accept excess charge easily. In this case,electrons are attracted to the sphere through a wire called the ground wire, because it supplies a conducting path to theground. The ground connection is broken before the charged rod is removed, leaving the sphere with an excess chargeopposite to that of the rod. Again, an opposite charge is achieved when charging by induction, and the charged rod losesnone of its excess charge.


Chapter 5 | Electric Charges and Fields 191




Figure 5.13 Charging by induction using a ground connection. (a) A positively charged rod is brought near a neutral metalsphere, polarizing it. (b) The sphere is grounded, allowing electrons to be attracted from Earth’s ample supply. (c) The groundconnection is broken. (d) The positive rod is removed, leaving the sphere with an induced negative charge.


5.3 | Coulomb's Law
Learning Objectives


By the end of this section, you will be able to:
• Describe the electric force, both qualitatively and quantitatively
• Calculate the force that charges exert on each other
• Determine the direction of the electric force for different source charges
• Correctly describe and apply the superposition principle for multiple source charges


Experiments with electric charges have shown that if two objects each have electric charge, then they exert an electric forceon each other. The magnitude of the force is linearly proportional to the net charge on each object and inversely proportionalto the square of the distance between them. (Interestingly, the force does not depend on the mass of the objects.) Thedirection of the force vector is along the imaginary line joining the two objects and is dictated by the signs of the chargesinvolved.
Let


• q1, q2 = the net electric charges of the two objects;
• r→ 12 = the vector displacement from q1 to q2 .


The electric force F→ on one of the charges is proportional to the magnitude of its own charge and the magnitude of the
other charge, and is inversely proportional to the square of the distance between them:


F ∝
q1q2
r12
2


.


This proportionality becomes an equality with the introduction of a proportionality constant. For reasons that will becomeclear in a later chapter, the proportionality constant that we use is actually a collection of constants. (We discuss this constantshortly.)
Coulomb’s Law
The electric force (or Coulomb force) between two electrically charged particles is equal to


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(5.1)
F


12(r) =
1


4πε0
|q1q2|
r12
2


r̂ 12


We use absolute value signs around the product q1q2 because one of the charges may be negative, but the magnitude
of the force is always positive. The unit vector r̂ points directly from the charge q1 toward q2 . If q1 and q2
have the same sign, the force vector on q2 points away from q1 ; if they have opposite signs, the force on q2 points
toward q1 (Figure 5.14).


Figure 5.14 The electrostatic force F→ between point charges q1 and q2
separated by a distance r is given by Coulomb’s law. Note that Newton’s third law(every force exerted creates an equal and opposite force) applies as usual—the force on
q1 is equal in magnitude and opposite in direction to the force it exerts on q2 . (a)
Like charges; (b) unlike charges.


It is important to note that the electric force is not constant; it is a function of the separation distance between the two
charges. If either the test charge or the source charge (or both) move, then r→ changes, and therefore so does the force. An
immediate consequence of this is that direct application of Newton’s laws with this force can be mathematically difficult,depending on the specific problem at hand. It can (usually) be done, but we almost always look for easier methods ofcalculating whatever physical quantity we are interested in. (Conservation of energy is the most common choice.)
Finally, the new constant ε0 in Coulomb’s law is called the permittivity of free space, or (better) the permittivity of
vacuum. It has a very important physical meaning that we will discuss in a later chapter; for now, it is simply an empiricalproportionality constant. Its numerical value (to three significant figures) turns out to be


ε0 = 8.85 × 10
−12 C2


N ·m2
.


These units are required to give the force in Coulomb’s law the correct units of newtons. Note that in Coulomb’s law, thepermittivity of vacuum is only part of the proportionality constant. For convenience, we often define a Coulomb’s constant:
ke = 14πε0


= 8.99 × 109 N ·m
2


C2
.


Example 5.1
The Force on the Electron in Hydrogen
A hydrogen atom consists of a single proton and a single electron. The proton has a charge of +e and the
electron has −e . In the “ground state” of the atom, the electron orbits the proton at most probable distance of
5.29 × 10−11 m (Figure 5.15). Calculate the electric force on the electron due to the proton.


Chapter 5 | Electric Charges and Fields 193




5.1


Figure 5.15 A schematic depiction of a hydrogen atom,showing the force on the electron. This depiction is only toenable us to calculate the force; the hydrogen atom does notreally look like this. Recall Figure 5.7.


Strategy
For the purposes of this example, we are treating the electron and proton as two point particles, each with anelectric charge, and we are told the distance between them; we are asked to calculate the force on the electron.We thus use Coulomb’s law.
Solution
Our two charges and the distance between them are,


q1 = +e = +1.602 × 10
−19 C


q2 = −e = −1.602 × 10
−19 C


r = 5.29 × 10−11 m.


The magnitude of the force on the electron is
F = 1


4πϵ0
|e|2


r2
= 1




⎝8.85 × 10


−12 C2


N ·m2




⎝1.602 × 10


−19 C⎞⎠
2



⎝5.29 × 10


−11 m⎞⎠
2


= 8.25 × 10−8 N.


As for the direction, since the charges on the two particles are opposite, the force is attractive; the force on theelectron points radially directly toward the proton, everywhere in the electron’s orbit. The force is thus expressedas
F


= ⎛⎝8.25 × 10
−8 N⎞⎠ r̂ .


Significance
This is a three-dimensional system, so the electron (and therefore the force on it) can be anywhere in animaginary spherical shell around the proton. In this “classical” model of the hydrogen atom, the electrostaticforce on the electron points in the inward centripetal direction, thus maintaining the electron’s orbit. But note thatthe quantum mechanical model of hydrogen (discussed in Quantum Mechanics (http://cnx.org/content/m58573/latest/) ) is utterly different.


Check Your Understanding What would be different if the electron also had a positive charge?


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Multiple Source Charges
The analysis that we have done for two particles can be extended to an arbitrary number of particles; we simply repeat theanalysis, two charges at a time. Specifically, we ask the question: Given N charges (which we refer to as source charge),what is the net electric force that they exert on some other point charge (which we call the test charge)? Note that we usethese terms because we can think of the test charge being used to test the strength of the force provided by the sourcecharges.
Like all forces that we have seen up to now, the net electric force on our test charge is simply the vector sum of eachindividual electric force exerted on it by each of the individual test charges. Thus, we can calculate the net force on the testchargeQ by calculating the force on it from each source charge, taken one at a time, and then adding all those forces together(as vectors). This ability to simply add up individual forces in this way is referred to as the principle of superposition, andis one of the more important features of the electric force. In mathematical form, this becomes


(5.2)
F


(r) = 1
4πε0


Q∑
i = 1


N
qi
ri
2
r̂ i.


In this expression, Q represents the charge of the particle that is experiencing the electric force F→ , and is located at r→
from the origin; the qi ’s are the N source charges, and the vectors r→ i = ri r̂ i are the displacements from the position
of the ith charge to the position of Q. Each of the N unit vectors points directly from its associated source charge toward thetest charge. All of this is depicted in Figure 5.16. Please note that there is no physical difference between Q and qi ; the
difference in labels is merely to allow clear discussion, with Q being the charge we are determining the force on.


Figure 5.16 The eight source charges each apply a force on thesingle test charge Q. Each force can be calculated independentlyof the other seven forces. This is the essence of the superpositionprinciple.


(Note that the force vector F→ i does not necessarily point in the same direction as the unit vector r̂ i ; it may point in
the opposite direction, − r̂ i . The signs of the source charge and test charge determine the direction of the force on the test
charge.)
There is a complication, however. Just as the source charges each exert a force on the test charge, so too (by Newton’s thirdlaw) does the test charge exert an equal and opposite force on each of the source charges. As a consequence, each sourcecharge would change position. However, by Equation 5.2, the force on the test charge is a function of position; thus, asthe positions of the source charges change, the net force on the test charge necessarily changes, which changes the force,


Chapter 5 | Electric Charges and Fields 195




which again changes the positions. Thus, the entire mathematical analysis quickly becomes intractable. Later, we will learntechniques for handling this situation, but for now, we make the simplifying assumption that the source charges are fixedin place somehow, so that their positions are constant in time. (The test charge is allowed to move.) With this restriction inplace, the analysis of charges is known as electrostatics, where “statics” refers to the constant (that is, static) positions ofthe source charges and the force is referred to as an electrostatic force.
Example 5.2


The Net Force from Two Source Charges
Three different, small charged objects are placed as shown in Figure 5.17. The charges q1 and q3 are fixed
in place; q2 is free to move. Given q1 = 2e , q2 = −3e , and q3 = −5e , and that d = 2.0 × 10−7 m , what is
the net force on the middle charge q2 ?


Figure 5.17 Source charges q1 and q3 each apply a force
on q2 .


Strategy
We use Coulomb’s law again. The way the question is phrased indicates that q2 is our test charge, so that q1 and
q3 are source charges. The principle of superposition says that the force on q2 from each of the other charges
is unaffected by the presence of the other charge. Therefore, we write down the force on q2 from each and add
them together as vectors.
Solution
We have two source charges (q1 and q3), a test charge (q2), distances (r21 and r23), and we are asked to
find a force. This calls for Coulomb’s law and superposition of forces. There are two forces:


F


= F


21 + F


23 =
1


4πε0





⎢q2q1
r21
2


j
^


+



⎜−


q2q3
r23
2


i
^⎞







⎥.


We can’t add these forces directly because they don’t point in the same direction: F→ 12 points only in
the −x-direction, while F→ 13 points only in the +y-direction. The net force is obtained from applying the
Pythagorean theorem to its x- and y-components:


F = Fx
2 + Fy


2


where


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5.2


Fx = −F23 = −
1


4πε0


q2q3
r23
2


= −

⎝8.99 × 10


9 N ·m2


C2




⎝4.806 × 10


−19 C⎞⎠

⎝8.01 × 10


−19 C⎞⎠



⎝4.00 × 10


−7 m⎞⎠
2


= −2.16 × 10−14 N


and
Fy = F21 =


1
4πε0


q2q1
r21
2


=

⎝8.99 × 10


9 N ·m2


C2




⎝4.806 × 10


−19 C⎞⎠

⎝3.204 × 10


−19 C⎞⎠



⎝2.00 × 10


−7 m⎞⎠
2


= 3.46 × 10−14 N.


We find that
F = Fx


2 + Fy
2 = 4.08 × 10−14 N


at an angle of
ϕ = tan−1




Fy
Fx

⎠ = tan


−1⎛

3.46 × 10−14 N
−2.16 × 10−14 N



⎠ = −58°,


that is, 58° above the −x-axis, as shown in the diagram.
Significance
Notice that when we substituted the numerical values of the charges, we did not include the negative sign ofeither q2 or q3 . Recall that negative signs on vector quantities indicate a reversal of direction of the vector in
question. But for electric forces, the direction of the force is determined by the types (signs) of both interactingcharges; we determine the force directions by considering whether the signs of the two charges are the same orare opposite. If you also include negative signs from negative charges when you substitute numbers, you run therisk of mathematically reversing the direction of the force you are calculating. Thus, the safest thing to do is tocalculate just the magnitude of the force, using the absolute values of the charges, and determine the directionsphysically.
It’s also worth noting that the only new concept in this example is how to calculate the electric forces; everythingelse (getting the net force from its components, breaking the forces into their components, finding the directionof the net force) is the same as force problems you have done earlier.


Check Your Understanding What would be different if q1 were negative?


5.4 | Electric Field
Learning Objectives


By the end of this section, you will be able to:
• Explain the purpose of the electric field concept
• Describe the properties of the electric field
• Calculate the field of a collection of source charges of either sign


Chapter 5 | Electric Charges and Fields 197




As we showed in the preceding section, the net electric force on a test charge is the vector sum of all the electric forcesacting on it, from all of the various source charges, located at their various positions. But what if we use a different testcharge, one with a different magnitude, or sign, or both? Or suppose we have a dozen different test charges we wish to tryat the same location? We would have to calculate the sum of the forces from scratch. Fortunately, it is possible to define aquantity, called the electric field, which is independent of the test charge. It only depends on the configuration of the sourcecharges, and once found, allows us to calculate the force on any test charge.
Defining a Field
Suppose we have N source charges q1, q2, q3 ,…, qN located at positions r→ 1, r→ 2, r→ 3 ,…, r→ N , applying N
electrostatic forces on a test charge Q. The net force on Q is (see Equation 5.2)


F


= F


1 + F


2 + F


3 + ⋯ + F


N


= 1
4πε0





⎜Qq1
r1
2


r̂ 1 +
Qq2
r2
2


r̂ 2 +
Qq3
r3
2


r̂ 3 + ⋯ +
QqN
r1
2


r̂ N





= Q



⎢ 1
4πε0





⎜q1
r1
2
r̂ 1 +


q2
r2
2
r̂ 2 +


q3
r3
2
r̂ 3 + ⋯ +


qN
r1
2
r̂ N







⎥.


We can rewrite this as


(5.3)F→ = Q E→


where
E


≡ 1
4πε0





⎜q1
r1
2
r̂ 1 +


q2
r2
2
r̂ 2 +


q3
r3
2
r̂ 3 + ⋯ +


qN
r1
2
r̂ N





or, more compactly,


(5.4)
E


(P) ≡ 1
4πε0



i = 1


N
qi
ri
2
r̂ i.


This expression is called the electric field at position P = P(x, y, z) of the N source charges. Here, P is the location of the
point in space where you are calculating the field and is relative to the positions r→ i of the source charges (Figure 5.18).
Note that we have to impose a coordinate system to solve actual problems.


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Figure 5.18 Each of these eight source charges creates its ownelectric field at every point in space; shown here are the field vectorsat an arbitrary point P. Like the electric force, the net electric fieldobeys the superposition principle.


Notice that the calculation of the electric field makes no reference to the test charge. Thus, the physically useful approach isto calculate the electric field and then use it to calculate the force on some test charge later, if needed. Different test chargesexperience different forces Equation 5.3, but it is the same electric field Equation 5.4. That being said, recall that thereis no fundamental difference between a test charge and a source charge; these are merely convenient labels for the system ofinterest. Any charge produces an electric field; however, just as Earth’s orbit is not affected by Earth’s own gravity, a chargeis not subject to a force due to the electric field it generates. Charges are only subject to forces from the electric fields ofother charges.
In this respect, the electric field E→ of a point charge is similar to the gravitational field g→ of Earth; once we have
calculated the gravitational field at some point in space, we can use it any time we want to calculate the resulting forceon any mass we choose to place at that point. In fact, this is exactly what we do when we say the gravitational field of
Earth (near Earth’s surface) has a value of 9.81 m/s2, and then we calculate the resulting force (i.e., weight) on different
masses. Also, the general expression for calculating g→ at arbitrary distances from the center of Earth (i.e., not just near
Earth’s surface) is very similar to the expression for E→ : g→ = GM


r2
r̂ , where G is a proportionality constant, playing


the same role for g→ as 1
4πε0


does for E→ . The value of g→ is calculated once and is then used in an endless number
of problems.
To push the analogy further, notice the units of the electric field: From F = QE , the units of E are newtons per coulomb,
N/C, that is, the electric field applies a force on each unit charge. Now notice the units of g: From w = mg , the units of
g are newtons per kilogram, N/kg, that is, the gravitational field applies a force on each unit mass. We could say that thegravitational field of Earth, near Earth’s surface, has a value of 9.81 N/kg.
The Meaning of “Field”
Recall from your studies of gravity that the word “field” in this context has a precise meaning. A field, in physics, is aphysical quantity whose value depends on (is a function of) position, relative to the source of the field. In the case of the
electric field, Equation 5.4 shows that the value of E→ (both the magnitude and the direction) depends on where in space
the point P is located, measured from the locations r→ i of the source charges qi .
In addition, since the electric field is a vector quantity, the electric field is referred to as a vector field. (The gravitationalfield is also a vector field.) In contrast, a field that has only a magnitude at every point is a scalar field. The temperature in


Chapter 5 | Electric Charges and Fields 199




a room is an example of a scalar field. It is a field because the temperature, in general, is different at different locations inthe room, and it is a scalar field because temperature is a scalar quantity.
Also, as you did with the gravitational field of an object with mass, you should picture the electric field of a charge-bearingobject (the source charge) as a continuous, immaterial substance that surrounds the source charge, filling all of space—inprinciple, to ±∞ in all directions. The field exists at every physical point in space. To put it another way, the electric charge
on an object alters the space around the charged object in such a way that all other electrically charged objects in spaceexperience an electric force as a result of being in that field. The electric field, then, is the mechanism by which the electricproperties of the source charge are transmitted to and through the rest of the universe. (Again, the range of the electric forceis infinite.)
We will see in subsequent chapters that the speed at which electrical phenomena travel is the same as the speed of light.There is a deep connection between the electric field and light.
Superposition
Yet another experimental fact about the field is that it obeys the superposition principle. In this context, that means that wecan (in principle) calculate the total electric field of many source charges by calculating the electric field of only q1 at
position P, then calculate the field of q2 at P, while—and this is the crucial idea—ignoring the field of, and indeed even
the existence of, q1. We can repeat this process, calculating the field of each individual source charge, independently of
the existence of any of the other charges. The total electric field, then, is the vector sum of all these fields. That, in essence,is what Equation 5.4 says.
In the next section, we describe how to determine the shape of an electric field of a source charge distribution and how tosketch it.
The Direction of the Field
Equation 5.4 enables us to determine the magnitude of the electric field, but we need the direction also. We use theconvention that the direction of any electric field vector is the same as the direction of the electric force vector that the fieldwould apply to a positive test charge placed in that field. Such a charge would be repelled by positive source charges (theforce on it would point away from the positive source charge) but attracted to negative charges (the force points toward thenegative source).


Direction of the Electric Field
By convention, all electric fields E→ point away from positive source charges and point toward negative source
charges.


Add charges to the Electric Field of Dreams (https://openstaxcollege.org/l/21elefiedream) and see howthey react to the electric field. Turn on a background electric field and adjust the direction and magnitude.


Example 5.3
The E-field of an Atom
In an ionized helium atom, the most probable distance between the nucleus and the electron is
r = 26.5 × 10−12 m . What is the electric field due to the nucleus at the location of the electron?
Strategy
Note that although the electron is mentioned, it is not used in any calculation. The problem asks for an electricfield, not a force; hence, there is only one charge involved, and the problem specifically asks for the field due tothe nucleus. Thus, the electron is a red herring; only its distance matters. Also, since the distance between the twoprotons in the nucleus is much, much smaller than the distance of the electron from the nucleus, we can treat thetwo protons as a single charge +2e (Figure 5.19).


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Figure 5.19 A schematic representation of a helium atom.Again, helium physically looks nothing like this, but this sort ofdiagram is helpful for calculating the electric field of thenucleus.


Solution
The electric field is calculated by


E


= 1
4πε0



i = 1


N
qi
ri
2
r̂ i.


Since there is only one source charge (the nucleus), this expression simplifies to
E


= 1
4πε0


q


r2
r̂ .


Here q = 2e = 2⎛⎝1.6 × 10−19 C⎞⎠ (since there are two protons) and r is given; substituting gives


E


= 1




⎝8.85 × 10


−12 C2


N ·m2



2⎛⎝1.6 × 10
−19 C⎞⎠



⎝26.5 × 10


−12 m⎞⎠
2
r̂ = 4.1 × 1012 N


C
r̂ .


The direction of E→ is radially away from the nucleus in all directions. Why? Because a positive test charge
placed in this field would accelerate radially away from the nucleus (since it is also positively charged), and again,the convention is that the direction of the electric field vector is defined in terms of the direction of the force itwould apply to positive test charges.


Example 5.4
The E-Field above Two Equal Charges
(a) Find the electric field (magnitude and direction) a distance z above the midpoint between two equal charges
+q that are a distance d apart (Figure 5.20). Check that your result is consistent with what you’d expect when
z ≫ d .
(b) The same as part (a), only this time make the right-hand charge −q instead of +q .


Chapter 5 | Electric Charges and Fields 201




Figure 5.20 Finding the field of two identical source chargesat the point P. Due to the symmetry, the net field at P is entirelyvertical. (Notice that this is not true away from the midlinebetween the charges.)


Strategy
We add the two fields as vectors, per Equation 5.4. Notice that the system (and therefore the field) issymmetrical about the vertical axis; as a result, the horizontal components of the field vectors cancel. Thissimplifies the math. Also, we take care to express our final answer in terms of only quantities that are given in theoriginal statement of the problem: q, z, d, and constants (π, ε0).
Solution


a. By symmetry, the horizontal (x)-components of E→ cancel (Figure 5.21);
Ex = 14πε0


q


r2
sin θ − 1


4πε0


q


r2
sin θ = 0 .


Figure 5.21 Note that the horizontal components of theelectric fields from the two charges cancel each other out, whilethe vertical components add together.


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The vertical (z)-component is given by
Ez = 14πε0


q


r2
cos θ + 1


4πε0


q


r2
cos θ = 1


4πε0


2q


r2
cos θ.


Since none of the other components survive, this is the entire electric field, and it points in the k^
direction. Notice that this calculation uses the principle of superposition; we calculate the fields of thetwo charges independently and then add them together.What we want to do now is replace the quantities in this expression that we don’t know (such as r), orcan’t easily measure (such as cos θ) with quantities that we do know, or can measure. In this case, by
geometry,


r2 = z2 + ⎛⎝
d
2



2


and
cos θ = zr =


z



⎣z


2 + ⎛⎝
d
2



2⎤


1/2
.


Thus, substituting,
E


(z) = 1
4πε0


2q



⎣z


2 + ⎛⎝
d
2



2⎤


2
z



⎣z


2 + ⎛⎝
d
2



2⎤


1/2
k
^
.


Simplifying, the desired answer is
(5.5)


E


(z) = 1
4πε0


2qz



⎣z


2 + ⎛⎝
d
2



2⎤


3/2
k
^
.


b. If the source charges are equal and opposite, the vertical components cancel because
Ez = 14πε0


q


r2
cos θ − 1


4πε0


q


r2
cos θ = 0


and we get, for the horizontal component of E→ ,
E


(z) = 1
4πε0


q


r2
sin θ i


^
− 1


4πε0


−q


r2
sin θ i


^


= 1
4πε0


2q


r2
sin θ i


^


= 1
4πε0


2q



⎣z


2 + ⎛⎝
d
2



2⎤


2




d
2




⎣z


2 + ⎛⎝
d
2



2⎤


1/2
i
^
.


This becomes
(5.6)


E


(z) = 1
4πε0


qd



⎣z


2 + ⎛⎝
d
2



2⎤


3/2
i
^
.


Chapter 5 | Electric Charges and Fields 203




5.3


Significance
It is a very common and very useful technique in physics to check whether your answer is reasonable byevaluating it at extreme cases. In this example, we should evaluate the field expressions for the cases d = 0 ,
z ≫ d , and z → ∞ , and confirm that the resulting expressions match our physical expectations. Let’s do so:
Let’s start with Equation 5.5, the field of two identical charges. From far away (i.e., z ≫ d), the two source
charges should “merge” and we should then “see” the field of just one charge, of size 2q. So, let z ≫ d; then we
can neglect d2 in Equation 5.5 to obtain


lim
d → 0


E


= 1
4πε0


2qz



⎣z


2⎤


3/2
k
^


= 1
4πε0


2qz


z3
k
^


= 1
4πε0



⎝2q⎞⎠
z2


k
^
,


which is the correct expression for a field at a distance z away from a charge 2q.
Next, we consider the field of equal and opposite charges, Equation 5.6. It can be shown (via a Taylorexpansion) that for d ≪ z ≪ ∞ , this becomes


(5.7)
E


(z) = 1
4πε0


qd


z3
i
^
,


which is the field of a dipole, a system that we will study in more detail later. (Note that the units of E→ are still
correct in this expression, since the units of d in the numerator cancel the unit of the “extra” z in the denominator.)If z is very large (z → ∞) , then E → 0 , as it should; the two charges “merge” and so cancel out.


Check Your Understanding What is the electric field due to a single point particle?


Try this simulation of electric field hockey (https://openstaxcollege.org/l/21elefielhocke) to get thecharge in the goal by placing other charges on the field.


5.5 | Calculating Electric Fields of Charge Distributions
Learning Objectives


By the end of this section, you will be able to:
• Explain what a continuous source charge distribution is and how it is related to the concept ofquantization of charge
• Describe line charges, surface charges, and volume charges
• Calculate the field of a continuous source charge distribution of either sign


The charge distributions we have seen so far have been discrete: made up of individual point particles. This is in contrastwith a continuous charge distribution, which has at least one nonzero dimension. If a charge distribution is continuousrather than discrete, we can generalize the definition of the electric field. We simply divide the charge into infinitesimalpieces and treat each piece as a point charge.
Note that because charge is quantized, there is no such thing as a “truly” continuous charge distribution. However, in most


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practical cases, the total charge creating the field involves such a huge number of discrete charges that we can safely ignorethe discrete nature of the charge and consider it to be continuous. This is exactly the kind of approximation we make whenwe deal with a bucket of water as a continuous fluid, rather than a collection of H2O molecules.
Our first step is to define a charge density for a charge distribution along a line, across a surface, or within a volume, asshown in Figure 5.22.


Figure 5.22 The configuration of charge differential elements for a (a) line charge, (b) sheet ofcharge, and (c) a volume of charge. Also note that (d) some of the components of the total electric fieldcancel out, with the remainder resulting in a net electric field.


Definitions of charge density:
• λ ≡ charge per unit length ( linear charge density); units are coulombs per meter (C/m)
• σ ≡ charge per unit area ( surface charge density); units are coulombs per square meter (C/m2)
• ρ ≡ charge per unit volume ( volume charge density); units are coulombs per cubic meter (C/m3)


Then, for a line charge, a surface charge, and a volume charge, the summation in Equation 5.4 becomes an integral and
qi is replaced by dq = λdl , σdA , or ρdV , respectively:


(5.8)
Point charge: E



(P) = 1


4πε0

i = 1


N ⎛

qi
r2

⎠ r̂


(5.9)
Line charge: E



(P) = 1


4πε0

⌡line




λdl
r2

⎠ r̂


(5.10)
Surface charge: E



(P) = 1


4πε0

⌡surface




σdA
r2

⎠ r̂


(5.11)
Volume charge: E



(P) = 1


4πε0

⌡volume




ρdV


r2

⎠ r̂


The integrals are generalizations of the expression for the field of a point charge. They implicitly include and assume theprinciple of superposition. The “trick” to using them is almost always in coming up with correct expressions for dl, dA, ordV, as the case may be, expressed in terms of r, and also expressing the charge density function appropriately. It may beconstant; it might be dependent on location.
Note carefully the meaning of r in these equations: It is the distance from the charge element ⎛⎝qi, λdl, σdA, ρdV ⎞⎠ to the
location of interest, P(x, y, z) (the point in space where you want to determine the field). However, don’t confuse this with


Chapter 5 | Electric Charges and Fields 205




the meaning of r̂ ; we are using it and the vector notation E→ to write three integrals at once. That is, Equation 5.9 is
actually


Ex (P) = 14πε0

⌡line




λdl
r2

⎠x, Ey (P) =


1
4πε0

⌡line




λdl
r2

⎠y, Ez (P) =


1
4πε0

⌡line




λdl
r2

⎠z.


Example 5.5
Electric Field of a Line Segment
Find the electric field a distance z above the midpoint of a straight line segment of length L that carries a uniformline charge density λ .
Strategy
Since this is a continuous charge distribution, we conceptually break the wire segment into differential pieces oflength dl, each of which carries a differential amount of charge dq = λdl . Then, we calculate the differential
field created by two symmetrically placed pieces of the wire, using the symmetry of the setup to simplify thecalculation (Figure 5.23). Finally, we integrate this differential field expression over the length of the wire (halfof it, actually, as we explain below) to obtain the complete electric field expression.


Figure 5.23 A uniformly charged segment of wire. Theelectric field at point P can be found by applying thesuperposition principle to symmetrically placed charge elementsand integrating.


Solution
Before we jump into it, what do we expect the field to “look like” from far away? Since it is a finite line segment,from far away, it should look like a point charge. We will check the expression we get to see if it meets thisexpectation.
The electric field for a line charge is given by the general expression


E


(P) = 1
4πε0

⌡line


λdl
r2


r̂ .


The symmetry of the situation (our choice of the two identical differential pieces of charge) implies the horizontal(x)-components of the field cancel, so that the net field points in the z-direction. Let’s check this formally.
The total field E→ (P) is the vector sum of the fields from each of the two charge elements (call them E→ 1 and
E


2 , for now):
E


(P) = E


1 + E


2 = E1x i
^


+ E1z k
^


+ E2x

⎝− i


^⎞
⎠+ E2z k


^
.


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5.4


Because the two charge elements are identical and are the same distance away from the point P where we want tocalculate the field, E1x = E2x, so those components cancel. This leaves
E


(P) = E1z k
^


+ E2z k
^


= E1 cos θk
^


+ E2 cos θk
^
.


These components are also equal, so we have
E


(P) = 1
4πε0


λdl
r2


cos θk
^


+ 1
4πε0


λdl
r2


cos θk
^


= 1
4πε0

⌡0


L/2
2λdx
r2


cos θk
^


where our differential line element dl is dx, in this example, since we are integrating along a line of charge that
lies on the x-axis. (The limits of integration are 0 to L


2
, not −L


2
to +L


2
, because we have constructed the net


field from two differential pieces of charge dq. If we integrated along the entire length, we would pick up anerroneous factor of 2.)
In principle, this is complete. However, to actually calculate this integral, we need to eliminate all the variablesthat are not given. In this case, both r and θ change as we integrate outward to the end of the line charge, so those
are the variables to get rid of. We can do that the same way we did for the two point charges: by noticing that


r = ⎛⎝z
2 + x2⎞⎠


1/2


and
cos θ = zr =


z

⎝z


2 + x2⎞⎠
1/2


.


Substituting, we obtain
E


(P) = 1
4πε0

⌡0


L/2


2λdx

⎝z


2 + x2⎞⎠


z

⎝z


2 + x2⎞⎠
1/2


k
^


= 1
4πε0







0


L/2


2λz

⎝z


2 + x2⎞⎠
3/2


dxk
^


= 2λz
4πε0





⎢ x


z2 z2 + x2





⎥|0
L/2


k
^


which simplifies to
(5.12)


E


(z) = 1
4πε0


λL


z z2 + L
2


4


k
^
.


Significance
Notice, once again, the use of symmetry to simplify the problem. This is a very common strategy for calculatingelectric fields. The fields of nonsymmetrical charge distributions have to be handled with multiple integrals andmay need to be calculated numerically by a computer.


Check Your Understanding How would the strategy used above change to calculate the electric field ata point a distance z above one end of the finite line segment?


Chapter 5 | Electric Charges and Fields 207




Example 5.6
Electric Field of an Infinite Line of Charge
Find the electric field a distance z above the midpoint of an infinite line of charge that carries a uniform linecharge density λ .
Strategy
This is exactly like the preceding example, except the limits of integration will be −∞ to +∞ .
Solution
Again, the horizontal components cancel out, so we wind up with


E


(P) = 1
4πε0

⌡−∞



λdx
r2


cos θk
^


where our differential line element dl is dx, in this example, since we are integrating along a line of charge thatlies on the x-axis. Again,
cos θ = zr =


z

⎝z


2 + x2⎞⎠
1/2


.


Substituting, we obtain
E


(P) = 1
4πε0

⌡−∞




λdx

⎝z


2 + x2⎞⎠


z

⎝z


2 + x2⎞⎠
1/2


k
^


= 1
4πε0







−∞




λz

⎝z


2 + x2⎞⎠
3/2


dxk
^


= λz
4πε0





⎢ x


z2 z2 + x2





⎥|−∞


k
^
,


which simplifies to
E


(z) = 1
4πε0



z k
^
.


Significance
Our strategy for working with continuous charge distributions also gives useful results for charges with infinitedimension.


In the case of a finite line of charge, note that for z ≫ L , z2 dominates the L in the denominator, so that Equation 5.12
simplifies to


E


≈ 1
4πε0


λL
z2


k
^
.


If you recall that λL = q , the total charge on the wire, we have retrieved the expression for the field of a point charge, as
expected.
In the limit L → ∞ , on the other hand, we get the field of an infinite straight wire, which is a straight wire whose length
is much, much greater than either of its other dimensions, and also much, much greater than the distance at which the fieldis to be calculated:


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(5.13)
E


(z) = 1
4πε0



z k
^
.


An interesting artifact of this infinite limit is that we have lost the usual 1/r2 dependence that we are used to. This will
become even more intriguing in the case of an infinite plane.
Example 5.7


Electric Field due to a Ring of Charge
A ring has a uniform charge density λ , with units of coulomb per unit meter of arc. Find the electric potential at
a point on the axis passing through the center of the ring.
Strategy
We use the same procedure as for the charged wire. The difference here is that the charge is distributed on a circle.We divide the circle into infinitesimal elements shaped as arcs on the circle and use polar coordinates shown inFigure 5.24.


Figure 5.24 The system and variable for calculating theelectric field due to a ring of charge.


Solution
The electric field for a line charge is given by the general expression


E


(P) = 1
4πε0

⌡line


λdl
r2


r̂ .


A general element of the arc between θ and θ + dθ is of length Rdθ and therefore contains a charge equal to
λRdθ. The element is at a distance of r = z2 + R2 from P, the angle is cos ϕ = z


z2 + R2
, and therefore the


electric field is


Chapter 5 | Electric Charges and Fields 209




E


(P) = 1
4πε0

⌡line


λdl
r2


r̂ = 1
4πε0

⌡0



λRdθ
z2 + R2


z


z2 + R2


= 1
4πε0


λRz

⎝z


2 + R2⎞⎠
3/2


ẑ∫
0



dθ = 1


4πε0
2πλRz



⎝z


2 + R2⎞⎠
3/2




= 1
4πε0


qtot z



⎝z


2 + R2⎞⎠
3/2


ẑ .


Significance
As usual, symmetry simplified this problem, in this particular case resulting in a trivial integral. Also, when wetake the limit of z>>R , we find that


E


≈ 1
4πε0


qtot
z2


ẑ ,


as we expect.


Example 5.8
The Field of a Disk
Find the electric field of a circular thin disk of radius R and uniform charge density at a distance z above the centerof the disk (Figure 5.25)


Figure 5.25 A uniformly charged disk. As in the line chargeexample, the field above the center of this disk can be calculatedby taking advantage of the symmetry of the charge distribution.


Strategy
The electric field for a surface charge is given by


E


(P) = 1
4πε0

⌡surface


σdA
r2


r̂ .


To solve surface charge problems, we break the surface into symmetrical differential “stripes” that match theshape of the surface; here, we’ll use rings, as shown in the figure. Again, by symmetry, the horizontal components


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5.5


cancel and the field is entirely in the vertical (k^ ) direction. The vertical component of the electric field is
extracted by multiplying by cos θ , so


E


(P) = 1
4πε0

⌡surface


σdA
r2


cos θ k
^
.


As before, we need to rewrite the unknown factors in the integrand in terms of the given quantities. In this case,
dA = 2πr ′dr′


r2 = r′2 + z2


cos θ = z

⎝r′


2 + z2⎞⎠
1/2


.


(Please take note of the two different “r’s” here; r is the distance from the differential ring of charge to the pointP where we wish to determine the field, whereas r′ is the distance from the center of the disk to the differential
ring of charge.) Also, we already performed the polar angle integral in writing down dA.
Solution
Substituting all this in, we get


E


(P) = E


(z) = 1
4πε0







0


R


σ(2πr′dr′)z

⎝r′


2 + z2⎞⎠
3/2


k
^


= 1
4πε0


(2πσz)



⎜1z −


1
R2 + z2





⎟k
^


or, more simply,
(5.14)


E


(z) = 1
4πε0





⎜2πσ − 2πσz


R2 + z2





⎟k
^
.


Significance
Again, it can be shown (via a Taylor expansion) that when z ≫ R , this reduces to


E


(z) ≈ 1
4πε0


σπR2


z2
k
^
,


which is the expression for a point charge Q = σπR2.


Check Your Understanding How would the above limit change with a uniformly charged rectangleinstead of a disk?


As R → ∞ , Equation 5.14 reduces to the field of an infinite plane, which is a flat sheet whose area is much, much
greater than its thickness, and also much, much greater than the distance at which the field is to be calculated:


(5.15)
E


= σ
2ε0


k
^
.


Note that this field is constant. This surprising result is, again, an artifact of our limit, although one that we will make use ofrepeatedly in the future. To understand why this happens, imagine being placed above an infinite plane of constant charge.


Chapter 5 | Electric Charges and Fields 211




5.6


Does the plane look any different if you vary your altitude? No—you still see the plane going off to infinity, no matter howfar you are from it. It is important to note that Equation 5.15 is because we are above the plane. If we were below, the
field would point in the −k^ direction.
Example 5.9


The Field of Two Infinite Planes
Find the electric field everywhere resulting from two infinite planes with equal but opposite charge densities(Figure 5.26).


Figure 5.26 Two charged infinite planes. Note the direction ofthe electric field.


Strategy
We already know the electric field resulting from a single infinite plane, so we may use the principle ofsuperposition to find the field from two.
Solution
The electric field points away from the positively charged plane and toward the negatively charged plane. Sincethe σ are equal and opposite, this means that in the region outside of the two planes, the electric fields cancel
each other out to zero.
However, in the region between the planes, the electric fields add, and we get


E


= σε0
i
^


for the electric field. The i^ is because in the figure, the field is pointing in the +x-direction.
Significance
Systems that may be approximated as two infinite planes of this sort provide a useful means of creating uniformelectric fields.


Check Your Understanding What would the electric field look like in a system with two parallelpositively charged planes with equal charge densities?


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5.6 | Electric Field Lines
Learning Objectives


By the end of this section, you will be able to:
• Explain the purpose of an electric field diagram
• Describe the relationship between a vector diagram and a field line diagram
• Explain the rules for creating a field diagram and why these rules make physical sense
• Sketch the field of an arbitrary source charge


Now that we have some experience calculating electric fields, let’s try to gain some insight into the geometry of electricfields. As mentioned earlier, our model is that the charge on an object (the source charge) alters space in the region around itin such a way that when another charged object (the test charge) is placed in that region of space, that test charge experiencesan electric force. The concept of electric field lines, and of electric field line diagrams, enables us to visualize the way inwhich the space is altered, allowing us to visualize the field. The purpose of this section is to enable you to create sketchesof this geometry, so we will list the specific steps and rules involved in creating an accurate and useful sketch of an electricfield.
It is important to remember that electric fields are three-dimensional. Although in this book we include some pseudo-three-dimensional images, several of the diagrams that you’ll see (both here, and in subsequent chapters) will be two-dimensionalprojections, or cross-sections. Always keep in mind that in fact, you’re looking at a three-dimensional phenomenon.
Our starting point is the physical fact that the electric field of the source charge causes a test charge in that field to experiencea force. By definition, electric field vectors point in the same direction as the electric force that a (hypothetical) positive testcharge would experience, if placed in the field (Figure 5.27)


Figure 5.27 The electric field of a positive point charge. A large number of field vectors are shown. Like all vectorarrows, the length of each vector is proportional to the magnitude of the field at each point. (a) Field in twodimensions; (b) field in three dimensions.


