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Fundamentals of Physics II

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Fundamentals of Physics II
Electromagnetism, Optics, and Quantum Mechanics
Expanded Edition
R. SHANKAR
Yale UNIVERSITY PRESS New Haven and London

First edition 2016. Expanded edition 2020. Copyright © 2016, 2020 by Yale University. All rights reserved. This book may not be reproduced, in whole or in part, including illustrations, in any form (beyond that copying permitted by Sections 107 and 108 of the U.S. Copyright Law and except by reviewers for the public press), without written permission from the publishers.
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10 9 8 7 6 5 4 3 2 1

To Stella, Vesper, Matteo, Roro, Devi, Eddie, and the rest of g3.

Deep and original, but also humble and generous, the physicist Josiah Willard Gibbs spent much of his life at Yale University. His father was a professor of sacred languages at Yale, and Gibbs received his bachelor’s and doctorate degrees from the university before teaching there until his death in 1903. The sculptor Lee Lawrie created the memorial bronze tablet pictured above, which was installed in Yale’s Sloane Physics Laboratory in 1912. It now resides in the entrance to the J.W. Gibbs Laboratories, Yale University. His life and work continue to inspire the author as they do everyone who is familiar with them.

Contents
Preface to the Expanded Edition Preface to the First Edition 1. Electrostatics I
1.1 Review of F = ma 1.2 Enter electricity 1.3 Coulomb’s law 1.4 Properties of charge
1.4.1 Superposition principle 1.5 Verifying Coulomb’s law 1.6 The ratio of gravitational to electric forces 1.7 Coulomb’s law for continuous charge density 2. The Electric Field 2.1 Review of key ideas 2.2 Digression on nuclear forces 2.3 The electric field E 2.4 Visualizing the field 2.5 Field of a dipole
2.5.1 Far field of dipole: general case 2.6 Response to a field
2.6.1 Dipole in a uniform field 3. Gauss’s Law I
3.1 Field of an infinite line charge 3.2 Field of an infinite sheet of charge 3.3 Spherical charge distribution: Gauss’s law 3.4 Digression on the area vector dA
3.4.1 Composition of areas

3.4.2 An application of the area vector 3.5 Gauss’s law through pictures
3.5.1 Continuous charge density
4. Gauss’s Law II: Applications 4.1 Applications of Gauss’s law 4.2 Field inside a shell 4.3 Field of an infinite charged wire, redux 4.4 Field of an infinite plane, redux 4.5 Conductors 4.5.1 Field inside a perfect conductor is zero 4.5.2 The net charge on a conductor will reside at the surface 4.5.3 A conductor with a hole inside 4.5.4 Field on the surface of a conductor
5. The Coulomb Potential 5.1 Conservative forces and potential energy 5.2 Is the electrostatic field conservative? 5.3 Path independence through pictures 5.4 Potential and field of a dipole
6. Conductors and Capacitors 6.1 Cases where computing V from E is easier 6.2 Visualizing V 6.3 Equipotentials 6.4 Method of images 6.4.1 Proof of uniqueness (optional section) 6.4.2 Additional properties of the potential V(r) 6.5 Capacitors 6.6 Energy stored in a capacitor 6.7 Energy of a charge distribution
7. Circuits and Currents 7.1 Energy in the electric field 7.2 Circuits and conductivity

7.3 Circuits 7.4 The battery and the emf Ɛ 7.5 The RC circuit with a battery 7.6 Miscellaneous circuits
8. Magnetism I 8.1 Experiments pointing to magnetism 8.2 Examples of the Lorentz force, the cyclotron 8.3 Lorentz force on current-carrying wires 8.4 The magnetic dipole 8.5 The DC motor
9. Magnetism II: Biot-Savart Law 9.1 Practice with Biot-Savart: field of a loop 9.2 Microscopic description of a bar magnet 9.3 Magnetic field of an infinite wire 9.4 Ampère’s law 9.5 Maxwell’s equations (static case)
10. Ampère II, Faraday, and Lenz 10.1 Field of an infinite wire, redux 10.2 Field of a solenoid 10.3 Faraday and Lenz 10.4 Optional digression on Faraday’s law
11. More Faraday 11.1 Betatron 11.2 Generators 11.3 Inductance 11.4 Mutual inductance 11.5 Self-inductance 11.6 Energy in the magnetic field
12. AC Circuits 12.1 Review of inductors 12.2 The LC circuit

12.2.1 Driven LC circuit 12.3 The LCR circuit
12.3.1 Review of complex numbers 12.3.2 Solving the LCR equation 12.3.3 Visualizing Z 12.4 Complex form of Ohm’s law
13. LCR Circuits and Displacement Current 13.1 Analysis of LCR results 13.1.1 Transients and the complementary solution 13.2 Power of the complex numbers 13.3 Displacement current
14. Electromagnetic Waves 14.1 The wave equation 14.2 Restricted Maxwell equations in vacuum 14.2.1 Maxwell equations involving infinitesimal cubes 14.2.2 Maxwell equations involving infinitesimal loops 14.3 The wave! 14.4 Sinusoidal solution to the wave equation 14.5 Energy in the electromagnetic wave 14.6 Origin of electromagnetic waves 14.7 Maxwell equations—the general case (optional) 14.7.1 Maxwell equations involving infinitesimal cubes 14.7.2 Maxwell equations involving infinitesimal loops 14.7.3 Consequences for the restricted E and B 14.8 From microscopic to macroscopic (optional) 14.8.1 Maxwell equations involving cubes 14.8.2 Maxwell equations involving loops
15. Electromagnetism and Relativity 15.1 Magnetism from Coulomb’s law and relativity 15.2 Relativistic invariance of electrodynamics 15.3 Review of Lorentz transformations

15.3.1 Implications for Newtonian mechanics 15.4 Scalar and vector fields 15.5 The derivative operator 15.6 Lorentz scalars and vectors 15.7 The four-current J
15.7.1 Charge conservation and the four-current J 15.8 The four-potential A
15.8.1 Gauge invariance 15.9 Wave equation for the four-vector A
15.9.1 Why work with V and A? 15.10 The electromagnetic tensor Ƒ
15.10.1 Tensors 15.10.2 The electromagnetic field tensor Ƒ
16. Optics I: Geometric Optics Revisited 16.1 Geometric or ray optics 16.2 Brief history of c 16.3 Some highlights of geometric optics 16.4 The law of reflection from Fermat’s principle 16.5 Snell’s law from Fermat’s principle 16.6 Reflection off a curved surface by Fermat 16.7 Elliptical mirrors and Fermat’s principle 16.8 Parabolic mirrors
17. Optics II: More Mirrors and Lenses 17.1 Spherical approximations to parabolic mirrors 17.2 Image formation: geometric optics 17.2.1 A midlife crisis 17.3 Image formation by Fermat’s principle 17.4 Tricky cases 17.4.1 Fermat’s principle for virtual focal points 17.4.2 Ray optics for virtual images 17.5 Lenses à la Fermat

17.6 Principle of least action 17.7 The eye
18. Wave Theory of Light 18.1 Interference of waves 18.2 Adding waves using real numbers 18.3 Adding waves with complex numbers 18.4 Analysis of interference 18.5 Diffraction grating 18.6 Single-slit diffraction 18.7 Understanding reflection and crystal diffraction 18.8 Light incident on an oil slick 18.8.1 Normal incidence 18.8.2 Oblique incidence
19. Quantum Mechanics: The Main Experiment 19.1 Double-slit experiment with light 19.2 Trouble with Maxwell 19.3 Digression on photons 19.3.1 Photoelectric effect 19.3.2 Compton effect 19.4 Matter waves 19.5 Photons versus electrons 19.6 The Heisenberg uncertainty principle 19.6.1 There are no states of well-defined position and momentum 19.6.2 Heisenberg microscope 19.7 Let there be light 19.8 The wave function ψ 19.9 Collapse of the wave function 19.10 Summary
20. The Wave Function and Its Interpretation 20.1 Probability in classical and quantum mechanics 20.2 Getting to know ψ

20.3 Statistical concepts: mean and uncertainty
21. Quantization and Measurement 21.1 More on momentum states 21.2 Single-valuedness and quantization of momentum 21.2.1 Quantization 21.2.2 The integral of ψp(x) 21.3 Measurement postulate: momentum 21.3.1 An example solvable by inspection 21.3.2 Using a normalized ψ 21.4 Finding A(p) by computation 21.5 More on Fourier’s theorems 21.6 Measurement postulate: general 21.7 More than one variable
22. States of Definite Energy 22.1 Free particle on a ring 22.1.1 Analysis of energy levels: degeneracy 22.2 Thinking inside the box 22.2.1 Particle in a well 22.2.2 The box: an exact solution 22.3 Energy measurement in the box
23. Scattering and Dynamics 23.1 Quantum scattering 23.1.1 Scattering for E > V0 23.1.2 Scattering for E < V0 23.2 Tunneling 23.3 Quantum dynamics 23.3.1 A solution of the time-dependent Schrödinger equation 23.3.2 Derivation of the particular solution ψE(x, t) 23.4 Special properties of the product solution 23.5 General solution for time evolution 23.5.1 Time evolution: a more complicated example

24. Summary and Outlook 24.1 Postulates: first pass 24.2 Refining the postulates 24.2.1 Toward a compact set of postulates 24.2.2 Eigenvalue problem 24.2.3 The Dirac delta function and the operator X 24.3 Postulates: final 24.4 Many particles, bosons, and fermions 24.4.1 Identical versus indistinguishable 24.4.2 Implications for atomic structure 24.5 Energy-time uncertainty principle 24.6 What next?
Exercises Problem Set 1, for Chapters 1 and 2 Problem Set 2, for Chapters 3 and 4 Problem Set 3, for Chapters 5 and 6 Problem Set 4, for Chapter 7 Problem Set 5, for Chapters 8 and 9 Problem Set 6, for Chapter 10 Problem Set 7, for Chapters 11, 12, and 13 Problem Set 8, for Chapters 14 and 15 Problem Set 9, for Chapters 16, 17, and 18 Problem Set 10, for Chapters 19 and 20 Problem Set 11, for Chapters 21 and 22 Problem Set 12, for Chapters 23 and 24
Answers to Exercises Problem Set 1, for Chapters 1 and 2 Problem Set 2, for Chapters 3 and 4 Problem Set 3, for Chapters 5 and 6 Problem Set 4, for Chapter 7 Problem Set 5, for Chapters 8 and 9

Problem Set 6, for Chapter 10 Problem Set 7, for Chapters 11, 12, and 13 Problem Set 8, for Chapters 14 and 15 Problem Set 9, for Chapters 16, 17, and 18 Problem Set 10, for Chapters 19 and 20 Problem Set 11, for Chapters 21 and 22 Problem Set 12, for Chapters 23 and 24
Constants Index

Preface to the Expanded Edition
Besides the correction of errors, many pointed out by Robert Zuidema, whom I am pleased to thank here, the biggest change has been the inclusion of many exercises and answers (but not solutions) at the end of the book. As always it has been a pleasure to work with Yale University Press, especially Joseph Calamia and Ann-Marie Imbornoni, and with copy editor Liz Casey and our LaTeX compositor, Newgen North America. Thanks to MaryEllen Oliver for an exceptional job of proofreading both volumes of the expanded edition.

Preface to the First Edition
This is the companion volume to Fundamentals of Physics I: Mechanics, Relativity, and Thermodynamics. It is the second half of an introductory course taught at Yale and covers electromagnetism, optics, and quantum mechanics. Like Volume I, it is based on the lectures given at Yale to a diverse class. The two volumes could be used for a year-long course in introductory physics that covers all the major topics. It may also be used for self-study. Some instructors may prescribe it as a supplement to another text.
The chapters in the book more or less follow those lectures with a few minor modifications. The style preserves the classroom atmosphere. Often I introduce the questions asked by the students or the answers they give when I believe they will be of value to the reader. The problem sets and exams, without which one cannot learn or be sure one has learned the physics, may be found along with their solutions at the Yale website, http://oyc.yale.edu/physics, free and open to all. The lectures may also be found at venues YouTube, iTunes (https://itunes.apple.com/us/itunesu/physics-video/id341651848?mt=10), and Academic Earth, to name a few.
In the lectures I sometimes refer to my Basic Training in Mathematics, published by Springer and intended for anyone who wants to master the undergraduate mathematics needed for the physical sciences.
This book, like its predecessor, owes its existence to many people. Peter Salovey, now president, then dean of Yale College, persuaded me to be part of the first batch of Open Yale Courses, funded by the Hewlett Foundation. Diana E. E. Kleiner, Dunham Professor, History of Art and Classics, encouraged and guided me in many ways. She was also the one who persuaded me to write both these books. At Yale University Press, Joe Calamia has been an invaluable guide, making countless suggestions to improve the book’s contents. He has also lent his name to many subatomic particles that appear in this book. Once again Ann-Marie Imbornoni skillfully shepherded the book through various stages of production. I am delighted that Liz Casey was once again able to apply her editorial magic to