We’ve plotted many field vectors in the figure, which are distributed uniformly around the source charge. Since the electricfield is a vector, the arrows that we draw correspond at every point in space to both the magnitude and the direction of thefield at that point. As always, the length of the arrow that we draw corresponds to the magnitude of the field vector at thatpoint. For a point source charge, the length decreases by the square of the distance from the source charge. In addition, the


Chapter 5 | Electric Charges and Fields 213




direction of the field vector is radially away from the source charge, because the direction of the electric field is defined bythe direction of the force that a positive test charge would experience in that field. (Again, keep in mind that the actual fieldis three-dimensional; there are also field lines pointing out of and into the page.)
This diagram is correct, but it becomes less useful as the source charge distribution becomes more complicated. Forexample, consider the vector field diagram of a dipole (Figure 5.28).


Figure 5.28 The vector field of a dipole. Even with just twoidentical charges, the vector field diagram becomes difficult tounderstand.


There is a more useful way to present the same information. Rather than drawing a large number of increasingly smallervector arrows, we instead connect all of them together, forming continuous lines and curves, as shown in Figure 5.29.


Figure 5.29 (a) The electric field line diagram of a positive point charge. (b) The field line diagramof a dipole. In both diagrams, the magnitude of the field is indicated by the field line density. Thefield vectors (not shown here) are everywhere tangent to the field lines.


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Although it may not be obvious at first glance, these field diagrams convey the same information about the electric field asdo the vector diagrams. First, the direction of the field at every point is simply the direction of the field vector at that samepoint. In other words, at any point in space, the field vector at each point is tangent to the field line at that same point. Thearrowhead placed on a field line indicates its direction.
As for the magnitude of the field, that is indicated by the field line density—that is, the number of field lines per unitarea passing through a small cross-sectional area perpendicular to the electric field. This field line density is drawn to beproportional to the magnitude of the field at that cross-section. As a result, if the field lines are close together (that is, thefield line density is greater), this indicates that the magnitude of the field is large at that point. If the field lines are far apartat the cross-section, this indicates the magnitude of the field is small. Figure 5.30 shows the idea.


Figure 5.30 Electric field lines passing through imaginary areas. Since the number oflines passing through each area is the same, but the areas themselves are different, thefield line density is different. This indicates different magnitudes of the electric field atthese points.


In Figure 5.30, the same number of field lines passes through both surfaces (S and S′), but the surface S is larger than
surface S′ . Therefore, the density of field lines (number of lines per unit area) is larger at the location of S′ , indicating that
the electric field is stronger at the location of S′ than at S. The rules for creating an electric field diagram are as follows.


Problem-Solving Strategy: Drawing Electric Field Lines
1. Electric field lines either originate on positive charges or come in from infinity, and either terminate onnegative charges or extend out to infinity.
2. The number of field lines originating or terminating at a charge is proportional to the magnitude of that charge.A charge of 2q will have twice as many lines as a charge of q.
3. At every point in space, the field vector at that point is tangent to the field line at that same point.


Chapter 5 | Electric Charges and Fields 215




4. The field line density at any point in space is proportional to (and therefore is representative of) the magnitudeof the field at that point in space.
5. Field lines can never cross. Since a field line represents the direction of the field at a given point, if two fieldlines crossed at some point, that would imply that the electric field was pointing in two different directions ata single point. This in turn would suggest that the (net) force on a test charge placed at that point would pointin two different directions. Since this is obviously impossible, it follows that field lines must never cross.


Always keep in mind that field lines serve only as a convenient way to visualize the electric field; they are not physicalentities. Although the direction and relative intensity of the electric field can be deduced from a set of field lines, the linescan also be misleading. For example, the field lines drawn to represent the electric field in a region must, by necessity, bediscrete. However, the actual electric field in that region exists at every point in space.
Field lines for three groups of discrete charges are shown in Figure 5.31. Since the charges in parts (a) and (b) have thesame magnitude, the same number of field lines are shown starting from or terminating on each charge. In (c), however, wedraw three times as many field lines leaving the +3q charge as entering the −q . The field lines that do not terminate at
−q emanate outward from the charge configuration, to infinity.


Figure 5.31 Three typical electric field diagrams. (a) A dipole. (b) Two identical charges. (c) Two charges with oppositesigns and different magnitudes. Can you tell from the diagram which charge has the larger magnitude?


The ability to construct an accurate electric field diagram is an important, useful skill; it makes it much easier to estimate,predict, and therefore calculate the electric field of a source charge. The best way to develop this skill is with softwarethat allows you to place source charges and then will draw the net field upon request. We strongly urge you to search theInternet for a program. Once you’ve found one you like, run several simulations to get the essential ideas of field diagramconstruction. Then practice drawing field diagrams, and checking your predictions with the computer-drawn diagrams.
One example of a field-line drawing program (https://openstaxcollege.org/l/21fieldlindrapr) is fromthe PhET “Charges and Fields” simulation.


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5.7 | Electric Dipoles
Learning Objectives


By the end of this section, you will be able to:
• Describe a permanent dipole
• Describe an induced dipole
• Define and calculate an electric dipole moment
• Explain the physical meaning of the dipole moment


Earlier we discussed, and calculated, the electric field of a dipole: two equal and opposite charges that are “close” to eachother. (In this context, “close” means that the distance d between the two charges is much, much less than the distance ofthe field point P, the location where you are calculating the field.) Let’s now consider what happens to a dipole when it is
placed in an external field E→ . We assume that the dipole is a permanent dipole; it exists without the field, and does not
break apart in the external field.
Rotation of a Dipole due to an Electric Field
For now, we deal with only the simplest case: The external field is uniform in space. Suppose we have the situation depicted
in Figure 5.32, where we denote the distance between the charges as the vector d→ , pointing from the negative charge
to the positive charge. The forces on the two charges are equal and opposite, so there is no net force on the dipole. However,there is a torque:


τ→ =



⎜ d


2
× F



+





⎟ +



⎜− d




2
× F









=







⎜ d


2





⎟ × ⎛⎝+q E


→ ⎞
⎠+



⎜− d




2





⎟ × ⎛⎝−q E


→ ⎞






= q d


× E


.


Figure 5.32 A dipole in an external electric field. (a) The net force on the dipole is zero, but the net torque is not. As a result,the dipole rotates, becoming aligned with the external field. (b) The dipole moment is a convenient way to characterize this
effect. The d→ points in the same direction as p→ .


The quantity q d→ (the magnitude of each charge multiplied by the vector distance between them) is a property of the
dipole; its value, as you can see, determines the torque that the dipole experiences in the external field. It is useful, therefore,to define this product as the so-called dipole moment of the dipole:


(5.16)p→ ≡ q d→ .


Chapter 5 | Electric Charges and Fields 217




We can therefore write


(5.17)τ→ = p→ × E→ .


Recall that a torque changes the angular velocity of an object, the dipole, in this case. In this situation, the effect is to rotate
the dipole (that is, align the direction of p→ ) so that it is parallel to the direction of the external field.
Induced Dipoles
Neutral atoms are, by definition, electrically neutral; they have equal amounts of positive and negative charge. Furthermore,since they are spherically symmetrical, they do not have a “built-in” dipole moment the way most asymmetrical moleculesdo. They obtain one, however, when placed in an external electric field, because the external field causes oppositely directedforces on the positive nucleus of the atom versus the negative electrons that surround the nucleus. The result is a new chargedistribution of the atom, and therefore, an induced dipole moment (Figure 5.33).


Figure 5.33 A dipole is induced in a neutral atom by an external electric field. The induceddipole moment is aligned with the external field.


An important fact here is that, just as for a rotated polar molecule, the result is that the dipole moment ends up alignedparallel to the external electric field. Generally, the magnitude of an induced dipole is much smaller than that of an inherentdipole. For both kinds of dipoles, notice that once the alignment of the dipole (rotated or induced) is complete, the net effect
is to decrease the total electric field E→ total = E→ external + E→ dipole in the regions outside the dipole charges (Figure
5.34). By “outside” we mean further from the charges than they are from each other. This effect is crucial for capacitors, asyou will see in Capacitance.


Figure 5.34 The net electric field is the vector sum of thefield of the dipole plus the external field.


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Recall that we found the electric field of a dipole in Equation 5.7. If we rewrite it in terms of the dipole moment we get:
E


(z) = 1
4πε0


p→


z3
.


The form of this field is shown in Figure 5.34. Notice that along the plane perpendicular to the axis of the dipole andmidway between the charges, the direction of the electric field is opposite that of the dipole and gets weaker the further fromthe axis one goes. Similarly, on the axis of the dipole (but outside it), the field points in the same direction as the dipole,again getting weaker the further one gets from the charges.


Chapter 5 | Electric Charges and Fields 219




charging by induction
conduction electron
conductor
continuous charge distribution
coulomb
Coulomb force
Coulomb’s law
dipole
dipole moment
electric charge
electric field
electric force
electron
electrostatic attraction
electrostatic force
electrostatic repulsion
electrostatics
field line
field line density
induced dipole
infinite plane
infinite straight wire
insulator
ion
law of conservation of charge
linear charge density
neutron
permanent dipole


CHAPTER 5 REVIEW
KEY TERMS


process by which an electrically charged object brought near a neutral object creates a chargeseparation in that object
electron that is free to move away from its atomic orbit


material that allows electrons to move separately from their atomic orbits; object with properties that allowcharges to move about freely within it
total source charge composed of so large a number of elementary charges that it mustbe treated as continuous, rather than discrete


SI unit of electric charge
another term for the electrostatic force
mathematical equation calculating the electrostatic force vector between two charged particles


two equal and opposite charges that are fixed close to each other
property of a dipole; it characterizes the combination of distance between the opposite charges, and themagnitude of the charges
physical property of an object that causes it to be attracted toward or repelled from another chargedobject; each charged object generates and is influenced by a force called an electric force


physical phenomenon created by a charge; it “transmits” a force between a two charges
noncontact force observed between electrically charged objects


particle surrounding the nucleus of an atom and carrying the smallest unit of negative charge
phenomenon of two objects with opposite charges attracting each other


amount and direction of attraction or repulsion between two charged bodies; the assumption is thatthe source charges remain motionless
phenomenon of two objects with like charges repelling each other


study of charged objects which are not in motion
smooth, usually curved line that indicates the direction of the electric field


number of field lines per square meter passing through an imaginary area; its purpose is to indicate thefield strength at different points in space
typically an atom, or a spherically symmetric molecule; a dipole created due to opposite forcesdisplacing the positive and negative charges


flat sheet in which the dimensions making up the area are much, much greater than its thickness, and alsomuch, much greater than the distance at which the field is to be calculated; its field is constant
straight wire whose length is much, much greater than either of its other dimensions, and alsomuch, much greater than the distance at which the field is to be calculated


material that holds electrons securely within their atomic orbits
atom or molecule with more or fewer electrons than protons


net electric charge of a closed system is constant
amount of charge in an element of a charge distribution that is essentially one-dimensional (thewidth and height are much, much smaller than its length); its units are C/m


neutral particle in the nucleus of an atom, with (nearly) the same mass as a proton
typically a molecule; a dipole created by the arrangement of the charged particles from which thedipole is created


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permittivity of vacuum
polarization
principle of superposition
proton
static electricity
superposition
surface charge density
volume charge density


also called the permittivity of free space, and constant describing the strength of the electricforce in a vacuum
slight shifting of positive and negative charges to opposite sides of an object


useful fact that we can simply add up all of the forces due to charges acting on an object
particle in the nucleus of an atom and carrying a positive charge equal in magnitude to the amount of negativecharge carried by an electron


buildup of electric charge on the surface of an object; the arrangement of the charge remains constant(“static”)
concept that states that the net electric field of multiple source charges is the vector sum of the field ofeach source charge calculated individually


amount of charge in an element of a two-dimensional charge distribution (the thickness is
small); its units are C/m2


amount of charge in an element of a three-dimensional charge distribution; its units are C/m3


KEY EQUATIONS
Coulomb’s law F→ 12(r) = 14πε0


q1q2
r12
2


r̂ 12


Superposition of electric forces
F


(r) = 1
4πε0


Q∑
i = 1


N
qi
ri
2
r̂ i


Electric force due to an electric field F→ = Q E→
Electric field at point P


E


(P) ≡ 1
4πε0



i = 1


N
qi
ri
2
r̂ i


Field of an infinite wire
E


(z) = 1
4πε0



z k
^


Field of an infinite plane
E


= σ
2ε0


k
^


Dipole moment p→ ≡ q d→
Torque on dipole in external E-field τ→ = p→ × E→


SUMMARY
5.1 Electric Charge


• There are only two types of charge, which we call positive and negative. Like charges repel, unlike charges attract,and the force between charges decreases with the square of the distance.
• The vast majority of positive charge in nature is carried by protons, whereas the vast majority of negative charge iscarried by electrons. The electric charge of one electron is equal in magnitude and opposite in sign to the charge ofone proton.
• An ion is an atom or molecule that has nonzero total charge due to having unequal numbers of electrons and protons.
• The SI unit for charge is the coulomb (C), with protons and electrons having charges of opposite sign but equal
magnitude; the magnitude of this basic charge is e ≡ 1.602 × 10−19 C


Chapter 5 | Electric Charges and Fields 221




• Both positive and negative charges exist in neutral objects and can be separated by bringing the two objects intophysical contact; rubbing the objects together can remove electrons from the bonds in one object and place them onthe other object, increasing the charge separation.
• For macroscopic objects, negatively charged means an excess of electrons and positively charged means a depletionof electrons.
• The law of conservation of charge states that the net charge of a closed system is constant.


5.2 Conductors, Insulators, and Charging by Induction
• A conductor is a substance that allows charge to flow freely through its atomic structure.
• An insulator holds charge fixed in place.
• Polarization is the separation of positive and negative charges in a neutral object. Polarized objects have theirpositive and negative charges concentrated in different areas, giving them a charge distribution.


5.3 Coulomb's Law
• Coulomb’s law gives the magnitude of the force between point charges. It is


F


12(r) =
1


4πε0


q1q2
r12
2


r̂ 12


where q2 and q2 are two point charges separated by a distance r. This Coulomb force is extremely basic,
since most charges are due to point-like particles. It is responsible for all electrostatic effects and underlies mostmacroscopic forces.


5.4 Electric Field
• The electric field is an alteration of space caused by the presence of an electric charge. The electric field mediatesthe electric force between a source charge and a test charge.
• The electric field, like the electric force, obeys the superposition principle
• The field is a vector; by definition, it points away from positive charges and toward negative charges.


5.5 Calculating Electric Fields of Charge Distributions
• A very large number of charges can be treated as a continuous charge distribution, where the calculation of the fieldrequires integration. Common cases are:


◦ one-dimensional (like a wire); uses a line charge density λ
◦ two-dimensional (metal plate); uses surface charge density σ
◦ three-dimensional (metal sphere); uses volume charge density ρ


• The “source charge” is a differential amount of charge dq. Calculating dq depends on the type of source chargedistribution:
dq = λdl; dq = σdA; dq = ρdV .


• Symmetry of the charge distribution is usually key.
• Important special cases are the field of an “infinite” wire and the field of an “infinite” plane.


5.6 Electric Field Lines
• Electric field diagrams assist in visualizing the field of a source charge.
• The magnitude of the field is proportional to the field line density.
• Field vectors are everywhere tangent to field lines.


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5.7 Electric Dipoles
• If a permanent dipole is placed in an external electric field, it results in a torque that aligns it with the external field.
• If a nonpolar atom (or molecule) is placed in an external field, it gains an induced dipole that is aligned with theexternal field.
• The net field is the vector sum of the external field plus the field of the dipole (physical or induced).
• The strength of the polarization is described by the dipole moment of the dipole, p→ = q d→ .


CONCEPTUAL QUESTIONS
5.1 Electric Charge
1. There are very large numbers of charged particles inmost objects. Why, then, don’t most objects exhibit staticelectricity?
2. Why do most objects tend to contain nearly equalnumbers of positive and negative charges?
3. A positively charged rod attracts a small piece of cork.(a) Can we conclude that the cork is negatively charged?(b) The rod repels another small piece of cork. Can weconclude that this piece is positively charged?
4. Two bodies attract each other electrically. Do they bothhave to be charged? Answer the same question if the bodiesrepel one another.
5. How would you determine whether the charge on aparticular rod is positive or negative?


5.2 Conductors, Insulators, and Charging by
Induction
6. An eccentric inventor attempts to levitate a cork ball bywrapping it with foil and placing a large negative chargeon the ball and then putting a large positive charge onthe ceiling of his workshop. Instead, while attempting toplace a large negative charge on the ball, the foil flies off.Explain.
7. When a glass rod is rubbed with silk, it becomespositive and the silk becomes negative—yet both attractdust. Does the dust have a third type of charge that isattracted to both positive and negative? Explain.
8. Why does a car always attract dust right after it ispolished? (Note that car wax and car tires are insulators.)
9. Does the uncharged conductor shown below experiencea net electric force?


10. While walking on a rug, a person frequently becomescharged because of the rubbing between his shoes and therug. This charge then causes a spark and a slight shockwhen the person gets close to a metal object. Why are theseshocks so much more common on a dry day?
11. Compare charging by conduction to charging byinduction.
12. Small pieces of tissue are attracted to a charged comb.Soon after sticking to the comb, the pieces of tissue arerepelled from it. Explain.
13. Trucks that carry gasoline often have chains danglingfrom their undercarriages and brushing the ground. Why?
14. Why do electrostatic experiments work so poorly inhumid weather?
15. Why do some clothes cling together after beingremoved from the clothes dryer? Does this happen ifthey’re still damp?
16. Can induction be used to produce charge on aninsulator?


Chapter 5 | Electric Charges and Fields 223




17. Suppose someone tells you that rubbing quartz withcotton cloth produces a third kind of charge on the quartz.Describe what you might do to test this claim.
18. A handheld copper rod does not acquire a charge whenyou rub it with a cloth. Explain why.
19. Suppose you place a charge q near a large metal plate.(a) If q is attracted to the plate, is the plate necessarilycharged? (b) If q is repelled by the plate, is the platenecessarily charged?


5.3 Coulomb's Law
20. Would defining the charge on an electron to bepositive have any effect on Coulomb’s law?
21. An atomic nucleus contains positively charged protonsand uncharged neutrons. Since nuclei do stay together, whatmust we conclude about the forces between these nuclearparticles?
22. Is the force between two fixed charges influenced bythe presence of other charges?


5.4 Electric Field
23. When measuring an electric field, could we use anegative rather than a positive test charge?
24. During fair weather, the electric field due to the netcharge on Earth points downward. Is Earth chargedpositively or negatively?
25. If the electric field at a point on the line between twocharges is zero, what do you know about the charges?
26. Two charges lie along the x-axis. Is it true that thenet electric field always vanishes at some point (other thaninfinity) along the x-axis?


5.5 Calculating Electric Fields of Charge


Distributions
27. Give a plausible argument as to why the electric fieldoutside an infinite charged sheet is constant.
28. Compare the electric fields of an infinite sheet ofcharge, an infinite, charged conducting plate, and infinite,oppositely charged parallel plates.
29. Describe the electric fields of an infinite charged plateand of two infinite, charged parallel plates in terms of theelectric field of an infinite sheet of charge.
30. A negative charge is placed at the center of a ring ofuniform positive charge. What is the motion (if any) of thecharge? What if the charge were placed at a point on theaxis of the ring other than the center?


5.6 Electric Field Lines
31. If a point charge is released from rest in a uniformelectric field, will it follow a field line? Will it do so if theelectric field is not uniform?
32. Under what conditions, if any, will the trajectory of acharged particle not follow a field line?
33. How would you experimentally distinguish an electricfield from a gravitational field?
34. A representation of an electric field shows 10 fieldlines perpendicular to a square plate. How many field linesshould pass perpendicularly through the plate to depict afield with twice the magnitude?
35. What is the ratio of the number of electric field linesleaving a charge 10q and a charge q?


5.7 Electric Dipoles
36. What are the stable orientation(s) for a dipole in anexternal electric field? What happens if the dipole isslightly perturbed from these orientations?


PROBLEMS
5.1 Electric Charge
37. Common static electricity involves charges rangingfrom nanocoulombs to microcoulombs. (a) How manyelectrons are needed to form a charge of −2.00 nC? (b) Howmany electrons must be removed from a neutral object toleave a net charge of 0.500 µC ?


38. If 1.80 × 1020 electrons move through a pocket
calculator during a full day’s operation, how manycoulombs of charge moved through it?
39. To start a car engine, the car battery moves
3.75 × 1021 electrons through the starter motor. How
many coulombs of charge were moved?


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40. A certain lightning bolt moves 40.0 C of charge. Howmany fundamental units of charge is this?
41. A 2.5-g copper penny is given a charge of
−2.0 × 10−9 C . (a) How many excess electrons are on the
penny? (b) By what percent do the excess electrons changethe mass of the penny?
42. A 2.5-g copper penny is given a charge of
4.0 × 10−9 C . (a) How many electrons are removed from
the penny? (b) If no more than one electron is removedfrom an atom, what percent of the atoms are ionized by thischarging process?


5.2 Conductors, Insulators, and Charging by
Induction
43. Suppose a speck of dust in an electrostatic precipitator
has 1.0000 × 1012 protons in it and has a net charge of
−5.00 nC (a very large charge for a small speck). Howmany electrons does it have?
44. An amoeba has 1.00 × 1016 protons and a net charge
of 0.300 pC. (a) How many fewer electrons are there thanprotons? (b) If you paired them up, what fraction of theprotons would have no electrons?
45. A 50.0-g ball of copper has a net charge of 2.00 µC .
What fraction of the copper’s electrons has been removed?(Each copper atom has 29 protons, and copper has anatomic mass of 63.5.)
46. What net charge would you place on a 100-g piece
of sulfur if you put an extra electron on 1 in 1012 of its
atoms? (Sulfur has an atomic mass of 32.1 u.)
47. How many coulombs of positive charge are there in4.00 kg of plutonium, given its atomic mass is 244 and thateach plutonium atom has 94 protons?


5.3 Coulomb's Law
48. Two point particles with charges +3 µC and +5 µC
are held in place by 3-N forces on each charge inappropriate directions. (a) Draw a free-body diagram foreach particle. (b) Find the distance between the charges.
49. Two charges +3 µC and +12 µC are fixed 1 m
apart, with the second one to the right. Find the magnitudeand direction of the net force on a −2-nC charge whenplaced at the following locations: (a) halfway between thetwo (b) half a meter to the left of the +3 µC charge (c)


half a meter above the +12 µC charge in a direction
perpendicular to the line joining the two fixed charges
50. In a salt crystal, the distance between adjacent sodium
and chloride ions is 2.82 × 10−10 m. What is the force of
attraction between the two singly charged ions?
51. Protons in an atomic nucleus are typically 10−15 m
apart. What is the electric force of repulsion betweennuclear protons?
52. Suppose Earth and the Moon each carried a netnegative charge −Q. Approximate both bodies as pointmasses and point charges.
(a) What value of Q is required to balance the gravitationalattraction between Earth and the Moon?
(b) Does the distance between Earth and the Moon affectyour answer? Explain.
(c) How many electrons would be needed to produce thischarge?
53. Point charges q1 = 50 µC and q2 = −25 µC are
placed 1.0 m apart. What is the force on a third charge
q3 = 20 µC placed midway between q1 and q2 ?
54. Where must q3 of the preceding problem be placed
so that the net force on it is zero?
55. Two small balls, each of mass 5.0 g, are attached tosilk threads 50 cm long, which are in turn tied to the samepoint on the ceiling, as shown below. When the balls aregiven the same charge Q, the threads hang at 5.0° to the
vertical, as shown below. What is the magnitude of Q?What are the signs of the two charges?


Chapter 5 | Electric Charges and Fields 225




56. Point charges Q1 = 2.0 µC and Q2 = 4.0 µC are
located at r→ 1 = (4.0 i^ − 2.0 j^ + 5.0k^ )m and
r→ 2 = (8.0 i


^
+ 5.0 j


^
− 9.0k


^
)m . What is the force of


Q2 on Q1 ?
57. The net excess charge on two small spheres (smallenough to be treated as point charges) is Q. Show that theforce of repulsion between the spheres is greatest wheneach sphere has an excess charge Q/2. Assume that thedistance between the spheres is so large compared withtheir radii that the spheres can be treated as point charges.
58. Two small, identical conducting spheres repel eachother with a force of 0.050 N when they are 0.25 m apart.After a conducting wire is connected between the spheresand then removed, they repel each other with a force of0.060 N. What is the original charge on each sphere?
59. A charge q = 2.0 µC is placed at the point P shown
below. What is the force on q?


60. What is the net electric force on the charge located atthe lower right-hand corner of the triangle shown here?


61. Two fixed particles, each of charge 5.0 × 10−6 C,
are 24 cm apart. What force do they exert on a third particle
of charge −2.5 × 10−6 C that is 13 cm from each of
them?
62. The charges
q1 = 2.0 × 10


−7 C, q2 = −4.0 × 10
−7 C, and


q3 = −1.0 × 10
−7 C are placed at the corners of the


triangle shown below. What is the force on q1?


63. What is the force on the charge q at the lower-right-hand corner of the square shown here?


64. Point charges q1 = 10 µC and q2 = −30 µC are
fixed at r1 = ⎛⎝3.0 i^ − 4.0 j^⎞⎠m and
r2 =

⎝9.0 i


^
+ 6.0 j


^⎞
⎠m. What is the force of q2 on q1 ?


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5.4 Electric Field
65. A particle of charge 2.0 × 10−8 C experiences an
upward force of magnitude 4.0 × 10−6 N when it is
placed in a particular point in an electric field. (a) Whatis the electric field at that point? (b) If a charge
q = −1.0 × 10−8 C is placed there, what is the force on
it?
66. On a typical clear day, the atmospheric electric fieldpoints downward and has a magnitude of approximately100 N/C. Compare the gravitational and electric forces on
a small dust particle of mass 2.0 × 10−15 g that carries
a single electron charge. What is the acceleration (bothmagnitude and direction) of the dust particle?
67. Consider an electron that is 10−10 m from an alpha
particle (q = 3.2 × 10−19 C). (a) What is the electric
field due to the alpha particle at the location of the electron?(b) What is the electric field due to the electron at thelocation of the alpha particle? (c) What is the electric forceon the alpha particle? On the electron?
68. Each the balls shown below carries a charge q and hasa mass m. The length of each thread is l, and at equilibrium,the balls are separated by an angle 2θ . How does θ vary
with q and l? Show that θ satisfies
sin(θ)2 tan(θ) =


q2


16πε0gl
2m


.


69. What is the electric field at a point where the force on a
−2.0 × 10−6 −C charge is ⎛⎝4.0 i^ − 6.0 j^⎞⎠× 10−6 N?


70. A proton is suspended in the air by an electric fieldat the surface of Earth. What is the strength of this electric


field?
71. The electric field in a particular thundercloud is
2.0 × 105 N/C. What is the acceleration of an electron in
this field?
72. A small piece of cork whose mass is 2.0 g is given a
charge of 5.0 × 10−7 C. What electric field is needed to
place the cork in equilibrium under the combined electricand gravitational forces?
73. If the electric field is 100 N/C at a distance of 50 cm
from a point charge q, what is the value of q?
74. What is the electric field of a proton at the first Bohr
orbit for hydrogen (r = 5.29 × 10−11 m)? What is the
force on the electron in that orbit?
75. (a) What is the electric field of an oxygen nucleus at
a point that is 10−10 m from the nucleus? (b) What is the
force this electric field exerts on a second oxygen nucleusplaced at that point?
76. Two point charges, q1 = 2.0 × 10−7 C and
q2 = −6.0 × 10


−8 C, are held 25.0 cm apart. (a) What is
the electric field at a point 5.0 cm from the negative chargeand along the line between the two charges? (b)What is theforce on an electron placed at that point?
77. Point charges q1 = 50 µC and q2 = −25 µC are
placed 1.0 m apart. (a) What is the electric field at a pointmidway between them? (b) What is the force on a charge
q3 = 20 µC situated there?
78. Can you arrange the two point charges
q1 = −2.0 × 10


−6 C and q2 = 4.0 × 10−6 C along the
x-axis so that E = 0 at the origin?
79. Point charges q1 = q2 = 4.0 × 10−6 C are fixed on
the x-axis at x = −3.0 m and x = 3.0 m. What charge
q must be placed at the origin so that the electric fieldvanishes at x = 0, y = 3.0 m?


5.5 Calculating Electric Fields of Charge
Distributions
80. A thin conducting plate 1.0 m on the side is given a
charge of −2.0 × 10−6 C . An electron is placed 1.0 cm
above the center of the plate. What is the acceleration of the


Chapter 5 | Electric Charges and Fields 227




electron?
81. Calculate the magnitude and direction of the electricfield 2.0 m from a long wire that is charged uniformly at
λ = 4.0 × 10−6 C/m.


82. Two thin conducting plates, each 25.0 cm on a side,are situated parallel to one another and 5.0 mm apart. If
10−11 electrons are moved from one plate to the other,
what is the electric field between the plates?
83. The charge per unit length on the thin rod shownbelow is λ . What is the electric field at the point P? (Hint:
Solve this problem by first considering the electric field
d E
→ at P due to a small segment dx of the rod, which


contains charge dq = λdx . Then find the net field by
integrating d E→ over the length of the rod.)


84. The charge per unit length on the thin semicircularwire shown below is λ . What is the electric field at the
point P?


85. Two thin parallel conducting plates are placed 2.0 cmapart. Each plate is 2.0 cm on a side; one plate carries a netcharge of 8.0 µC, and the other plate carries a net charge
of −8.0 µC. What is the charge density on the inside
surface of each plate? What is the electric field between theplates?
86. A thin conducing plate 2.0 m on a side is given a totalcharge of −10.0 µC . (a) What is the electric field 1.0 cm
above the plate? (b) What is the force on an electron atthis point? (c) Repeat these calculations for a point 2.0 cmabove the plate. (d) When the electron moves from 1.0 to2,0 cm above the plate, how much work is done on it by theelectric field?
87. A total charge q is distributed uniformly along a thin,straight rod of length L (see below). What is the electricfield at P1? At P2?


88. Charge is distributed along the entire x-axis withuniform density λ. How much work does the electric field
of this charge distribution do on an electron that movesalong the y-axis from y = a to y = b?
89. Charge is distributed along the entire x-axis withuniform density λx and along the entire y-axis with
uniform density λy. Calculate the resulting electric field at
(a) r→ = a i^ + b j^ and (b) r→ = ck^ .
90. A rod bent into the arc of a circle subtends an angle
2θ at the center P of the circle (see below). If the rod is
charged uniformly with a total chargeQ, what is the electricfield at P?


91. A proton moves in the electric field
E


= 200 i
^


N/C. (a) What are the force on and the
acceleration of the proton? (b) Do the same calculation foran electron moving in this field.
92. An electron and a proton, each starting from rest, areaccelerated by the same uniform electric field of 200 N/C.Determine the distance and time for each particle to acquire
a kinetic energy of 3.2 × 10−16 J.
93. A spherical water droplet of radius 25 µm carries an
excess 250 electrons. What vertical electric field is neededto balance the gravitational force on the droplet at thesurface of the earth?
94. A proton enters the uniform electric field producedby the two charged plates shown below. The magnitude of
the electric field is 4.0 × 105 N/C, and the speed of the
proton when it enters is 1.5 × 107 m/s. What distance d
has the proton been deflected downward when it leaves theplates?


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95. Shown below is a small sphere of mass 0.25 g that
carries a charge of 9.0 × 10−10 C. The sphere is attached
to one end of a very thin silk string 5.0 cm long. The otherend of the string is attached to a large vertical conducting
plate that has a charge density of 30 × 10−6 C/m2. What
is the angle that the string makes with the vertical?


96. Two infinite rods, each carrying a uniform chargedensity λ, are parallel to one another and perpendicular
to the plane of the page. (See below.) What is the electricalfield at P1? At P2?


97. Positive charge is distributed with a uniform density
λ along the positive x-axis from r to ∞, along the
positive y-axis from r to ∞, and along a 90° arc of a
circle of radius r, as shown below. What is the electric fieldat O?


98. From a distance of 10 cm, a proton is projected with a
speed of v = 4.0 × 106 m/s directly at a large, positively
charged plate whose charge density is
σ = 2.0 × 10−5 C/m2. (See below.) (a) Does the proton
reach the plate? (b) If not, how far from the plate does itturn around?


99. A particle of mass m and charge −q moves along
a straight line away from a fixed particle of charge Q.When the distance between the two particles is r0, −q is
moving with a speed v0. (a) Use the work-energy theorem
to calculate the maximum separation of the charges. (b)What do you have to assume about v0 to make this
calculation? (c) What is the minimum value of v0 such that
−q escapes from Q?


5.6 Electric Field Lines
100. Which of the following electric field lines areincorrect for point charges? Explain why.


Chapter 5 | Electric Charges and Fields 229




101. In this exercise, you will practice drawing electricfield lines. Make sure you represent both the magnitudeand direction of the electric field adequately. Note that thenumber of lines into or out of charges is proportional to thecharges.
(a) Draw the electric field lines map for two charges
+20 µC and −20 µC situated 5 cm from each other.
(b) Draw the electric field lines map for two charges
+20 µC and +20 µC situated 5 cm from each other.
(c) Draw the electric field lines map for two charges
+20 µC and −30 µC situated 5 cm from each other.
102. Draw the electric field for a system of three particlesof charges +1 µC, +2 µC, and −3 µC fixed at the
corners of an equilateral triangle of side 2 cm.
103. Two charges of equal magnitude but opposite signmake up an electric dipole. A quadrupole consists of twoelectric dipoles are placed anti-parallel at two edges of a


square as shown.


Draw the electric field of the charge distribution.
104. Suppose the electric field of an isolated point charge
decreased with distance as 1/r2 + δ rather than as 1/r2 .
Show that it is then impossible to draw continous field linesso that their number per unit area is proportional to E.


5.7 Electric Dipoles
105. Consider the equal and opposite charges shownbelow. (a) Show that at all points on the x-axis for which
|x| ≫ a, E ≈ Qa/2πε0 x


3. (b) Show that at all points on
the y-axis for which |y| ≫ a, E ≈ Qa/πε0 y3.


106. (a) What is the dipole moment of the configurationshown above? If Q = 4.0 µC , (b) what is the torque on
this dipole with an electric field of 4.0 × 105 N/C i^ ? (c)
What is the torque on this dipole with an electric field of
−4.0 × 105 N/C i


^ ? (d) What is the torque on this dipole
with an electric field of ±4.0 × 105 N/C j^ ?
107. A water molecule consists of two hydrogen atomsbonded with one oxygen atom. The bond angle betweenthe two hydrogen atoms is 104° (see below). Calculate
the net dipole moment of a water molecule that is placedin a uniform, horizontal electric field of magnitude
2.3 × 10−8 N/C. (You are missing some information for
solving this problem; you will need to determine what


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information you need, and look it up.)


ADDITIONAL PROBLEMS
108. Point charges q1 = 2.0 µC and q1 = 4.0 µC are
located at r1 = ⎛⎝4.0 i^ − 2.0 j^ + 2.0k^⎞⎠m and
r2 =

⎝8.0 i


^
+ 5.0 j


^
− 9.0k


^⎞
⎠m . What is the force of


q2 on q1?


109. What is the force on the 5.0-µC charges shown
below?


110. What is the force on the 2.0-µC charge placed at the
center of the square shown below?


111. Four charged particles are positioned at the corners


of a parallelogram as shown below. If q = 5.0 µC and
Q = 8.0 µC, what is the net force on q?


112. A charge Q is fixed at the origin and a second chargeqmoves along the x-axis, as shown below. How much workis done on q by the electric force when q moves from
x1 to x2?


113. A charge q = −2.0 µC is released from rest when
it is 2.0 m from a fixed charge Q = 6.0 µC. What is the
kinetic energy of q when it is 1.0 m from Q?
114. What is the electric field at the midpoint M of thehypotenuse of the triangle shown below?


Chapter 5 | Electric Charges and Fields 231




115. Find the electric field at P for the chargeconfigurations shown below.


116. (a) What is the electric field at the lower-right-handcorner of the square shown below? (b) What is the force ona charge q placed at that point?


117. Point charges are placed at the four corners of a
rectangle as shown below: q1 = 2.0 × 10−6 C,
q2 = −2.0 × 10


−6 C, q3 = 4.0 × 10
−6 C, and


q4 = 1.0 × 10
−6 C. What is the electric field at P?


118. Three charges are positioned at the corners of aparallelogram as shown below. (a) If Q = 8.0 µC, what is
the electric field at the unoccupied corner? (b) What is theforce on a 5.0-µC charge placed at this corner?


119. A positive charge q is released from rest at the originof a rectangular coordinate system and moves under the
influence of the electric field E→ = E0 (1 + x/a) i^ . What
is the kinetic energy of q when it passes through x = 3a?
120. A particle of charge −q and mass m is placed at the
center of a uniformaly charged ring of total charge Q andradius R. The particle is displaced a small distance alongthe axis perpendicular to the plane of the ring and released.Assuming that the particle is constrained to move along theaxis, show that the particle oscillates in simple harmonic
motion with a frequency f = 1



qQ


4πε0mR
3
.


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121. Charge is distributed uniformly along the entirey-axis with a density λy and along the positive x-axis
from x = a to x = b with a density λx. What is the force
between the two distributions?
122. The circular arc shown below carries a charge perunit length λ = λ0 cos θ, where θ is measured from the
x-axis. What is the electric field at the origin?


123. Calculate the electric field due to a uniformlycharged rod of length L, aligned with the x-axis with oneend at the origin; at a point P on the z-axis.
124. The charge per unit length on the thin rod shownbelow is λ. What is the electric force on the point charge
q? Solve this problem by first considering the electric force
d F
→ on q due to a small segment dx of the rod, which


contains charge λdx. Then, find the net force by


integrating d F→ over the length of the rod.


125. The charge per unit length on the thin rod shown hereis λ. What is the electric force on the point charge q? (See
the preceding problem.)


126. The charge per unit length on the thin semicircularwire shown below is λ. What is the electric force on the
point charge q? (See the preceding problems.)


Chapter 5 | Electric Charges and Fields 233




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6 | GAUSS'S LAW


Figure 6.1 This chapter introduces the concept of flux, which relates a physical quantity and the area through which it isflowing. Although we introduce this concept with the electric field, the concept may be used for many other quantities, such asfluid flow. (credit: modification of work by “Alessandro”/Flickr)


Chapter Outline
6.1 Electric Flux
6.2 Explaining Gauss’s Law
6.3 Applying Gauss’s Law
6.4 Conductors in Electrostatic Equilibrium


Introduction
Flux is a general and broadly applicable concept in physics. However, in this chapter, we concentrate on the flux of theelectric field. This allows us to introduce Gauss’s law, which is particularly useful for finding the electric fields of chargedistributions exhibiting spatial symmetry. The main topics discussed here are


1. Electric flux. We define electric flux for both open and closed surfaces.
2. Gauss’s law. We derive Gauss’s law for an arbitrary charge distribution and examine the role of electric flux inGauss’s law.
3. Calculating electric fields with Gauss’s law. The main focus of this chapter is to explain how to use Gauss’s lawto find the electric fields of spatially symmetrical charge distributions. We discuss the importance of choosing aGaussian surface and provide examples involving the applications of Gauss’s law.
4. Electric fields in conductors. Gauss’s law provides useful insight into the absence of electric fields in conductingmaterials.


So far, we have found that the electrostatic field begins and ends at point charges and that the field of a point charge variesinversely with the square of the distance from that charge. These characteristics of the electrostatic field lead to an importantmathematical relationship known as Gauss’s law. This law is named in honor of the extraordinary German mathematicianand scientist Karl Friedrich Gauss (Figure 6.2). Gauss’s law gives us an elegantly simple way of finding the electricfield, and, as you will see, it can be much easier to use than the integration method described in the previous chapter.However, there is a catch—Gauss’s law has a limitation in that, while always true, it can be readily applied only for chargedistributions with certain symmetries.


Chapter 6 | Gauss's Law 235




Figure 6.2 Karl Friedrich Gauss (1777–1855) was a legendarymathematician of the nineteenth century. Although his majorcontributions were to the field of mathematics, he also didimportant work in physics and astronomy.


6.1 | Electric Flux
Learning Objectives


By the end of this section, you will be able to:
• Define the concept of flux
• Describe electric flux
• Calculate electric flux for a given situation


The concept of flux describes how much of something goes through a given area. More formally, it is the dot product of avector field (in this chapter, the electric field) with an area. You may conceptualize the flux of an electric field as a measureof the number of electric field lines passing through an area (Figure 6.3). The larger the area, the more field lines gothrough it and, hence, the greater the flux; similarly, the stronger the electric field is (represented by a greater density oflines), the greater the flux. On the other hand, if the area rotated so that the plane is aligned with the field lines, none willpass through and there will be no flux.


Figure 6.3 The flux of an electric field through the shadedarea captures information about the “number” of electric fieldlines passing through the area. The numerical value of theelectric flux depends on the magnitudes of the electric field andthe area, as well as the relative orientation of the area withrespect to the direction of the electric field.


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A macroscopic analogy that might help you imagine this is to put a hula hoop in a flowing river. As you change the angleof the hoop relative to the direction of the current, more or less of the flow will go through the hoop. Similarly, the amountof flow through the hoop depends on the strength of the current and the size of the hoop. Again, flux is a general concept;we can also use it to describe the amount of sunlight hitting a solar panel or the amount of energy a telescope receives froma distant star, for example.
To quantify this idea, Figure 6.4(a) shows a planar surface S1 of area A1 that is perpendicular to the uniform electric field
E


= E ŷ . If N field lines pass through S1 , then we know from the definition of electric field lines (Electric Charges
and Fields) that N/A1 ∝ E, or N ∝ EA1.
The quantity EA1 is the electric flux through S1 . We represent the electric flux through an open surface like S1 by the
symbol Φ . Electric flux is a scalar quantity and has an SI unit of newton-meters squared per coulomb (N ·m2 /C ). Notice
that N ∝ EA1 may also be written as N ∝ Φ , demonstrating that electric flux is a measure of the number of field lines
crossing a surface.


Figure 6.4 (a) A planar surface S1 of area A1 is perpendicular to the electric field E j^ . N field lines cross
surface S1 . (b) A surface S2 of area A2 whose projection onto the xz-plane is S1 .The same number of field lines
cross each surface.


Now consider a planar surface that is not perpendicular to the field. How would we represent the electric flux? Figure6.4(b) shows a surface S2 of area A2 that is inclined at an angle θ to the xz-plane and whose projection in that plane is
S1 (area A1 ). The areas are related by A2 cos θ = A1. Because the same number of field lines crosses both S1 and S2 ,
the fluxes through both surfaces must be the same. The flux through S2 is therefore Φ = EA1 = EA2 cos θ. Designating
n̂ 2 as a unit vector normal to S2 (see Figure 6.4(b)), we obtain


Φ = E


· n̂ 2 A2.


Check out this video (https://openstaxcollege.org/l/21fluxsizeangl) to observe what happens to the flux asthe area changes in size and angle, or the electric field changes in strength.


Area Vector
For discussing the flux of a vector field, it is helpful to introduce an area vector A→ . This allows us to write the last
equation in a more compact form. What should the magnitude of the area vector be? What should the direction of the areavector be? What are the implications of how you answer the previous question?
The area vector of a flat surface of area A has the following magnitude and direction:


• Magnitude is equal to area (A)


Chapter 6 | Gauss's Law 237




• Direction is along the normal to the surface ( n̂ ); that is, perpendicular to the surface.
Since the normal to a flat surface can point in either direction from the surface, the direction of the area vector of an opensurface needs to be chosen, as shown in Figure 6.5.


Figure 6.5 The direction of the area vector of an open surfaceneeds to be chosen; it could be either of the two cases displayedhere. The area vector of a part of a closed surface is defined topoint from the inside of the closed space to the outside. This rulegives a unique direction.


Since n̂ is a unit normal to a surface, it has two possible directions at every point on that surface (Figure 6.6(a)). For an
open surface, we can use either direction, as long as we are consistent over the entire surface. Part (c) of the figure showsseveral cases.


Figure 6.6 (a) Two potential normal vectors arise at every point on a surface. (b) The outward normalis used to calculate the flux through a closed surface. (c) Only S3 has been given a consistent set of
normal vectors that allows us to define the flux through the surface.


However, if a surface is closed, then the surface encloses a volume. In that case, the direction of the normal vector at any


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point on the surface points from the inside to the outside. On a closed surface such as that of Figure 6.6(b), n̂ is chosen
to be the outward normal at every point, to be consistent with the sign convention for electric charge.
Electric Flux
Now that we have defined the area vector of a surface, we can define the electric flux of a uniform electric field through aflat area as the scalar product of the electric field and the area vector, as defined in Products of Vectors (http://cnx.org/content/m58280/latest/) :


(6.1)Φ = E→ · A→ (uniform E→ , flat su face).


Figure 6.7 shows the electric field of an oppositely charged, parallel-plate system and an imaginary box between the plates.The electric field between the plates is uniform and points from the positive plate toward the negative plate. A calculationof the flux of this field through various faces of the box shows that the net flux through the box is zero. Why does the fluxcancel out here?


Figure 6.7 Electric flux through a cube, placed between twocharged plates. Electric flux through the bottom face (ABCD) is
negative, because E→ is in the opposite direction to the normal to
the surface. The electric flux through the top face (FGHK) is positive,because the electric field and the normal are in the same direction.The electric flux through the other faces is zero, since the electricfield is perpendicular to the normal vectors of those faces. The netelectric flux through the cube is the sum of fluxes through the sixfaces. Here, the net flux through the cube is equal to zero. Themagnitude of the flux through rectangle BCKF is equal to themagnitudes of the flux through both the top and bottom faces.


The reason is that the sources of the electric field are outside the box. Therefore, if any electric field line enters the volumeof the box, it must also exit somewhere on the surface because there is no charge inside for the lines to land on. Therefore,quite generally, electric flux through a closed surface is zero if there are no sources of electric field, whether positive ornegative charges, inside the enclosed volume. In general, when field lines leave (or “flow out of”) a closed surface, Φ is
positive; when they enter (or “flow into”) the surface, Φ is negative.
Any smooth, non-flat surface can be replaced by a collection of tiny, approximately flat surfaces, as shown in Figure 6.8.


Chapter 6 | Gauss's Law 239




If we divide a surface S into small patches, then we notice that, as the patches become smaller, they can be approximated byflat surfaces. This is similar to the way we treat the surface of Earth as locally flat, even though we know that globally, it isapproximately spherical.


Figure 6.8 A surface is divided into patches to find the flux.


To keep track of the patches, we can number them from 1 through N . Now, we define the area vector for each patch as
the area of the patch pointed in the direction of the normal. Let us denote the area vector for the ith patch by δ A→ i. (We
have used the symbol δ to remind us that the area is of an arbitrarily small patch.) With sufficiently small patches, we may
approximate the electric field over any given patch as uniform. Let us denote the average electric field at the location of the
ith patch by E→ i.


E


i = average electric field ver the ith patch.


Therefore, we can write the electric flux Φi through the area of the ith patch as
Φi = E



i · δ A



i (ith patch).


The flux through each of the individual patches can be constructed in this manner and then added to give us an estimate ofthe net flux through the entire surface S, which we denote simply as Φ .
Φ = ∑


i = 1


N


Φi = ∑
i = 1


N


E


i · δ A


i (N patch estimate).


This estimate of the flux gets better as we decrease the size of the patches. However, when you use smaller patches, youneed more of them to cover the same surface. In the limit of infinitesimally small patches, they may be considered to have
area dA and unit normal n̂ . Since the elements are infinitesimal, they may be assumed to be planar, and E→ i may be
taken as constant over any element. Then the flux dΦ through an area dA is given by dΦ = E→ · n̂ dA. It is positive
when the angle between E→ i and n̂ is less than 90° and negative when the angle is greater than 90° . The net flux is the
sum of the infinitesimal flux elements over the entire surface. With infinitesimally small patches, you need infinitely many
patches, and the limit of the sum becomes a surface integral. With ∫


S
representing the integral over S,


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(6.2)
Φ = ∫


S
E


· n̂ dA = ⌠
⌡S


E


· d A


(open surface).


In practical terms, surface integrals are computed by taking the antiderivatives of both dimensions defining the area, withthe edges of the surface in question being the bounds of the integral.
To distinguish between the flux through an open surface like that of Figure 6.4 and the flux through a closed surface (onethat completely bounds some volume), we represent flux through a closed surface by


(6.3)Φ = ∮
S
E


· n̂ dA = ∮
S
E


· d A


(closed surface)


where the circle through the integral symbol simply means that the surface is closed, and we are integrating over the entirething. If you only integrate over a portion of a closed surface, that means you are treating a subset of it as an open surface.
Example 6.1


Flux of a Uniform Electric Field
A constant electric field of magnitude E0 points in the direction of the positive z-axis (Figure 6.9). What is the
electric flux through a rectangle with sides a and b in the (a) xy-plane and in the (b) xz-plane?


Figure 6.9 Calculating the flux of E0 through a rectangular
surface.


Strategy
Apply the definition of flux: Φ = E→ · A→ (uniform E→ ) , where the definition of dot product is crucial.
Solution


a. In this case, Φ = E→ 0 · A→ = E0 A = E0ab.
b. Here, the direction of the area vector is either along the positive y-axis or toward the negative y-axis.Therefore, the scalar product of the electric field with the area vector is zero, giving zero flux.


Significance
The relative directions of the electric field and area can cause the flux through the area to be zero.


Chapter 6 | Gauss's Law 241




Example 6.2
Flux of a Uniform Electric Field through a Closed Surface
A constant electric field of magnitude E0 points in the direction of the positive z-axis (Figure 6.10). What is
the net electric flux through a cube?


Figure 6.10 Calculating the flux of E0 through a closed cubic
surface.


Strategy
Apply the definition of flux: Φ = E→ · A→ (uniform E→ ) , noting that a closed surface eliminates the ambiguity
in the direction of the area vector.
Solution
Through the top face of the cube, Φ = E→ 0 · A→ = E0 A.
Through the bottom face of the cube, Φ = E→ 0 · A→ = −E0 A, because the area vector here points downward.
Along the other four sides, the direction of the area vector is perpendicular to the direction of the electric field.Therefore, the scalar product of the electric field with the area vector is zero, giving zero flux.
The net flux is Φnet = E0 A − E0 A + 0 + 0 + 0 + 0 = 0 .
Significance
The net flux of a uniform electric field through a closed surface is zero.


Example 6.3
Electric Flux through a Plane, Integral Method
A uniform electric field E→ of magnitude 10 N/C is directed parallel to the yz-plane at 30° above the xy-plane,
as shown in Figure 6.11. What is the electric flux through the plane surface of area 6.0 m2 located in the


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6.1


xz-plane? Assume that n̂ points in the positive y-direction.


Figure 6.11 The electric field produces a net electric fluxthrough the surface S.


Strategy
Apply Φ = ∫


S
E


· n̂ dA , where the direction and magnitude of the electric field are constant.
Solution
The angle between the uniform electric field E→ and the unit normal n̂ to the planar surface is 30° . Since
both the direction and magnitude are constant, E comes outside the integral. All that is left is a surface integralover dA, which is A. Therefore, using the open-surface equation, we find that the electric flux through the surfaceis


Φ = ∫
S
E


· n̂ dA = EA cos θ


= (10 N/C)(6.0 m2)( cos 30°) = 52 N ·m2 /C.


Significance
Again, the relative directions of the field and the area matter, and the general equation with the integral willsimplify to the simple dot product of area and electric field.


Check Your Understanding What angle should there be between the electric field and the surfaceshown in Figure 6.11 in the previous example so that no electric flux passes through the surface?


Chapter 6 | Gauss's Law 243




6.2


Example 6.4
Inhomogeneous Electric Field
What is the total flux of the electric field E→ = cy2 k^ through the rectangular surface shown in Figure 6.12?


Figure 6.12 Since the electric field is not constant over thesurface, an integration is necessary to determine the flux.


Strategy
Apply Φ = ∫


S
E


· n̂ dA . We assume that the unit normal n̂ to the given surface points in the positive
z-direction, so n̂ = k^ . Since the electric field is not uniform over the surface, it is necessary to divide the surface
into infinitesimal strips along which E→ is essentially constant. As shown in Figure 6.12, these strips are
parallel to the x-axis, and each strip has an area dA = b dy.
Solution
From the open surface integral, we find that the net flux through the rectangular surface is


Φ = ∫
S
E


· n̂ dA = ∫
0


a
(cy2 k


^
) · k


^
(b dy)


= cb⌠
⌡0


a
y2 dy = 1


3
a3bc.


Significance
For a non-constant electric field, the integral method is required.


Check Your Understanding If the electric field in Example 6.4 is E→ = mxk^ , what is the flux
through the rectangular area?


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6.2 | Explaining Gauss’s Law
Learning Objectives


By the end of this section, you will be able to:
• State Gauss’s law
• Explain the conditions under which Gauss’s law may be used
• Apply Gauss’s law in appropriate systems


We can now determine the electric flux through an arbitrary closed surface due to an arbitrary charge distribution. We foundthat if a closed surface does not have any charge inside where an electric field line can terminate, then any electric fieldline entering the surface at one point must necessarily exit at some other point of the surface. Therefore, if a closed surfacedoes not have any charges inside the enclosed volume, then the electric flux through the surface is zero. Now, what happensto the electric flux if there are some charges inside the enclosed volume? Gauss’s law gives a quantitative answer to thisquestion.
To get a feel for what to expect, let’s calculate the electric flux through a spherical surface around a positive point chargeq, since we already know the electric field in such a situation. Recall that when we place the point charge at the origin of acoordinate system, the electric field at a point P that is at a distance r from the charge at the origin is given by


E


P =
1


4πε0
1
r2


r̂ ,


where r̂ is the radial vector from the charge at the origin to the point P. We can use this electric field to find the flux
through the spherical surface of radius r, as shown in Figure 6.13.


Figure 6.13 A closed spherical surface surrounding a pointcharge q.


Then we apply Φ = ∫
S
E


· n̂ dA to this system and substitute known values. On the sphere, n̂ = r̂ and r = R , so for
an infinitesimal area dA,


dΦ = E


· n̂ dA = 1
4πε0


q


R2
r̂ · r̂ dA = 1


4πε0


q


R2
dA.


We now find the net flux by integrating this flux over the surface of the sphere:
Φ = 1


4πε0


q


R2

S
dA = 1


4πε0


q


R2
(4πR2) =


q
ε0


.


where the total surface area of the spherical surface is 4πR2. This gives the flux through the closed spherical surface at
radius r as


(6.4)Φ = qε0.


Chapter 6 | Gauss's Law 245




A remarkable fact about this equation is that the flux is independent of the size of the spherical surface. This can be directly
attributed to the fact that the electric field of a point charge decreases as 1/r2 with distance, which just cancels the r2 rate
of increase of the surface area.
Electric Field Lines Picture
An alternative way to see why the flux through a closed spherical surface is independent of the radius of the surface is tolook at the electric field lines. Note that every field line from q that pierces the surface at radius R1 also pierces the surface
at R2 (Figure 6.14).


Figure 6.14 Flux through spherical surfaces of radii R1 and
R2 enclosing a charge q are equal, independent of the size of
the surface, since all E-field lines that pierce one surface fromthe inside to outside direction also pierce the other surface in thesame direction.


Therefore, the net number of electric field lines passing through the two surfaces from the inside to outside direction isequal. This net number of electric field lines, which is obtained by subtracting the number of lines in the direction fromoutside to inside from the number of lines in the direction from inside to outside gives a visual measure of the electric fluxthrough the surfaces.
You can see that if no charges are included within a closed surface, then the electric flux through it must be zero. A typicalfield line enters the surface at dA1 and leaves at dA2. Every line that enters the surface must also leave that surface. Hence
the net “flow” of the field lines into or out of the surface is zero (Figure 6.15(a)). The same thing happens if charges ofequal and opposite sign are included inside the closed surface, so that the total charge included is zero (part (b)). A surfacethat includes the same amount of charge has the same number of field lines crossing it, regardless of the shape or size of thesurface, as long as the surface encloses the same amount of charge (part (c)).


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Figure 6.15 Understanding the flux in terms of field lines. (a) The electric flux through a closed surface due to a chargeoutside that surface is zero. (b) Charges are enclosed, but because the net charge included is zero, the net flux through theclosed surface is also zero. (c) The shape and size of the surfaces that enclose a charge does not matter because all surfacesenclosing the same charge have the same flux.
Statement of Gauss’s Law
Gauss’s law generalizes this result to the case of any number of charges and any location of the charges in the space inside
the closed surface. According to Gauss’s law, the flux of the electric field E→ through any closed surface, also called a
Gaussian surface, is equal to the net charge enclosed (qenc) divided by the permittivity of free space (ε0) :


ΦClosed Surface =
qenc
ε0


.


This equation holds for charges of either sign, because we define the area vector of a closed surface to point outward. If theenclosed charge is negative (see Figure 6.16(b)), then the flux through either S or S ' is negative.


Figure 6.16 The electric flux through any closed surface surrounding a point charge q isgiven by Gauss’s law. (a) Enclosed charge is positive. (b) Enclosed charge is negative.


The Gaussian surface does not need to correspond to a real, physical object; indeed, it rarely will. It is a mathematicalconstruct that may be of any shape, provided that it is closed. However, since our goal is to integrate the flux over it, wetend to choose shapes that are highly symmetrical.


Chapter 6 | Gauss's Law 247




If the charges are discrete point charges, then we just add them. If the charge is described by a continuous distribution, thenwe need to integrate appropriately to find the total charge that resides inside the enclosed volume. For example, the fluxthrough the Gaussian surface S of Figure 6.17 is Φ = (q1 + q2 + q5)/ε0. Note that qenc is simply the sum of the point
charges. If the charge distribution were continuous, we would need to integrate appropriately to compute the total chargewithin the Gaussian surface.


Figure 6.17 The flux through the Gaussian surface shown,due to the charge distribution, is Φ = (q1 + q2 + q5)/ε0.


Recall that the principle of superposition holds for the electric field. Therefore, the total electric field at any point, includingthose on the chosen Gaussian surface, is the sum of all the electric fields present at this point. This allows us to write Gauss’slaw in terms of the total electric field.
Gauss’s Law
The flux Φ of the electric field E→ through any closed surface S (a Gaussian surface) is equal to the net charge
enclosed (qenc) divided by the permittivity of free space (ε0) :


(6.5)Φ = ∮
S
E


· n̂ dA = qencε0
.


To use Gauss’s law effectively, you must have a clear understanding of what each term in the equation represents. The field
E
→ is the total electric field at every point on the Gaussian surface. This total field includes contributions from charges
both inside and outside the Gaussian surface. However, qenc is just the charge inside the Gaussian surface. Finally, the
Gaussian surface is any closed surface in space. That surface can coincide with the actual surface of a conductor, or it canbe an imaginary geometric surface. The only requirement imposed on a Gaussian surface is that it be closed (Figure 6.18).


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Figure 6.18 A Klein bottle partially filled with a liquid. Couldthe Klein bottle be used as a Gaussian surface?


Example 6.5
Electric Flux through Gaussian Surfaces
Calculate the electric flux through each Gaussian surface shown in Figure 6.19.


Figure 6.19 Various Gaussian surfaces and charges.


Chapter 6 | Gauss's Law 249




Strategy
From Gauss’s law, the flux through each surface is given by qenc/ε0, where qenc is the charge enclosed by that
surface.
Solution
For the surfaces and charges shown, we find


a. Φ = 2.0 µCε0 = 2.3 × 105 N ·m2 /C.
b. Φ = −2.0 µCε0 = −2.3 × 105 N ·m2 /C.
c. Φ = 2.0 µCε0 = 2.3 × 105 N ·m2 /C.
d. Φ = −4.0 µC + 6.0 µC − 1.0 µCε0 = 1.1 × 105 N ·m2 /C.
e. Φ = 4.0 µC + 6.0 µC − 10.0 µCε0 = 0.


Significance
In the special case of a closed surface, the flux calculations become a sum of charges. In the next section, this willallow us to work with more complex systems.


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6.3 Check Your Understanding Calculate the electric flux through the closed cubical surface for eachcharge distribution shown in Figure 6.20.


Figure 6.20 A cubical Gaussian surface with various charge distributions.


Use this simulation (https://openstaxcollege.org/l/21gaussimulat) to adjust the magnitude of the chargeand the radius of the Gaussian surface around it. See how this affects the total flux and the magnitude of the electricfield at the Gaussian surface.


6.3 | Applying Gauss’s Law
Learning Objectives


By the end of this section, you will be able to:
• Explain what spherical, cylindrical, and planar symmetry are
• Recognize whether or not a given system possesses one of these symmetries
• Apply Gauss’s law to determine the electric field of a system with one of these symmetries


Gauss’s law is very helpful in determining expressions for the electric field, even though the law is not directly about theelectric field; it is about the electric flux. It turns out that in situations that have certain symmetries (spherical, cylindrical,or planar) in the charge distribution, we can deduce the electric field based on knowledge of the electric flux. In these


Chapter 6 | Gauss's Law 251




systems, we can find a Gaussian surface S over which the electric field has constant magnitude. Furthermore, if E→ is
parallel to n̂ everywhere on the surface, then E→ · n̂ = E. (If E→ and n̂ are antiparallel everywhere on the surface,
then E→ · n̂ = −E. ) Gauss’s law then simplifies to


(6.6)Φ = ∮
S
E


· n̂ dA = E∮
S
dA = EA =


qenc
ε0


,


where A is the area of the surface. Note that these symmetries lead to the transformation of the flux integral into a productof the magnitude of the electric field and an appropriate area. When you use this flux in the expression for Gauss’s law, youobtain an algebraic equation that you can solve for the magnitude of the electric field, which looks like
E~


qenc
ε0 area


.


The direction of the electric field at the field point P is obtained from the symmetry of the charge distribution and the type
of charge in the distribution. Therefore, Gauss’s law can be used to determine E→ . Here is a summary of the steps we will
follow:


Problem-Solving Strategy: Gauss’s Law
1. Identify the spatial symmetry of the charge distribution. This is an important first step that allows us to choosethe appropriate Gaussian surface. As examples, an isolated point charge has spherical symmetry, and an infiniteline of charge has cylindrical symmetry.
2. Choose a Gaussian surface with the same symmetry as the charge distribution and identify its consequences.


With this choice, E→ · n̂ is easily determined over the Gaussian surface.
3. Evaluate the integral ∮


S
E


· n̂ dA over the Gaussian surface, that is, calculate the flux through the surface.
The symmetry of the Gaussian surface allows us to factor E→ · n̂ outside the integral.


4. Determine the amount of charge enclosed by the Gaussian surface. This is an evaluation of the right-handside of the equation representing Gauss’s law. It is often necessary to perform an integration to obtain the netenclosed charge.
5. Evaluate the electric field of the charge distribution. The field may now be found using the results of steps 3and 4.


Basically, there are only three types of symmetry that allow Gauss’s law to be used to deduce the electric field. They are
• A charge distribution with spherical symmetry
• A charge distribution with cylindrical symmetry
• A charge distribution with planar symmetry


To exploit the symmetry, we perform the calculations in appropriate coordinate systems and use the right kind of Gaussiansurface for that symmetry, applying the remaining four steps.
Charge Distribution with Spherical Symmetry
A charge distribution has spherical symmetry if the density of charge depends only on the distance from a point in spaceand not on the direction. In other words, if you rotate the system, it doesn’t look different. For instance, if a sphere ofradius R is uniformly charged with charge density ρ0 then the distribution has spherical symmetry (Figure 6.21(a)). On
the other hand, if a sphere of radius R is charged so that the top half of the sphere has uniform charge density ρ1 and the
bottom half has a uniform charge density ρ2 ≠ ρ1, then the sphere does not have spherical symmetry because the charge


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density depends on the direction (Figure 6.21(b)). Thus, it is not the shape of the object but rather the shape of the chargedistribution that determines whether or not a system has spherical symmetry.
Figure 6.21(c) shows a sphere with four different shells, each with its own uniform charge density. Although this is asituation where charge density in the full sphere is not uniform, the charge density function depends only on the distancefrom the center and not on the direction. Therefore, this charge distribution does have spherical symmetry.


Figure 6.21 Illustrations of spherically symmetrical and nonsymmetrical systems. Different shadings indicatedifferent charge densities. Charges on spherically shaped objects do not necessarily mean the charges aredistributed with spherical symmetry. The spherical symmetry occurs only when the charge density does notdepend on the direction. In (a), charges are distributed uniformly in a sphere. In (b), the upper half of the spherehas a different charge density from the lower half; therefore, (b) does not have spherical symmetry. In (c), thecharges are in spherical shells of different charge densities, which means that charge density is only a functionof the radial distance from the center; therefore, the system has spherical symmetry.


One good way to determine whether or not your problem has spherical symmetry is to look at the charge density functionin spherical coordinates, ρ⎛⎝r, θ, ϕ⎞⎠ . If the charge density is only a function of r, that is ρ = ρ(r) , then you have spherical
symmetry. If the density depends on θ or ϕ , you could change it by rotation; hence, you would not have spherical
symmetry.
Consequences of symmetry
In all spherically symmetrical cases, the electric field at any point must be radially directed, because the charge and, hence,the field must be invariant under rotation. Therefore, using spherical coordinates with their origins at the center of thespherical charge distribution, we can write down the expected form of the electric field at a point P located at a distance rfrom the center:


(6.7)Spherical symmetry: E→ P = EP(r) r̂ ,
where r̂ is the unit vector pointed in the direction from the origin to the field point P. The radial component EP of the
electric field can be positive or negative. When EP > 0, the electric field at P points away from the origin, and when
EP < 0, the electric field at P points toward the origin.
Gaussian surface and flux calculations
We can now use this form of the electric field to obtain the flux of the electric field through the Gaussian surface. Forspherical symmetry, the Gaussian surface is a closed spherical surface that has the same center as the center of the chargedistribution. Thus, the direction of the area vector of an area element on the Gaussian surface at any point is parallel to thedirection of the electric field at that point, since they are both radially directed outward (Figure 6.22).


Chapter 6 | Gauss's Law 253




Figure 6.22 The electric field at any point of the sphericalGaussian surface for a spherically symmetrical chargedistribution is parallel to the area element vector at that point,giving flux as the product of the magnitude of electric field andthe value of the area. Note that the radius R of the chargedistribution and the radius r of the Gaussian surface are differentquantities.


The magnitude of the electric field E→ must be the same everywhere on a spherical Gaussian surface concentric with the
distribution. For a spherical surface of radius r,


Φ = ∮
S
E


P · n̂ dA = EP∮
S
dA = EP 4πr


2.


Using Gauss’s law
According to Gauss’s law, the flux through a closed surface is equal to the total charge enclosed within the closed surfacedivided by the permittivity of vacuum ε0 . Let qenc be the total charge enclosed inside the distance r from the origin, which
is the space inside the Gaussian spherical surface of radius r. This gives the following relation for Gauss’s law:


4πr2E =
qenc
ε0


.


Hence, the electric field at point P that is a distance r from the center of a spherically symmetrical charge distribution hasthe following magnitude and direction:
(6.8)Magnitude: E(r) = 1


4πε0


qenc
r2


Direction: radial from O to P or from P to O.
The direction of the field at point P depends on whether the charge in the sphere is positive or negative. For a net positivecharge enclosed within the Gaussian surface, the direction is from O to P, and for a net negative charge, the direction isfrom P to O. This is all we need for a point charge, and you will notice that the result above is identical to that for a pointcharge. However, Gauss’s law becomes truly useful in cases where the charge occupies a finite volume.
Computing enclosed charge
The more interesting case is when a spherical charge distribution occupies a volume, and asking what the electric fieldinside the charge distribution is thus becomes relevant. In this case, the charge enclosed depends on the distance r of thefield point relative to the radius of the charge distribution R, such as that shown in Figure 6.23.


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Figure 6.23 A spherically symmetrical charge distribution and the Gaussian surface used for finding thefield (a) inside and (b) outside the distribution.


If point P is located outside the charge distribution—that is, if r ≥ R—then the Gaussian surface containing P encloses
all charges in the sphere. In this case, qenc equals the total charge in the sphere. On the other hand, if point P is within
the spherical charge distribution, that is, if r < R, then the Gaussian surface encloses a smaller sphere than the sphere of
charge distribution. In this case, qenc is less than the total charge present in the sphere. Referring to Figure 6.23, we can
write qenc as


qenc =




qtot(total charge) if r ≥ R


qwithin r < R(only charge within r < R) if r < R
.


The field at a point outside the charge distribution is also called E→ out , and the field at a point inside the charge distribution
is called E→ in. Focusing on the two types of field points, either inside or outside the charge distribution, we can now write
the magnitude of the electric field as


(6.9)P outside sphere Eout = 14πε0
qtot
r2


(6.10)P inside sphere Ein = 14πε0
qwithin r < R


r2
.


Note that the electric field outside a spherically symmetrical charge distribution is identical to that of a point charge at thecenter that has a charge equal to the total charge of the spherical charge distribution. This is remarkable since the chargesare not located at the center only. We now work out specific examples of spherical charge distributions, starting with thecase of a uniformly charged sphere.
Example 6.6


Uniformly Charged Sphere
A sphere of radius R, such as that shown in Figure 6.23, has a uniform volume charge density ρ0 . Find the
electric field at a point outside the sphere and at a point inside the sphere.
Strategy
Apply the Gauss’s law problem-solving strategy, where we have already worked out the flux calculation.
Solution
The charge enclosed by the Gaussian surface is given by


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qenc = ∫ ρ0dV = ∫
0


r
ρ04πr′


2dr′ = ρ0


4
3
πr3⎞⎠.


The answer for electric field amplitude can then be written down immediately for a point outside the sphere,labeled Eout, and a point inside the sphere, labeled Ein.
Eout =


1
4πε0


qtot
r2


, qtot =
4
3
πR3 ρ0,


Ein =
qenc


4πε0 r
2
=


ρ0 r
3ε0


, since qenc = 43
πr3 ρ0.


It is interesting to note that the magnitude of the electric field increases inside the material as you go out, since theamount of charge enclosed by the Gaussian surface increases with the volume. Specifically, the charge enclosed
grows ∝ r3 , whereas the field from each infinitesimal element of charge drops off ∝ 1/r2 with the net result
that the electric field within the distribution increases in strength linearly with the radius. The magnitude of theelectric field outside the sphere decreases as you go away from the charges, because the included charge remainsthe same but the distance increases. Figure 6.24 displays the variation of the magnitude of the electric field withdistance from the center of a uniformly charged sphere.


Figure 6.24 Electric field of a uniformly charged, non-conducting sphere increases inside the sphere to a maximum at
the surface and then decreases as 1/r2 . Here, ER = ρ0R3ε0 . The
electric field is due to a spherical charge distribution of uniformcharge density and total charge Q as a function of distance fromthe center of the distribution.


The direction of the electric field at any point P is radially outward from the origin if ρ0 is positive, and inward
(i.e., toward the center) if ρ0 is negative. The electric field at some representative space points are displayed in
Figure 6.25 whose radial coordinates r are r = R/2 , r = R , and r = 2R .


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Figure 6.25 Electric field vectors inside and outside a uniformly chargedsphere.


Significance
Notice that Eout has the same form as the equation of the electric field of an isolated point charge. In determining
the electric field of a uniform spherical charge distribution, we can therefore assume that all of the charge insidethe appropriate spherical Gaussian surface is located at the center of the distribution.


Example 6.7
Non-Uniformly Charged Sphere
A non-conducting sphere of radius R has a non-uniform charge density that varies with the distance from its centeras given by


ρ(r) = arn (r ≤ R; n ≥ 0),


where a is a constant. We require n ≥ 0 so that the charge density is not undefined at r = 0 . Find the electric
field at a point outside the sphere and at a point inside the sphere.
Strategy
Apply the Gauss’s law strategy given above, where we work out the enclosed charge integrals separately for casesinside and outside the sphere.
Solution
Since the given charge density function has only a radial dependence and no dependence on direction, we have aspherically symmetrical situation. Therefore, the magnitude of the electric field at any point is given above andthe direction is radial. We just need to find the enclosed charge qenc, which depends on the location of the field
point.


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A note about symbols: We use r′ for locating charges in the charge distribution and r for locating the field
point(s) at the Gaussian surface(s). The letter R is used for the radius of the charge distribution.
As charge density is not constant here, we need to integrate the charge density function over the volume enclosedby the Gaussian surface. Therefore, we set up the problem for charges in one spherical shell, say between r′
and r′ + dr′, as shown in Figure 6.26. The volume of charges in the shell of infinitesimal width is equal to
the product of the area of surface 4πr′2 and the thickness dr′ . Multiplying the volume with the density at this
location, which is ar′n , gives the charge in the shell:


dq = ar′n 4πr′2dr′.


Figure 6.26 Spherical symmetry with non-uniform chargedistribution. In this type of problem, we need four radii: R is theradius of the charge distribution, r is the radius of the Gaussiansurface, r′ is the inner radius of the spherical shell, and
r′ + dr′ is the outer radius of the spherical shell. The spherical
shell is used to calculate the charge enclosed within theGaussian surface. The range for r′ is from 0 to r for the field at
a point inside the charge distribution and from 0 to R for thefield at a point outside the charge distribution. If r > R , then
the Gaussian surface encloses more volume than the chargedistribution, but the additional volume does not contribute to
qenc .


(a) Field at a point outside the charge distribution. In this case, the Gaussian surface, which contains the fieldpoint P, has a radius r that is greater than the radius R of the charge distribution, r > R . Therefore, all charges of
the charge distribution are enclosed within the Gaussian surface. Note that the space between r′ = R and r′ = r
is empty of charges and therefore does not contribute to the integral over the volume enclosed by the Gaussiansurface:


qenc = ∫ dq = ∫
0


R
ar′n 4πr′2dr′ = 4πa


n + 3
Rn + 3.


This is used in the general result for E→ out above to obtain the electric field at a point outside the charge
distribution as


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6.4


E


out =


aRn + 3


ε0(n + 3)


1
r2


r̂ ,


where r̂ is a unit vector in the direction from the origin to the field point at the Gaussian surface.
(b) Field at a point inside the charge distribution. The Gaussian surface is now buried inside the chargedistribution, with r < R . Therefore, only those charges in the distribution that are within a distance r of the center
of the spherical charge distribution count in renc :


qenc = ⌠⌡0


r
ar′n4πr′2 dr′ = 4πa


n + 3
rn + 3.


Now, using the general result above for E→ in, we find the electric field at a point that is a distance r from the
center and lies within the charge distribution as


E


in =



a
ε0(n + 3)



⎦r


n + 1 r̂ ,


where the direction information is included by using the unit radial vector.
Check Your Understanding Check that the electric fields for the sphere reduce to the correct values fora point charge.


Charge Distribution with Cylindrical Symmetry
A charge distribution has cylindrical symmetry if the charge density depends only upon the distance r from the axis of acylinder and must not vary along the axis or with direction about the axis. In other words, if your system varies if you rotateit around the axis, or shift it along the axis, you do not have cylindrical symmetry.
Figure 6.27 shows four situations in which charges are distributed in a cylinder. A uniform charge density ρ0. in an
infinite straight wire has a cylindrical symmetry, and so does an infinitely long cylinder with constant charge density ρ0.
An infinitely long cylinder that has different charge densities along its length, such as a charge density ρ1 for z > 0 and
ρ2 ≠ ρ1 for z < 0 , does not have a usable cylindrical symmetry for this course. Neither does a cylinder in which charge
density varies with the direction, such as a charge density ρ1 for 0 ≤ θ < π and ρ2 ≠ ρ1 for π ≤ θ < 2π . A system with
concentric cylindrical shells, each with uniform charge densities, albeit different in different shells, as in Figure 6.27(d),does have cylindrical symmetry if they are infinitely long. The infinite length requirement is due to the charge densitychanging along the axis of a finite cylinder. In real systems, we don’t have infinite cylinders; however, if the cylindricalobject is considerably longer than the radius from it that we are interested in, then the approximation of an infinite cylinderbecomes useful.