the manuscript, greatly improving not only the punctuation, syntax, and grammar but also the clarity. She made sure my intended sense was captured by the words used.
I thank Professor Ganpathy Murthy (University of Kentucky) and Branislav Djordjevic (George Mason University) for thoughtful comments. My very special thanks go to Phil Nelson of the University of Pennsylvania for his detailed and insightful comments on many parts of the book.
The writing of this book started a year ago and ended August 2015 at the Aspen Center for Physics (ACP). I am most grateful for the climate provided by the ACP where both the scientist and author in me found intellectual nourishment. The ACP is supported by the National Science Foundation (NSF) Grant number 1066293.
A large portion of the book was written at the Kavli Institute for Theoretical Physics (KITP) in Santa Barbara, where I was fortunate to receive a Simons Distinguished Visiting Scholar award for Fall 2014. The KITP is supported in part by the National Science Foundation under Grant number NSF PHY11–25915. I am especially grateful to Professor Lars Bildsten for making this possible.
The day I find I cannot write books at either of these marvelous places, I will switch to another line of work.
Barry Bradlyn and Alexey Shkarin were two exceptional graduate students who proofread the book, caught bugs, and suggested stylistic changes.
My family, all three generations of it, was very supportive as always. The final check was provided by Stella, who left many unsolicited notes in the margins and inside using her crayons. She is responsible for all remaining errors.

CHAPTER 1
Electrostatics I
We begin the second half of the course with an introduction to a new force: electromagnetism. Then we will study optics. We will conclude with a study of quantum mechanics. Now, quantum mechanics is not like a new force. It’s a whole different ball game. It’s not about what forces are acting on this or that object that determine its trajectory. The question there is: should we be even thinking about particles going on any trajectory? The answer will be negative. You will find out that most of the cherished ideas from Newtonian mechanics get overthrown. But the good news is that you need quantum mechanics only to study very small things like atoms or molecules. Of course, the big question is, where do you draw the line? How small is small? Some people even ask me, “Do you need quantum mechanics to describe the human brain?” And the answer is, “Yes, if it is small enough.” I’ve gone to parties where after a few minutes of talking to a person I’m thinking, “Okay, this person’s brain needs a full-fledged quantum mechanical treatment.” But most of the time everything is macroscopic, and you can describe it with Newtonian mechanics and classical electrodynamics.
1.1 Review of F = ma
Before we start with electromagnetism, let us recall the interplay between the ideas of force, mass, acceleration, and F = ma discussed at length in the prequel to this book, referred to as Volume I. The only thing everyone knows from the nursery is that a stands for acceleration, and we all know how to measure it. You find the position now and the position slightly later, take the difference, divide by the time, and get the velocity. Even though velocity requires two successive position measurements, we talk of velocity “right now,” because you can make those two successive measurements arbitrarily close to each other, and in the limit in which the time difference between them goes to zero, you can talk about the velocity right now. If the

speedometer in your car points to 60 miles per hour, that’s your velocity right now. Likewise, find the velocity now, find the velocity a little later, divide the difference by time, and you get the acceleration. It’s also an instantaneous quantity. If you step on the gas and feel the seat pushing you, that reflects your acceleration right now.
Given that we know how to measure acceleration, how should we determine the mass of anything? First of all you need an arbitrarily chosen standard of mass. The Bureau of Standards has a block of some material that defines a kilogram. Using that, can you find the value of another mass? Surely you know that using a weighing scale is not the correct answer because that measures the weight of the object due to earth’s gravity, while the mass of an object is defined anywhere, even far from the earth. Now you might say, “Well, take a known force and divide by the acceleration it produces,” but we haven’t talked about how to measure the force either. All you have is this equation F = ma.
One correct option is to use F = ma itself, as follows. Take a spring, attach one end to a wall, hook the known 1 kg mass to the other end, pull it by some amount, release it, and measure a1, the acceleration. Now pick any object whose mass you want, say an elephant. You detach that 1 kg mass, attach the elephant, pull the spring by the same amount, and measure aE, the acceleration of the elephant. Since you pulled the spring by the same amount, the force is the same in both cases. You don’t know and don’t have to know what it is, just that it is the same. Therefore we know
which determines mE, the mass of the elephant. So imagine that the masses of all objects can be determined by this
process. Can we now use F = ma to find the trajectory of bodies? No, we still need to know what forces will be acting on a body in any given situation. We need to know the F in F = ma in the given context. Newton does not tell you that in general. For example, for the spring, you have to determine what force it exerts when it’s pulled by various amounts. To this end, you pull it by some amount x, attach it to a known mass, find the acceleration, and then the product ma gives the force as a function of x, namely Hooke’s law F = −kx. So this is an example of your finding out the left-hand side of Newton’s law by measuring the acceleration of known

masses. Newton did, however, give the left-hand side in one famous case, the law of universal gravity between masses M and m in terms of their separation r and the gravitational constant G:
Using this law we have been able to do some very impressive celestial mechanics, right up to the present.
Unlike the spring force, there is no real contact between the earth and the object that it is pulling, whether it be the apple or the moon. This is an example of action at a distance. It was a great abstraction to believe that things can reach out and pull (or push) other things without touching them. Gravity was the first formally described force of this kind.
Remember the distinction between F =ma and F = −kx. The first is always true and relates the force on a body to the acceleration it produces, but does not tell us what force F will act in any given situation. It is our job to find out every time what forces might be acting on a body. If it’s connected to a spring, we have to study the spring experimentally to find out that F = −kx.
So F = ma is good for three things: to define mass, to determine the forces acting on bodies of known mass by seeing how they accelerate, and to find the acceleration of bodies given the forces.
Every time a body accelerates, we must be able to relate its acceleration to the sum of all the forces acting on it. But now and then we will not be able to do this. We can either abandon F = ma, or, putting our faith in the correctness of F = ma, provided all forces are included, we can go on to discover and characterize the new force behind the discrepancy.
1.2 Enter electricity
Now I will describe an experiment that reveals a new force. I take a comb and vigorously brush my hair and then touch a small piece of paper with the comb. I find it sticks to the comb and I can lift it. But when I shake the comb vigorously, the paper falls down. What can we learn from this?
Clearly, the force between the comb and paper is not the force of gravity, because gravity doesn’t care if you comb your hair or not. We may concede that there is a new force, but we may conclude it is feeble compared to

gravity because it eventually yields to gravity when we shake the comb. It would be a mistake to think so. In fact, this new force is roughly 1040 times stronger than the gravitational force, as determined by a criterion that I will explain shortly. But first, let us grasp this fact intuitively.
Look at Figure 1.1. You see me holding the comb, which is holding up the piece of paper. What is trying to pull it down? The entire planet! The Himalayas are pulling it down, the Pacific Ocean is pulling it down, even the Loch Ness Monster is pulling it down. Everything is pulling it down. I am one of these people generally convinced the world is acting against me, but this time I’m right. Everything is against me and my comb, and yet we are able to triumph against all of that. And that is how you compare the electric force with the gravitational force. It takes the entire planet to compensate whatever force I created between the comb and the piece of paper. Later we will see how the number 1040 quantifies this fact.
Something has happened to the comb when I rubbed it against my hair, something that allowed it to attract the paper. We describe that condition of the comb by saying “The comb is charged.” If the comb is briefly dipped in water and removed, we find it no longer attracts the piece of paper. We say the comb is now discharged.
Figure 1.1 The comb is pulling the paper electrostatically, and all of the world is pulling the other way gravitationally.