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Figure 6.27 To determine whether a given charge distribution has cylindrical symmetry,look at the cross-section of an “infinitely long” cylinder. If the charge density does notdepend on the polar angle of the cross-section or along the axis, then you have cylindricalsymmetry. (a) Charge density is constant in the cylinder; (b) upper half of the cylinder has adifferent charge density from the lower half; (c) left half of the cylinder has a differentcharge density from the right half; (d) charges are constant in different cylindrical rings, butthe density does not depend on the polar angle. Cases (a) and (d) have cylindrical symmetry,whereas (b) and (c) do not.
Consequences of symmetry
In all cylindrically symmetrical cases, the electric field E→ P at any point P must also display cylindrical symmetry.
Cylindrical symmetry: E→ P = EP(r) r̂ ,
where r is the distance from the axis and r̂ is a unit vector directed perpendicularly away from the axis (Figure 6.28).


Figure 6.28 The electric field in a cylindrically symmetricalsituation depends only on the distance from the axis. Thedirection of the electric field is pointed away from the axis forpositive charges and toward the axis for negative charges.
Gaussian surface and flux calculation
To make use of the direction and functional dependence of the electric field, we choose a closed Gaussian surface in theshape of a cylinder with the same axis as the axis of the charge distribution. The flux through this surface of radius s andheight L is easy to compute if we divide our task into two parts: (a) a flux through the flat ends and (b) a flux through thecurved surface (Figure 6.29).


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Figure 6.29 The Gaussian surface in the case of cylindricalsymmetry. The electric field at a patch is either parallel orperpendicular to the normal to the patch of the Gaussian surface.


The electric field is perpendicular to the cylindrical side and parallel to the planar end caps of the surface. The flux throughthe cylindrical part is

⌡S


E


· n̂ dA = E∫
S
dA = E(2πrL),


whereas the flux through the end caps is zero because E→ · n̂ = 0 there. Thus, the flux is

S
E


· n̂ dA = E(2πrL) + 0 + 0 = 2πrLE.


Using Gauss’s law
According to Gauss’s law, the flux must equal the amount of charge within the volume enclosed by this surface, divided bythe permittivity of free space. When you do the calculation for a cylinder of length L, you find that qenc of Gauss’s law is
directly proportional to L. Let us write it as charge per unit length (λenc) times length L:


qenc = λencL.


Hence, Gauss’s law for any cylindrically symmetrical charge distribution yields the following magnitude of the electric fielda distance s away from the axis:
Magnitude: E(r) = λenc


2πε0
1
r .


The charge per unit length λenc depends on whether the field point is inside or outside the cylinder of charge distribution,
just as we have seen for the spherical distribution.
Computing enclosed charge
Let R be the radius of the cylinder within which charges are distributed in a cylindrically symmetrical way. Let the fieldpoint P be at a distance s from the axis. (The side of the Gaussian surface includes the field point P.) When r > R (that is,
when P is outside the charge distribution), the Gaussian surface includes all the charge in the cylinder of radius R and lengthL. When r < R (P is located inside the charge distribution), then only the charge within a cylinder of radius s and length L
is enclosed by the Gaussian surface:


λencL =




(total charge) if r ≥ R


(only charge within r < R) if r < R
.


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Example 6.8
Uniformly Charged Cylindrical Shell
A very long non-conducting cylindrical shell of radius R has a uniform surface charge density σ0. Find the
electric field (a) at a point outside the shell and (b) at a point inside the shell.
Strategy
Apply the Gauss’s law strategy given earlier, where we treat the cases inside and outside the shell separately.
Solutiona. Electric field at a point outside the shell. For a point outside the cylindrical shell, the Gaussian surfaceis the surface of a cylinder of radius r > R and length L, as shown in Figure 6.30. The charge enclosed


by the Gaussian cylinder is equal to the charge on the cylindrical shell of length L. Therefore, λenc is
given by


λenc =
σ02π RL


L
= 2π Rσ0.


Figure 6.30 A Gaussian surface surrounding a cylindricalshell.


Hence, the electric field at a point P outside the shell at a distance r away from the axis is
E


=
2πRσ0
2πεo


1
r r̂ =


Rσ0
εo


1
r r̂ (r > R)


where r̂ is a unit vector, perpendicular to the axis and pointing away from it, as shown in the figure.
The electric field at P points in the direction of r̂ given in Figure 6.30 if σ0 > 0 and in the opposite
direction to r̂ if σ0 < 0 .


b. Electric field at a point inside the shell. For a point inside the cylindrical shell, the Gaussian surface isa cylinder whose radius r is less than R (Figure 6.31). This means no charges are included inside theGaussian surface:
λenc = 0.


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6.5


Figure 6.31 A Gaussian surface within a cylindrical shell.


This gives the following equation for the magnitude of the electric field Ein at a point whose r is less
than R of the shell of charges.


Ein 2πrL = 0 (r < R),


This gives us
Ein = 0 (r < R).


Significance
Notice that the result inside the shell is exactly what we should expect: No enclosed charge means zero electricfield. Outside the shell, the result becomes identical to a wire with uniform charge Rσ0.


Check Your Understanding A thin straight wire has a uniform linear charge density λ0. Find the
electric field at a distance d from the wire, where d is much less than the length of the wire.


Charge Distribution with Planar Symmetry
A planar symmetry of charge density is obtained when charges are uniformly spread over a large flat surface. In planarsymmetry, all points in a plane parallel to the plane of charge are identical with respect to the charges.
Consequences of symmetry
We take the plane of the charge distribution to be the xy-plane and we find the electric field at a space point P withcoordinates (x, y, z). Since the charge density is the same at all (x, y)-coordinates in the z = 0 plane, by symmetry, the
electric field at P cannot depend on the x- or y-coordinates of point P, as shown in Figure 6.32. Therefore, the electricfield at P can only depend on the distance from the plane and has a direction either toward the plane or away from the plane.That is, the electric field at P has only a nonzero z-component.
Uniform charges in xy plane: E→ = E(z) ẑ
where z is the distance from the plane and ẑ is the unit vector normal to the plane. Note that in this system,
E(z) = E(−z), although of course they point in opposite directions.


Chapter 6 | Gauss's Law 263




Figure 6.32 The components of the electric field parallel to aplane of charges cancel out the two charges locatedsymmetrically from the field point P. Therefore, the field at anypoint is pointed vertically from the plane of charges. For anypoint P and charge q1, we can always find a q2 with this
effect.


Gaussian surface and flux calculation
In the present case, a convenient Gaussian surface is a box, since the expected electric field points in one direction only. Tokeep the Gaussian box symmetrical about the plane of charges, we take it to straddle the plane of the charges, such that oneface containing the field point P is taken parallel to the plane of the charges. In Figure 6.33, sides I and II of the Gaussiansurface (the box) that are parallel to the infinite plane have been shaded. They are the only surfaces that give rise to nonzeroflux because the electric field and the area vectors of the other faces are perpendicular to each other.


Figure 6.33 A thin charged sheet and the Gaussian box forfinding the electric field at the field point P. The normal to eachface of the box is from inside the box to outside. On two faces ofthe box, the electric fields are parallel to the area vectors, and onthe other four faces, the electric fields are perpendicular to thearea vectors.


Let A be the area of the shaded surface on each side of the plane and EP be the magnitude of the electric field at point
P. Since sides I and II are at the same distance from the plane, the electric field has the same magnitude at points in theseplanes, although the directions of the electric field at these points in the two planes are opposite to each other.
Magnitude at I or II: E(z) = EP.
If the charge on the plane is positive, then the direction of the electric field and the area vectors are as shown in Figure6.33. Therefore, we find for the flux of electric field through the box


(6.11)Φ = ∮
S
E


P · n̂ dA = EP A + EP A + 0 + 0 + 0 + 0 = 2EP A


where the zeros are for the flux through the other sides of the box. Note that if the charge on the plane is negative, thedirections of electric field and area vectors for planes I and II are opposite to each other, and we get a negative sign for theflux. According to Gauss’s law, the flux must equal qenc /ε0 . From Figure 6.33, we see that the charges inside the volume


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enclosed by the Gaussian box reside on an area A of the xy-plane. Hence,
(6.12)qenc = σ0 A.


Using the equations for the flux and enclosed charge in Gauss’s law, we can immediately determine the electric field at apoint at height z from a uniformly charged plane in the xy-plane:
E


P =
σ0
2ε0


n̂ .


The direction of the field depends on the sign of the charge on the plane and the side of the plane where the field point P is
located. Note that above the plane, n̂ = + ẑ , while below the plane, n̂ = − ẑ .
You may be surprised to note that the electric field does not actually depend on the distance from the plane; this is an effectof the assumption that the plane is infinite. In practical terms, the result given above is still a useful approximation for finiteplanes near the center.
6.4 | Conductors in Electrostatic Equilibrium


Learning Objectives
By the end of this section, you will be able to:
• Describe the electric field within a conductor at equilibrium
• Describe the electric field immediately outside the surface of a charged conductor atequilibrium
• Explain why if the field is not as described in the first two objectives, the conductor is not atequilibrium


So far, we have generally been working with charges occupying a volume within an insulator. We now study what happenswhen free charges are placed on a conductor. Generally, in the presence of a (generally external) electric field, the freecharge in a conductor redistributes and very quickly reaches electrostatic equilibrium. The resulting charge distribution andits electric field have many interesting properties, which we can investigate with the help of Gauss’s law and the concept ofelectric potential.
The Electric Field inside a Conductor Vanishes
If an electric field is present inside a conductor, it exerts forces on the free electrons (also called conduction electrons),which are electrons in the material that are not bound to an atom. These free electrons then accelerate. However, movingcharges by definition means nonstatic conditions, contrary to our assumption. Therefore, when electrostatic equilibrium isreached, the charge is distributed in such a way that the electric field inside the conductor vanishes.
If you place a piece of a metal near a positive charge, the free electrons in the metal are attracted to the external positivecharge and migrate freely toward that region. The region the electrons move to then has an excess of electrons over theprotons in the atoms and the region from where the electrons have migrated has more protons than electrons. Consequently,the metal develops a negative region near the charge and a positive region at the far end (Figure 6.34). As we saw inthe preceding chapter, this separation of equal magnitude and opposite type of electric charge is called polarization. If youremove the external charge, the electrons migrate back and neutralize the positive region.


Chapter 6 | Gauss's Law 265




Figure 6.34 Polarization of a metallic sphere by an externalpoint charge +q . The near side of the metal has an opposite
surface charge compared to the far side of the metal. The sphereis said to be polarized. When you remove the external charge,the polarization of the metal also disappears.


The polarization of the metal happens only in the presence of external charges. You can think of this in terms of electricfields. The external charge creates an external electric field. When the metal is placed in the region of this electric field,the electrons and protons of the metal experience electric forces due to this external electric field, but only the conductionelectrons are free to move in the metal over macroscopic distances. The movement of the conduction electrons leads to thepolarization, which creates an induced electric field in addition to the external electric field (Figure 6.35). The net electricfield is a vector sum of the fields of +q and the surface charge densities −σA and +σB. This means that the net field
inside the conductor is different from the field outside the conductor.


Figure 6.35 In the presence of an external charge q, thecharges in a metal redistribute. The electric field at any point hasthree contributions, from +q and the induced charges −σA
and +σB. Note that the surface charge distribution will not be
uniform in this case.


The redistribution of charges is such that the sum of the three contributions at any point P inside the conductor is
E


P = E


q + E


B + E


A = 0


.


Now, thanks to Gauss’s law, we know that there is no net charge enclosed by a Gaussian surface that is solely within thevolume of the conductor at equilibrium. That is, qenc = 0 and hence
(6.13)E→ net = 0→ (at points inside a conductor).Charge on a Conductor


An interesting property of a conductor in static equilibrium is that extra charges on the conductor end up on the outersurface of the conductor, regardless of where they originate. Figure 6.36 illustrates a system in which we bring an externalpositive charge inside the cavity of a metal and then touch it to the inside surface. Initially, the inside surface of the cavity isnegatively charged and the outside surface of the conductor is positively charged. When we touch the inside surface of thecavity, the induced charge is neutralized, leaving the outside surface and the whole metal charged with a net positive charge.


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Figure 6.36 Electric charges on a conductor migrate to the outside surface nomatter where you put them initially.


To see why this happens, note that the Gaussian surface in Figure 6.37 (the dashed line) follows the contour of the actualsurface of the conductor and is located an infinitesimal distance within it. Since E = 0 everywhere inside a conductor,

s
E


· n̂dA = 0.


Thus, from Gauss’ law, there is no net charge inside the Gaussian surface. But the Gaussian surface lies just below theactual surface of the conductor; consequently, there is no net charge inside the conductor. Any excess charge must lie on itssurface.


Figure 6.37 The dashed line represents a Gaussian surfacethat is just beneath the actual surface of the conductor.


This particular property of conductors is the basis for an extremely accurate method developed by Plimpton and Lawtonin 1936 to verify Gauss’s law and, correspondingly, Coulomb’s law. A sketch of their apparatus is shown in Figure 6.38.Two spherical shells are connected to one another through an electrometer E, a device that can detect a very slight amountof charge flowing from one shell to the other. When switch S is thrown to the left, charge is placed on the outer shell by thebattery B. Will charge flow through the electrometer to the inner shell?
No. Doing so would mean a violation of Gauss’s law. Plimpton and Lawton did not detect any flow and, knowing the
sensitivity of their electrometer, concluded that if the radial dependence in Coulomb’s law were 1/r2 + δ , δ would be less
than 2 × 10−9 [1]. More recent measurements place δ at less than 3 × 10−16 [2], a number so small that the validity of
Coulomb’s law seems indisputable.


1. S. Plimpton and W. Lawton. 1936. “A Very Accurate Test of Coulomb’s Law of Force between Charges.” PhysicalReview 50, No. 11: 1066, doi:10.1103/PhysRev.50.10662. E. Williams, J. Faller, and H. Hill. 1971. “New Experimental Test of Coulomb’s Law: A Laboratory Upper Limit onthe Photon Rest Mass.” Physical Review Letters 26 , No. 12: 721, doi:10.1103/PhysRevLett.26.721


Chapter 6 | Gauss's Law 267




Figure 6.38 A representation of the apparatus used by Plimpton and Lawton.Any transfer of charge between the spheres is detected by the electrometer E.
The Electric Field at the Surface of a Conductor
If the electric field had a component parallel to the surface of a conductor, free charges on the surface would move, asituation contrary to the assumption of electrostatic equilibrium. Therefore, the electric field is always perpendicular to thesurface of a conductor.
At any point just above the surface of a conductor, the surface charge density δ and the magnitude of the electric field E
are related by


(6.14)E = σε0.


To see this, consider an infinitesimally small Gaussian cylinder that surrounds a point on the surface of the conductor, asin Figure 6.39. The cylinder has one end face inside and one end face outside the surface. The height and cross-sectionalarea of the cylinder are δ and ΔA , respectively. The cylinder’s sides are perpendicular to the surface of the conductor, and
its end faces are parallel to the surface. Because the cylinder is infinitesimally small, the charge density σ is essentially
constant over the surface enclosed, so the total charge inside the Gaussian cylinder is σΔA . Now E is perpendicular to the
surface of the conductor outside the conductor and vanishes within it, because otherwise, the charges would accelerate, andwe would not be in equilibrium. Electric flux therefore crosses only the outer end face of the Gaussian surface and may bewritten as EΔA , since the cylinder is assumed to be small enough that E is approximately constant over that area. From
Gauss’ law,


EΔA = σΔAε0
.


Thus,
E = σε0


.


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Figure 6.39 An infinitesimally small cylindrical Gaussian surface surrounds point P, which is on the surface of
the conductor. The field E→ is perpendicular to the surface of the conductor outside the conductor and vanishes
within it.


Example 6.9
Electric Field of a Conducting Plate
The infinite conducting plate in Figure 6.40 has a uniform surface charge density σ . Use Gauss’ law to find the
electric field outside the plate. Compare this result with that previously calculated directly.


Figure 6.40 A side view of an infinite conducting plate andGaussian cylinder with cross-sectional area A.


Chapter 6 | Gauss's Law 269




Strategy
For this case, we use a cylindrical Gaussian surface, a side view of which is shown.
Solution
The flux calculation is similar to that for an infinite sheet of charge from the previous chapter with one major
exception: The left face of the Gaussian surface is inside the conductor where E→ = 0→ , so the total flux
through the Gaussian surface is EA rather than 2EA. Then from Gauss’ law,


EA = σAε0


and the electric field outside the plate is
E = σε0


.


Significance
This result is in agreement with the result from the previous section, and consistent with the rule stated above.


Example 6.10
Electric Field between Oppositely Charged Parallel Plates
Two large conducting plates carry equal and opposite charges, with a surface charge density σ of magnitude
6.81 × 10−7 C/m2, as shown in Figure 6.41. The separation between the plates is l = 6.50 mm . What is the
electric field between the plates?


Figure 6.41 The electric field between oppositely chargedparallel plates. A test charge is released at the positive plate.


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Strategy
Note that the electric field at the surface of one plate only depends on the charge on that plate. Thus, apply
E = σ/ε0 with the given values.
Solution
The electric field is directed from the positive to the negative plate, as shown in the figure, and its magnitude isgiven by


E = σε0
= 6.81 × 10


−7 C/m2


8.85 × 10−12 C2 /N m2
= 7.69 × 104 N/C.


Significance
This formula is applicable to more than just a plate. Furthermore, two-plate systems will be important later.


Example 6.11
A Conducting Sphere
The isolated conducting sphere (Figure 6.42) has a radius R and an excess charge q. What is the electric fieldboth inside and outside the sphere?


Figure 6.42 An isolated conducting sphere.


Strategy
The sphere is isolated, so its surface change distribution and the electric field of that distribution are spherically
symmetrical. We can therefore represent the field as E→ = E(r) r̂ . To calculate E(r), we apply Gauss’s law over
a closed spherical surface S of radius r that is concentric with the conducting sphere.
Solution
Since r is constant and n̂ = r̂ on the sphere,



S
E


· n̂ dA = E(r)∮
S
dA = E(r) 4πr2.


For r < R , S is within the conductor, so qenc = 0, and Gauss’s law gives
E(r) = 0,


Chapter 6 | Gauss's Law 271




6.6


as expected inside a conductor. If r > R , S encloses the conductor so qenc = q. From Gauss’s law,
E(r) 4πr2 =


q
ε0


.


The electric field of the sphere may therefore be written as
E


= 0


(r < R),


E


= 1
4πε0


q


r2
r̂ (r ≥ R).


Significance
Notice that in the region r ≥ R , the electric field due to a charge q placed on an isolated conducting sphere of
radius R is identical to the electric field of a point charge q located at the center of the sphere. The differencebetween the charged metal and a point charge occurs only at the space points inside the conductor. For a pointcharge placed at the center of the sphere, the electric field is not zero at points of space occupied by the sphere,but a conductor with the same amount of charge has a zero electric field at those points (Figure 6.43). However,there is no distinction at the outside points in space where r > R , and we can replace the isolated charged
spherical conductor by a point charge at its center with impunity.


Figure 6.43 Electric field of a positively charged metalsphere. The electric field inside is zero, and the electric fieldoutside is same as the electric field of a point charge at thecenter, although the charge on the metal sphere is at the surface.


Check Your Understanding How will the system above change if there are charged objects external tothe sphere?


For a conductor with a cavity, if we put a charge +q inside the cavity, then the charge separation takes place in the
conductor, with −q amount of charge on the inside surface and a +q amount of charge at the outside surface (Figure
6.44(a)). For the same conductor with a charge +q outside it, there is no excess charge on the inside surface; both the
positive and negative induced charges reside on the outside surface (Figure 6.44(b)).


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Figure 6.44 (a) A charge inside a cavity in a metal. The distribution of chargesat the outer surface does not depend on how the charges are distributed at theinner surface, since the E-field inside the body of the metal is zero. Thatmagnitude of the charge on the outer surface does depend on the magnitude of thecharge inside, however. (b) A charge outside a conductor containing an innercavity. The cavity remains free of charge. The polarization of charges on theconductor happens at the surface.


If a conductor has two cavities, one of them having a charge +qa inside it and the other a charge −qb, the polarization
of the conductor results in −qa on the inside surface of the cavity a, +qb on the inside surface of the cavity b, and
qa − qb on the outside surface (Figure 6.45). The charges on the surfaces may not be uniformly spread out; their spread
depends upon the geometry. The only rule obeyed is that when the equilibrium has been reached, the charge distribution ina conductor is such that the electric field by the charge distribution in the conductor cancels the electric field of the externalcharges at all space points inside the body of the conductor.


Figure 6.45 The charges induced by two equal and oppositecharges in two separate cavities of a conductor. If the net chargeon the cavity is nonzero, the external surface becomes chargedto the amount of the net charge.


Chapter 6 | Gauss's Law 273




area vector
cylindrical symmetry
electric flux
flux
free electrons
Gaussian surface
planar symmetry
spherical symmetry


CHAPTER 6 REVIEW
KEY TERMS


vector with magnitude equal to the area of a surface and direction perpendicular to the surface
system only varies with distance from the axis, not direction


dot product of the electric field and the area through which it is passing
quantity of something passing through a given area


also called conduction electrons, these are the electrons in a conductor that are not bound to anyparticular atom, and hence are free to move around
any enclosed (usually imaginary) surface
system only varies with distance from a plane


system only varies with the distance from the origin, not in direction


KEY EQUATIONS
Definition of electric flux, for uniform electric field Φ = E→ · A→ → EA cos θ
Electric flux through an open surface Φ = ∫


S
E


· n̂ dA = ∫
S


E


· d A


Electric flux through a closed surface Φ = ∮
S


E


· n̂ dA = ∮
S


E


· d A


Gauss’s law Φ = ∮
S


E


· n̂ dA = qencε0


Gauss’s Law for systems with symmetry Φ = ∮
S


E


· n̂ dA = E∮
S
dA = EA =


qenc
ε0


The magnitude of the electric field just outside the surfaceof a conductor E =
σ
ε0


SUMMARY
6.1 Electric Flux


• The electric flux through a surface is proportional to the number of field lines crossing that surface. Note that thismeans the magnitude is proportional to the portion of the field perpendicular to the area.
• The electric flux is obtained by evaluating the surface integral


Φ = ∮
S
E


· n̂ dA = ∮
S
E


· d A


,


where the notation used here is for a closed surface S.
6.2 Explaining Gauss’s Law


• Gauss’s law relates the electric flux through a closed surface to the net charge within that surface,
Φ = ∮


S
E


· n̂ dA = qencε0
,


where qenc is the total charge inside the Gaussian surface S.


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• All surfaces that include the same amount of charge have the same number of field lines crossing it, regardless ofthe shape or size of the surface, as long as the surfaces enclose the same amount of charge.
6.3 Applying Gauss’s Law


• For a charge distribution with certain spatial symmetries (spherical, cylindrical, and planar), we can find a Gaussian
surface over which E→ · n̂ = E , where E is constant over the surface. The electric field is then determined with
Gauss’s law.


• For spherical symmetry, the Gaussian surface is also a sphere, and Gauss’s law simplifies to 4πr2E = qencε0 .
• For cylindrical symmetry, we use a cylindrical Gaussian surface, and find that Gauss’s law simplifies to


2πrLE =
qenc
ε0


.
• For planar symmetry, a convenient Gaussian surface is a box penetrating the plane, with two faces parallel to the
plane and the remainder perpendicular, resulting in Gauss’s law being 2AE = qencε0 .


6.4 Conductors in Electrostatic Equilibrium
• The electric field inside a conductor vanishes.
• Any excess charge placed on a conductor resides entirely on the surface of the conductor.
• The electric field is perpendicular to the surface of a conductor everywhere on that surface.
• The magnitude of the electric field just above the surface of a conductor is given by E = σε0 .


CONCEPTUAL QUESTIONS
6.1 Electric Flux
1. Discuss how would orient a planar surface of area A ina uniform electric field of magnitude E0 to obtain (a) the
maximum flux and (b) the minimum flux through the area.
2. What are the maximum and minimum values of the fluxin the preceding question?
3. The net electric flux crossing a closed surface is alwayszero. True or false?
4. The net electric flux crossing an open surface is neverzero. True or false?


6.2 Explaining Gauss’s Law
5. Two concentric spherical surfaces enclose a pointcharge q. The radius of the outer sphere is twice that ofthe inner one. Compare the electric fluxes crossing the twosurfaces.
6. Compare the electric flux through the surface of a cubeof side length a that has a charge q at its center to the fluxthrough a spherical surface of radius a with a charge q at itscenter.


7. (a) If the electric flux through a closed surface is zero,is the electric field necessarily zero at all points on thesurface? (b) What is the net charge inside the surface?
8. Discuss how Gauss’s law would be affected if the
electric field of a point charge did not vary as 1/r2.
9. Discuss the similarities and differences between thegravitational field of a point mass m and the electric field ofa point charge q.
10. Discuss whether Gauss’s law can be applied to otherforces, and if so, which ones.
11. Is the term E→ in Gauss’s law the electric field
produced by just the charge inside the Gaussian surface?
12. Reformulate Gauss’s law by choosing the unit normalof the Gaussian surface to be the one directed inward.


6.3 Applying Gauss’s Law
13. Would Gauss’s law be helpful for determining theelectric field of two equal but opposite charges a fixeddistance apart?


Chapter 6 | Gauss's Law 275




14. Discuss the role that symmetry plays in the applicationof Gauss’s law. Give examples of continuous chargedistributions in which Gauss’s law is useful and not usefulin determining the electric field.
15. Discuss the restrictions on the Gaussian surface usedto discuss planar symmetry. For example, is its lengthimportant? Does the cross-section have to be square? Mustthe end faces be on opposite sides of the sheet?


6.4 Conductors in Electrostatic Equilibrium
16. Is the electric field inside a metal always zero?
17. Under electrostatic conditions, the excess charge on aconductor resides on its surface. Does this mean that all theconduction electrons in a conductor are on the surface?
18. A charge q is placed in the cavity of a conductoras shown below. Will a charge outside the conductorexperience an electric field due to the presence of q?


19. The conductor in the preceding figure has an excesscharge of – 5.0 µC . If a 2.0-µC point charge is placed
in the cavity, what is the net charge on the surface of thecavity and on the outer surface of the conductor?


PROBLEMS
6.1 Electric Flux
20. A uniform electric field of magnitude 1.1 × 104 N/C
is perpendicular to a square sheet with sides 2.0 m long.What is the electric flux through the sheet?
21. Calculate the flux through the sheet of the previousproblem if the plane of the sheet is at an angle of 60° to the
field. Find the flux for both directions of the unit normal tothe sheet.
22. Find the electric flux through a rectangular area
3 cm × 2 cm between two parallel plates where there is
a constant electric field of 30 N/C for the followingorientations of the area: (a) parallel to the plates, (b)perpendicular to the plates, and (c) the normal to the areamaking a 30° angle with the direction of the electric field.
Note that this angle can also be given as 180° + 30°.
23. The electric flux through a square-shaped area of side5 cm near a large charged sheet is found to be
3 × 10−5 N ·m2 /C when the area is parallel to the plate.
Find the charge density on the sheet.
24. Two large rectangular aluminum plates of area
150 cm2 face each other with a separation of 3 mm
between them. The plates are charged with equal amount ofopposite charges, ±20 µC . The charges on the plates face
each other. Find the flux through a circle of radius 3 cm


between the plates when the normal to the circle makes anangle of 5° with a line perpendicular to the plates. Note
that this angle can also be given as 180° + 5°.
25. A square surface of area 2 cm2 is in a space of
uniform electric field of magnitude 103 N/C . The amount
of flux through it depends on how the square is orientedrelative to the direction of the electric field. Find theelectric flux through the square, when the normal to itmakes the following angles with electric field: (a) 30° , (b)
90° , and (c) 0° . Note that these angles can also be given
as 180° + θ .
26. A vector field is pointed along the z-axis,
v→ = α


x2 + y2
ẑ . (a) Find the flux of the vector field


through a rectangle in the xy-plane between a < x < b and
c < y < d . (b) Do the same through a rectangle in the
yz-plane between a < z < b and c < y < d . (Leave your
answer as an integral.)
27. Consider the uniform electric field
E


= (4.0 j
^


+ 3.0 k
^
) × 103 N/C. What is its electric


flux through a circular area of radius 2.0 m that lies in thexy-plane?
28. Repeat the previous problem, given that the circular


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area is (a) in the yz-plane and (b) 45° above the xy-plane.
29. An infinite charged wire with charge per unit length λ
lies along the central axis of a cylindrical surface of radiusr and length l. What is the flux through the surface due tothe electric field of the charged wire?


6.2 Explaining Gauss’s Law
30. Determine the electric flux through each surfacewhose cross-section is shown below.


31. Find the electric flux through the closed surface whosecross-sections are shown below.


32. A point charge q is located at the center of a cubewhose sides are of length a. If there are no other charges inthis system, what is the electric flux through one face of thecube?
33. A point charge of 10 µC is at an unspecified location


Chapter 6 | Gauss's Law 277




inside a cube of side 2 cm. Find the net electric flux thoughthe surfaces of the cube.
34. A net flux of 1.0 × 104 N ·m2 /C passes inward
through the surface of a sphere of radius 5 cm. (a) Howmuch charge is inside the sphere? (b) How precisely can wedetermine the location of the charge from this information?
35. A charge q is placed at one of the corners of a cube ofside a, as shown below. Find the magnitude of the electricflux through the shaded face due to q. Assume q > 0 .


36. The electric flux through a cubical box 8.0 cm on
a side is 1.2 × 103 N ·m2 /C. What is the total charge
enclosed by the box?
37. The electric flux through a spherical surface is
4.0 × 104 N ·m2 /C. What is the net charge enclosed by
the surface?
38. A cube whose sides are of length d is placed in a
uniform electric field of magnitude E = 4.0 × 103 N/C
so that the field is perpendicular to two opposite faces ofthe cube. What is the net flux through the cube?
39. Repeat the previous problem, assuming that theelectric field is directed along a body diagonal of the cube.
40. A total charge 5.0 × 10−6 C is distributed uniformly
throughout a cubical volume whose edges are 8.0 cm long.(a) What is the charge density in the cube? (b) What isthe electric flux through a cube with 12.0-cm edges thatis concentric with the charge distribution? (c) Do the samecalculation for cubes whose edges are 10.0 cm long and 5.0cm long. (d) What is the electric flux through a sphericalsurface of radius 3.0 cm that is also concentric with thecharge distribution?


6.3 Applying Gauss’s Law
41. Recall that in the example of a uniform charged
sphere, ρ0 = Q/(43πR3). Rewrite the answers in terms of


the total charge Q on the sphere.
42. Suppose that the charge density of the spherical chargedistribution shown in Figure 6.23 is ρ(r) = ρ0 r/R for
r ≤ R and zero for r > R. Obtain expressions for the
electric field both inside and outside the distribution.
43. A very long, thin wire has a uniform linear chargedensity of 50 µC/m. What is the electric field at a distance
2.0 cm from the wire?
44. A charge of −30 µC is distributed uniformly
throughout a spherical volume of radius 10.0 cm.Determine the electric field due to this charge at a distanceof (a) 2.0 cm, (b) 5.0 cm, and (c) 20.0 cm from the centerof the sphere.
45. Repeat your calculations for the preceding problem,given that the charge is distributed uniformly over thesurface of a spherical conductor of radius 10.0 cm.
46. A total charge Q is distributed uniformly throughouta spherical shell of inner and outer radii r1 and r2,
respectively. Show that the electric field due to the chargeis
E


= 0


(r ≤ r1);


E


= Q


4πε0 r
2





⎜ r


3 − r1
3


r2
3 − r1


3





⎟ r̂ (r1 ≤ r ≤ r2);


E


= Q


4πε0 r
2
r̂ (r ≥ r2).


47. When a charge is placed on a metal sphere, it ends upin equilibrium at the outer surface. Use this information todetermine the electric field of +3.0 µC charge put on a
5.0-cm aluminum spherical ball at the following two pointsin space: (a) a point 1.0 cm from the center of the ball (aninside point) and (b) a point 10 cm from the center of theball (an outside point).
48. A large sheet of charge has a uniform charge density
of 10 µC/m2 . What is the electric field due to this charge
at a point just above the surface of the sheet?
49. Determine if approximate cylindrical symmetry holdsfor the following situations. State why or why not. (a) A300-cm long copper rod of radius 1 cm is charged with+500 nC of charge and we seek electric field at a point 5cm from the center of the rod. (b) A 10-cm long copper rodof radius 1 cm is charged with +500 nC of charge and weseek electric field at a point 5 cm from the center of the rod.(c) A 150-cm wooden rod is glued to a 150-cm plastic rod


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to make a 300-cm long rod, which is then painted with acharged paint so that one obtains a uniform charge density.The radius of each rod is 1 cm, and we seek an electric fieldat a point that is 4 cm from the center of the rod. (d) Samerod as (c), but we seek electric field at a point that is 500cm from the center of the rod.
50. A long silver rod of radius 3 cm has a charge of
−5 µC/cm on its surface. (a) Find the electric field at a
point 5 cm from the center of the rod (an outside point). (b)Find the electric field at a point 2 cm from the center of therod (an inside point).
51. The electric field at 2 cm from the center of longcopper rod of radius 1 cm has a magnitude 3 N/C anddirected outward from the axis of the rod. (a) How muchcharge per unit length exists on the copper rod? (b) Whatwould be the electric flux through a cube of side 5 cmsituated such that the rod passes through opposite sides ofthe cube perpendicularly?
52. A long copper cylindrical shell of inner radius 2 cmand outer radius 3 cm surrounds concentrically a chargedlong aluminum rod of radius 1 cm with a charge densityof 4 pC/m. All charges on the aluminum rod reside at itssurface. The inner surface of the copper shell has exactlyopposite charge to that of the aluminum rod while theouter surface of the copper shell has the same charge asthe aluminum rod. Find the magnitude and direction of theelectric field at points that are at the following distancesfrom the center of the aluminum rod: (a) 0.5 cm, (b) 1.5 cm,(c) 2.5 cm, (d) 3.5 cm, and (e) 7 cm.
53. Charge is distributed uniformly with a density ρ
throughout an infinitely long cylindrical volume of radiusR. Show that the field of this charge distribution is directedradially with respect to the cylinder and that
E =


ρr
2ε0


(r ≤ R);


E =
ρR2


2ε0 r
(r ≥ R).


54. Charge is distributed throughout a very longcylindrical volume of radius R such that the charge densityincreases with the distance r from the central axis of thecylinder according to ρ = αr, where α is a constant.
Show that the field of this charge distribution is directedradially with respect to the cylinder and that
E = αr


2


3ε0
(r ≤ R);


E = αR
3


3ε0 r
(r ≥ R).


55. The electric field 10.0 cm from the surface of a copperball of radius 5.0 cm is directed toward the ball’s center
and has magnitude 4.0 × 102 N/C. How much charge is
on the surface of the ball?
56. Charge is distributed throughout a spherical shell ofinner radius r1 and outer radius r2 with a volume density
given by ρ = ρ0 r1/r, where ρ0 is a constant. Determine
the electric field due to this charge as a function of r, thedistance from the center of the shell.
57. Charge is distributed throughout a spherical volume of
radius R with a density ρ = αr2, where α is a constant.
Determine the electric field due to the charge at points bothinside and outside the sphere.
58. Consider a uranium nucleus to be sphere of radius
R = 7.4 × 10−15 m with a charge of 92e distributed
uniformly throughout its volume. (a) What is the electric
force exerted on an electron when it is 3.0 × 10−15m
from the center of the nucleus? (b) What is the accelerationof the electron at this point?
59. The volume charge density of a spherical chargedistribution is given by ρ(r) = ρ0 e−αr, where ρ0 and
α are constants. What is the electric field produced by this
charge distribution?


6.4 Conductors in Electrostatic Equilibrium
60. An uncharged conductor with an internal cavity isshown in the following figure. Use the closed surface Salong with Gauss’ law to show that when a charge q isplaced in the cavity a total charge –q is induced on the innersurface of the conductor. What is the charge on the outersurface of the conductor?


Figure 6.46 A charge inside a cavity of a metal. Charges atthe outer surface do not depend on how the charges aredistributed at the inner surface since E field inside the body ofthe metal is zero.
61. An uncharged spherical conductor S of radius R hastwo spherical cavities A and B of radii a and b, respectively


Chapter 6 | Gauss's Law 279




as shown below. Two point charges +qa and +qb are
placed at the center of the two cavities by using non-conducting supports. In addition, a point charge +q0 is
placed outside at a distance r from the center of the sphere.(a) Draw approximate charge distributions in the metalalthough metal sphere has no net charge. (b) Draw electricfield lines. Draw enough lines to represent all distinctlydifferent places.


62. A positive point charge is placed at the angle bisectorof two uncharged plane conductors that make an angle of
45°. See below. Draw the electric field lines.


63. A long cylinder of copper of radius 3 cm is chargedso that it has a uniform charge per unit length on its surfaceof 3 C/m. (a) Find the electric field inside and outsidethe cylinder. (b) Draw electric field lines in a planeperpendicular to the rod.
64. An aluminum spherical ball of radius 4 cm is chargedwith 5 µC of charge. A copper spherical shell of inner
radius 6 cm and outer radius 8 cm surrounds it. A totalcharge of −8 µC is put on the copper shell. (a) Find the
electric field at all points in space, including points insidethe aluminum and copper shell when copper shell andaluminum sphere are concentric. (b) Find the electric fieldat all points in space, including points inside the aluminumand copper shell when the centers of copper shell andaluminum sphere are 1 cm apart.
65. A long cylinder of aluminum of radius R meters ischarged so that it has a uniform charge per unit length on itssurface of λ . (a) Find the electric field inside and outside
the cylinder. (b) Plot electric field as a function of distancefrom the center of the rod.


66. At the surface of any conductor in electrostaticequilibrium, E = σ/ε0. Show that this equation is
consistent with the fact that E = kq/r2 at the surface of a
spherical conductor.
67. Two parallel plates 10 cm on a side are given equal and
opposite charges of magnitude 5.0 × 10−9 C. The plates
are 1.5 mm apart. What is the electric field at the center ofthe region between the plates?
68. Two parallel conducting plates, each of cross-sectional
area 400 cm2 , are 2.0 cm apart and uncharged. If
1.0 × 1012 electrons are transferred from one plate to the
other, what are (a) the charge density on each plate? (b) Theelectric field between the plates?
69. The surface charge density on a long straight metallicpipe is σ . What is the electric field outside and inside the
pipe? Assume the pipe has a diameter of 2a.