Figure 1.2 Top: A negatively charged rod imparts some negative charge to the sphere upon contact and they repel each other. Second: The negatively charged rod attracts a neutral sphere by polarizing it without touching it. Third: A charged rod polarizes a dielectric. The light region in the middle is the overlap of rectangles with positive (dotted boundary) and negative charges (solid boundary). The light region is neutral and the edges carry the uncanceled charges. Bottom: The charged spheres attract because they have been charged oppositely.
I am going to describe in detail the microscopic theory that can explain this experiment and many more, qualitatively and quantitatively. But, first, let us consider, at a qualitative level, a few more such experiments (depicted in Figure 1.2) and their explanations in terms of these ideas.
Experiment 1: Wearing silk gloves, take an aluminum rod and rub it against a passing furry animal, say a Yeti. Briefly touch an uncharged metallic sphere, isolated from everything else. The rod and sphere will repel.
Experiment 2: Move the charged rod near an isolated and uncharged sphere. Before they touch, they will attract each other. The same happens when the metallic sphere is replaced by a piece of paper.
Experiment 3: Take two more uncharged spheres. Repeat the previous two experiments after rubbing the rod on a piece of polyester and you will find the same results.

Experiment 4: Take two uncharged spheres; touch one with the rod that has been rubbed against Yeti fur and the other with a rod that has been rubbed against polyester. This time the two spheres will attract each other.
Experiment 5: Connect the spheres with a wire and they no longer attract.
Now we turn to the underlying theory, which is a result of centuries of investigation. We first consider some qualitative facts.
The most important idea is that everything is made of atoms. The atom has a nucleus consisting of protons and neutrons. The nucleus is surrounded by some very light particles called electrons. Normally the number of electrons and protons in an atom is equal. Two protons will repel each other, as will two electrons, but the proton and electron will attract each other. A neutron will not interact with, i.e., attract or repel, another neutron, proton, or electron. (Here I refer to electrical interactions, not nuclear interactions. These are much stronger, but significant only at very short distances [≃ 10−15 m]. The neutron fully participates in nuclear interactions and the electron does not.) Objects like electrons and protons that take part in electrical interactions are said to be charged or to carry a charge, while neutrons are said to be (electrically) neutral.
Just as mass is the reason particles experience the force of gravity, charge is the reason they interact electrically. But there are differences. There are no gravitationally neutral particles—everything has a positive mass. Second, gravity is always attractive but the electric forces can go either way. This is described by saying there are two kinds of charge, positive and negative, which can cancel each other out just like positive and negative numbers can. By convention the proton has positive charge and the electron has negative charge. Like charges repel and unlike charges attract. A system made of an equal amount of positive and negative charges will appear neutral, at least from a distance, when the internal structure is irrelevant. There is no such way to neutralize gravity.
An atom with an equal number of protons and electrons is neutral. This is due to a remarkable fact that the electron and proton have exactly equal and opposite charges. This equality is quite a mystery since the two particles are otherwise very dissimilar: the proton is about 1836 times as heavy and experiences forces that the electron does not.

Here is how we understand Experiments 1 through 5 in terms of the preceding facts.
Experiment 1: Upon rubbing, electrons flow from the Yeti to the rod, leaving the rod negative and Yeti positive. The protons stay where they are. The silk gloves keep the electrons in the rod from jumping on to your body and then to the ground: silk is an insulator. When the rod touches the sphere, some electrons migrate to the sphere in order to get away from each other. The sphere and the rod are both negatively charged and repel each other.
Experiment 2: When the negatively charged rod goes near the neutral metallic sphere, the electrons in the sphere are repelled by the extra electrons in the rod, and they preferentially occupy the far side, leaving a positive region near the rod. The positive region is attracted to the rod and the negative one repelled by it, but the attraction wins since the positive part is closer. Such free motion of electrons can take place in a conductor. If we replace the metallic sphere by a piece of paper, it too gets attracted, but by a more complicated mechanism. The paper is a dielectric. The electrons in it are not free to run off to one end, because paper is also an insulator, but they can move a little from their orbits centered on the nuclei if coaxed. Think of a rectangular piece of paper as made of two superposed layers, one positive (bounded by the dotted line in the figure) and made of the nuclei and one negative, made of the electrons (bounded by the solid line). Initially the two layers overlap completely and neutralize each other everywhere. When the negative rod comes near one edge, the electronic layer is displaced by a tiny amount (of atomic dimensions) away from the rod, while the positive nuclei stay put. The bulk of the paper (solid region of overlap in the middle) is still neutral, but the edge near the rod has a strip of unbalanced protons and the edge far from it has a strip of unbalanced electrons. This process, in which the positive and negative charges are displaced relative to each other by a small amount, is called polarization. Again, the attraction of the nearby positive strip beats the repulsion of the distant negative one.
Experiment 3: When the rod is rubbed against polyester, the electrons flow the other way: from the rod to the polyester, leaving the rod positively

charged. We can repeat the arguments from Experiments 1 and 2, simply reversing the sign of all charges.
Experiment 4: Now one sphere is positively charged (by polyester) and one is negatively charged (thanks to Yeti) and they attract.
Experiment 5: The wire, a conductor, allows electrons to flow from the negative to the positive sphere till both become neutral (assuming they had equal and opposite charges).
Observe that in all cases, it is only the electrons that do the moving. Consider in particular Experiment 3, when a positively charged rod touches the neutral sphere, and both end up positive. The protons do not flow from the rod to the sphere. Instead, the rod starts out with a deficit of electrons it lost to the polyester. It is hungry for electrons, some of which it takes from the sphere when it touches it. The sphere then becomes positive and the rod slightly less positive. It is as if positive charge had migrated from the rod to the sphere. Likewise, the electric current in a wire is assigned a direction conventionally associated with the flow of positive charge, while in reality it is the electrons that are moving in the opposite direction. We will run into one exception: within a cell or battery, current is carried by positive and negative ions (non-neutral atoms with an excess or deficit of electrons).
1.3 Coulomb’s law
We now progress from a qualitative description of charges to a quantitative one. How do we measure or quantify q, the charge? What precisely is the force between two static charges q1 and q2 as a function of their positions? All the answers are contained in one formula called Coulomb’s law, after Charles-Augustine de Coulomb (1736–1806). Even though only Coulomb’s name is on it, his work was the culmination of many previous efforts. He did, however, give the law its final and direct verification, which is why the unit of charge, denoted by C, is called a coulomb.
Coulomb’s law says that the force between two charges q1 and q2, located at points r1 and r2 (as shown in Figure 1.3), is

Figure 1.3 The forces between two charges q1 and q2 located at r1 and r2. The force F12 acts on q1 due to q2 and is equal and opposite to F21, similarly defined.
In the formula F21 (= −F12) is the force on charge 2, due to charge 1. The figure corresponds to the case when the charges are of the same sign and hence repel. If they are of opposite sign, the forces will be reversed and describe mutual attraction.
We will spend considerable time unearthing the numerous implications of this formula.
First, notice that regardless of how q is measured, the formula shows the charges pushing each other away if they have the same sign and attracting if they have opposite signs.
Next, the formula defines a charge of one coulomb: if two charges, 1 coulomb each, are separated by 1 meter, the repulsive force between them will be 9 · 109 N.
That’s an enormous force (the weight of about 10,000,000 adults), and normally you don’t run into 1 coulomb of unneutralized charge. A coulomb arises more commonly when we consider currents: an ampere (denoted by A) is the flow of one coulomb per second and that is not unusual.

(Remember the wire is still neutral: the flowing electrons are neutralized by a static nuclear background.)
In these units the charge of the proton, denoted by e, is 1.6· 10−19 C and that of the electron is −e.
1.4 Properties of charge
Now we consider two fundamental facts about charge that are not part of Coulomb’s law: it is conserved and it is quantized.
As you know, “conserved” is a physics term for saying “does not change with time.” Electrical charge may migrate from body to body or place to place, but the total charge is conserved, provided you keep track of the signs. In a chemical reaction or in particle accelerators where all kinds of new particles are produced in a collision, the total charge of the final products always equals the total charge of the incoming products.
Charge is not merely conserved: it is conserved locally. I will illustrate what I mean by considering a conservation law that is not local. Suppose I say the number of students in the class is conserved. That means that if you count them at any time, you will get the same number. But suppose Joe suddenly disappears from the back of the room and instantaneously reappears at the front. The number of Joes is conserved. This is, however, not local conservation because Joe disappears in one part of the world and appears in another, without following an interpolating trajectory. Such nonlocal conservation laws do not seem to exist and do not interest us, since they cannot survive relativity: the disappearance and reappearance of Joe, simultaneous in one frame, need not be simultaneous in another frame. There we could have a period with no Joe anywhere or two Joes. If you want conservation laws that hold in all frames, they have to be local.
The conservation of electrical charge is local. So charge doesn’t just disappear at one place and reappear somewhere else; it moves around. As it moves we can follow this motion continuously. We can employ this notion to restate the local conservation of charge as follows. Suppose you mark off a closed region of space and (i) count all the charge inside and (ii) keep track of all charge entering or leaving the region via the boundary. You will find that the increase (decrease) of the enclosed charge is precisely accounted for by the charge flowing in (out) across the boundary. This would not have been the case with Joe: if you had counted the number of

Joes inside a region in the back of the class and another region in front of the class, both numbers would have jumped abruptly with no accompanying flow of Joe at either boundary.
The conservation of charge had been assumed from Coulomb’s time and played a big part in the explanation of the electrostatic experiments described earlier.
The second feature of charge is that it is quantized. That means the electrical charge does not take a continuum of possible values, unlike, say, the x-coordinate of any object, which can be any number you like. All the charges we have ever seen are integral (positive or negative) multiples of a basic unit of charge, e = 1.6 · 10−19 C. (Quarks are an exception, but they are always trapped inside particles like protons and neutrons. Their charge is also quantized but as a fraction of e. For example, the proton is made up of 2 quarks of charge and one of charge and a cloud of quarkantiquark pairs of net charge 0.)
Paul Dirac (1902–1984) has provided a possible explanation of charge quantization using two ideas I have not discussed yet: magnetic monopoles and quantum mechanics. I nonetheless digress here to describe Dirac’s work because by the time I cover these two topics, you may have forgotten the question we are discussing. Briefly, a magnetic monopole, if it existed, would possess an attribute called magnetic charge that comes in two signs, just like electric charge. Monopoles of like charge would repel and monopoles of unlike charge would attract with an inverse square law. All the magnetic phenomena we see, like with bar magnets, are associated with magnetic dipoles, which have net zero magnetic charge and are actually produced by electric currents. A magnetic monopole will be like a bar magnet with just the north pole, something we have not seen yet. So far we have not had direct and reproducible evidence of even a single monopole, let alone a macroscopic manifestation in the form of a magnet with just one pole. Some grand unified theories, however, predict magnetic monopoles. They are expected to be fairly heavy and to interact more strongly than electric charges. Dirac showed that if quantum theory is to consistently describe the interaction of electric charges with monopoles, all electric charges have to be multiples of some basic unit, inversely proportional to the monopole’s magnetic charge. Thus even a single monopole, anywhere in the universe, guarantees electric charge quantization. If you believe that

anything that can exist will exist, you can hope that one day these monopoles will be seen.
Let us briefly consider a few facts you may have known but not wondered about at any length.
Every electron, anywhere in the universe, is identical to every other one: it has exactly the same charge and exactly the same mass. Now, you might say, “Look, that’s a tautology, because if it hadn’t the same charge and the same mass, you would simply call it something else.” But what makes my sentence non-empty is that there are many, many, many electrons that are absolutely identical. This never happens macroscopically. Even identical twins are not identical, and cars that are supposed to be identical are not. But at the microscopic level, elementary particles like electrons are identical to other electrons anywhere in the universe, even if they were produced in collisions in different parts of the universe. That is a mystery, at least in classical mechanics, though relativistic quantum field theory gives an explanation. (Relativistic quantum field theory is a description of fields, like the electromagnetic field, satisfying the laws of relativity and quantum mechanics. It forms the basis of all modern particle theory.) The fact that they are absolutely identical makes our life easy, because if every particle were different from every other particle, we could not make many useful predictions. For example, assuming that the hydrogen atom on a receding galaxy is identical to the hydrogen atom on the earth and observing that the light from it has a shifted frequency, we deduce the galaxy’s velocity from the Doppler shift, instead of simply saying the “hydrogen” in the other galaxy is a different atom. This identity of atoms and molecules is also why structures like DNA are stable and reproducible.
Why is the charge of the electron exactly equal and opposite to the charge of the proton, given that they have very different masses and non-electric interactions? The standard model of strong, weak, and electromagnetic interaction can explain this based on a consistency condition called “anomaly cancellation.” This equality of charge is the key to the neutrality of atoms and the reason behind our existence. It is also why we can detect gravity despite its relative weakness, a point we will explore in greater detail shortly.
1.4.1 Superposition principle