70. A point charge q = −5.0 × 10−12 C is placed at the
center of a spherical conducting shell of inner radius 3.5 cmand outer radius 4.0 cm. The electric field just above thesurface of the conductor is directed radially outward andhas magnitude 8.0 N/C. (a) What is the charge density onthe inner surface of the shell? (b) What is the charge densityon the outer surface of the shell? (c) What is the net chargeon the conductor?
71. A solid cylindrical conductor of radius a is surroundedby a concentric cylindrical shell of inner radius b. Thesolid cylinder and the shell carry charges +Q and –Q,


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respectively. Assuming that the length L of both conductorsis much greater than a or b, determine the electric field as afunction of r, the distance from the common central axis of
the cylinders, for (a) r < a; (b) a < r < b; and (c) r > b.


ADDITIONAL PROBLEMS
72. A vector field E→ (not necessarily an electric field;
note units) is given by E→ = 3x2 k^ . Calculate

S
E


· n̂ da, where S is the area shown below. Assume
that n̂ = k^ .


73. Repeat the preceding problem, with
E


= 2x i
^


+ 3x2 k
^
.


74. A circular area S is concentric with the origin, has
radius a, and lies in the yz-plane. Calculate ∫


S
E


· n̂ dA


for E→ = 3z2 i^ .
75. (a) Calculate the electric flux through the open
hemispherical surface due to the electric field E→ = E0 k^
(see below). (b) If the hemisphere is rotated by 90° around
the x-axis, what is the flux through it?


76. Suppose that the electric field of an isolated point


charge were proportional to 1/r2 + σ rather than 1/r2.
Determine the flux that passes through the surface of asphere of radius R centered at the charge. Would Gauss’slaw remain valid?
77. The electric field in a region is given by
E


= a/(b + cx) i
^
, where a = 200 N ·m/C,


b = 2.0 m, and c = 2.0. What is the net charge enclosed
by the shaded volume shown below?


78. Two equal and opposite charges of magnitude Q arelocated on the x-axis at the points +a and –a, as shownbelow. What is the net flux due to these charges througha square surface of side 2a that lies in the yz-plane andis centered at the origin? (Hint: Determine the flux dueto each charge separately, then use the principle ofsuperposition. You may be able to make a symmetryargument.)


79. A fellow student calculated the flux through the squarefor the system in the preceding problem and got 0. What


Chapter 6 | Gauss's Law 281




went wrong?
80. A 10 cm × 10 cm piece of aluminum foil of 0.1 mm
thickness has a charge of 20 µC that spreads on both
wide side surfaces evenly. You may ignore the charges onthe thin sides of the edges. (a) Find the charge density.(b) Find the electric field 1 cm from the center, assumingapproximate planar symmetry.
81. Two 10 cm × 10 cm pieces of aluminum foil of
thickness 0.1 mm face each other with a separation of 5mm. One of the foils has a charge of +30 µC and the other
has −30 µC . (a) Find the charge density at all surfaces,
i.e., on those facing each other and those facing away. (b)Find the electric field between the plates near the centerassuming planar symmetry.
82. Two large copper plates facing each other have charge
densities ±4.0 C/m2 on the surface facing the other plate,
and zero in between the plates. Find the electric fluxthrough a 3 cm × 4 cm rectangular area between the
plates, as shown below, for the following orientations of thearea. (a) If the area is parallel to the plates, and (b) if thearea is tilted θ = 30° from the parallel direction. Note, this
angle can also be θ = 180° + 30°.


83. The infinite slab between the planes defined by
z = −a/2 and z = a/2 contains a uniform volume charge
density ρ (see below). What is the electric field produced
by this charge distribution, both inside and outside thedistribution?


84. A total charge Q is distributed uniformly throughout


a spherical volume that is centered at O1 and has a radius
R. Without disturbing the charge remaining, charge isremoved from the spherical volume that is centered at O2
(see below). Show that the electric field everywhere in theempty region is given by
E


= Q r


4πε0R
3
,


where r→ is the displacement vector directed from
O1 toO2.


85. A non-conducting spherical shell of inner radius a1
and outer radius b1 is uniformly charged with charged
density ρ1 inside another non-conducting spherical shell
of inner radius a2 and outer radius b2 that is also
uniformly charged with charge density ρ2 . See below.
Find the electric field at space point P at a distance rfrom the common center such that (a) r > b2, (b)
a2 < r < b2, (c) b1 < r < a2, (d) a1 < r < b1, and
(e) r < a1 .


86. Two non-conducting spheres of radii R1 and R2
are uniformly charged with charge densities ρ1 and ρ2,
respectively. They are separated at center-to-center distancea (see below). Find the electric field at point P located


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at a distance r from the center of sphere 1 and is in thedirection θ from the line joining the two spheres assuming
their charge densities are not affected by the presence of theother sphere. (Hint: Work one sphere at a time and use thesuperposition principle.)


87. A disk of radius R is cut in a non-conducting largeplate that is uniformly charged with charge density σ
(coulomb per square meter). See below. Find the electricfield at a height h above the center of the disk.
(h > > R, h < < l or w). (Hint: Fill the hole with
±σ.)


88. Concentric conducting spherical shells carry chargesQ and –Q, respectively (see below). The inner shell hasnegligible thickness. Determine the electric field for (a)
r < a; (b) a < r < b; (c) b < r < c; and (d) r > c.


89. Shown below are two concentric conducting sphericalshells of radii R1 and R2 , each of finite thickness much
less than either radius. The inner and outer shell carry netcharges q1 and q2, respectively, where both q1 and q2
are positive. What is the electric field for (a) r < R1; (b)
R1 < r < R2; and (c) r > R2? (d) What is the net charge
on the inner surface of the inner shell, the outer surface ofthe inner shell, the inner surface of the outer shell, and theouter surface of the outer shell?


90. A point charge of q = 5.0 × 10−8 C is placed at
the center of an uncharged spherical conducting shell ofinner radius 6.0 cm and outer radius 9.0 cm. Find theelectric field at (a) r = 4.0 cm , (b) r = 8.0 cm , and (c)
r = 12.0 cm . (d) What are the charges induced on the
inner and outer surfaces of the shell?


CHALLENGE PROBLEMS
91. The Hubble Space Telescope can measure the energyflux from distant objects such as supernovae and stars.Scientists then use this data to calculate the energy emittedby that object. Choose an interstellar object which scientists


have observed the flux at the Hubble with (for example,Vega[3]), find the distance to that object and the size ofHubble’s primary mirror, and calculate the total energyflux. (Hint: The Hubble intercepts only a small part of the
3. http://adsabs.harvard.edu/abs/2004AJ....127.3508B


Chapter 6 | Gauss's Law 283




total flux.)
92. Re-derive Gauss’s law for the gravitational field, with
g→ directed positively outward.


93. An infinite plate sheet of charge of surface chargedensity σ is shown below. What is the electric field at
a distance x from the sheet? Compare the result of thiscalculation with that of worked out in the text.


94. A spherical rubber balloon carries a total charge Qdistributed uniformly over its surface. At t = 0 , the radius
of the balloon is R. The balloon is then slowly inflated until


its radius reaches 2R at the time t0. Determine the electric
field due to this charge as a function of time (a) at thesurface of the balloon, (b) at the surface of radius R, and (c)at the surface of radius 2R. Ignore any effect on the electricfield due to the material of the balloon and assume that theradius increases uniformly with time.
95. Find the electric field of a large conducting platecontaining a net charge q. Let A be area of one side ofthe plate and h the thickness of the plate (see below). Thecharge on the metal plate will distribute mostly on the twoplanar sides and very little on the edges if the plate is thin.


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7 | ELECTRIC POTENTIAL


Figure 7.1 The energy released in a lightning strike is an excellent illustration of the vast quantities of energy that may bestored and released by an electric potential difference. In this chapter, we calculate just how much energy can be released in alightning strike and how this varies with the height of the clouds from the ground. (credit: Anthony Quintano)


Chapter Outline
7.1 Electric Potential Energy
7.2 Electric Potential and Potential Difference
7.3 Calculations of Electric Potential
7.4 Determining Field from Potential
7.5 Equipotential Surfaces and Conductors
7.6 Applications of Electrostatics


Introduction
In Electric Charges and Fields, we just scratched the surface (or at least rubbed it) of electrical phenomena. Two termscommonly used to describe electricity are its energy and voltage, which we show in this chapter is directly related to thepotential energy in a system.
We know, for example, that great amounts of electrical energy can be stored in batteries, are transmitted cross-country viacurrents through power lines, and may jump from clouds to explode the sap of trees. In a similar manner, at the molecularlevel, ions cross cell membranes and transfer information.
We also know about voltages associated with electricity. Batteries are typically a few volts, the outlets in your homefrequently produce 120 volts, and power lines can be as high as hundreds of thousands of volts. But energy and voltageare not the same thing. A motorcycle battery, for example, is small and would not be very successful in replacing a muchlarger car battery, yet each has the same voltage. In this chapter, we examine the relationship between voltage and electricalenergy, and begin to explore some of the many applications of electricity.


Chapter 7 | Electric Potential 285




7.1 | Electric Potential Energy
Learning Objectives


By the end of this section, you will be able to:
• Define the work done by an electric force
• Define electric potential energy
• Apply work and potential energy in systems with electric charges


When a free positive charge q is accelerated by an electric field, it is given kinetic energy (Figure 7.2). The process isanalogous to an object being accelerated by a gravitational field, as if the charge were going down an electrical hill where itselectric potential energy is converted into kinetic energy, although of course the sources of the forces are very different. Letus explore the work done on a charge q by the electric field in this process, so that we may develop a definition of electricpotential energy.


Figure 7.2 A charge accelerated by an electric field isanalogous to a mass going down a hill. In both cases, potentialenergy decreases as kinetic energy increases, – ΔU = ΔK .
Work is done by a force, but since this force is conservative, wecan write W = – ΔU .


The electrostatic or Coulomb force is conservative, which means that the work done on q is independent of the path taken, aswe will demonstrate later. This is exactly analogous to the gravitational force. When a force is conservative, it is possible todefine a potential energy associated with the force. It is usually easier to work with the potential energy (because it dependsonly on position) than to calculate the work directly.
To show this explicitly, consider an electric charge +q fixed at the origin and move another charge +Q toward q in such
a manner that, at each instant, the applied force F→ exactly balances the electric force F→ e on Q (Figure 7.3). The
work done by the applied force F→ on the charge Q changes the potential energy of Q. We call this potential energy the
electrical potential energy of Q.


Figure 7.3 Displacement of “test” charge Q in the presence offixed “source” charge q.


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The work W12 done by the applied force F→ when the particle moves from P1 to P2 may be calculated by
W12 = ∫


P1


P2
F


· d l


.


Since the applied force F→ balances the electric force F→ e on Q, the two forces have equal magnitude and opposite
directions. Therefore, the applied force is


F


= − Fe


= −
kqQ


r2
r̂ ,


where we have defined positive to be pointing away from the origin and r is the distance from the origin. The directions ofboth the displacement and the applied force in the system in Figure 7.3 are parallel, and thus the work done on the systemis positive.
We use the letterU to denote electric potential energy, which has units of joules (J). When a conservative force does negativework, the system gains potential energy. When a conservative force does positive work, the system loses potential energy,
ΔU = −W. In the system in Figure 7.3, the Coulomb force acts in the opposite direction to the displacement; therefore,
the work is negative. However, we have increased the potential energy in the two-charge system.
Example 7.1


Kinetic Energy of a Charged Particle
A +3.0-nC charge Q is initially at rest a distance of 10 cm ( r1 ) from a +5.0-nC charge q fixed at the origin
(Figure 7.4). Naturally, the Coulomb force accelerates Q away from q, eventually reaching 15 cm ( r2 ).


Figure 7.4 The charge Q is repelled by q, thus having workdone on it and gaining kinetic energy.
a. What is the work done by the electric field between r1 and r2 ?
b. How much kinetic energy does Q have at r2 ?


Strategy
Calculate the work with the usual definition. Since Q started from rest, this is the same as the kinetic energy.
Solution
Integrating force over distance, we obtain


W12 = ∫
r1


r2
F


· d r→ = ⌠
⌡r1


r2
kqQ


r2
dr =

⎣−


kqQ
r

⎦r1


r2


= kqQ⎡⎣
−1
r2


+ 1r1



= ⎛⎝8.99 × 10
9 Nm2 /C2⎞⎠



⎝5.0 × 10


−9 C⎞⎠

⎝3.0 × 10


−9 C⎞⎠



−1
0.15 m


+ 1
0.10 m





= 4.5 × 10−7 J.


This is also the value of the kinetic energy at r2.
Significance
Charge Q was initially at rest; the electric field of q did work on Q, so now Q has kinetic energy equal to the workdone by the electric field.


Chapter 7 | Electric Potential 287




7.1 Check Your Understanding If Q has a mass of 4.00 µg, what is the speed of Q at r2?


In this example, the work W done to accelerate a positive charge from rest is positive and results from a loss in U, or anegative ΔU . A value for U can be found at any point by taking one point as a reference and calculating the work needed
to move a charge to the other point.


Electric Potential Energy
Work W done to accelerate a positive charge from rest is positive and results from a loss in U, or a negative ΔU .
Mathematically,


(7.1)W = −ΔU.
Gravitational potential energy and electric potential energy are quite analogous. Potential energy accounts for work done bya conservative force and gives added insight regarding energy and energy transformation without the necessity of dealingwith the force directly. It is much more common, for example, to use the concept of electric potential energy than to dealwith the Coulomb force directly in real-world applications.
In polar coordinates with q at the origin and Q located at r, the displacement element vector is d l→ = r̂ dr and thus the
work becomes


W12 = −kqQ

⌡r1


r2
1
r2


r̂ · r̂ dr = kqQ 1r2
− kqQ 1r1


.


Notice that this result only depends on the endpoints and is otherwise independent of the path taken. To explore this further,compare path P1 to P2 with path P1P3P4P2 in Figure 7.5.


Figure 7.5 Two paths for displacement P1 to P2. The work
on segments P1P3 and P4P2 are zero due to the electrical
force being perpendicular to the displacement along these paths.Therefore, work on paths P1P2 and P1P3P4P2 are equal.


The segments P1P3 and P4P2 are arcs of circles centered at q. Since the force on Q points either toward or away from
q, no work is done by a force balancing the electric force, because it is perpendicular to the displacement along these arcs.Therefore, the only work done is along segment P3P4, which is identical to P1P2.
One implication of this work calculation is that if we were to go around the path P1P3P4P2P1, the net work would be
zero (Figure 7.6). Recall that this is how we determine whether a force is conservative or not. Hence, because the electric
force is related to the electric field by F→ = q E→ , the electric field is itself conservative. That is,


∮ E→ · d l→ = 0.


Note that Q is a constant.


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Figure 7.6 A closed path in an electric field. The net workaround this path is zero.


Another implication is that we may define an electric potential energy. Recall that the work done by a conservative forceis also expressed as the difference in the potential energy corresponding to that force. Therefore, the work Wref to bring a
charge from a reference point to a point of interest may be written as


Wref = ∫
rref


r
F


· d l


and, by Equation 7.1, the difference in potential energy (U2 − U1) of the test charge Q between the two points is
ΔU = −∫


rref


r
F


· d l


.


Therefore, we can write a general expression for the potential energy of two point charges (in spherical coordinates):
ΔU = −⌠


⌡rref


r
kqQ


r2
dr = −



⎣−


kqQ
r

⎦rref


r


= kqQ⎡⎣
1
r −


1
rref

⎦.


We may take the second term to be an arbitrary constant reference level, which serves as the zero reference:
U(r) = k


qQ
r − Uref.


A convenient choice of reference that relies on our common sense is that when the two charges are infinitely far apart,there is no interaction between them. (Recall the discussion of reference potential energy in Potential Energy andConservation of Energy (http://cnx.org/content/m58311/latest/) .) Taking the potential energy of this state to bezero removes the term Uref from the equation (just like when we say the ground is zero potential energy in a gravitational
potential energy problem), and the potential energy of Q when it is separated from q by a distance r assumes the form


(7.2)U(r) = kqQr (zero reference at r = ∞).


This formula is symmetrical with respect to q and Q, so it is best described as the potential energy of the two-charge system.
Example 7.2


Potential Energy of a Charged Particle
A +3.0-nC charge Q is initially at rest a distance of 10 cm ( r1 ) from a +5.0-nC charge q fixed at the origin
(Figure 7.7). Naturally, the Coulomb force accelerates Q away from q, eventually reaching 15 cm ( r2 ).


Chapter 7 | Electric Potential 289




7.2


Figure 7.7 The charge Q is repelled by q, thus having workdone on it and losing potential energy.


What is the change in the potential energy of the two-charge system from r1 to r2?
Strategy
Calculate the potential energy with the definition given above: ΔU12 = −∫


r1


r2
F


· d r→ . Since Q started from
rest, this is the same as the kinetic energy.
Solution
We have


ΔU12 = −∫
r1


r2
F


· d r→ = −⌠
⌡r1


r2
kqQ


r2
dr = −



⎣−


kqQ
r

⎦r1


r2


= kqQ⎡⎣
1
r2


− 1r1



= ⎛⎝8.99 × 10
9 Nm2 /C2⎞⎠



⎝5.0 × 10


−9 C⎞⎠

⎝3.0 × 10


−9 C⎞⎠



1
0.15 m


− 1
0.10 m





= −4.5 × 10−7 J.


Significance
The change in the potential energy is negative, as expected, and equal in magnitude to the change in kinetic energyin this system. Recall from Example 7.1 that the change in kinetic energy was positive.


Check Your Understanding What is the potential energy of Q relative to the zero reference at infinity at
r2 in the above example?


Due to Coulomb’s law, the forces due to multiple charges on a test charge Q superimpose; they may be calculatedindividually and then added. This implies that the work integrals and hence the resulting potential energies exhibit the samebehavior. To demonstrate this, we consider an example of assembling a system of four charges.
Example 7.3


Assembling Four Positive Charges
Find the amount of work an external agent must do in assembling four charges +2.0 µC,
+3.0 µC, + 4.0 µC, and +5.0 µC at the vertices of a square of side 1.0 cm, starting each charge from infinity
(Figure 7.8).


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Figure 7.8 How much work is needed to assemble this chargeconfiguration?


Strategy
We bring in the charges one at a time, giving them starting locations at infinity and calculating the work to bringthem in from infinity to their final location. We do this in order of increasing charge.
Solution
Step 1. First bring the +2.0-µC charge to the origin. Since there are no other charges at a finite distance from
this charge yet, no work is done in bringing it from infinity,


W1 = 0.


Step 2. While keeping the +2.0-µC charge fixed at the origin, bring the +3.0-µC charge to
(x, y, z) = (1.0 cm, 0, 0) (Figure 7.9). Now, the applied force must do work against the force exerted by the
+2.0-µC charge fixed at the origin. The work done equals the change in the potential energy of the +3.0-µC
charge:


W2 = k
q1q2
r12


=

⎝9.0 × 10


9 N ·m2


C2




⎝2.0 × 10


−6 C⎞⎠

⎝3.0 × 10


−6 C⎞⎠


1.0 × 10−2 m
= 5.4 J.


Figure 7.9 Step 2. Work W2 to bring the +3.0-µC charge
from infinity.


Step 3. While keeping the charges of +2.0 µC and +3.0 µC fixed in their places, bring in the +4.0-µC charge
to (x, y, z) = (1.0 cm, 1.0 cm, 0) (Figure 7.10). The work done in this step is


W3 = k
q1q3
r13


+ k
q2q3
r23


=

⎝9.0 × 10


9 N ·m2


C2







⎝2.0 × 10


−6 C⎞⎠

⎝4.0 × 10


−6 C⎞⎠


2 × 10−2 m
+

⎝3.0 × 10


−6 C⎞⎠

⎝4.0 × 10


−6 C⎞⎠


1.0 × 10−2 m





⎥ = 15.9 J.


Chapter 7 | Electric Potential 291




7.3


Figure 7.10 Step 3. The work W3 to bring the +4.0-µC
charge from infinity.


Step 4. Finally, while keeping the first three charges in their places, bring the +5.0-µC charge to
(x, y, z) = (0, 1.0 cm, 0) (Figure 7.11). The work done here is
W4 = kq4




q1
r14


+
q2
r24


+
q3
r34

⎦,


=

⎝9.0 × 10


9 N ·m2


C2



⎝5.0 × 10


−6 C⎞⎠





⎝2.0 × 10


−6 C⎞⎠


1.0 × 10−2 m
+

⎝3.0 × 10


−6 C⎞⎠


2 × 10−2 m
+

⎝4.0 × 10


−6 C⎞⎠


1.0 × 10−2 m





⎥ = 36.5 J.


Figure 7.11 Step 4. The work W4 to bring the +5.0-µC
charge from infinity.


Hence, the total work done by the applied force in assembling the four charges is equal to the sum of the work inbringing each charge from infinity to its final position:
WT = W1 +W2 +W3 +W4 = 0 + 5.4 J + 15.9 J + 36.5 J = 57.8 J.


Significance
The work on each charge depends only on its pairwise interactions with the other charges. No more complicatedinteractions need to be considered; the work on the third charge only depends on its interaction with the first andsecond charges, the interaction between the first and second charge does not affect the third.


Check Your Understanding Is the electrical potential energy of two point charges positive or negative ifthe charges are of the same sign? Opposite signs? How does this relate to the work necessary to bring thecharges into proximity from infinity?
Note that the electrical potential energy is positive if the two charges are of the same type, either positive or negative, andnegative if the two charges are of opposite types. This makes sense if you think of the change in the potential energy ΔU
as you bring the two charges closer or move them farther apart. Depending on the relative types of charges, you may have


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to work on the system or the system would do work on you, that is, your work is either positive or negative. If you haveto do positive work on the system (actually push the charges closer), then the energy of the system should increase. If youbring two positive charges or two negative charges closer, you have to do positive work on the system, which raises theirpotential energy. Since potential energy is proportional to 1/r, the potential energy goes up when r goes down between twopositive or two negative charges.
On the other hand, if you bring a positive and a negative charge nearer, you have to do negative work on the system (thecharges are pulling you), which means that you take energy away from the system. This reduces the potential energy. Sincepotential energy is negative in the case of a positive and a negative charge pair, the increase in 1/rmakes the potential energymore negative, which is the same as a reduction in potential energy.
The result from Example 7.1 may be extended to systems with any arbitrary number of charges. In this case, it is mostconvenient to write the formula as


(7.3)
W12 ⋯ N =


k
2

i


N



j


N qiq j
ri j


for i ≠ j.


The factor of 1/2 accounts for adding each pair of charges twice.
7.2 | Electric Potential and Potential Difference


Learning Objectives
By the end of this section, you will be able to:
• Define electric potential, voltage, and potential difference
• Define the electron-volt
• Calculate electric potential and potential difference from potential energy and electric field
• Describe systems in which the electron-volt is a useful unit
• Apply conservation of energy to electric systems


Recall that earlier we defined electric field to be a quantity independent of the test charge in a given system, which wouldnonetheless allow us to calculate the force that would result on an arbitrary test charge. (The default assumption in theabsence of other information is that the test charge is positive.) We briefly defined a field for gravity, but gravity is alwaysattractive, whereas the electric force can be either attractive or repulsive. Therefore, although potential energy is perfectlyadequate in a gravitational system, it is convenient to define a quantity that allows us to calculate the work on a charge
independent of the magnitude of the charge. Calculating the work directly may be difficult, since W = F→ · d→ and the
direction and magnitude of F→ can be complex for multiple charges, for odd-shaped objects, and along arbitrary paths. But
we do know that because F→ = q E→ , the work, and hence ΔU, is proportional to the test charge q. To have a physical
quantity that is independent of test charge, we define electric potential V (or simply potential, since electric is understood)to be the potential energy per unit charge:


Electric Potential
The electric potential energy per unit charge is


(7.4)V = Uq .


Since U is proportional to q, the dependence on q cancels. Thus, V does not depend on q. The change in potential energy
ΔU is crucial, so we are concerned with the difference in potential or potential difference ΔV between two points, where


Chapter 7 | Electric Potential 293




ΔV = VB − VA =
ΔU
q .


Electric Potential Difference
The electric potential difference between points A and B, VB − VA, is defined to be the change in potential energy
of a charge q moved from A to B, divided by the charge. Units of potential difference are joules per coulomb, given thename volt (V) after Alessandro Volta.


1 V = 1 J/C


The familiar term voltage is the common name for electric potential difference. Keep in mind that whenever a voltage isquoted, it is understood to be the potential difference between two points. For example, every battery has two terminals, andits voltage is the potential difference between them. More fundamentally, the point you choose to be zero volts is arbitrary.This is analogous to the fact that gravitational potential energy has an arbitrary zero, such as sea level or perhaps a lecturehall floor. It is worthwhile to emphasize the distinction between potential difference and electrical potential energy.
Potential Difference and Electrical Potential Energy
The relationship between potential difference (or voltage) and electrical potential energy is given by


(7.5)ΔV = ΔUq or ΔU = qΔV .


Voltage is not the same as energy. Voltage is the energy per unit charge. Thus, a motorcycle battery and a car battery can bothhave the same voltage (more precisely, the same potential difference between battery terminals), yet one stores much moreenergy than the other because ΔU = qΔV . The car battery can move more charge than the motorcycle battery, although
both are 12-V batteries.
Example 7.4


Calculating Energy
You have a 12.0-V motorcycle battery that can move 5000 C of charge, and a 12.0-V car battery that can move60,000 C of charge. How much energy does each deliver? (Assume that the numerical value of each charge isaccurate to three significant figures.)
Strategy
To say we have a 12.0-V battery means that its terminals have a 12.0-V potential difference. When such a batterymoves charge, it puts the charge through a potential difference of 12.0 V, and the charge is given a change inpotential energy equal to ΔU = qΔV . To find the energy output, we multiply the charge moved by the potential
difference.
Solution
For the motorcycle battery, q = 5000 C and ΔV = 12.0 V . The total energy delivered by the motorcycle
battery is


ΔUcycle = (5000 C)(12.0 V) = (5000 C)(12.0 J/C) = 6.00 × 10
4 J.


Similarly, for the car battery, q = 60,000 C and
ΔUcar = (60,000 C)(12.0 V) = 7.20 × 10


5 J.


Significance
Voltage and energy are related, but they are not the same thing. The voltages of the batteries are identical, but theenergy supplied by each is quite different. A car battery has a much larger engine to start than a motorcycle. Notealso that as a battery is discharged, some of its energy is used internally and its terminal voltage drops, such as


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7.4


when headlights dim because of a depleted car battery. The energy supplied by the battery is still calculated as inthis example, but not all of the energy is available for external use.
Check Your Understanding How much energy does a 1.5-V AAA battery have that can move 100 C?


Note that the energies calculated in the previous example are absolute values. The change in potential energy for the batteryis negative, since it loses energy. These batteries, like many electrical systems, actually move negative charge—electrons inparticular. The batteries repel electrons from their negative terminals (A) through whatever circuitry is involved and attractthem to their positive terminals (B), as shown in Figure 7.12. The change in potential is ΔV = VB − VA = + 12 V and
the charge q is negative, so that ΔU = qΔV is negative, meaning the potential energy of the battery has decreased when q
has moved from A to B.


Figure 7.12 A battery moves negative charge from itsnegative terminal through a headlight to its positive terminal.Appropriate combinations of chemicals in the battery separatecharges so that the negative terminal has an excess of negativecharge, which is repelled by it and attracted to the excesspositive charge on the other terminal. In terms of potential, thepositive terminal is at a higher voltage than the negativeterminal. Inside the battery, both positive and negative chargesmove.


Example 7.5
How Many Electrons Move through a Headlight Each Second?
When a 12.0-V car battery powers a single 30.0-W headlight, how many electrons pass through it each second?
Strategy
To find the number of electrons, we must first find the charge that moves in 1.00 s. The charge moved is relatedto voltage and energy through the equations ΔU = qΔV . A 30.0-W lamp uses 30.0 joules per second. Since the
battery loses energy, we have ΔU = −30 J and, since the electrons are going from the negative terminal to the
positive, we see that ΔV = +12.0 V.
Solution
To find the charge q moved, we solve the equation ΔU = qΔV :


q = ΔU
ΔV


.


Entering the values for ΔU and ΔV , we get


Chapter 7 | Electric Potential 295




7.5


q = −30.0 J
+12.0 V


= −30.0 J
+12.0 J/C


= −2.50 C.


The number of electrons ne is the total charge divided by the charge per electron. That is,
ne = −2.50 C


−1.60 × 10−19 C/e−
= 1.56 × 1019 electrons.


Significance
This is a very large number. It is no wonder that we do not ordinarily observe individual electrons with so manybeing present in ordinary systems. In fact, electricity had been in use for many decades before it was determinedthat the moving charges in many circumstances were negative. Positive charge moving in the opposite direction ofnegative charge often produces identical effects; this makes it difficult to determine which is moving or whetherboth are moving.


Check Your Understanding How many electrons would go through a 24.0-W lamp?


The Electron-Volt
The energy per electron is very small in macroscopic situations like that in the previous example—a tiny fraction of a joule.But on a submicroscopic scale, such energy per particle (electron, proton, or ion) can be of great importance. For example,even a tiny fraction of a joule can be great enough for these particles to destroy organic molecules and harm living tissue.The particle may do its damage by direct collision, or it may create harmful X-rays, which can also inflict damage. It isuseful to have an energy unit related to submicroscopic effects.
Figure 7.13 shows a situation related to the definition of such an energy unit. An electron is accelerated between twocharged metal plates, as it might be in an old-model television tube or oscilloscope. The electron gains kinetic energy thatis later converted into another form—light in the television tube, for example. (Note that in terms of energy, “downhill” forthe electron is “uphill” for a positive charge.) Since energy is related to voltage by ΔU = qΔV , we can think of the joule
as a coulomb-volt.


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Figure 7.13 A typical electron gun accelerates electrons using a potential difference between two separated metal plates.By conservation of energy, the kinetic energy has to equal the change in potential energy, so KE = qV . The energy of the
electron in electron-volts is numerically the same as the voltage between the plates. For example, a 5000-V potentialdifference produces 5000-eV electrons. The conceptual construct, namely two parallel plates with a hole in one, is shown in(a), while a real electron gun is shown in (b).


Electron-Volt
On the submicroscopic scale, it is more convenient to define an energy unit called the electron-volt (eV), which is theenergy given to a fundamental charge accelerated through a potential difference of 1 V. In equation form,


1 eV = (1.60 × 10−19 C)(1 V) = (1.60 × 10−19 C)(1 J/C) = 1.60 × 10−19 J.


An electron accelerated through a potential difference of 1 V is given an energy of 1 eV. It follows that an electronaccelerated through 50 V gains 50 eV. A potential difference of 100,000 V (100 kV) gives an electron an energy of 100,000eV (100 keV), and so on. Similarly, an ion with a double positive charge accelerated through 100 V gains 200 eV ofenergy. These simple relationships between accelerating voltage and particle charges make the electron-volt a simple andconvenient energy unit in such circumstances.
The electron-volt is commonly employed in submicroscopic processes—chemical valence energies and molecular andnuclear binding energies are among the quantities often expressed in electron-volts. For example, about 5 eV of energyis required to break up certain organic molecules. If a proton is accelerated from rest through a potential differenceof 30 kV, it acquires an energy of 30 keV (30,000 eV) and can break up as many as 6000 of these molecules
(30,000 eV ÷ 5 eV per molecule = 6000 molecules). Nuclear decay energies are on the order of 1 MeV (1,000,000 eV)
per event and can thus produce significant biological damage.
Conservation of Energy
The total energy of a system is conserved if there is no net addition (or subtraction) due to work or heat transfer. Forconservative forces, such as the electrostatic force, conservation of energy states that mechanical energy is a constant.


Chapter 7 | Electric Potential 297




7.6


Mechanical energy is the sum of the kinetic energy and potential energy of a system; that is, K + U = constant. A loss of
U for a charged particle becomes an increase in its K. Conservation of energy is stated in equation form as


K + U = constant


or
Ki + Ui = Kf + Uf


where i and f stand for initial and final conditions. As we have found many times before, considering energy can give usinsights and facilitate problem solving.
Example 7.6


Electrical Potential Energy Converted into Kinetic Energy
Calculate the final speed of a free electron accelerated from rest through a potential difference of 100 V. (Assumethat this numerical value is accurate to three significant figures.)
Strategy
We have a system with only conservative forces. Assuming the electron is accelerated in a vacuum, andneglecting the gravitational force (we will check on this assumption later), all of the electrical potential energyis converted into kinetic energy. We can identify the initial and final forms of energy to be
Ki = 0, Kf =


1
2
mv2, Ui = qV , Uf = 0.


Solution
Conservation of energy states that


Ki + Ui = Kf + Uf.


Entering the forms identified above, we obtain
qV = mv


2


2
.


We solve this for v:
v =


2qV
m .


Entering values for q, V, and m gives
v = 2(−1.60 × 10


−19 C)( − 100 J/C)


9.11 × 10−31 kg
= 5.93 × 106 m/s.


Significance
Note that both the charge and the initial voltage are negative, as in Figure 7.13. From the discussion of electriccharge and electric field, we know that electrostatic forces on small particles are generally very large comparedwith the gravitational force. The large final speed confirms that the gravitational force is indeed negligible here.The large speed also indicates how easy it is to accelerate electrons with small voltages because of their verysmall mass. Voltages much higher than the 100 V in this problem are typically used in electron guns. Thesehigher voltages produce electron speeds so great that effects from special relativity must be taken into account andhence are reserved for a later chapter (Relativity (http://cnx.org/content/m58555/latest/) ). That is why weconsider a low voltage (accurately) in this example.


Check Your Understanding How would this example change with a positron? A positron is identical toan electron except the charge is positive.


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Voltage and Electric Field
So far, we have explored the relationship between voltage and energy. Now we want to explore the relationship between
voltage and electric field. We will start with the general case for a non-uniform E→ field. Recall that our general formula
for the potential energy of a test charge q at point P relative to reference point R is


UP = −∫
R


P
F


· d l


.


When we substitute in the definition of electric field ( E→ = F→ /q), this becomes
UP = −q∫


R


P
E


· d l


.


Applying our definition of potential (V = U/q) to this potential energy, we find that, in general,


(7.6)
VP = −∫


R


P
E


· d l


.


From our previous discussion of the potential energy of a charge in an electric field, the result is independent of the pathchosen, and hence we can pick the integral path that is most convenient.
Consider the special case of a positive point charge q at the origin. To calculate the potential caused by q at a distance r fromthe origin relative to a reference of 0 at infinity (recall that we did the same for potential energy), let P = r and R = ∞,
with d l→ = d r→ = r̂ dr and use E→ = kq


r2
r̂ . When we evaluate the integral


VP = −∫
R


P
E


· d l


for this system, we have
Vr = −⌠


⌡∞


r
kq


r2
r̂ · r̂ dr,


which simplifies to
Vr = −⌠


⌡∞


r
kq


r2
dr =


kq
r −


kq
∞ =


kq
r .


This result,
Vr =


kq
r


is the standard form of the potential of a point charge. This will be explored further in the next section.
To examine another interesting special case, suppose a uniform electric field E→ is produced by placing a potential
difference (or voltage) ΔV across two parallel metal plates, labeled A and B (Figure 7.14). Examining this situation will
tell us what voltage is needed to produce a certain electric field strength. It will also reveal a more fundamental relationshipbetween electric potential and electric field.


Chapter 7 | Electric Potential 299




Figure 7.14 The relationship between V and E for parallelconducting plates is E = V / d . (Note that ΔV = VAB in
magnitude. For a charge that is moved from plate A at higherpotential to plate B at lower potential, a minus sign needs to beincluded as follows: −ΔV = VA − VB = VAB. )


From a physicist’s point of view, either ΔV or E→ can be used to describe any interaction between charges. However,
ΔV is a scalar quantity and has no direction, whereas E→ is a vector quantity, having both magnitude and direction. (Note
that the magnitude of the electric field, a scalar quantity, is represented by E.) The relationship between ΔV and E→ is
revealed by calculating the work done by the electric force in moving a charge from point A to point B. But, as noted earlier,arbitrary charge distributions require calculus. We therefore look at a uniform electric field as an interesting special case.
The work done by the electric field in Figure 7.14 to move a positive charge q from A, the positive plate, higher potential,to B, the negative plate, lower potential, is


W = −ΔU = −qΔV .


The potential difference between points A and B is
−ΔV = −(VB − VA) = VA − VB = VAB.


Entering this into the expression for work yields
W = qVAB.


Work is W = F→ · d→ = Fd cos θ ; here cos θ = 1 , since the path is parallel to the field. Thus, W = Fd . Since
F = qE , we see that W = qEd .
Substituting this expression for work into the previous equation gives


qEd = qVAB.


The charge cancels, so we obtain for the voltage between points A and B


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VAB = Ed


E =
VAB
d





⎬(uniform E-field on y)


where d is the distance from A to B, or the distance between the plates in Figure 7.14. Note that this equation implies thatthe units for electric field are volts per meter. We already know the units for electric field are newtons per coulomb; thus,the following relation among units is valid:
1 N / C = 1 V / m.


Furthermore, we may extend this to the integral form. Substituting Equation 7.5 into our definition for the potentialdifference between points A and B, we obtain
VAB = VB − VA = −∫


R


B
E


· d l


+ ∫
R


A
E


· d l


which simplifies to
VB − VA = −∫


A


B
E


· d l


.


As a demonstration, from this we may calculate the potential difference between two points (A and B) equidistant from apoint charge q at the origin, as shown in Figure 7.15.


Figure 7.15 The arc for calculating the potential differencebetween two points that are equidistant from a point charge atthe origin.


To do this, we integrate around an arc of the circle of constant radius r between A and B, which means we let
d l


= rφ̂dφ, while using E→ = kq
r2


r̂ . Thus,


(7.7)
ΔVAB = VB − VA = −∫


A


B
E


· d l


for this system becomes
VB − VA = −



⌡A


B
kq


r2
r̂ · r φ̂dφ.


However, r̂ · φ̂ = 0 and therefore
VB − VA = 0.


This result, that there is no difference in potential along a constant radius from a point charge, will come in handy when wemap potentials.


Chapter 7 | Electric Potential 301




Example 7.7
What Is the Highest Voltage Possible between Two Plates?
Dry air can support a maximum electric field strength of about 3.0 × 106 V/m. Above that value, the field
creates enough ionization in the air to make the air a conductor. This allows a discharge or spark that reduces thefield. What, then, is the maximum voltage between two parallel conducting plates separated by 2.5 cm of dry air?
Strategy
We are given the maximum electric field E between the plates and the distance d between them. We can use theequation VAB = Ed to calculate the maximum voltage.
Solution
The potential difference or voltage between the plates is


VAB = Ed.


Entering the given values for E and d gives
VAB = (3.0 × 10


6 V/m)(0.025 m) = 7.5 × 104 V


or
VAB = 75 kV.


(The answer is quoted to only two digits, since the maximum field strength is approximate.)
Significance
One of the implications of this result is that it takes about 75 kV to make a spark jump across a 2.5-cm (1-in.)gap, or 150 kV for a 5-cm spark. This limits the voltages that can exist between conductors, perhaps on a powertransmission line. A smaller voltage can cause a spark if there are spines on the surface, since sharp pointshave larger field strengths than smooth surfaces. Humid air breaks down at a lower field strength, meaning thata smaller voltage will make a spark jump through humid air. The largest voltages can be built up with staticelectricity on dry days (Figure 7.16).


Figure 7.16 A spark chamber is used to trace the paths of high-energy particles. Ionization created by the particles asthey pass through the gas between the plates allows a spark to jump. The sparks are perpendicular to the plates,following electric field lines between them. The potential difference between adjacent plates is not high enough tocause sparks without the ionization produced by particles from accelerator experiments (or cosmic rays). This form ofdetector is now archaic and no longer in use except for demonstration purposes. (credit b: modification of work by JackCollins)


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Example 7.8
Field and Force inside an Electron Gun
An electron gun (Figure 7.13) has parallel plates separated by 4.00 cm and gives electrons 25.0 keV of energy.(a) What is the electric field strength between the plates? (b) What force would this field exert on a piece of plasticwith a 0.500-µC charge that gets between the plates?
Strategy
Since the voltage and plate separation are given, the electric field strength can be calculated directly from
the expression E = VAB


d
. Once we know the electric field strength, we can find the force on a charge by


using F→ = q E→ . Since the electric field is in only one direction, we can write this equation in terms of the
magnitudes, F = qE .
Solutiona. The expression for the magnitude of the electric field between two uniform metal plates is


E =
VAB
d


.


Since the electron is a single charge and is given 25.0 keV of energy, the potential difference must be 25.0kV. Entering this value for VAB and the plate separation of 0.0400 m, we obtain
E = 25.0 kV


0.0400 m
= 6.25 × 105 V/m.


b. The magnitude of the force on a charge in an electric field is obtained from the equation
F = qE.


Substituting known values gives
F = (0.500 × 10−6 C)(6.25 × 105 V/m) = 0.313 N.


Significance
Note that the units are newtons, since 1 V/m = 1 N/C . Because the electric field is uniform between the plates,
the force on the charge is the same no matter where the charge is located between the plates.


Example 7.9
Calculating Potential of a Point Charge
Given a point charge q = + 2.0 nC at the origin, calculate the potential difference between point P1 a distance
a = 4.0 cm from q, and P2 a distance b = 12.0 cm from q, where the two points have an angle of φ = 24°
between them (Figure 7.17).


Chapter 7 | Electric Potential 303




7.7


Figure 7.17 Find the difference in potential between P1 and
P2 .


Strategy
Do this in two steps. The first step is to use VB − VA = −∫


A


B
E


· d l
→ and let A = a = 4.0 cm and


B = b = 12.0 cm, with d l→ = d r→ = r̂ dr and E→ = kq
r2


r̂ . Then perform the integral. The second step is
to integrate VB − VA = −∫


A


B
E


· d l
→ around an arc of constant radius r, which means we let d l→ = rφ̂dφ


with limits 0 ≤ φ ≤ 24°, still using E→ = kq
r2


r̂ . Then add the two results together.
Solution
For the first part, VB − VA = −∫


A


B
E


· d l
→ for this system becomes Vb − Va = −⌠⌡a


b
kq


r2
r̂ · r̂ dr which


computes to
ΔV = −⌠


⌡a


b
kq


r2
dr = kq⎡⎣


1
a −


1
b



= ⎛⎝8.99 × 10
9 Nm2 /C2⎞⎠



⎝2.0 × 10


−9 C⎞⎠



1
0.040 m


− 1
0.12 m



⎦ = 300 V.


For the second step, VB − VA = −∫
A


B
E


· d l
→ becomes ΔV = −⌠


⌡0


24°
kq


r2
r̂ · r φ̂dφ , but r̂ · φ̂ = 0 and


therefore ΔV = 0. Adding the two parts together, we get 300 V.
Significance
We have demonstrated the use of the integral form of the potential difference to obtain a numerical result. Noticethat, in this particular system, we could have also used the formula for the potential due to a point charge at thetwo points and simply taken the difference.


Check Your Understanding From the examples, how does the energy of a lightning strike vary with theheight of the clouds from the ground? Consider the cloud-ground system to be two parallel plates.


Before presenting problems involving electrostatics, we suggest a problem-solving strategy to follow for this topic.
Problem-Solving Strategy: Electrostatics


1. Examine the situation to determine if static electricity is involved; this may concern separated stationary


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charges, the forces among them, and the electric fields they create.
2. Identify the system of interest. This includes noting the number, locations, and types of charges involved.
3. Identify exactly what needs to be determined in the problem (identify the unknowns). A written list is useful.Determine whether the Coulomb force is to be considered directly—if so, it may be useful to draw a free-bodydiagram, using electric field lines.
4. Make a list of what is given or can be inferred from the problem as stated (identify the knowns). It is importantto distinguish the Coulomb force F from the electric field E, for example.
5. Solve the appropriate equation for the quantity to be determined (the unknown) or draw the field lines asrequested.
6. Examine the answer to see if it is reasonable: Does it make sense? Are units correct and the numbers involvedreasonable?


7.3 | Calculations of Electric Potential
Learning Objectives


By the end of this section, you will be able to:
• Calculate the potential due to a point charge
• Calculate the potential of a system of multiple point charges
• Describe an electric dipole
• Define dipole moment
• Calculate the potential of a continuous charge distribution


Point charges, such as electrons, are among the fundamental building blocks of matter. Furthermore, spherical chargedistributions (such as charge on a metal sphere) create external electric fields exactly like a point charge. The electricpotential due to a point charge is, thus, a case we need to consider.
We can use calculus to find the work needed to move a test charge q from a large distance away to a distance of r from apoint charge q. Noting the connection between work and potential W = −qΔV , as in the last section, we can obtain the
following result.


Electric Potential V of a Point Charge
The electric potential V of a point charge is given by


(7.8)
V =


kq
r (point charge)


where k is a constant equal to 9.0 × 109 N ·m2 /C2.


The potential at infinity is chosen to be zero. Thus, V for a point charge decreases with distance, whereas E→ for a point
charge decreases with distance squared:


E = Fqt
=


kq


r2
.


Recall that the electric potential V is a scalar and has no direction, whereas the electric field E→ is a vector. To find the
voltage due to a combination of point charges, you add the individual voltages as numbers. To find the total electric field,you must add the individual fields as vectors, taking magnitude and direction into account. This is consistent with the fact
that V is closely associated with energy, a scalar, whereas E→ is closely associated with force, a vector.


Chapter 7 | Electric Potential 305




Example 7.10
What Voltage Is Produced by a Small Charge on a Metal Sphere?
Charges in static electricity are typically in the nanocoulomb (nC) to microcoulomb (µC) range. What is the
voltage 5.00 cm away from the center of a 1-cm-diameter solid metal sphere that has a –3.00-nC static charge?
Strategy
As we discussed in Electric Charges and Fields, charge on a metal sphere spreads out uniformly andproduces a field like that of a point charge located at its center. Thus, we can find the voltage using the equation
V =


kq
r .


Solution
Entering known values into the expression for the potential of a point charge, we obtain


V = k
q
r =

⎝8.99 × 10


9 N ·m2 /C2⎞⎠


−3.00 × 10−9 C
5.00 × 10−2 m



⎠ = −539 V.


Significance
The negative value for voltage means a positive charge would be attracted from a larger distance, since thepotential is lower (more negative) than at larger distances. Conversely, a negative charge would be repelled, asexpected.


Example 7.11
What Is the Excess Charge on a Van de Graaff Generator?
A demonstration Van de Graaff generator has a 25.0-cm-diameter metal sphere that produces a voltage of 100 kVnear its surface (Figure 7.18). What excess charge resides on the sphere? (Assume that each numerical valuehere is shown with three significant figures.)


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7.8


Figure 7.18 The voltage of this demonstration Van de Graaffgenerator is measured between the charged sphere and ground.Earth’s potential is taken to be zero as a reference. The potentialof the charged conducting sphere is the same as that of an equalpoint charge at its center.


Strategy
The potential on the surface is the same as that of a point charge at the center of the sphere, 12.5 cm away. (Theradius of the sphere is 12.5 cm.) We can thus determine the excess charge using the equation


V =
kq
r .


Solution
Solving for q and entering known values gives


q = rV
k


=
(0.125 m)⎛⎝100 × 10


3 V⎞⎠


8.99 × 109 N ·m2 /C2
= 1.39 × 10−6 C = 1.39 µC.


Significance
This is a relatively small charge, but it produces a rather large voltage. We have another indication here that it isdifficult to store isolated charges.


Check Your Understanding What is the potential inside the metal sphere in Example 7.10?


The voltages in both of these examples could be measured with a meter that compares the measured potential with groundpotential. Ground potential is often taken to be zero (instead of taking the potential at infinity to be zero). It is the potentialdifference between two points that is of importance, and very often there is a tacit assumption that some reference point,such as Earth or a very distant point, is at zero potential. As noted earlier, this is analogous to taking sea level as h = 0
when considering gravitational potential energy Ug = mgh .


Chapter 7 | Electric Potential 307




Systems of Multiple Point Charges
Just as the electric field obeys a superposition principle, so does the electric potential. Consider a system consisting of Ncharges q1, q2, …, qN. What is the net electric potential V at a space point P from these charges? Each of these charges
is a source charge that produces its own electric potential at point P, independent of whatever other changes may be doing.Let V1, V2, …, VN be the electric potentials at P produced by the charges q1, q2, …, qN, respectively. Then, the net
electric potential VP at that point is equal to the sum of these individual electric potentials. You can easily show this by
calculating the potential energy of a test charge when you bring the test charge from the reference point at infinity to pointP:


VP = V1 + V2 +⋯+ VN = ∑
1


N


Vi.


Note that electric potential follows the same principle of superposition as electric field and electric potential energy. Toshow this more explicitly, note that a test charge qi at the point P in space has distances of r1, r2, …, rN from the N
charges fixed in space above, as shown in Figure 7.19. Using our formula for the potential of a point charge for each ofthese (assumed to be point) charges, we find that


(7.9)
VP = ∑


1


N


k
qi
ri


= k∑
1


N
qi
ri
.


Therefore, the electric potential energy of the test charge is
UP = qtVP = qt k∑


1


N
qi
ri
,


which is the same as the work to bring the test charge into the system, as found in the first section of the chapter.


Figure 7.19 Notation for direct distances from charges to aspace point P.
The Electric Dipole
An electric dipole is a system of two equal but opposite charges a fixed distance apart. This system is used to model manyreal-world systems, including atomic and molecular interactions. One of these systems is the water molecule, under certaincircumstances. These circumstances are met inside a microwave oven, where electric fields with alternating directions makethe water molecules change orientation. This vibration is the same as heat at the molecular level.
Example 7.12


Electric Potential of a Dipole
Consider the dipole in Figure 7.20 with the charge magnitude of q = 3.0 nC and separation distance
d = 4.0 cm. What is the potential at the following locations in space? (a) (0, 0, 1.0 cm); (b) (0, 0, –5.0 cm); (c)
(3.0 cm, 0, 2.0 cm).


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7.9


Figure 7.20 A general diagram of an electric dipole, and thenotation for the distances from the individual charges to a pointP in space.


Strategy
Apply VP = k∑


1


N
qi
ri


to each of these three points.
Solution


a. VP = k∑
1


N
qi
ri


= (9.0 × 109 N ·m2 /C2)⎛⎝
3.0 nC
0.010 m


− 3.0 nC
0.030 m



⎠ = 1.8 × 10


3 V


b. VP = k∑
1


N
qi
ri


= (9.0 × 109 N ·m2 /C2)⎛⎝
3.0 nC
0.070 m


− 3.0 nC
0.030 m



⎠ = −5.1 × 10


2 V


c. VP = k∑
1


N
qi
ri


= (9.0 × 109 N ·m2 /C2)⎛⎝
3.0 nC
0.030 m


− 3.0 nC
0.050 m



⎠ = 3.6 × 10


2 V


Significance
Note that evaluating potential is significantly simpler than electric field, due to potential being a scalar instead ofa vector.


Check Your Understanding What is the potential on the x-axis? The z-axis?


Now let us consider the special case when the distance of the point P from the dipole is much greater than the distancebetween the charges in the dipole, r ≫ d; for example, when we are interested in the electric potential due to a polarized
molecule such as a water molecule. This is not so far (infinity) that we can simply treat the potential as zero, but the distanceis great enough that we can simplify our calculations relative to the previous example.
We start by noting that in Figure 7.21 the potential is given by


Chapter 7 | Electric Potential 309




VP = V+ + V− = k


q
r+



q
r−



where
r± = x


2 + ⎛⎝z ∓
d
2



2
.


Figure 7.21 A general diagram of an electric dipole, and thenotation for the distances from the individual charges to a point Pin space.


This is still the exact formula. To take advantage of the fact that r ≫ d, we rewrite the radii in terms of polar coordinates,
with x = r sin θ and z = r cos θ . This gives us


r± = r
2 sin2 θ + ⎛⎝r cos θ ∓


d
2



2
.


We can simplify this expression by pulling r out of the root,
r± = r sin


2 θ + ⎛⎝cos θ ∓
d
2r



2


and then multiplying out the parentheses
r± = r sin


2 θ + cos2 θ ∓ cos θdr +


d
2r



2
= r 1 ∓ cos θdr +




d
2r



2
.


The last term in the root is small enough to be negligible (remember r ≫ d, and hence (d/r)2 is extremely small,
effectively zero to the level we will probably be measuring), leaving us with


r± = r 1 ∓ cos θ
d
r .


Using the binomial approximation (a standard result from the mathematics of series, when α is small)
1


1 ∓ α
≈ 1 ± α


2


and substituting this into our formula for VP , we get
VP = k




q
r

⎝1 +


d cos θ
2r

⎠−


q
r

⎝1 −


d cos θ
2r



⎦ = k


qd cos θ


r2
.


This may be written more conveniently if we define a new quantity, the electric dipole moment,


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(7.10)p→ = q d→ ,


where these vectors point from the negative to the positive charge. Note that this has magnitude qd. This quantity allows usto write the potential at point P due to a dipole at the origin as


(7.11)
VP = k


p→ · r̂
r2


.


A diagram of the application of this formula is shown in Figure 7.22.


Figure 7.22 The geometry for the application of the potentialof a dipole.


There are also higher-order moments, for quadrupoles, octupoles, and so on. You will see these in future classes.
Potential of Continuous Charge Distributions
We have been working with point charges a great deal, but what about continuous charge distributions? Recall fromEquation 7.9 that


VP = k∑
qi
ri
.


We may treat a continuous charge distribution as a collection of infinitesimally separated individual points. This yields theintegral


(7.12)
VP = k∫


dq
r


for the potential at a point P. Note that r is the distance from each individual point in the charge distribution to the point P.As we saw in Electric Charges and Fields, the infinitesimal charges are given by
dq =




λ dl (one dimension)
σ dA (two dimensions)
ρ dV (three dimensions)


where λ is linear charge density, σ is the charge per unit area, and ρ is the charge per unit volume.


Chapter 7 | Electric Potential 311




Example 7.13
Potential of a Line of Charge
Find the electric potential of a uniformly charged, nonconducting wire with linear density λ (coulomb/meter)
and length L at a point that lies on a line that divides the wire into two equal parts.
Strategy
To set up the problem, we choose Cartesian coordinates in such a way as to exploit the symmetry in the problemas much as possible. We place the origin at the center of the wire and orient the y-axis along the wire so that theends of the wire are at y = ± L/2 . The field point P is in the xy-plane and since the choice of axes is up to us,
we choose the x-axis to pass through the field point P, as shown in Figure 7.23.


Figure 7.23 We want to calculate the electric potential due toa line of charge.


Solution
Consider a small element of the charge distribution between y and y + dy . The charge in this cell is dq = λ dy
and the distance from the cell to the field point P is x2 + y2. Therefore, the potential becomes


VP = k



⎮dqr = k⌠


⌡−L/2


L/2
λdy


x2 + y2
= kλ⎡⎣ln



⎝y + y


2 + x2⎞⎠

⎦−L/2


L/2


= kλ



⎢ln



⎜⎛⎝
L
2

⎠+


L
2



2
+ x2



⎟− ln



⎜⎛⎝−


L
2

⎠+

⎝−


L
2



2
+ x2









= kλln



⎢ L + L


2 + 4x2


−L + L2 + 4x2





⎥.


Significance
Note that this was simpler than the equivalent problem for electric field, due to the use of scalar quantities. Recallthat we expect the zero level of the potential to be at infinity, when we have a finite charge. To examine this, wetake the limit of the above potential as x approaches infinity; in this case, the terms inside the natural log approachone, and hence the potential approaches zero in this limit. Note that we could have done this problem equivalentlyin cylindrical coordinates; the only effect would be to substitute r for x and z for y.


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Example 7.14
Potential Due to a Ring of Charge
A ring has a uniform charge density λ , with units of coulomb per unit meter of arc. Find the electric potential at
a point on the axis passing through the center of the ring.
Strategy
We use the same procedure as for the charged wire. The difference here is that the charge is distributed on a circle.We divide the circle into infinitesimal elements shaped as arcs on the circle and use cylindrical coordinates shownin Figure 7.24.


Figure 7.24 We want to calculate the electric potential due toa ring of charge.


Solution
A general element of the arc between θ and θ + dθ is of length Rdθ and therefore contains a charge equal to
λRdθ. The element is at a distance of z2 + R2 from P, and therefore the potential is


VP = k∫
dq
r = k

⌡0



λRdθ


z2 + R2
= kλR


z2 + R2


0



dθ = 2πkλR


z2 + R2
= k


qtot


z2 + R2
.


Significance
This result is expected because every element of the ring is at the same distance from point P. The net potential
at P is that of the total charge placed at the common distance, z2 + R2 .


Example 7.15
Potential Due to a Uniform Disk of Charge
A disk of radius R has a uniform charge density σ , with units of coulomb meter squared. Find the electric
potential at any point on the axis passing through the center of the disk.
Strategy
We divide the disk into ring-shaped cells, and make use of the result for a ring worked out in the previousexample, then integrate over r in addition to θ . This is shown in Figure 7.25.


Chapter 7 | Electric Potential 313




Figure 7.25 We want to calculate the electric potential due toa disk of charge.


Solution
An infinitesimal width cell between cylindrical coordinates r and r + dr shown in Figure 7.25 will be a ring of
charges whose electric potential dVP at the field point has the following expression


dVP = k
dq


z2 + r2


where
dq = σ2πrdr.


The superposition of potential of all the infinitesimal rings that make up the disk gives the net potential at pointP. This is accomplished by integrating from r = 0 to r = R :
VP = ∫ dVP = k2πσ⌠


⌡0


R
r dr


z2 + r2
,


= k2πσ⎛⎝ z
2 + R2 − z2⎞⎠.


Significance
The basic procedure for a disk is to first integrate around θ and then over r. This has been demonstrated for
uniform (constant) charge density. Often, the charge density will vary with r, and then the last integral will givedifferent results.


Example 7.16
Potential Due to an Infinite Charged Wire
Find the electric potential due to an infinitely long uniformly charged wire.
Strategy
Since we have already worked out the potential of a finite wire of length L in Example 7.7, we might wonder iftaking L → ∞ in our previous result will work:


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7.10


VP = limL → ∞
kλln



⎜ L + L


2 + 4x2


−L + L2 + 4x2





⎟.


However, this limit does not exist because the argument of the logarithm becomes [2/0] as L → ∞ , so this way
of finding V of an infinite wire does not work. The reason for this problem may be traced to the fact that thecharges are not localized in some space but continue to infinity in the direction of the wire. Hence, our (unspoken)assumption that zero potential must be an infinite distance from the wire is no longer valid.
To avoid this difficulty in calculating limits, let us use the definition of potential by integrating over the electricfield from the previous section, and the value of the electric field from this charge configuration from the previouschapter.
Solution
We use the integral


VP = −∫
R


P
E


· d l


where R is a finite distance from the line of charge, as shown in Figure 7.26.


Figure 7.26 Points of interest for calculating the potential ofan infinite line of charge.


With this setup, we use E→ P = 2kλ1s ŝ and d l→ = d s→ to obtain
VP − VR = −∫


R


P
2kλ1sds = −2kλln


sP
sR


.


Now, if we define the reference potential VR = 0 at sR = 1m, this simplifies to
VP = −2kλ ln sP.


Note that this form of the potential is quite usable; it is 0 at 1 m and is undefined at infinity, which is why wecould not use the latter as a reference.
Significance
Although calculating potential directly can be quite convenient, we just found a system for which this strategydoes not work well. In such cases, going back to the definition of potential in terms of the electric field may offera way forward.


Check Your Understanding What is the potential on the axis of a nonuniform ring of charge, where thecharge density is λ(θ) = λ cos θ ?


Chapter 7 | Electric Potential 315




7.4 | Determining Field from Potential
Learning Objectives


By the end of this section, you will be able to:
• Explain how to calculate the electric field in a system from the given potential
• Calculate the electric field in a given direction from a given potential
• Calculate the electric field throughout space from a given potential


Recall that we were able, in certain systems, to calculate the potential by integrating over the electric field. As you mayalready suspect, this means that we may calculate the electric field by taking derivatives of the potential, although going
from a scalar to a vector quantity introduces some interesting wrinkles. We frequently need E→ to calculate the force in a
system; since it is often simpler to calculate the potential directly, there are systems in which it is useful to calculate V and
then derive E→ from it.
In general, regardless of whether the electric field is uniform, it points in the direction of decreasing potential, because
the force on a positive charge is in the direction of E→ and also in the direction of lower potential V. Furthermore, the
magnitude of E→ equals the rate of decrease of V with distance. The faster V decreases over distance, the greater the
electric field. This gives us the following result.


Relationship between Voltage and Uniform Electric Field
In equation form, the relationship between voltage and uniform electric field is


E = − ΔV
Δs


where Δs is the distance over which the change in potential ΔV takes place. The minus sign tells us that E points in
the direction of decreasing potential. The electric field is said to be the gradient (as in grade or slope) of the electricpotential.


For continually changing potentials, ΔV and Δs become infinitesimals, and we need differential calculus to determine
the electric field. As shown in Figure 7.27, if we treat the distance Δs as very small so that the electric field is essentially
constant over it, we find that


Es = − dVds
.


Figure 7.27 The electric field component along the
displacement Δs is given by E = − ΔV


Δs
. Note that A and B


are assumed to be so close together that the field is constantalong Δs .


Therefore, the electric field components in the Cartesian directions are given by


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(7.13)Ex = − ∂V∂ x , Ey = − ∂V∂ y , Ez = − ∂V∂z .


This allows us to define the “grad” or “del” vector operator, which allows us to compute the gradient in one step. InCartesian coordinates, it takes the form


(7.14)



= i
^ ∂
∂ x


+ j
^ ∂
∂ y


+ k
^ ∂
∂z


.


With this notation, we can calculate the electric field from the potential with


(7.15)E→ = − ∇→ V ,


a process we call calculating the gradient of the potential.
If we have a system with either cylindrical or spherical symmetry, we only need to use the del operator in the appropriatecoordinates:


(7.16)Cylindrical: ∇→ = r̂ ∂
∂r


+ φ̂1r

∂φ


+ ẑ ∂
∂z


(7.17)
Spherical: ∇



= r̂ ∂


∂r
+ θ


^ 1
r



∂θ


+ φ̂ 1
r sin θ



∂φ


Example 7.17
Electric Field of a Point Charge
Calculate the electric field of a point charge from the potential.
Strategy
The potential is known to be V = kqr , which has a spherical symmetry. Therefore, we use the spherical del
operator in the formula E→ = − ∇→ V .
Solution
Performing this calculation gives us


E


= −

⎝ r̂



∂r


+ θ
^ 1
r



∂θ


+ φ̂ 1
r sin θ



∂φ

⎠k
q
r = −kq



⎝ r̂



∂r


1
r + θ


^ 1
r



∂θ


1
r + φ̂


1
r sin θ



∂φ


1
r

⎠.


This equation simplifies to
E


= −kq

⎝ r̂


−1
r2


+ θ
^
0 + φ̂0



⎠ = k


q


r2


as expected.


Chapter 7 | Electric Potential 317




Significance
We not only obtained the equation for the electric field of a point particle that we’ve seen before, we also have a
demonstration that E→ points in the direction of decreasing potential, as shown in Figure 7.28.


Figure 7.28 Electric field vectors inside and outside a uniformly chargedsphere.


Example 7.18
Electric Field of a Ring of Charge
Use the potential found in Example 7.8 to calculate the electric field along the axis of a ring of charge (Figure7.29).


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7.11


Figure 7.29 We want to calculate the electric field from theelectric potential due to a ring charge.


Strategy
In this case, we are only interested in one dimension, the z-axis. Therefore, we use Ez = − ∂V∂z
with the potential V = k qtot


z2 + R2
found previously.


Solution
Taking the derivative of the potential yields


Ez = − ∂∂z
kqtot


z2 + R2
= k


qtot z



⎝z


2 + R2⎞⎠
3/2


.


Significance
Again, this matches the equation for the electric field found previously. It also demonstrates a system in whichusing the full del operator is not necessary.


Check Your Understanding Which coordinate system would you use to calculate the electric field of adipole?


7.5 | Equipotential Surfaces and Conductors
Learning Objectives


By the end of this section, you will be able to:
• Define equipotential surfaces and equipotential lines
• Explain the relationship between equipotential lines and electric field lines
• Map equipotential lines for one or two point charges
• Describe the potential of a conductor
• Compare and contrast equipotential lines and elevation lines on topographic maps


We can represent electric potentials (voltages) pictorially, just as we drew pictures to illustrate electric fields. This is not


Chapter 7 | Electric Potential 319




surprising, since the two concepts are related. Consider Figure 7.30, which shows an isolated positive point charge andits electric field lines, which radiate out from a positive charge and terminate on negative charges. We use blue arrows torepresent the magnitude and direction of the electric field, and we use green lines to represent places where the electricpotential is constant. These are called equipotential surfaces in three dimensions, or equipotential lines in two dimensions.The term equipotential is also used as a noun, referring to an equipotential line or surface. The potential for a point chargeis the same anywhere on an imaginary sphere of radius r surrounding the charge. This is true because the potential for apoint charge is given by V = kq/r and thus has the same value at any point that is a given distance r from the charge. An
equipotential sphere is a circle in the two-dimensional view of Figure 7.30. Because the electric field lines point radiallyaway from the charge, they are perpendicular to the equipotential lines.


Figure 7.30 An isolated point charge Q with its electric fieldlines in blue and equipotential lines in green. The potential is thesame along each equipotential line, meaning that no work isrequired to move a charge anywhere along one of those lines.Work is needed to move a charge from one equipotential line toanother. Equipotential lines are perpendicular to electric fieldlines in every case. For a three-dimensional version, explore thefirst media link.


It is important to note that equipotential lines are always perpendicular to electric field lines. No work is required to movea charge along an equipotential, since ΔV = 0 . Thus, the work is
W = −ΔU = −qΔV = 0.


Work is zero if the direction of the force is perpendicular to the displacement. Force is in the same direction as E, so motionalong an equipotential must be perpendicular to E. More precisely, work is related to the electric field by
W = F



⋅ d


= q E


⋅ d


= qEd cos θ = 0.


Note that in this equation, E and F symbolize the magnitudes of the electric field and force, respectively. Neither q nor Eis zero; d is also not zero. So cos θ must be 0, meaning θ must be 90º . In other words, motion along an equipotential is
perpendicular to E.
One of the rules for static electric fields and conductors is that the electric field must be perpendicular to the surface of anyconductor. This implies that a conductor is an equipotential surface in static situations. There can be no voltage differenceacross the surface of a conductor, or charges will flow. One of the uses of this fact is that a conductor can be fixed at whatwe consider zero volts by connecting it to the earth with a good conductor—a process called grounding. Grounding can bea useful safety tool. For example, grounding the metal case of an electrical appliance ensures that it is at zero volts relativeto Earth.
Because a conductor is an equipotential, it can replace any equipotential surface. For example, in Figure 7.30, a chargedspherical conductor can replace the point charge, and the electric field and potential surfaces outside of it will be unchanged,confirming the contention that a spherical charge distribution is equivalent to a point charge at its center.
Figure 7.31 shows the electric field and equipotential lines for two equal and opposite charges. Given the electric field


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lines, the equipotential lines can be drawn simply by making them perpendicular to the electric field lines. Conversely,given the equipotential lines, as in Figure 7.32(a), the electric field lines can be drawn by making them perpendicular tothe equipotentials, as in Figure 7.32(b).


Figure 7.31 The electric field lines and equipotential lines for two equal butopposite charges. The equipotential lines can be drawn by making themperpendicular to the electric field lines, if those are known. Note that thepotential is greatest (most positive) near the positive charge and least (mostnegative) near the negative charge. For a three-dimensional version, explore thefirst media link.


Figure 7.32 (a) These equipotential lines might be measured with a voltmeter in a laboratory experiment. (b) Thecorresponding electric field lines are found by drawing them perpendicular to the equipotentials. Note that these fields areconsistent with two equal negative charges. For a three-dimensional version, play with the first media link.


To improve your intuition, we show a three-dimensional variant of the potential in a system with two opposing charges.Figure 7.33 displays a three-dimensional map of electric potential, where lines on the map are for equipotential surfaces.The hill is at the positive charge, and the trough is at the negative charge. The potential is zero far away from the charges.Note that the cut off at a particular potential implies that the charges are on conducting spheres with a finite radius.


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Figure 7.33 Electric potential map of two opposite charges ofequal magnitude on conducting spheres. The potential isnegative near the negative charge and positive near the positivecharge.


A two-dimensional map of the cross-sectional plane that contains both charges is shown in Figure 7.34. The line that isequidistant from the two opposite charges corresponds to zero potential, since at the points on the line, the positive potentialfrom the positive charge cancels the negative potential from the negative charge. Equipotential lines in the cross-sectionalplane are closed loops, which are not necessarily circles, since at each point, the net potential is the sum of the potentialsfrom each charge.


Figure 7.34 A cross-section of the electric potential map oftwo opposite charges of equal magnitude. The potential isnegative near the negative charge and positive near the positivecharge.
View this simulation (https://openstaxcollege.org/l/21equipsurelec) to observe and modify theequipotential surfaces and electric fields for many standard charge configurations. There’s a lot to explore.


One of the most important cases is that of the familiar parallel conducting plates shown in Figure 7.35. Between the plates,the equipotentials are evenly spaced and parallel. The same field could be maintained by placing conducting plates at theequipotential lines at the potentials shown.


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Figure 7.35 The electric field and equipotential lines betweentwo metal plates. Note that the electric field is perpendicular tothe equipotentials and hence normal to the plates at their surfaceas well as in the center of the region between them.


Consider the parallel plates in Figure 7.2. These have equipotential lines that are parallel to the plates in the space betweenand evenly spaced. An example of this (with sample values) is given in Figure 7.35. We could draw a similar set ofequipotential isolines for gravity on the hill shown in Figure 7.2. If the hill has any extent at the same slope, the isolinesalong that extent would be parallel to each other. Furthermore, in regions of constant slope, the isolines would be evenlyspaced. An example of real topographic lines is shown in Figure 7.36.


Figure 7.36 A topographical map along a ridge has roughly parallel elevation lines, similar to the equipotential lines in Figure7.35. (a) A topographical map of Devil’s Tower, Wyoming. Lines that are close together indicate very steep terrain. (b) Aperspective photo of Devil’s Tower shows just how steep its sides are. Notice the top of the tower has the same shape as thecenter of the topographical map.


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Example 7.19
Calculating Equipotential Lines
You have seen the equipotential lines of a point charge in Figure 7.30. How do we calculate them? For example,if we have a +10-nC charge at the origin, what are the equipotential surfaces at which the potential is (a) 100 V,
(b) 50 V, (c) 20 V, and (d) 10 V?
Strategy
Set the equation for the potential of a point charge equal to a constant and solve for the remaining variable(s).Then calculate values as needed.
Solution
In V = kqr , let V be a constant. The only remaining variable is r; hence, r = k qV = constant . Thus, the
equipotential surfaces are spheres about the origin. Their locations are:


a. r = k q
V


= ⎛⎝8.99 × 10
9 Nm2 /C2⎞⎠



⎝10 × 10


−9 C⎞⎠
100 V


= 0.90 m ;


b. r = k q
V


= ⎛⎝8.99 × 10
9 Nm2 /C2⎞⎠



⎝10 × 10


−9 C⎞⎠
50 V


= 1.8 m ;


c. r = k q
V


= ⎛⎝8.99 × 10
9 Nm2 /C2⎞⎠



⎝10 × 10


−9 C⎞⎠
20 V


= 4.5 m ;


d. r = k q
V


= ⎛⎝8.99 × 10
9 Nm2 /C2⎞⎠



⎝10 × 10


−9 C⎞⎠
10 V


= 9.0 m .
Significance
This means that equipotential surfaces around a point charge are spheres of constant radius, as shown earlier, withwell-defined locations.


Example 7.20
Potential Difference between Oppositely Charged Parallel Plates
Two large conducting plates carry equal and opposite charges, with a surface charge density σ of magnitude
6.81 × 10−7 C/m2, as shown in Figure 7.37. The separation between the plates is l = 6.50 mm . (a) What
is the electric field between the plates? (b) What is the potential difference between the plates? (c) What is thedistance between equipotential planes which differ by 100 V?


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Figure 7.37 The electric field between oppositely chargedparallel plates. A portion is released at the positive plate.


Strategy
(a) Since the plates are described as “large” and the distance between them is not, we will approximate each ofthem as an infinite plane, and apply the result from Gauss’s law in the previous chapter.
(b) Use ΔVAB = −∫


A


B
E


· d l
→ .


(c) Since the electric field is constant, find the ratio of 100 V to the total potential difference; then calculate thisfraction of the distance.
Solutiona. The electric field is directed from the positive to the negative plate as shown in the figure, and itsmagnitude is given by


E = σε0
= 6.81 × 10


−7 C/m2


8.85 × 10−12 C2 /N ·m2
= 7.69 × 104 V/m.


b. To find the potential difference ΔV between the plates, we use a path from the negative to the positive
plate that is directed against the field. The displacement vector d l→ and the electric field E→ are
antiparallel so E→ · d l→ = −E dl. The potential difference between the positive plate and the negative
plate is then


ΔV = −∫ E · dl = E∫ dl = El = (7.69 × 104 V/m)(6.50 × 10−3 m) = 500 V.
c. The total potential difference is 500 V, so 1/5 of the distance between the plates will be the distancebetween 100-V potential differences. The distance between the plates is 6.5 mm, so there will be 1.3 mmbetween 100-V potential differences.


Significance
You have now seen a numerical calculation of the locations of equipotentials between two charged parallel plates.


Chapter 7 | Electric Potential 325




7.12 Check Your Understanding What are the equipotential surfaces for an infinite line charge?


Distribution of Charges on Conductors
In Example 7.19 with a point charge, we found that the equipotential surfaces were in the form of spheres, with the pointcharge at the center. Given that a conducting sphere in electrostatic equilibrium is a spherical equipotential surface, weshould expect that we could replace one of the surfaces in Example 7.19 with a conducting sphere and have an identicalsolution outside the sphere. Inside will be rather different, however.


Figure 7.38 An isolated conducting sphere.


To investigate this, consider the isolated conducting sphere of Figure 7.38 that has a radius R and an excess charge q. Tofind the electric field both inside and outside the sphere, note that the sphere is isolated, so its surface change distribution
and the electric field of that distribution are spherically symmetric. We can therefore represent the field as E→ = E(r) r̂ .
To calculate E(r), we apply Gauss’s law over a closed spherical surface S of radius r that is concentric with the conducting
sphere. Since r is constant and n̂ = r̂ on the sphere,



S


E


· n̂ da = E(r)∮ da = E(r) 4πr2.


For r < R , S is within the conductor, so recall from our previous study of Gauss’s law that qenc = 0 and Gauss’s law
gives E(r) = 0 , as expected inside a conductor at equilibrium. If r > R , S encloses the conductor so qenc = q. From
Gauss’s law,


E(r) 4πr2 =
q
ε0


.


The electric field of the sphere may therefore be written as
E = 0 (r < R),


E = 1
4πε0


q


r2
r̂ (r ≥ R).


As expected, in the region r ≥ R, the electric field due to a charge q placed on an isolated conducting sphere of radius R
is identical to the electric field of a point charge q located at the center of the sphere.
To find the electric potential inside and outside the sphere, note that for r ≥ R, the potential must be the same as that of
an isolated point charge q located at r = 0 ,


V(r) = 1
4πrε0


q
r (r ≥ R)


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simply due to the similarity of the electric field.
For r < R, E = 0, so V(r) is constant in this region. Since V(R) = q/4πε0R,


V(r) = 1
4πrε0


q
R
(r < R).


We will use this result to show that
σ1R1 = σ2R2,


for two conducting spheres of radii R1 and R2 , with surface charge densities σ1 and σ2 respectively, that are connected
by a thin wire, as shown in Figure 7.39. The spheres are sufficiently separated so that each can be treated as if it wereisolated (aside from the wire). Note that the connection by the wire means that this entire system must be an equipotential.


Figure 7.39 Two conducting spheres are connected by a thin conducting wire.


We have just seen that the electrical potential at the surface of an isolated, charged conducting sphere of radius R is
V = 1


4πrε0


q
R
.