We now pass to an application of Coulomb’s law when there are three charges q1, q2, and q3 at r1, r2, and r3 as shown in Figure 1.4. What will be the force on q3 due to the other two? Most students answer right away that it is the vector sum of F31 and F32 in our notation, i.e., the sum of the force q1 by itself would exert on q3 and what q2 would exert by itself. While this is indeed correct, it is not simply a consequence of Coulomb’s law. The law only says what happens when we have just one pair of charges, while the students’ answer assumes that F31, the force on q3 due to q1, is unaffected by the presence of q2. This is not a logical necessity or a consequence of Coulomb’s law, and it is not even true if effects of relativistic quantum mechanics are included. We then find that when there are three charges present, certain new forces appear that cannot be described in terms of the pairwise “two-body” interactions. In other words, studying pairs of particles in isolation will not tell us everything we need to know when more than one pair is present. However, in classical electrodynamics, which we focus on here, we may add the force q1 would have exerted on q3 in the absence of q2 to the force q2 would have exerted on q3 in the absence of q1 to find the force on q3 when all three are present. This is called the superposition principle. I repeat: this is not a logical consequence of Coulomb’s law, but an empirically established feature of classical electrodynamics that simplifies our life enormously.
Figure 1.4 The force on q3 due to charges q1 and q2 is the sum of the forces each would have exerted on q3 in the absence of the other. This is the superposition principle.
1.5 Verifying Coulomb’s law

Suppose I give you Coulomb’s law and ask you to verify it. How will you confirm the dependence of the force on q1, q2, and r = |r1 − r2|? Think about it, before reading the answers given by my class. An idea my class generated was that we keep q1 and q2 fixed and vary r, and measure the mutual force as a function of r. Here are two ideas my students came up with for measuring the force. One was to connect the charges to the two ends of a spring and watch how much it expanded (or contracted) to balance the electrical force. The other was to tie one of the charges down, let the other accelerate, and use F = ma to find the force (as a function of the starting separation r).
I should point out that I accepted any procedure that was right in principle and did not require that they corresponded to what experimentalists, who are devilishly clever, would employ in practice.
Notice that to confirm the 1/r2 dependence, we don’t have to know what q1 and q2 are, as long as we keep them fixed. Double (or triple) the r and see if the force (measured as described above) falls to one fourth (one ninth) of the initial. Of course, you need to consider a lot of values of r to truly nail down the r dependence.
Next, you want to verify that the force goes as the first power of q1 and the first power of q2. Consider the following suggestion: “Take two metal spheres, put a fixed charge on one (this will be the fixed q1) and vary the charge on the other (q2), and track the force. For example, if you halve q2, the force should drop to half the old value.” To halve the charge on sphere 2, you cannot simply say “Halve the number of electrons dumped on it,” because the existence of electrons was unknown at Coulomb’s time, and you have to play by the rules of that pre–atomic theory period. After some discussion the following acceptable strategy was generated.
Take two charged spheres numbered 1 and 2, and find the force between them. Do not touch number 1. Take sphere number 2 and bring it in contact with an identical uncharged sphere and separate them. By symmetry, they should each end up with Even though we did not know what q2 was, we know we have halved it in the process. Put 2 at the old location and see if the force has halved.
We physicists love these symmetry arguments, which transcend physics and border on philosophy: When two identical spheres are made to share

some charge, there is no reason why nature would not give each exactly half the total.
Returning to the spheres, by another splitting, you can get a sphere with
charge . By making a sphere with share its charge with an identical one carrying you can get one with and so on.
That’s how we can verify that the force depends linearly on q2. Of course, it must then also depend linearly on q1, because it’s up to us to decide which one we want to call q2.
Here is another challenge. I give you a charged sphere and I want you to find how much charge it carries, in coulombs. What will you do? When a student said: “Put it in the vicinity of a reference charge and then measure the acceleration,” I asked her how to get a known reference charge. Her answer (correct, but by no means unique) was as follows. Take these two identical spheres, each with the same unknown charge q (say by making them share 2q equally), place them at a known separation, say a meter, and measure the force needed to keep them where they are. Use Coulomb’s law to extract q2.
If you constantly think about how you would measure anything you work with, you’ll understand physics more deeply and also find solving problems a lot easier. If instead you are busy pushing symbols around and chasing factors of 2π, you will eventually be lost.
1.6 The ratio of gravitational to electric forces
Recall the claim that Fg/Fe, the ratio of gravitational to electric forces, is of the order 10−40. We have to specify how we got this number. Our task is not like selling toothpaste where one can glibly say it makes teeth 3.14 times whiter: that is a different game, not subject to any rules.
We do have to explain how we come up with 10−40. It turns out the answer does depend slightly on what comparison method we choose. There will be some variations, but they will be tiny compared to the enormous ratio, i.e., the number of zeros may range from 37 to 43 depending on the comparison method.
Consider two particles of mass m1 and m2 and charges q1 and q2, a distance r apart. We find