Now, the spheres are connected by a conductor and are therefore at the same potential; hence
1


4πrε0


q1
R1


= 1
4πrε0


q2
R2


,


and
q1
R1


=
q2
R2


.


The net charge on a conducting sphere and its surface charge density are related by q = σ(4πR2). Substituting this equation
into the previous one, we find


σ1R1 = σ2R2.


Obviously, two spheres connected by a thin wire do not constitute a typical conductor with a variable radius of curvature.Nevertheless, this result does at least provide a qualitative idea of how charge density varies over the surface of a conductor.The equation indicates that where the radius of curvature is large (points B and D in Figure 7.40), σ and E are small.
Similarly, the charges tend to be denser where the curvature of the surface is greater, as demonstrated by the chargedistribution on oddly shaped metal (Figure 7.40). The surface charge density is higher at locations with a small radius ofcurvature than at locations with a large radius of curvature.


Chapter 7 | Electric Potential 327




Figure 7.40 The surface charge density and the electric fieldof a conductor are greater at regions with smaller radii ofcurvature.


A practical application of this phenomenon is the lightning rod, which is simply a grounded metal rod with a sharp endpointing upward. As positive charge accumulates in the ground due to a negatively charged cloud overhead, the electric
field around the sharp point gets very large. When the field reaches a value of approximately 3.0 × 106 N/C (the dielectric
strength of the air), the free ions in the air are accelerated to such high energies that their collisions with air moleculesactually ionize the molecules. The resulting free electrons in the air then flow through the rod to Earth, thereby neutralizingsome of the positive charge. This keeps the electric field between the cloud and the ground from getting large enough toproduce a lightning bolt in the region around the rod.
An important application of electric fields and equipotential lines involves the heart. The heart relies on electrical signalsto maintain its rhythm. The movement of electrical signals causes the chambers of the heart to contract and relax. Whena person has a heart attack, the movement of these electrical signals may be disturbed. An artificial pacemaker and adefibrillator can be used to initiate the rhythm of electrical signals. The equipotential lines around the heart, the thoracicregion, and the axis of the heart are useful ways of monitoring the structure and functions of the heart. An electrocardiogram(ECG) measures the small electric signals being generated during the activity of the heart.


Play around with this simulation (https://openstaxcollege.org/l/21pointcharsim) to move point chargesaround on the playing field and then view the electric field, voltages, equipotential lines, and more.


7.6 | Applications of Electrostatics
Learning Objectives


By the end of this section, you will be able to:
• Describe some of the many practical applications of electrostatics, including several printingtechnologies
• Relate these applications to Newton’s second law and the electric force


The study of electrostatics has proven useful in many areas. This module covers just a few of the many applications ofelectrostatics.


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The Van de Graaff Generator
Van de Graaff generators (or Van de Graaffs) are not only spectacular devices used to demonstrate high voltage due tostatic electricity—they are also used for serious research. The first was built by Robert Van de Graaff in 1931 (based onoriginal suggestions by Lord Kelvin) for use in nuclear physics research. Figure 7.41 shows a schematic of a large researchversion. Van de Graaffs use both smooth and pointed surfaces, and conductors and insulators to generate large static chargesand, hence, large voltages.
A very large excess charge can be deposited on the sphere because it moves quickly to the outer surface. Practical limitsarise because the large electric fields polarize and eventually ionize surrounding materials, creating free charges thatneutralize excess charge or allow it to escape. Nevertheless, voltages of 15 million volts are well within practical limits.


Figure 7.41 Schematic of Van de Graaff generator. A battery(A) supplies excess positive charge to a pointed conductor, thepoints of which spray the charge onto a moving insulating beltnear the bottom. The pointed conductor (B) on top in the largesphere picks up the charge. (The induced electric field at thepoints is so large that it removes the charge from the belt.) Thiscan be done because the charge does not remain inside theconducting sphere but moves to its outside surface. An ionsource inside the sphere produces positive ions, which areaccelerated away from the positive sphere to high velocities.
Xerography
Most copy machines use an electrostatic process called xerography—a word coined from the Greek words xeros for dryand graphos for writing. The heart of the process is shown in simplified form in Figure 7.42.


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Figure 7.42 Xerography is a dry copying process based on electrostatics. The major steps in theprocess are the charging of the photoconducting drum, transfer of an image, creating a positive chargeduplicate, attraction of toner to the charged parts of the drum, and transfer of toner to the paper. Notshown are heat treatment of the paper and cleansing of the drum for the next copy.


A selenium-coated aluminum drum is sprayed with positive charge from points on a device called a corotron. Selenium isa substance with an interesting property—it is a photoconductor. That is, selenium is an insulator when in the dark and aconductor when exposed to light.
In the first stage of the xerography process, the conducting aluminum drum is grounded so that a negative charge is inducedunder the thin layer of uniformly positively charged selenium. In the second stage, the surface of the drum is exposed to theimage of whatever is to be copied. In locations where the image is light, the selenium becomes conducting, and the positivecharge is neutralized. In dark areas, the positive charge remains, so the image has been transferred to the drum.
The third stage takes a dry black powder, called toner, and sprays it with a negative charge so that it is attracted to thepositive regions of the drum. Next, a blank piece of paper is given a greater positive charge than on the drum so that it willpull the toner from the drum. Finally, the paper and electrostatically held toner are passed through heated pressure rollers,which melt and permanently adhere the toner to the fibers of the paper.
Laser Printers
Laser printers use the xerographic process to make high-quality images on paper, employing a laser to produce an image onthe photoconducting drum as shown in Figure 7.43. In its most common application, the laser printer receives output froma computer, and it can achieve high-quality output because of the precision with which laser light can be controlled. Manylaser printers do significant information processing, such as making sophisticated letters or fonts, and in the past may havecontained a computer more powerful than the one giving them the raw data to be printed.


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Figure 7.43 In a laser printer, a laser beam is scanned across a photoconducting drum,leaving a positively charged image. The other steps for charging the drum andtransferring the image to paper are the same as in xerography. Laser light can be veryprecisely controlled, enabling laser printers to produce high-quality images.
Ink Jet Printers and Electrostatic Painting
The ink jet printer, commonly used to print computer-generated text and graphics, also employs electrostatics. A nozzlemakes a fine spray of tiny ink droplets, which are then given an electrostatic charge (Figure 7.44).
Once charged, the droplets can be directed, using pairs of charged plates, with great precision to form letters and images onpaper. Ink jet printers can produce color images by using a black jet and three other jets with primary colors, usually cyan,magenta, and yellow, much as a color television produces color. (This is more difficult with xerography, requiring multipledrums and toners.)


Figure 7.44 The nozzle of an ink-jet printer produces small inkdroplets, which are sprayed with electrostatic charge. Various computer-driven devices are then used to direct the droplets to the correct positionson a page.


Electrostatic painting employs electrostatic charge to spray paint onto oddly shaped surfaces. Mutual repulsion of likecharges causes the paint to fly away from its source. Surface tension forms drops, which are then attracted by unlikecharges to the surface to be painted. Electrostatic painting can reach hard-to-get-to places, applying an even coat in acontrolled manner. If the object is a conductor, the electric field is perpendicular to the surface, tending to bring the dropsin perpendicularly. Corners and points on conductors will receive extra paint. Felt can similarly be applied.
Smoke Precipitators and Electrostatic Air Cleaning
Another important application of electrostatics is found in air cleaners, both large and small. The electrostatic part of theprocess places excess (usually positive) charge on smoke, dust, pollen, and other particles in the air and then passes the airthrough an oppositely charged grid that attracts and retains the charged particles (Figure 7.45)
Large electrostatic precipitators are used industrially to remove over 99% of the particles from stack gas emissions


Chapter 7 | Electric Potential 331




associated with the burning of coal and oil. Home precipitators, often in conjunction with the home heating and airconditioning system, are very effective in removing polluting particles, irritants, and allergens.


Figure 7.45 (a) Schematic of an electrostatic precipitator. Air is passed through grids of opposite charge. The first grid chargesairborne particles, while the second attracts and collects them. (b) The dramatic effect of electrostatic precipitators is seen by theabsence of smoke from this power plant. (credit b: modification of work by “Cmdalgleish”/Wikimedia Commons)


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electric dipole
electric dipole moment


electric potential
electric potential difference
electric potential energy
electron-volt
electrostatic precipitators
equipotential line
equipotential surface
grounding
ink jet printer
photoconductor
Van de Graaff generator
voltage
xerography


CHAPTER 7 REVIEW
KEY TERMS


system of two equal but opposite charges a fixed distance apart
quantity defined as p→ = q d→ for all dipoles, where the vector points from the negative to


positive charge
potential energy per unit charge


the change in potential energy of a charge q moved between two points, divided by thecharge.
potential energy stored in a system of charged objects due to the charges


energy given to a fundamental charge accelerated through a potential difference of one volt
filters that apply charges to particles in the air, then attract those charges to a filter,removing them from the airstream


two-dimensional representation of an equipotential surface
surface (usually in three dimensions) on which all points are at the same potential


process of attaching a conductor to the earth to ensure that there is no potential difference between it andEarth
small ink droplets sprayed with an electric charge are controlled by electrostatic plates to create images onpaper


substance that is an insulator until it is exposed to light, when it becomes a conductor
machine that produces a large amount of excess charge, used for experiments with highvoltage


change in potential energy of a charge moved from one point to another, divided by the charge; units of potentialdifference are joules per coulomb, known as volt
dry copying process based on electrostatics


KEY EQUATIONS
Potential energy of a two-charge system U(r) = kqQr
Work done to assemble a system of charges


W12 ⋯ N =
k
2

i


N



j


N qiq j
ri j


for i ≠ j


Potential difference ΔV = ΔUq or ΔU = qΔV
Electric potential


V = Uq = − ∫
R


P
E


⋅ d l


Potential difference between two points
ΔVAB = VB − VA = −∫


A


B
E


· d l


Electric potential of a point charge
V =


kq
r


Electric potential of a system of point charges
VP = k∑


1


N
qi
ri


Chapter 7 | Electric Potential 333




Electric dipole moment p→ = q d→
Electric potential due to a dipole


VP = k
p→ · r̂
r2


Electric potential of a continuous charge distribution
VP = k∫


dq
r


Electric field components Ex = − ∂V∂ x , Ey = − ∂V∂ y , Ez = − ∂V∂z
Del operator in Cartesian coordinates





= i
^ ∂
∂ x


+ j
^ ∂
∂ y


+ k
^ ∂
∂z


Electric field as gradient of potential E→ = − ∇→ V
Del operator in cylindrical coordinates ∇→ = r̂ ∂


∂r
+ φ̂1r



∂φ


+ ẑ ∂
∂z


Del operator in spherical coordinates



= r̂ ∂
∂r


+ θ
^ 1
r



∂θ


+ φ̂ 1
r sin θ



∂φ


SUMMARY
7.1 Electric Potential Energy


• The work done to move a charge from point A to B in an electric field is path independent, and the work around aclosed path is zero. Therefore, the electric field and electric force are conservative.
• We can define an electric potential energy, which between point charges is U(r) = kqQr , with the zero reference
taken to be at infinity.


• The superposition principle holds for electric potential energy; the potential energy of a system of multiple chargesis the sum of the potential energies of the individual pairs.
7.2 Electric Potential and Potential Difference


• Electric potential is potential energy per unit charge.
• The potential difference between points A and B, VB − VA, that is, the change in potential of a charge q moved
from A to B, is equal to the change in potential energy divided by the charge.


• Potential difference is commonly called voltage, represented by the symbol ΔV :
ΔV = ΔUq or ΔU = qΔV .


• An electron-volt is the energy given to a fundamental charge accelerated through a potential difference of 1 V. Inequation form,
1 eV = ⎛⎝1.60 × 10


−19 C⎞⎠(1 V) =

⎝1.60 × 10


−19 C⎞⎠(1 J/C) = 1.60 × 10
−19 J.


7.3 Calculations of Electric Potential
• Electric potential is a scalar whereas electric field is a vector.
• Addition of voltages as numbers gives the voltage due to a combination of point charges, allowing us to use the
principle of superposition: VP = k∑


1


N
qi
ri
.


• An electric dipole consists of two equal and opposite charges a fixed distance apart, with a dipole moment


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p→ = q d
→ .


• Continuous charge distributions may be calculated with VP = k∫ dqr .


7.4 Determining Field from Potential
• Just as we may integrate over the electric field to calculate the potential, we may take the derivative of the potentialto calculate the electric field.
• This may be done for individual components of the electric field, or we may calculate the entire electric field vectorwith the gradient operator.


7.5 Equipotential Surfaces and Conductors
• An equipotential surface is the collection of points in space that are all at the same potential. Equipotential lines arethe two-dimensional representation of equipotential surfaces.
• Equipotential surfaces are always perpendicular to electric field lines.
• Conductors in static equilibrium are equipotential surfaces.
• Topographic maps may be thought of as showing gravitational equipotential lines.


7.6 Applications of Electrostatics
• Electrostatics is the study of electric fields in static equilibrium.
• In addition to research using equipment such as a Van de Graaff generator, many practical applications ofelectrostatics exist, including photocopiers, laser printers, ink jet printers, and electrostatic air filters.


CONCEPTUAL QUESTIONS
7.1 Electric Potential Energy
1. Would electric potential energy be meaningful if theelectric field were not conservative?
2. Why do we need to be careful about work done on thesystem versus work done by the system in calculations?
3. Does the order in which we assemble a system of pointcharges affect the total work done?


7.2 Electric Potential and Potential Difference
4. Discuss how potential difference and electric fieldstrength are related. Give an example.
5. What is the strength of the electric field in a regionwhere the electric potential is constant?
6. If a proton is released from rest in an electric field,will it move in the direction of increasing or decreasingpotential? Also answer this question for an electron and aneutron. Explain why.
7. Voltage is the common word for potential difference.Which term is more descriptive, voltage or potential


difference?
8. If the voltage between two points is zero, can a testcharge be moved between them with zero net work beingdone? Can this necessarily be done without exerting aforce? Explain.
9. What is the relationship between voltage and energy?More precisely, what is the relationship between potentialdifference and electric potential energy?
10. Voltages are always measured between two points.Why?
11. How are units of volts and electron-volts related? Howdo they differ?
12. Can a particle move in a direction of increasingelectric potential, yet have its electric potential energydecrease? Explain


7.3 Calculations of Electric Potential
13. Compare the electric dipole moments of charges ±Q
separated by a distance d and charges ±Q/2 separated by
a distance d/2.


Chapter 7 | Electric Potential 335




14. Would Gauss’s law be helpful for determining theelectric field of a dipole? Why?
15. In what region of space is the potential due to auniformly charged sphere the same as that of a pointcharge? In what region does it differ from that of a pointcharge?
16. Can the potential of a nonuniformly charged sphere bethe same as that of a point charge? Explain.


7.4 Determining Field from Potential
17. If the electric field is zero throughout a region, mustthe electric potential also be zero in that region?
18. Explain why knowledge of E→ (x, y, z) is not
sufficient to determine V(x,y,z). What about the other wayaround?


7.5 Equipotential Surfaces and Conductors
19. If two points are at the same potential, are there anyelectric field lines connecting them?
20. Suppose you have a map of equipotential surfacesspaced 1.0 V apart. What do the distances between thesurfaces in a particular region tell you about the strength of
the E→ in that region?


21. Is the electric potential necessarily constant over thesurface of a conductor?
22. Under electrostatic conditions, the excess charge on aconductor resides on its surface. Does this mean that all ofthe conduction electrons in a conductor are on the surface?
23. Can a positively charged conductor be at a negativepotential? Explain.
24. Can equipotential surfaces intersect?


7.6 Applications of Electrostatics
25. Why are the metal support rods for satellite networkdishes generally grounded?
26. (a) Why are fish reasonably safe in an electrical storm?(b) Why are swimmers nonetheless ordered to get out of thewater in the same circumstance?
27. What are the similarities and differences between theprocesses in a photocopier and an electrostaticprecipitator?
28. About what magnitude of potential is used to chargethe drum of a photocopy machine? A web search for“xerography” may be of use.


PROBLEMS
7.1 Electric Potential Energy
29. Consider a charge Q1( + 5.0 µC) fixed at a site with
another charge Q2 (charge +3.0 µC , mass 6.0 µg)
moving in the neighboring space. (a) Evaluate the potentialenergy of Q2 when it is 4.0 cm from Q1. (b) If Q2 starts
from rest from a point 4.0 cm from Q1, what will be its
speed when it is 8.0 cm from Q1 ? (Note: Q1 is held fixed
in its place.)
30. Two charges Q1( + 2.00 µC) and Q2( + 2.00 µC)
are placed symmetrically along the x-axis at
x = ± 3.00 cm . Consider a charge Q3 of charge
+4.00 µC and mass 10.0 mg moving along the y-axis.
If Q3 starts from rest at y = 2.00 cm, what is its speed
when it reaches y = 4.00 cm?


31. To form a hydrogen atom, a proton is fixed at a pointand an electron is brought from far away to a distance of
0.529 × 10−10 m, the average distance between proton
and electron in a hydrogen atom. How much work is done?
32. (a) What is the average power output of a heartdefibrillator that dissipates 400 J of energy in 10.0 ms?(b) Considering the high-power output, why doesn’t thedefibrillator produce serious burns?


7.2 Electric Potential and Potential Difference
33. Find the ratio of speeds of an electron and a negativehydrogen ion (one having an extra electron) acceleratedthrough the same voltage, assuming non-relativistic finalspeeds. Take the mass of the hydrogen ion to be
1.67 × 10−27 kg.


34. An evacuated tube uses an accelerating voltage of 40kV to accelerate electrons to hit a copper plate and produce


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X-rays. Non-relativistically, what would be the maximumspeed of these electrons?
35. Show that units of V/m and N/C for electric fieldstrength are indeed equivalent.
36. What is the strength of the electric field betweentwo parallel conducting plates separated by 1.00 cm andhaving a potential difference (voltage) between them of
1.50 × 104 V ?
37. The electric field strength between two parallel
conducting plates separated by 4.00 cm is 7.50 × 104 V .
(a) What is the potential difference between the plates? (b)The plate with the lowest potential is taken to be zero volts.What is the potential 1.00 cm from that plate and 3.00 cmfrom the other?
38. The voltage across a membrane forming a cell wall is80.0 mV and the membrane is 9.00 nm thick. What is theelectric field strength? (The value is surprisingly large, butcorrect.) You may assume a uniform electric field.
39. Two parallel conducting plates are separated by 10.0cm, and one of them is taken to be at zero volts. (a) Whatis the electric field strength between them, if the potential8.00 cm from the zero volt plate (and 2.00 cm from theother) is 450 V? (b) What is the voltage between theplates?
40. Find the maximum potential difference between twoparallel conducting plates separated by 0.500 cm of air,given the maximum sustainable electric field strength in air
to be 3.0 × 106 V/m .
41. An electron is to be accelerated in a uniform electric
field having a strength of 2.00 × 106 V/m. (a) What
energy in keV is given to the electron if it is acceleratedthrough 0.400 m? (b) Over what distance would it have tobe accelerated to increase its energy by 50.0 GeV?
42. Use the definition of potential difference in terms ofelectric field to deduce the formula for potential differencebetween r = ra and r = rb for a point charge located at
the origin. Here r is the spherical radial coordinate.
43. The electric field in a region is pointed away fromthe z-axis and the magnitude depends upon the distance sfrom the axis. The magnitude of the electric field is given as
E = αs where α is a constant. Find the potential difference
between points P1 and P2 , explicitly stating the path over
which you conduct the integration for the line integral.


44. Singly charged gas ions are accelerated from restthrough a voltage of 13.0 V. At what temperature will theaverage kinetic energy of gas molecules be the same as thatgiven these ions?


7.3 Calculations of Electric Potential
45. A 0.500-cm-diameter plastic sphere, used in a staticelectricity demonstration, has a uniformly distributed40.0-pC charge on its surface. What is the potential near itssurface?
46. How far from a 1.00-µC point charge is the potential
100 V? At what distance is it 2.00 × 102 V?
47. If the potential due to a point charge is 5.00 × 102 V
at a distance of 15.0 m, what are the sign and magnitude ofthe charge?
48. In nuclear fission, a nucleus splits roughly in half. (a)
What is the potential 2.00 × 10−14 m from a fragment
that has 46 protons in it? (b) What is the potential energy inMeV of a similarly charged fragment at this distance?
49. A research Van de Graaff generator has a 2.00-m-diameter metal sphere with a charge of 5.00 mC on it. (a)What is the potential near its surface? (b) At what distancefrom its center is the potential 1.00 MV? (c) An oxygenatom with three missing electrons is released near the Vande Graaff generator. What is its energy in MeV when theatom is at the distance found in part b?
50. An electrostatic paint sprayer has a 0.200-m-diametermetal sphere at a potential of 25.0 kV that repels paintdroplets onto a grounded object.
(a) What charge is on the sphere? (b) What charge must a0.100-mg drop of paint have to arrive at the object with aspeed of 10.0 m/s?
51. (a) What is the potential between two points situated10 cm and 20 cm from a 3.0-µC point charge? (b) To what
location should the point at 20 cm be moved to increase thispotential difference by a factor of two?
52. Find the potential at points P1, P2, P3, and P4 in


Chapter 7 | Electric Potential 337




the diagram due to the two given charges.


53. Two charges –2.0 µC and +2.0 µC are separated by
4.0 cm on the z-axis symmetrically about origin, with thepositive one uppermost. Two space points of interest
P1 and P2 are located 3.0 cm and 30 cm from origin at
an angle 30° with respect to the z-axis. Evaluate electric
potentials at P1 and P2 in two ways: (a) Using the exact
formula for point charges, and (b) using the approximatedipole potential formula.
54. (a) Plot the potential of a uniformly charged 1-mrod with 1 C/m charge as a function of the perpendiculardistance from the center. Draw your graph from
s = 0.1 m to s = 1.0 m . (b) On the same graph, plot the
potential of a point charge with a 1-C charge at the origin.(c) Which potential is stronger near the rod? (d) Whathappens to the difference as the distance increases?Interpret your result.


7.4 Determining Field from Potential
55. Throughout a region, equipotential surfaces are givenby z = constant . The surfaces are equally spaced with
V = 100 V for z = 0.00 m, V = 200 V for
z = 0.50 m, V = 300 V for z = 1.00 m. What is the
electric field in this region?
56. In a particular region, the electric potential is given
by V = −xy2 z + 4xy. What is the electric field in this
region?
57. Calculate the electric field of an infinite line charge,throughout space.


7.5 Equipotential Surfaces and Conductors
58. Two very large metal plates are placed 2.0 cm apart,with a potential difference of 12 V between them. Considerone plate to be at 12 V, and the other at 0 V. (a) Sketchthe equipotential surfaces for 0, 4, 8, and 12 V. (b) Nextsketch in some electric field lines, and confirm that they are


perpendicular to the equipotential lines.
59. A very large sheet of insulating material has had anexcess of electrons placed on it to a surface charge density
of –3.00 nC/m2 . (a) As the distance from the sheet
increases, does the potential increase or decrease? Can youexplain why without any calculations? Does the locationof your reference point matter? (b) What is the shape ofthe equipotential surfaces? (c) What is the spacing betweensurfaces that differ by 1.00 V?
60. A metallic sphere of radius 2.0 cm is charged with
+5.0-µC charge, which spreads on the surface of the
sphere uniformly. The metallic sphere stands on aninsulated stand and is surrounded by a larger metallicspherical shell, of inner radius 5.0 cm and outer radius 6.0cm. Now, a charge of −5.0-µC is placed on the inside
of the spherical shell, which spreads out uniformly on theinside surface of the shell. If potential is zero at infinity,what is the potential of (a) the spherical shell, (b) thesphere, (c) the space between the two, (d) inside the sphere,and (e) outside the shell?


61. Two large charged plates of charge density
±30 µC/m2 face each other at a separation of 5.0 mm. (a)
Find the electric potential everywhere. (b) An electron isreleased from rest at the negative plate; with what speedwill it strike the positive plate?
62. A long cylinder of aluminum of radius R meters ischarged so that it has a uniform charge per unit length onits surface of λ .
(a) Find the electric field inside and outside the cylinder. (b)Find the electric potential inside and outside the cylinder.(c) Plot electric field and electric potential as a function ofdistance from the center of the rod.


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63. Two parallel plates 10 cm on a side are given equal and
opposite charges of magnitude 5.0 × 10−9 C. The plates
are 1.5 mm apart. What is the potential difference betweenthe plates?
64. The surface charge density on a long straight metallicpipe is σ . What is the electric potential outside and inside
the pipe? Assume the pipe has a diameter of 2a.


65. Concentric conducting spherical shells carry chargesQ and –Q, respectively. The inner shell has negligiblethickness. What is the potential difference between theshells?


66. Shown below are two concentric spherical shells ofnegligible thicknesses and radii R1 and R2. The inner
and outer shell carry net charges q1 and q2, respectively,
where both q1 and q2 are positive. What is the electric


potential in the regions (a) r < R1, (b) R1 < r < R2,
and (c) r > R2?


67. A solid cylindrical conductor of radius a is surroundedby a concentric cylindrical shell of inner radius b. The solidcylinder and the shell carry charges Q and –Q, respectively.Assuming that the length L of both conductors is muchgreater than a or b, what is the potential difference betweenthe two conductors?


7.6 Applications of Electrostatics
68. (a) What is the electric field 5.00 m from the centerof the terminal of a Van de Graaff with a 3.00-mC charge,noting that the field is equivalent to that of a point charge atthe center of the terminal? (b) At this distance, what forcedoes the field exert on a 2.00-µC charge on the Van de
Graaff’s belt?
69. (a) What is the direction and magnitude of an electricfield that supports the weight of a free electron near thesurface of Earth? (b) Discuss what the small value forthis field implies regarding the relative strength of thegravitational and electrostatic forces.
70. A simple and common technique for acceleratingelectrons is shown in Figure 7.46, where there is auniform electric field between two plates. Electrons arereleased, usually from a hot filament, near the negativeplate, and there is a small hole in the positive plate thatallows the electrons to continue moving. (a) Calculate theacceleration of the electron if the field strength is
2.50 × 104 N/C . (b) Explain why the electron will not be
pulled back to the positive plate once it moves through thehole.


Chapter 7 | Electric Potential 339




Figure 7.46 Parallel conducting plates with opposite chargeson them create a relatively uniform electric field used toaccelerate electrons to the right. Those that go through the holecan be used to make a TV or computer screen glow or toproduce X- rays.
71. In a Geiger counter, a thin metallic wire at the centerof a metallic tube is kept at a high voltage with respect tothe metal tube. Ionizing radiation entering the tube knockselectrons off gas molecules or sides of the tube that thenaccelerate towards the center wire, knocking off even moreelectrons. This process eventually leads to an avalanchethat is detectable as a current. A particular Geiger counterhas a tube of radius R and the inner wire of radius a isat a potential of V0 volts with respect to the outer metal
tube. Consider a point P at a distance s from the centerwire and far away from the ends. (a) Find a formula forthe electric field at a point P inside using the infinite wireapproximation. (b) Find a formula for the electric potentialat a point P inside. (c) Use
V0 = 900 V, a = 3.00 mm, R = 2.00 cm, and find the
value of the electric field at a point 1.00 cm from the center.


72. The practical limit to an electric field in air is about
3.00 × 106 N/C . Above this strength, sparking takes
place because air begins to ionize. (a) At this electric fieldstrength, how far would a proton travel before hitting thespeed of light (ignore relativistic effects)? (b) Is it practicalto leave air in particle accelerators?
73. To form a helium atom, an alpha particle that containstwo protons and two neutrons is fixed at one location,and two electrons are brought in from far away, one at
a time. The first electron is placed at 0.600 × 10−10 m
from the alpha particle and held there while the second
electron is brought to 0.600 × 10−10 m from the alpha
particle on the other side from the first electron. See thefinal configuration below. (a) How much work is done ineach step? (b) What is the electrostatic energy of the alphaparticle and two electrons in the final configuration?


74. Find the electrostatic energy of eight equal charges

⎝+3 µC⎞⎠ each fixed at the corners of a cube of side 2 cm.
75. The probability of fusion occurring is greatlyenhanced when appropriate nuclei are brought closetogether, but mutual Coulomb repulsion must be overcome.This can be done using the kinetic energy of high-temperature gas ions or by accelerating the nuclei towardone another. (a) Calculate the potential energy of two singly
charged nuclei separated by 1.00 × 10−12 m. (b) At what
temperature will atoms of a gas have an average kineticenergy equal to this needed electrical potential energy?
76. A bare helium nucleus has two positive charges and a
mass of 6.64 × 10 – 27 kg . (a) Calculate its kinetic energy
in joules at 2.00% of the speed of light. (b) What is this in
electron-volts? (c) What voltage would be needed to obtainthis energy?
77. An electron enters a region between two large parallelplates made of aluminum separated by a distance of 2.0 cmand kept at a potential difference of 200 V. The electronenters through a small hole in the negative plate and movestoward the positive plate. At the time the electron is near
the negative plate, its speed is 4.0 × 105 m/s. Assume the
electric field between the plates to be uniform, and find thespeed of electron at (a) 0.10 cm, (b) 0.50 cm, (c) 1.0 cm,and (d) 1.5 cm from the negative plate, and (e) immediatelybefore it hits the positive plate.


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78. How far apart are two conducting plates that have an
electric field strength of 4.50 × 103 V/m between them,
if their potential difference is 15.0 kV?
79. (a) Will the electric field strength between two parallelconducting plates exceed the breakdown strength of dry
air, which is 3.00 × 106 V/m , if the plates are separated
by 2.00 mm and a potential difference of 5.0 × 103 V is
applied? (b) How close together can the plates be with thisapplied voltage?
80. Membrane walls of living cells have surprisingly largeelectric fields across them due to separation of ions. Whatis the voltage across an 8.00-nm-thick membrane if theelectric field strength across it is 5.50 MV/m? You mayassume a uniform electric field.
81. A double charged ion is accelerated to an energyof 32.0 keV by the electric field between two parallel


conducting plates separated by 2.00 cm. What is the electricfield strength between the plates?
82. The temperature near the center of the Sun is thought
to be 15 million degrees Celsius (1.5 × 107 °C) (or
kelvin). Through what voltage must a singly charged ion beaccelerated to have the same energy as the average kineticenergy of ions at this temperature?
83. A lightning bolt strikes a tree, moving 20.0 C of
charge through a potential difference of 1.00 × 102 MV.
(a) What energy was dissipated? (b) What mass of watercould be raised from 15 °C to the boiling point and then
boiled by this energy? (c) Discuss the damage that could becaused to the tree by the expansion of the boiling steam.
84. What is the potential 0.530 × 10−10 m from a proton
(the average distance between the proton and electron in ahydrogen atom)?
85. (a) A sphere has a surface uniformly charged with 1.00C. At what distance from its center is the potential 5.00MV? (b) What does your answer imply about the practicalaspect of isolating such a large charge?
86. What are the sign and magnitude of a point charge thatproduces a potential of –2.00 V at a distance of 1.00 mm?
87. In one of the classic nuclear physics experiments atthe beginning of the twentieth century, an alpha particlewas accelerated toward a gold nucleus, and its path wassubstantially deflected by the Coulomb interaction. If theenergy of the doubly charged alpha nucleus was 5.00 MeV,how close to the gold nucleus (79 protons) could it comebefore being deflected?


ADDITIONAL PROBLEMS
88. A 12.0-V battery-operated bottle warmer heats 50.0
g of glass, 2.50 × 102 g of baby formula, and
2.00 × 102 g of aluminum from 20.0 °C to 90.0 °C . (a)
How much charge is moved by the battery? (b) How manyelectrons per second flow if it takes 5.00 min to warmthe formula? (Hint: Assume that the specific heat of babyformula is about the same as the specific heat of water.)
89. A battery-operated car uses a 12.0-V system. Findthe charge the batteries must be able to move in orderto accelerate the 750 kg car from rest to 25.0 m/s, make
it climb a 2.00 × 102 -m high hill, and finally cause it
to travel at a constant 25.0 m/s while climbing with
5.00 × 102 -N force for an hour.


90. (a) Find the voltage near a 10.0 cm diameter metalsphere that has 8.00 C of excess positive charge on it.(b) What is unreasonable about this result? (c) Whichassumptions are responsible?
91. A uniformly charged ring of radius 10 cm is placedon a nonconducting table. It is found that 3.0 cm above thecenter of the half-ring the potential is –3.0 V with respectto zero potential at infinity. How much charge is in the half-ring?
92. A glass ring of radius 5.0 cm is painted with a chargedpaint such that the charge density around the ring variescontinuously given by the following function of the polar
angle θ, λ = ⎛⎝3.0 × 10−6 C/m⎞⎠ cos2 θ. Find the potential


Chapter 7 | Electric Potential 341




at a point 15 cm above the center.
93. A CD disk of radius ( R = 3.0 cm ) is sprayed with
a charged paint so that the charge varies continually withradial distance r from the center in the following manner:
σ = − (6.0 C/m)r / R .
Find the potential at a point 4 cm above the center.
94. (a) What is the final speed of an electron acceleratedfrom rest through a voltage of 25.0 MV by a negativelycharged Van de Graff terminal? (b) What is unreasonableabout this result? (c) Which assumptions are responsible?
95. A large metal plate is charged uniformly to a density
of σ = 2.0 × 10−9 C/m2 . How far apart are the
equipotential surfaces that represent a potential differenceof 25 V?
96. Your friend gets really excited by the idea of makinga lightning rod or maybe just a sparking toy by connectingtwo spheres as shown in Figure 7.39, and making R2
so small that the electric field is greater than the dielectricstrength of air, just from the usual 150 V/m electric fieldnear the surface of the Earth. If R1 is 10 cm, how small
does R2 need to be, and does this seem practical? (Hint:
recall the calculation for electric field at the surface of aconductor from Gauss’s Law.)
97. (a) Find x > > L limit of the potential of a finite
uniformly charged rod and show that it coincides with thatof a point charge formula. (b) Why would you expect thisresult?
98. A small spherical pith ball of radius 0.50 cm is paintedwith a silver paint and then −10 µC of charge is placed
on it. The charged pith ball is put at the center of a goldspherical shell of inner radius 2.0 cm and outer radius 2.2cm. (a) Find the electric potential of the gold shell withrespect to zero potential at infinity. (b) How much chargeshould you put on the gold shell if you want to make itspotential 100 V?
99. Two parallel conducting plates, each of cross-sectional
area 400 cm2 , are 2.0 cm apart and uncharged. If
1.0 × 1012 electrons are transferred from one plate to the


other, (a) what is the potential difference between theplates? (b) What is the potential difference between thepositive plate and a point 1.25 cm from it that is betweenthe plates?
100. A point charge of q = 5.0 × 10−8 C is placed at
the center of an uncharged spherical conducting shell ofinner radius 6.0 cm and outer radius 9.0 cm. Find theelectric potential at (a) r = 4.0 cm, (b) r = 8.0 cm, (c)
r = 12.0 cm.


101. Earth has a net charge that produces an electric fieldof approximately 150 N/C downward at its surface. (a)What is the magnitude and sign of the excess charge, notingthe electric field of a conducting sphere is equivalent to apoint charge at its center? (b) What acceleration will thefield produce on a free electron near Earth’s surface? (c)What mass object with a single extra electron will have itsweight supported by this field?
102. Point charges of 25.0 µC and 45.0 µC are placed
0.500 m apart.
(a) At what point along the line between them is the electricfield zero?
(b) What is the electric field halfway between them?
103. What can you say about two charges q1 and q2 , if
the electric field one-fourth of the way from q1 to q2 is
zero?
104. Calculate the angular velocity ω of an electron
orbiting a proton in the hydrogen atom, given the radius of
the orbit is 0.530 × 10−10 m . You may assume that the
proton is stationary and the centripetal force is supplied byCoulomb attraction.
105. An electron has an initial velocity of
5.00 × 106 m/s in a uniform 2.00 × 105 -N/C electric
field. The field accelerates the electron in the directionopposite to its initial velocity. (a) What is the directionof the electric field? (b) How far does the electron travelbefore coming to rest? (c) How long does it take theelectron to come to rest? (d) What is the electron’s velocitywhen it returns to its starting point?


CHALLENGE PROBLEMS
106. Three Na+ and three Cl− ions are placed
alternately and equally spaced around a circle of radius 50nm. Find the electrostatic energy stored.


107. Look up (presumably online, or by dismantling anold device and making measurements) the magnitude of thepotential deflection plates (and the space between them) inan ink jet printer. Then look up the speed with which the ink


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comes out the nozzle. Can you calculate the typical mass ofan ink drop?
108. Use the electric field of a finite sphere with constantvolume charge density to calculate the electric potential,


throughout space. Then check your results by calculatingthe electric field from the potential.
109. Calculate the electric field of a dipole throughoutspace from the potential.


Chapter 7 | Electric Potential 343




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8 | CAPACITANCE


Figure 8.1 The tree-like branch patterns in this clear Plexiglas® block are known as a Lichtenberg figure, named for theGerman physicist Georg Christof Lichtenberg (1742–1799), who was the first to study these patterns. The “branches” are createdby the dielectric breakdown produced by a strong electric field. (credit: modification of work by Bert Hickman)


Chapter Outline
8.1 Capacitors and Capacitance
8.2 Capacitors in Series and in Parallel
8.3 Energy Stored in a Capacitor
8.4 Capacitor with a Dielectric
8.5 Molecular Model of a Dielectric


Introduction
Capacitors are important components of electrical circuits in many electronic devices, including pacemakers, cell phones,and computers. In this chapter, we study their properties, and, over the next few chapters, we examine their function incombination with other circuit elements. By themselves, capacitors are often used to store electrical energy and release itwhen needed; with other circuit components, capacitors often act as part of a filter that allows some electrical signals to passwhile blocking others. You can see why capacitors are considered one of the fundamental components of electrical circuits.
8.1 | Capacitors and Capacitance


Learning Objectives
By the end of this section, you will be able to:
• Explain the concepts of a capacitor and its capacitance
• Describe how to evaluate the capacitance of a system of conductors


A capacitor is a device used to store electrical charge and electrical energy. It consists of at least two electrical conductorsseparated by a distance. (Note that such electrical conductors are sometimes referred to as “electrodes,” but more correctly,


Chapter 8 | Capacitance 345




they are “capacitor plates.”) The space between capacitors may simply be a vacuum, and, in that case, a capacitor is thenknown as a “vacuum capacitor.” However, the space is usually filled with an insulating material known as a dielectric. (Youwill learn more about dielectrics in the sections on dielectrics later in this chapter.) The amount of storage in a capacitor isdetermined by a property called capacitance, which you will learn more about a bit later in this section.
Capacitors have applications ranging from filtering static from radio reception to energy storage in heart defibrillators.Typically, commercial capacitors have two conducting parts close to one another but not touching, such as those in Figure8.2. Most of the time, a dielectric is used between the two plates. When battery terminals are connected to an initiallyuncharged capacitor, the battery potential moves a small amount of charge of magnitude Q from the positive plate to thenegative plate. The capacitor remains neutral overall, but with charges +Q and −Q residing on opposite plates.


Figure 8.2 Both capacitors shown here were initially uncharged before beingconnected to a battery. They now have charges of +Q and −Q (respectively) on
their plates. (a) A parallel-plate capacitor consists of two plates of opposite chargewith area A separated by distance d. (b) A rolled capacitor has a dielectric materialbetween its two conducting sheets (plates).


A system composed of two identical parallel-conducting plates separated by a distance is called a parallel-plate capacitor(Figure 8.3). The magnitude of the electrical field in the space between the parallel plates is E = σ/ε0 , where σ denotes
the surface charge density on one plate (recall that σ is the charge Q per the surface area A). Thus, the magnitude of the
field is directly proportional to Q.


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Figure 8.3 The charge separation in a capacitor shows that thecharges remain on the surfaces of the capacitor plates. Electricalfield lines in a parallel-plate capacitor begin with positivecharges and end with negative charges. The magnitude of theelectrical field in the space between the plates is in directproportion to the amount of charge on the capacitor.


Capacitors with different physical characteristics (such as shape and size of their plates) store different amounts of chargefor the same applied voltage V across their plates. The capacitance C of a capacitor is defined as the ratio of the maximumcharge Q that can be stored in a capacitor to the applied voltage V across its plates. In other words, capacitance is the largestamount of charge per volt that can be stored on the device:


(8.1)C = Q
V
.


The SI unit of capacitance is the farad (F), named after Michael Faraday (1791–1867). Since capacitance is the charge perunit voltage, one farad is one coulomb per one volt, or
1 F = 1C


1V
.


By definition, a 1.0-F capacitor is able to store 1.0 C of charge (a very large amount of charge) when the potentialdifference between its plates is only 1.0 V. One farad is therefore a very large capacitance. Typical capacitance values range
from picofarads (1 pF = 10−12 F) to millifarads (1 mF = 10−3 F) , which also includes microfarads ( 1 µF = 10−6 F ).
Capacitors can be produced in various shapes and sizes (Figure 8.4).


Chapter 8 | Capacitance 347




Figure 8.4 These are some typical capacitors used inelectronic devices. A capacitor’s size is not necessarily related toits capacitance value.
Calculation of Capacitance
We can calculate the capacitance of a pair of conductors with the standard approach that follows.


Problem-Solving Strategy: Calculating Capacitance
1. Assume that the capacitor has a charge Q.
2. Determine the electrical field E→ between the conductors. If symmetry is present in the arrangement of


conductors, you may be able to use Gauss’s law for this calculation.
3. Find the potential difference between the conductors from


(8.2)
VB − VA = −∫


A


B


E


· d l


,


where the path of integration leads from one conductor to the other. The magnitude of the potential differenceis then V = |VB − VA| .
4. With V known, obtain the capacitance directly from Equation 8.1.


To show how this procedure works, we now calculate the capacitances of parallel-plate, spherical, and cylindricalcapacitors. In all cases, we assume vacuum capacitors (empty capacitors) with no dielectric substance in the space betweenconductors.
Parallel-Plate Capacitor
The parallel-plate capacitor (Figure 8.5) has two identical conducting plates, each having a surface area A, separated by adistance d. When a voltage V is applied to the capacitor, it stores a charge Q, as shown. We can see how its capacitance maydepend on A and d by considering characteristics of the Coulomb force. We know that force between the charges increaseswith charge values and decreases with the distance between them. We should expect that the bigger the plates are, the morecharge they can store. Thus, C should be greater for a larger value of A. Similarly, the closer the plates are together, thegreater the attraction of the opposite charges on them. Therefore, C should be greater for a smaller d.


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Figure 8.5 In a parallel-plate capacitor with plates separatedby a distance d, each plate has the same surface area A.


We define the surface charge density σ on the plates as
σ = Q


A
.


We know from previous chapters that when d is small, the electrical field between the plates is fairly uniform (ignoring edgeeffects) and that its magnitude is given by
E = σε0


,


where the constant ε0 is the permittivity of free space, ε0 = 8.85 × 10−12 F/m. The SI unit of F/m is equivalent to
C2 /N ·m2. Since the electrical field E→ between the plates is uniform, the potential difference between the plates is


V = Ed = σdε0
= Qd


ε0 A
.


Therefore Equation 8.1 gives the capacitance of a parallel-plate capacitor as


(8.3)C = Q
V


= Q
Qd/ε0A


= ε0
A
d
.


Notice from this equation that capacitance is a function only of the geometry and what material fills the space between theplates (in this case, vacuum) of this capacitor. In fact, this is true not only for a parallel-plate capacitor, but for all capacitors:The capacitance is independent of Q or V. If the charge changes, the potential changes correspondingly so that Q/V remainsconstant.


Chapter 8 | Capacitance 349




8.1


8.2


Example 8.1
Capacitance and Charge Stored in a Parallel-Plate Capacitor
(a) What is the capacitance of an empty parallel-plate capacitor with metal plates that each have an area of
1.00 m2 , separated by 1.00 mm? (b) How much charge is stored in this capacitor if a voltage of 3.00 × 103 V
is applied to it?
Strategy
Finding the capacitance C is a straightforward application of Equation 8.3. Once we find C, we can find thecharge stored by using Equation 8.1.
Solutiona. Entering the given values into Equation 8.3 yields


C = ε0
A
d


= ⎛⎝8.85 × 10
−12 F


m



1.00 m2


1.00 × 10−3 m
= 8.85 × 10−9 F = 8.85 nF.


This small capacitance value indicates how difficult it is to make a device with a large capacitance.
b. Inverting Equation 8.1 and entering the known values into this equation gives


Q = CV = (8.85 × 10−9 F)(3.00 × 103 V) = 26.6 µC.


Significance
This charge is only slightly greater than those found in typical static electricity applications. Since air breaksdown (becomes conductive) at an electrical field strength of about 3.0 MV/m, no more charge can be stored onthis capacitor by increasing the voltage.


Example 8.2
A 1-F Parallel-Plate Capacitor
Suppose you wish to construct a parallel-plate capacitor with a capacitance of 1.0 F. What area must you use foreach plate if the plates are separated by 1.0 mm?
Solution
Rearranging Equation 8.3, we obtain


A = Cdε0
= (1.0 F)(1.0 × 10


−3 m)
8.85 × 10−12 F/m


= 1.1 × 108 m2.


Each square plate would have to be 10 km across. It used to be a common prank to ask a student to go to thelaboratory stockroom and request a 1-F parallel-plate capacitor, until stockroom attendants got tired of the joke.
Check Your Understanding The capacitance of a parallel-plate capacitor is 2.0 pF. If the area of each


plate is 2.4 cm2 , what is the plate separation?


Check Your Understanding Verify that σ/V and ε0/d have the same physical units.


Spherical Capacitor
A spherical capacitor is another set of conductors whose capacitance can be easily determined (Figure 8.6). It consists oftwo concentric conducting spherical shells of radii R1 (inner shell) and R2 (outer shell). The shells are given equal and


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opposite charges +Q and −Q , respectively. From symmetry, the electrical field between the shells is directed radially
outward. We can obtain the magnitude of the field by applying Gauss’s law over a spherical Gaussian surface of radius rconcentric with the shells. The enclosed charge is +Q ; therefore we have



S


E


· n̂dA = E(4πr2) = Qε0
.


Thus, the electrical field between the conductors is
E


= 1
4πε0


Q


r2
r̂ .


We substitute this E→ into Equation 8.2 and integrate along a radial path between the shells:
V = ∫


R1


R2
E


· d l


= ⌠
⌡R1


R2⎛


1
4πε0


Q


r2


⎠ · ( r̂  dr) =


Q
4πε0

⌡R1


R2
dr
r2


= Q
4πε0




1
R1


− 1
R2



⎠.


In this equation, the potential difference between the plates is V = −(V2 − V1) = V1 − V2 . We substitute this result into
Equation 8.1 to find the capacitance of a spherical capacitor:


(8.4)
C = Q


V
= 4πε0


R1R2
R2 − R1


.


Figure 8.6 A spherical capacitor consists of two concentricconducting spheres. Note that the charges on a conductor resideon its surface.


Example 8.3
Capacitance of an Isolated Sphere
Calculate the capacitance of a single isolated conducting sphere of radius R1 and compare it with Equation 8.4
in the limit as R2 → ∞ .
Strategy
We assume that the charge on the sphere is Q, and so we follow the four steps outlined earlier. We also assumethe other conductor to be a concentric hollow sphere of infinite radius.


Chapter 8 | Capacitance 351




8.3


Solution
On the outside of an isolated conducting sphere, the electrical field is given by Equation 8.2. The magnitude ofthe potential difference between the surface of an isolated sphere and infinity is


V = ∫
R1


+∞
E


· d l


= Q
4πε0

⌡R1


+∞
1
r2


r̂ · ( r̂ dr) = Q
4πε0

⌡R1


+∞
dr
r2


= 1
4πε0


Q
R1


.


The capacitance of an isolated sphere is therefore
C = Q


V
= Q


4πε0R1
Q


= 4πε0R1.


Significance
The same result can be obtained by taking the limit of Equation 8.4 as R2 → ∞ . A single isolated sphere is
therefore equivalent to a spherical capacitor whose outer shell has an infinitely large radius.


Check Your Understanding The radius of the outer sphere of a spherical capacitor is five times theradius of its inner shell. What are the dimensions of this capacitor if its capacitance is 5.00 pF?


Cylindrical Capacitor
A cylindrical capacitor consists of two concentric, conducting cylinders (Figure 8.7). The inner cylinder, of radius R1 ,
may either be a shell or be completely solid. The outer cylinder is a shell of inner radius R2 . We assume that the length of
each cylinder is l and that the excess charges +Q and −Q reside on the inner and outer cylinders, respectively.


Figure 8.7 A cylindrical capacitor consists of two concentric, conducting cylinders. Here, the chargeon the outer surface of the inner cylinder is positive (indicated by + ) and the charge on the inner
surface of the outer cylinder is negative (indicated by − ).


With edge effects ignored, the electrical field between the conductors is directed radially outward from the common axis ofthe cylinders. Using the Gaussian surface shown in Figure 8.7, we have

S


E


· n̂ dA = E(2πrl) = Qε0
.


Therefore, the electrical field between the cylinders is


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8.4


(8.5)E→ = 1
2πε0


Q
r l


r̂ .


Here r̂ is the unit radial vector along the radius of the cylinder. We can substitute into Equation 8.2 and find the potential
difference between the cylinders:


V = ∫
R1


R2
E


· d l


p =
Q


2πε0 l

R1


R2
1
r r̂ · ( r̂ dr) =


Q
2πε0 l



R1


R2
dr
r =


Q
2πε0 l


lnr|R1
R2 = Q


2πε0 l
ln
R2
R1


.


Thus, the capacitance of a cylindrical capacitor is


(8.6)
C = Q


V
=


2πε0 l
ln(R2/R1)


.


As in other cases, this capacitance depends only on the geometry of the conductor arrangement. An important applicationof Equation 8.6 is the determination of the capacitance per unit length of a coaxial cable, which is commonly used totransmit time-varying electrical signals. A coaxial cable consists of two concentric, cylindrical conductors separated by aninsulating material. (Here, we assume a vacuum between the conductors, but the physics is qualitatively almost the samewhen the space between the conductors is filled by a dielectric.) This configuration shields the electrical signal propagatingdown the inner conductor from stray electrical fields external to the cable. Current flows in opposite directions in the innerand the outer conductors, with the outer conductor usually grounded. Now, from Equation 8.6, the capacitance per unitlength of the coaxial cable is given by
C
l
=


2πε0
ln(R2/R1)


.


In practical applications, it is important to select specific values of C/l. This can be accomplished with appropriate choicesof radii of the conductors and of the insulating material between them.
Check Your Understanding When a cylindrical capacitor is given a charge of 0.500 nC, a potentialdifference of 20.0 V is measured between the cylinders. (a) What is the capacitance of this system? (b) If thecylinders are 1.0 m long, what is the ratio of their radii?


Several types of practical capacitors are shown in Figure 8.4. Common capacitors are often made of two small pieces ofmetal foil separated by two small pieces of insulation (see Figure 8.2(b)). The metal foil and insulation are encased in aprotective coating, and two metal leads are used for connecting the foils to an external circuit. Some common insulatingmaterials are mica, ceramic, paper, and Teflon™ non-stick coating.
Another popular type of capacitor is an electrolytic capacitor. It consists of an oxidized metal in a conducting paste.The main advantage of an electrolytic capacitor is its high capacitance relative to other common types of capacitors. Forexample, capacitance of one type of aluminum electrolytic capacitor can be as high as 1.0 F. However, you must be carefulwhen using an electrolytic capacitor in a circuit, because it only functions correctly when the metal foil is at a higherpotential than the conducting paste. When reverse polarization occurs, electrolytic action destroys the oxide film. This typeof capacitor cannot be connected across an alternating current source, because half of the time, ac voltage would have thewrong polarity, as an alternating current reverses its polarity (see Alternating-Current Circuts on alternating-currentcircuits).
A variable air capacitor (Figure 8.8) has two sets of parallel plates. One set of plates is fixed (indicated as “stator”), andthe other set of plates is attached to a shaft that can be rotated (indicated as “rotor”). By turning the shaft, the cross-sectionalarea in the overlap of the plates can be changed; therefore, the capacitance of this system can be tuned to a desired value.Capacitor tuning has applications in any type of radio transmission and in receiving radio signals from electronic devices.Any time you tune your car radio to your favorite station, think of capacitance.


Chapter 8 | Capacitance 353




Figure 8.8 In a variable air capacitor, capacitance can be tuned by changing the effective areaof the plates. (credit: modification of work by Robbie Sproule)


The symbols shown in Figure 8.9 are circuit representations of various types of capacitors. We generally use the symbolshown in Figure 8.9(a). The symbol in Figure 8.9(c) represents a variable-capacitance capacitor. Notice the similarity ofthese symbols to the symmetry of a parallel-plate capacitor. An electrolytic capacitor is represented by the symbol in partFigure 8.9(b), where the curved plate indicates the negative terminal.


Figure 8.9 This shows three different circuit representationsof capacitors. The symbol in (a) is the most commonly used one.The symbol in (b) represents an electrolytic capacitor. Thesymbol in (c) represents a variable-capacitance capacitor.


An interesting applied example of a capacitor model comes from cell biology and deals with the electrical potential in theplasma membrane of a living cell (Figure 8.10). Cell membranes separate cells from their surroundings but allow someselected ions to pass in or out of the cell. The potential difference across a membrane is about 70 mV. The cell membranemay be 7 to 10 nm thick. Treating the cell membrane as a nano-sized capacitor, the estimate of the smallest electrical field
strength across its ‘plates’ yields the value E = V


d
= 70 × 10


−3V
10 × 10−9m


= 7 × 106 V/m > 3MV/m .
This magnitude of electrical field is great enough to create an electrical spark in the air.


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Figure 8.10 The semipermeable membrane of a biological cell has differentconcentrations of ions on its interior surface than on its exterior. Diffusion moves
the K+ (potassium) and Cl– (chloride) ions in the directions shown, until the
Coulomb force halts further transfer. In this way, the exterior of the membraneacquires a positive charge and its interior surface acquires a negative charge,creating a potential difference across the membrane. The membrane is normallyimpermeable to Na+ (sodium ions).


Visit the PhET Explorations: Capacitor Lab (https://openstaxcollege.org/l/21phetcapacitor) toexplore how a capacitor works. Change the size of the plates and add a dielectric to see the effect on capacitance.Change the voltage and see charges built up on the plates. Observe the electrical field in the capacitor. Measure thevoltage and the electrical field.


8.2 | Capacitors in Series and in Parallel
Learning Objectives


By the end of this section, you will be able to:
• Explain how to determine the equivalent capacitance of capacitors in series and in parallelcombinations
• Compute the potential difference across the plates and the charge on the plates for a capacitorin a network and determine the net capacitance of a network of capacitors


Several capacitors can be connected together to be used in a variety of applications. Multiple connections of capacitorsbehave as a single equivalent capacitor. The total capacitance of this equivalent single capacitor depends both on theindividual capacitors and how they are connected. Capacitors can be arranged in two simple and common types ofconnections, known as series and parallel, for which we can easily calculate the total capacitance. These two basiccombinations, series and parallel, can also be used as part of more complex connections.
The Series Combination of Capacitors
Figure 8.11 illustrates a series combination of three capacitors, arranged in a row within the circuit. As for any capacitor,the capacitance of the combination is related to the charge and voltage by using Equation 8.1. When this seriescombination is connected to a battery with voltage V, each of the capacitors acquires an identical charge Q. To explain,first note that the charge on the plate connected to the positive terminal of the battery is +Q and the charge on the plate
connected to the negative terminal is −Q . Charges are then induced on the other plates so that the sum of the charges on
all plates, and the sum of charges on any pair of capacitor plates, is zero. However, the potential drop V1 = Q/C1 on one


Chapter 8 | Capacitance 355




capacitor may be different from the potential drop V2 = Q/C2 on another capacitor, because, generally, the capacitors may
have different capacitances. The series combination of two or three capacitors resembles a single capacitor with a smallercapacitance. Generally, any number of capacitors connected in series is equivalent to one capacitor whose capacitance(called the equivalent capacitance) is smaller than the smallest of the capacitances in the series combination. Charge on thisequivalent capacitor is the same as the charge on any capacitor in a series combination: That is, all capacitors of a seriescombination have the same charge. This occurs due to the conservation of charge in the circuit. When a charge Q in a seriescircuit is removed from a plate of the first capacitor (which we denote as −Q ), it must be placed on a plate of the second
capacitor (which we denote as +Q), and so on.


Figure 8.11 (a) Three capacitors are connected in series. The magnitude of the charge on each plateis Q. (b) The network of capacitors in (a) is equivalent to one capacitor that has a smaller capacitancethan any of the individual capacitances in (a), and the charge on its plates is Q.


We can find an expression for the total (equivalent) capacitance by considering the voltages across the individual capacitors.The potentials across capacitors 1, 2, and 3 are, respectively, V1 = Q/C1 , V2 = Q/C2 , and V3 = Q/C3 . These potentials
must sum up to the voltage of the battery, giving the following potential balance:


V = V1 + V2 + V3.


Potential V is measured across an equivalent capacitor that holds charge Q and has an equivalent capacitance CS . Entering
the expressions for V1 , V2 , and V3 , we get


Q
CS


= Q
C1


+ Q
C2


+ Q
C3


.


Canceling the charge Q, we obtain an expression containing the equivalent capacitance, CS , of three capacitors connected
in series:


1
CS


= 1
C1


+ 1
C2


+ 1
C3


.


This expression can be generalized to any number of capacitors in a series network.
Series Combination
For capacitors connected in a series combination, the reciprocal of the equivalent capacitance is the sum of reciprocalsof individual capacitances:


(8.7)1
CS


= 1
C1


+ 1
C2


+ 1
C3


+⋯.


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Example 8.4
Equivalent Capacitance of a Series Network
Find the total capacitance for three capacitors connected in series, given their individual capacitances are
1.000 µF , 5.000 µF , and 8.000 µF .
Strategy
Because there are only three capacitors in this network, we can find the equivalent capacitance by usingEquation 8.7 with three terms.
Solution
We enter the given capacitances into Equation 8.7:


1
CS


= 1
C1


+ 1
C2


+ 1
C3


= 1
1.000 µF


+ 1
5.000 µF


+ 1
8.000 µF


1
CS


= 1.325
µF


.


Now we invert this result and obtain CS = µF1.325 = 0.755 µF.
Significance
Note that in a series network of capacitors, the equivalent capacitance is always less than the smallest individualcapacitance in the network.


The Parallel Combination of Capacitors
A parallel combination of three capacitors, with one plate of each capacitor connected to one side of the circuit and the otherplate connected to the other side, is illustrated in Figure 8.12(a). Since the capacitors are connected in parallel, they allhave the same voltage V across their plates. However, each capacitor in the parallel network may store a different charge.To find the equivalent capacitance CP of the parallel network, we note that the total charge Q stored by the network is the
sum of all the individual charges:


Q = Q1 + Q2 + Q3.


On the left-hand side of this equation, we use the relation Q = CPV , which holds for the entire network. On the right-
hand side of the equation, we use the relations Q1 = C1V , Q2 = C2V , and Q3 = C3V for the three capacitors in the
network. In this way we obtain


CPV = C1V + C2V + C3V .


This equation, when simplified, is the expression for the equivalent capacitance of the parallel network of three capacitors:
CP = C1 + C2 + C3.


This expression is easily generalized to any number of capacitors connected in parallel in the network.
Parallel Combination
For capacitors connected in a parallel combination, the equivalent (net) capacitance is the sum of all individualcapacitances in the network,


(8.8)CP = C1 + C2 + C3 +⋯.


Chapter 8 | Capacitance 357




Figure 8.12 (a) Three capacitors are connected in parallel. Eachcapacitor is connected directly to the battery. (b) The charge on theequivalent capacitor is the sum of the charges on the individualcapacitors.


Example 8.5
Equivalent Capacitance of a Parallel Network
Find the net capacitance for three capacitors connected in parallel, given their individual capacitances are
1.0 µF, 5.0 µF, and 8.0 µF.


Strategy
Because there are only three capacitors in this network, we can find the equivalent capacitance by usingEquation 8.8 with three terms.
Solution
Entering the given capacitances into Equation 8.8 yields


CP = C1 + C2 + C3 = 1.0 µF + 5.0 µF + 8.0 µF


CP = 14.0 µF.


Significance
Note that in a parallel network of capacitors, the equivalent capacitance is always larger than any of the individualcapacitances in the network.


Capacitor networks are usually some combination of series and parallel connections, as shown in Figure 8.13. To find thenet capacitance of such combinations, we identify parts that contain only series or only parallel connections, and find theirequivalent capacitances. We repeat this process until we can determine the equivalent capacitance of the entire network.The following example illustrates this process.


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Figure 8.13 (a) This circuit contains both series and parallel connections of capacitors. (b) C1 and C2 are in series;
their equivalent capacitance is CS. (c) The equivalent capacitance CS is connected in parallel with C3. Thus, the
equivalent capacitance of the entire network is the sum of CS and C3.


Example 8.6
Equivalent Capacitance of a Network
Find the total capacitance of the combination of capacitors shown in Figure 8.13. Assume the capacitances areknown to three decimal places (C1 = 1.000 µF, C2 = 5.000 µF, C3 = 8.000 µF). Round your answer to three
decimal places.
Strategy
We first identify which capacitors are in series and which are in parallel. Capacitors C1 and C2 are in series.
Their combination, labeled CS, is in parallel with C3.
Solution
Since C1 andC2 are in series, their equivalent capacitance CS is obtained with Equation 8.7:


1
CS


= 1
C1


+ 1
C2


= 1
1.000 µF


+ 1
5.000 µF


= 1.200
µF


⇒ CS = 0.833 µF.


Capacitance CS is connected in parallel with the third capacitance C3 , so we use Equation 8.8 to find the
equivalent capacitance C of the entire network:


C = CS + C3 = 0.833 µF + 8.000 µF = 8.833 µF.


Example 8.7
Network of Capacitors
Determine the net capacitance C of the capacitor combination shown in Figure 8.14 when the capacitances are
C1 = 12.0 µF, C2 = 2.0 µF, and C3 = 4.0 µF . When a 12.0-V potential difference is maintained across the
combination, find the charge and the voltage across each capacitor.


Chapter 8 | Capacitance 359




Figure 8.14 (a) A capacitor combination. (b) An equivalent two-capacitor combination.


Strategy
We first compute the net capacitance C23 of the parallel connection C2 and C3 . Then C is the net capacitance
of the series connection C1 and C23 . We use the relation C = Q/V to find the charges Q1 ,Q2 , and Q3 , and
the voltages V1 , V2 , and V3 , across capacitors 1, 2, and 3, respectively.
Solution
The equivalent capacitance for C2 and C3 is


C23 = C2 + C3 = 2.0 µF + 4.0 µF = 6.0 µF.


The entire three-capacitor combination is equivalent to two capacitors in series,
1
C


= 1
12.0 µF


+ 1
6.0 µF


= 1
4.0 µF


⇒ C = 4.0 µF.


Consider the equivalent two-capacitor combination in Figure 8.14(b). Since the capacitors are in series, theyhave the same charge, Q1 = Q23 . Also, the capacitors share the 12.0-V potential difference, so
12.0 V = V1 + V23 =


Q1
C1


+
Q23
C23


=
Q1


12.0 µF
+


Q1
6.0 µF


⇒ Q1 = 48.0 µC.


Now the potential difference across capacitor 1 is
V1 =


Q1
C1


=
48.0 µC
12.0 µF


= 4.0 V.


Because capacitors 2 and 3 are connected in parallel, they are at the same potential difference:
V2 = V3 = 12.0 V − 4.0 V = 8.0 V.


Hence, the charges on these two capacitors are, respectively,
Q2 = C2V2 = (2.0 µF)(8.0 V) = 16.0 µC,


Q3 = C3V3 = (4.0 µF)(8.0 V) = 32.0 µC.


Significance
As expected, the net charge on the parallel combination of C2 and C3 is Q23 = Q2 + Q3 = 48.0 µC.


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8.5 Check Your Understanding Determine the net capacitance C of each network of capacitors shownbelow. Assume that C1 = 1.0 pF , C2 = 2.0 pF , C3 = 4.0 pF , and C4 = 5.0 pF . Find the charge on each
capacitor, assuming there is a potential difference of 12.0 V across each network.


8.3 | Energy Stored in a Capacitor
Learning Objectives


By the end of this section, you will be able to:
• Explain how energy is stored in a capacitor
• Use energy relations to determine the energy stored in a capacitor network


Most of us have seen dramatizations of medical personnel using a defibrillator to pass an electrical current through apatient’s heart to get it to beat normally. Often realistic in detail, the person applying the shock directs another person to“make it 400 joules this time.” The energy delivered by the defibrillator is stored in a capacitor and can be adjusted to fit thesituation. SI units of joules are often employed. Less dramatic is the use of capacitors in microelectronics to supply energywhen batteries are charged (Figure 8.15). Capacitors are also used to supply energy for flash lamps on cameras.


Chapter 8 | Capacitance 361




Figure 8.15 The capacitors on the circuit board for an electronic device follow a labelingconvention that identifies each one with a code that begins with the letter “C.”


The energy UC stored in a capacitor is electrostatic potential energy and is thus related to the charge Q and voltage V
between the capacitor plates. A charged capacitor stores energy in the electrical field between its plates. As the capacitor isbeing charged, the electrical field builds up. When a charged capacitor is disconnected from a battery, its energy remains inthe field in the space between its plates.
To gain insight into how this energy may be expressed (in terms of Q and V), consider a charged, empty, parallel-platecapacitor; that is, a capacitor without a dielectric but with a vacuum between its plates. The space between its plates has avolume Ad, and it is filled with a uniform electrostatic field E. The total energy UC of the capacitor is contained within
this space. The energy density uE in this space is simply UC divided by the volume Ad. If we know the energy density,
the energy can be found as UC = uE(Ad) . We will learn in Electromagnetic Waves (after completing the study of
Maxwell’s equations) that the energy density uE in a region of free space occupied by an electrical field E depends only
on the magnitude of the field and is


(8.9)uE = 12ε0E2.


If we multiply the energy density by the volume between the plates, we obtain the amount of energy stored between the
plates of a parallel-plate capacitor: UC = uE(Ad) = 12ε0E2 Ad = 12ε0 V


2


d2
Ad = 1


2
V 2 ε0


A
d


= 1
2
V 2C .


In this derivation, we used the fact that the electrical field between the plates is uniform so that E = V /d and C = ε0 A/d.
Because C = Q/V , we can express this result in other equivalent forms:


(8.10)
UC =


1
2
V 2C = 1


2
Q2


C
= 1


2
QV .


The expression in Equation 8.10 for the energy stored in a parallel-plate capacitor is generally valid for all types ofcapacitors. To see this, consider any uncharged capacitor (not necessarily a parallel-plate type). At some instant, we connectit across a battery, giving it a potential difference V = q/C between its plates. Initially, the charge on the plates is Q = 0.
As the capacitor is being charged, the charge gradually builds up on its plates, and after some time, it reaches the value Q.To move an infinitesimal charge dq from the negative plate to the positive plate (from a lower to a higher potential), the
amount of work dW that must be done on dq is dW = Vdq = q


C
dq .


This work becomes the energy stored in the electrical field of the capacitor. In order to charge the capacitor to a charge Q,the total work required is


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8.6


W = ∫
0


W(Q)
dW = ∫


0


Q q
C
dq = 1


2
Q2


C
.


Since the geometry of the capacitor has not been specified, this equation holds for any type of capacitor. The total work Wneeded to charge a capacitor is the electrical potential energy UC stored in it, or UC = W . When the charge is expressed
in coulombs, potential is expressed in volts, and the capacitance is expressed in farads, this relation gives the energy injoules.
Knowing that the energy stored in a capacitor is UC = Q2/(2C) , we can now find the energy density uE stored in a
vacuum between the plates of a charged parallel-plate capacitor. We just have to divide UC by the volume Ad of space
between its plates and take into account that for a parallel-plate capacitor, we have E = σ/ε0 and C = ε0 A/d . Therefore,
we obtain


uE =
UC
Ad


= 1
2
Q2


C
1
Ad


= 1
2


Q2


ε0 A/d
1
Ad


= 1
2


1
ε0


Q
A



2
= σ


2


2ε0
=


(Eε0)
2


2ε0
=


ε0
2
E2.


We see that this expression for the density of energy stored in a parallel-plate capacitor is in accordance with the generalrelation expressed in Equation 8.9. We could repeat this calculation for either a spherical capacitor or a cylindricalcapacitor—or other capacitors—and in all cases, we would end up with the general relation given by Equation 8.9.
Example 8.8


Energy Stored in a Capacitor
Calculate the energy stored in the capacitor network in Figure 8.14(a) when the capacitors are fully charged andwhen the capacitances are C1 = 12.0 µF, C2 = 2.0 µF, and C3 = 4.0 µF, respectively.
Strategy
We use Equation 8.10 to find the energy U1 , U2 , and U3 stored in capacitors 1, 2, and 3, respectively. The
total energy is the sum of all these energies.
Solution
We identify C1 = 12.0 µF and V1 = 4.0 V , C2 = 2.0 µF and V2 = 8.0 V , C3 = 4.0 µF and V3 = 8.0 V.
The energies stored in these capacitors are


U1 =
1
2
C1V1


2 = 1
2
(12.0 µF)(4.0 V)2 = 96 µJ,


U2 =
1
2
C2V2


2 = 1
2
(2.0 µF)(8.0 V)2 = 64 µJ,


U3 =
1
2
C3V3


2 = 1
2
(4.0 µF)(8.0 V)2 = 130 µJ.


The total energy stored in this network is
UC = U1 + U2 + U3 = 96 µJ + 64 µJ + 130 µJ = 0.29 mJ.


Significance
We can verify this result by calculating the energy stored in the single 4.0-µF capacitor, which is found to be
equivalent to the entire network. The voltage across the network is 12.0 V. The total energy obtained in this way
agrees with our previously obtained result,UC = 12CV 2 = 12(4.0 µF)(12.0 V)2 = 0.29 mJ .


Check Your Understanding The potential difference across a 5.0-pF capacitor is 0.40 V. (a) What is theenergy stored in this capacitor? (b) The potential difference is now increased to 1.20 V. By what factor is thestored energy increased?
In a cardiac emergency, a portable electronic device known as an automated external defibrillator (AED) can be a lifesaver.


Chapter 8 | Capacitance 363




A defibrillator (Figure 8.16) delivers a large charge in a short burst, or a shock, to a person’s heart to correct abnormalheart rhythm (an arrhythmia). A heart attack can arise from the onset of fast, irregular beating of the heart—called cardiacor ventricular fibrillation. Applying a large shock of electrical energy can terminate the arrhythmia and allow the body’snatural pacemaker to resume its normal rhythm. Today, it is common for ambulances to carry AEDs. AEDs are also foundin many public places. These are designed to be used by lay persons. The device automatically diagnoses the patient’sheart rhythm and then applies the shock with appropriate energy and waveform. CPR (cardiopulmonary resuscitation) isrecommended in many cases before using a defibrillator.


Figure 8.16 Automated external defibrillators are found inmany public places. These portable units provide verbalinstructions for use in the important first few minutes for aperson suffering a cardiac attack.


Example 8.9
Capacitance of a Heart Defibrillator
A heart defibrillator delivers 4.00 × 102 J of energy by discharging a capacitor initially at 1.00 × 104 V. What
is its capacitance?
Strategy
We are given UC and V, and we are asked to find the capacitance C. We solve Equation 8.10 for C and
substitute.
Solution
Solving this expression for C and entering the given values yields C = 2UC


V 2
= 2 4.00 × 10


2 J
(1.00 × 104 V)2


= 8.00 µF.


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8.4 | Capacitor with a Dielectric
Learning Objectives


By the end of this section, you will be able to:
• Describe the effects a dielectric in a capacitor has on capacitance and other properties
• Calculate the capacitance of a capacitor containing a dielectric


As we discussed earlier, an insulating material placed between the plates of a capacitor is called a dielectric. Inserting adielectric between the plates of a capacitor affects its capacitance. To see why, let’s consider an experiment described inFigure 8.17. Initially, a capacitor with capacitance C0 when there is air between its plates is charged by a battery to
voltage V0 . When the capacitor is fully charged, the battery is disconnected. A charge Q0 then resides on the plates, and
the potential difference between the plates is measured to be V0 . Now, suppose we insert a dielectric that totally fills the
gap between the plates. If we monitor the voltage, we find that the voltmeter reading has dropped to a smaller value V. Wewrite this new voltage value as a fraction of the original voltage V0 , with a positive number κ , κ > 1 :


V = 1κV0.


The constant κ in this equation is called the dielectric constant of the material between the plates, and its value is
characteristic for the material. A detailed explanation for why the dielectric reduces the voltage is given in the next section.Different materials have different dielectric constants (a table of values for typical materials is provided in the next section).Once the battery becomes disconnected, there is no path for a charge to flow to the battery from the capacitor plates. Hence,the insertion of the dielectric has no effect on the charge on the plate, which remains at a value of Q0 . Therefore, we find
that the capacitance of the capacitor with a dielectric is


(8.11)
C =


Q0
V


=
Q0
V0/κ


= κ
Q0
V0


= κC0.


This equation tells us that the capacitance C0 of an empty (vacuum) capacitor can be increased by a factor of κ when we
insert a dielectric material to completely fill the space between its plates. Note that Equation 8.11 can also be used foran empty capacitor by setting κ = 1 . In other words, we can say that the dielectric constant of the vacuum is 1, which is a
reference value.


Chapter 8 | Capacitance 365




Figure 8.17 (a) When fully charged, a vacuum capacitor has a voltage V0 and charge Q0
(the charges remain on plate’s inner surfaces; the schematic indicates the sign of charge oneach plate). (b) In step 1, the battery is disconnected. Then, in step 2, a dielectric (that iselectrically neutral) is inserted into the charged capacitor. When the voltage across thecapacitor is now measured, it is found that the voltage value has decreased to V = V0/κ .
The schematic indicates the sign of the induced charge that is now present on the surfaces ofthe dielectric material between the plates.


The principle expressed by Equation 8.11 is widely used in the construction industry (Figure 8.18). Metal plates inan electronic stud finder act effectively as a capacitor. You place a stud finder with its flat side on the wall and move itcontinually in the horizontal direction. When the finder moves over a wooden stud, the capacitance of its plates changes,because wood has a different dielectric constant than a gypsum wall. This change triggers a signal in a circuit, and thus thestud is detected.


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Figure 8.18 An electronic stud finder is used to detect wooden studsbehind drywall.


The electrical energy stored by a capacitor is also affected by the presence of a dielectric. When the energy stored in anempty capacitor is U0 , the energy U stored in a capacitor with a dielectric is smaller by a factor of κ ,


(8.12)
U = 1


2
Q2


C
= 1


2
Q0


2


κC0
= 1κU0.


As a dielectric material sample is brought near an empty charged capacitor, the sample reacts to the electrical field of thecharges on the capacitor plates. Just as we learned in Electric Charges and Fields on electrostatics, there will be theinduced charges on the surface of the sample; however, they are not free charges like in a conductor, because a perfectinsulator does not have freely moving charges. These induced charges on the dielectric surface are of an opposite sign tothe free charges on the plates of the capacitor, and so they are attracted by the free charges on the plates. Consequently, thedielectric is “pulled” into the gap, and the work to polarize the dielectric material between the plates is done at the expenseof the stored electrical energy, which is reduced, in accordance with Equation 8.12.
Example 8.10


Inserting a Dielectric into an Isolated Capacitor
An empty 20.0-pF capacitor is charged to a potential difference of 40.0 V. The charging battery is thendisconnected, and a piece of Teflon™ with a dielectric constant of 2.1 is inserted to completely fill the spacebetween the capacitor plates (see Figure 8.17). What are the values of (a) the capacitance, (b) the charge of theplate, (c) the potential difference between the plates, and (d) the energy stored in the capacitor with and withoutdielectric?


Chapter 8 | Capacitance 367




8.7


Strategy
We identify the original capacitance C0 = 20.0 pF and the original potential difference V0 = 40.0 V between
the plates. We combine Equation 8.11 with other relations involving capacitance and substitute.
Solutiona. The capacitance increases to


C = κC0 = 2.1(20.0 pF) = 42.0 pF.


b. Without dielectric, the charge on the plates is
Q0 = C0V0 = (20.0 pF)(40.0 V) = 0.8 nC.


Since the battery is disconnected before the dielectric is inserted, the plate charge is unaffected by thedielectric and remains at 0.8 nC.
c. With the dielectric, the potential difference becomes


V = 1κV0 =
1
2.1


40.0 V = 19.0 V.