Fortunately the ratio does not depend on the separation r we choose for comparison since both forces fall as 1/r2. It does, however, depend on the charges and masses. If there were only one kind of particle (and its antiparticle) in the universe, we could plug in its mass and charge. But there are of course many. However, we can focus on the two key players out of which everything we see is made, the proton and electron.
If we take two electrons we get
If we take a proton and an electron, the ratio will be of order 10−40, and if we take two protons it will be of the order 10−36. Gravity is incredibly weaker than electricity, no matter how you slice it.
If gravity is so weak, how did anyone discover it? Suppose we knew only about electricity and didn’t know about gravitation. One way to find out that there is an extra force is to measure the force between two particles to a fantastic accuracy and find some discrepancy in the 40th decimal place. But that’s not how it was done, of course. Everyone seems to know the reason: the electric force, even though it’s very strong, comes with opposite charges. Consider the planet Earth. It has lots and lots of atoms and lots of charges in each atom, but every atom is neutral. The moon too has lots and lots of atoms, but they’re also neutral. So all the powerful electric forces amount to nothing, due to internal cancellations. But the mass of the electron does not cancel the mass of the proton in determining the mass of the atoms. So mass can never be hidden, whereas charge can be hidden. That’s the reason why, in spite of the incredible amount of electrical forces they’re potentially capable of exerting, the earth and moon see each other as neutral entities. In most cosmological calculations you can forget the electric force. The remaining (gravitational) force plays a dramatic role in the structure of the universe.
It is this feature of gravity, that mass cannot be hidden, that allowed us to infer existence of dark matter. Let us recall how we know of its existence in our own galaxy. If a star is orbiting the center of our galaxy, just by using

Newtonian gravity, by knowing the velocity of the object as it goes around, you can calculate how much mass is enclosed by the orbit. In case you forgot, for a circular orbit, the velocity at radius r is constrained by
where M is the enclosed mass. If you take orbits of bigger and bigger radius, you will find more and more enclosed mass, until you reach orbits as big as the visible galaxy. So far so good. But you find that as you consider bigger orbits, you still keep picking up more mass, out to some great distance. That is the dark matter halo of our galaxy. Dark matter is made of hard-to-detect particles, but its gravitational effects cannot be hidden. It occurs everywhere, even in galaxy clusters. Physicists around the world, including here at Yale, are trying to find dark matter. The problem is, we don’t know exactly what particles dark matter is made of. They are not any of the usual suspects, which would have interacted with other particles and been detected already. You have to build detectors that will detect that unknown species. And you’re hoping that one of these dark matter particles will collide with the stuff in your detector and trigger a reaction. Of course, there will be lots of reactions due to other particles. That’s called background. You’ve got to throw the background events out and hope that whatever is left over is due to dark matter. One diagnostic is that while normal particles will typically collide multiple times in the detector, we weep for joy if the dark matter particles collide even once. The particles that form dark matter are very interesting to astrophysicists and particle physicists, and there are many candidates.
1.7 Coulomb’s law for continuous charge density
We conclude with one final variant of Coulomb’s law. We have seen how to use the superposition principle to add up the pairwise forces on any one charge due to many others. But often we consider problems where the charges are continuous. (In real life everything is discrete, made of protons and electrons, but at some macroscopic scale, it will look like charge is continuous, just as water, which is made of molecules, appears to be a continuous fluid.) We tackle this variation just as we did the problem of gravity due to a continuous mass distribution: we replace the sums by integrals.

As an example, consider a circular wire of radius R with λ coulombs per meter, lying in the xy-plane with its center at the origin, as shown in Figure 1.5. I want to find the electric force it exerts on a point charge q located at a height z on the z-axis. I divide the loop into tiny segments of length dl. Consider the tiny highlighted segment of length dl perpendicular to the yzplane, as shown in Figure 1.5. It can be treated as a point charge λ dl. It exerts a force dF of magnitude
Figure 1.5 The electric force due to a loop in the xy-plane, on a charge located on the z-axis. The highlighted segment of length dl has charge λdl and exerts a force dF. We keep only the vertical part along the z-axis since the diametrically opposite segment dl∗ (shown by a dotted curve) will cancel the horizontal part.
The force vector lies in the yz-plane. We need keep only the vertical part, pointing up the z-axis, since the horizontal part will be canceled by the segment dl∗ at the diametrically opposite point, shown as a dotted curve. The total vertical force has a magnitude given by integration:

where the factor
projects out the vertical part of dF and 2π R is the integral over dl. Once you’ve done such a calculation you must think of ways to test the
result. Here are two good tests. First, if you set z = 0, i.e., find the force at the center of the loop, you should get zero since every segment that exerts a force toward the center is countered by the diametrically opposite one. This is true of our answer.
The second test is to go very far away from the loop, when it should look like a point charge λ · 2π R. How far is far? Any one length, say the diameter of my head, can be made to look impressively large or depressingly small by choosing the unit of length to be a micron or a light year. Only ratios of lengths can be described as large or small, and to be useful, the ratios should be relative to some intrinsic length in the problem. For my head to appear point-like I should be seen from a distance many times the size of my head. For the loop to appear point-like, it should be viewed from a distance z≫R. You may verify that in this limit the formula indeed reduces to the force between q and a point charge 2π Rλ, separated by a distance z.

CHAPTER 2
The Electric Field
I begin with a review of a subset of ideas from the last chapter that you will need going forward.
2.1 Review of key ideas
Several species of particles, such as protons and electrons, have an attribute called electric charge or simply charge. Others like the neutron do not. Objects with charge exert forces on other objects with charge. The coulomb is the unit for measuring charge. It is denoted by C and is defined by Coulomb’s law, which I repeat for convenience:
In the formula, q1 and q2 are charges of the particles located at r1 and r2, and F12 (= −F21) is the force on charge 1, due to charge 2.
If two charges, each equal to 1 C, are placed one meter apart, they will experience a force equal to 9 · 109 N. Once such a reference charge (or a known fraction of it) is given, any other charge may be measured using Coulomb’s law. (Here is a way to create a reference charge. We take two identical uncharged spheres, charge one by an unknown amount q and let it share its charge with the other. Each then has q/2 and the force between them at a known separation then determines q2/4.)

Charge can be positive or negative. From Coulomb’s law we see that like charges repel (i.e., F12 and F21 point away from each other) while opposite charges attract. The proton and electron have charges e = 1.6 · 10−19 C and −e respectively. The neutron has no charge. Finally, we need the superposition principle to go beyond a pair of charges. This principle allows us to compute the force on any one charge due to many others by adding their individual contributions. The force between a pair of charges is indifferent to the presence of other charges.
The total charge of a collection of charges is the algebraic sum of the charges of the constituents. As a result, an atom with an equal number of electrons and protons is electrically neutral. This is the reason we can detect the gravitation force between the earth and the moon: given their electrical neutrality, only gravity remains and is detectable despite being 1040 times weaker.
2.2 Digression on nuclear forces
Now for a brief digression. We can understand the atom as resulting from the attraction between the protons in the nucleus and the electrons. But what are the protons doing, so close to each other inside a nucleus of size 10−15 m? Why doesn’t the Coulomb repulsion make the nucleus explode? The answer, which you might already know, is that protons experience another force, the nuclear or strong force, which is attractive and much stronger than their Coulomb repulsion. If that is so, how did we manage to detect the relatively tiny electrical force hiding underneath this nuclear force? The answer has to do with the fact that the nuclear force Fn has a very different distance dependence compared to the electrical force Fe. It varies with distance roughly as
where r0 ≃ 10−15 m is called the range of the nuclear force. The electric force of course behaves as

where k includes q, ε0, etc. As a result of the different r-dependences, the ratio Fn/Fe, unlike Fg/Fe, is distance dependent:

Deep inside the nucleus, i.e.,

and

and the nuclear force dominates because A≫k. As we go to distances r≫r0, the exponential e−r/r0 completely suppresses the factor and the Coulomb repulsion wins. Of course, the crossover between the two forces is not abrupt, but occurs over the rough dimension of the nucleus.
Now for the role of the neutron in the nucleus. What are the neutrons doing here? Whereas they are nobodies with respect to Coulomb interactions, the attractive nuclear force between two neutrons or between a neutron and a proton is as strong as the nuclear force between two protons. (This is one reason protons and neutrons are collectively called nucleons.) As the nuclei get bigger, the exponential suppression of the attractive nuclear force really kicks in, while the Coulomb repulsion between protons lives on. So additional protons eventually cause instability, while neutrons contribute to the stability: they bring in nuclear attraction without the Coulomb repulsion that necessarily accompanies protons and tries to blow up the nucleus. There are far more neutrons than protons as the nuclei get
heavier. (For example, U has 92 protons and 143 neutrons.) But neutrons can only do so much: the laws of quantum mechanics force the added neutrons to have more and more kinetic energy, and beyond some size nuclei are unstable and decay into stable nuclei, say by emitting α particles, which are He nuclei, made of two protons and two neutrons.
This ends the brief digression on the complicated subject of nuclear physics.
2.3 The electric field E

Now for the main business of this chapter: the seminal notion of the electric field.
Let us rewrite the force on q2 due to q1 as follows:
What we have done is to write the force on q2 due to q1 as a product of q2 and E(r2), which is called the electric field at the location of q2.
Where does this cosmetic factorization of F21 into E(r2) and q2 lead us? First we will say that the interaction between q1 and q2 is a two-step process: Step 1. The charge q1 produces a field E(r2) at the location of q2 given by
Step 2. The charge q2 responds to the field by feeling a force F21 = q2E(r2). Thus we have split the simple Coulomb interaction into two parts: the
creation of the field by one charge and the response to that field of the other. Of course, we could just as well factorize F12 as E(r1), the field produced by q2 at the location of q1, times q1.
It will be a while before you can appreciate the cleverness behind this factorization. For now, just understand the terminology and the procedure.
Notice two things. Thing 1: While it takes two charges to feel a force, it takes only one charge to produce a field. A charge q at the origin produces the following field at point r:

where er = r/r is a unit vector in the radial direction, from the origin (where q is) to r where E is being computed.
Thing 2: The field due to q is non-zero everywhere, not just where there is another charge to feel the field.
We think of E(r) as a condition in space, produced by the presence of q. Something is different at r when q is around, compared to when it isn’t: with q present, any charge placed at r will feel a force, while without it, it will just sit there.
A field is a force waiting to happen: just put a test charge there and you will see it in action. The field of a charge is felt only by other charges.
If there are many charges, we invoke the superposition principle: the field at some r due to many charges will be the (vector) sum of the fields due to each one. You have to perform this possibly very complicated vector sum to calculate the field there. To measure it is easier: put a known test charge q at r, equate the force it experiences to qE. If q =1 C, the force and E are numerically equal but dimensionally different. This is why one says the field is the force on a unit charge.
Let us get some practice by computing E at the corner (a, a) of a square with charges q at the other three corners (0, 0), (a, 0), and (0, a) as shown in Figure 2.1. Once you get this, you can add twists: make the square into a rectangle, make the charges unequal and of different signs, and so forth.
Figure 2.1 The electric fields E1, E2, and E3 at (a, a) in terms of the unit vectors i and j, due to three equal charges q located at (0, a), (0, 0), and (a, 0). The total field at (a, a) is the vector sum of the three pieces.

The figure shows separately the three contributions E1, E2, and E3 at (a,a) due to three equal charges q located at (0,a), (0, 0), and (a,0):
I hope you can see why the field E2 is half as big as the other two, and points in the 45◦ direction, i.e., along i +j. The corresponding unit vector is obtained by dividing i + j by its length, It is easy to add the three pieces to get the total field at (a,a):
We will soon be doing numerous versions of this problem, computing the field due to various charge distributions, discrete and continuous. But at the outset I must warn you that Coulomb’s law, as stated, violates relativity. Suppose you and I hold two positive charges q1 and q2, and I am one lightyear away from you. You hold your charge in place by pushing against the repulsive force mine exerts. Now I suddenly move mine away from you by a bit. You will feel the reduced repulsion right away. I have managed to send you a signal instantaneously and this faster-than-light signaling is disallowed.
Does electrodynamics then violate relativity? No, we will see it is remarkably compatible with it. What happens in the complete theory is that if I wiggle my charge, the signal will reach you a year later, traveling at the speed of light. Until such time, the field at your location due to my charge will remain unaltered.
Of the two parts of the story, the computation of E in terms of the charges and the response of a test charge q to the field, only the former gets modified in the complete theory. The field at some spacetime point, say (r = 0, t = 0), will receive contributions from all other charges based not on what

they are doing now, but what they were doing at an earlier time. The amount by which we have to go back in time is just the time light would take to go from the source of the field to (r = 0, t = 0). A charge that was a light-year away a year ago will be contributing to the field at (0, 0). This is called the retarded interaction. We will discuss this briefly toward the end of chapter 15.
What is the role of Coulomb’s law then? In principle it is to be used when none of the charges is moving. In this case, the delay does not matter: since every charge knows where every other charge is, all the signals have arrived and are unchanging. In practice we also use Coulomb’s law provided the charges in question are near each other and moving at and retardation effects are negligible, as in most electrical circuits.
Remarkably, the second part of the story, the equation giving the response to the field, F = qE(r) remains unaltered in the final theory of electrodynamics. It is a local relation between the field at a spacetime point and the charge at that point. The field at any point could be a very complicated function of every charge in the history of the universe, but the response (of test charge q) to it depends only on its current value at the location of the charge q. It does not care what went into producing E. So the field concept is essential to making electromagnetic theory compatible with relativity.
2.4 Visualizing the field
Let us go back now to the simplest problem in the world: the electric field due to one charge. The formula is very simple. Let’s put that charge q at the origin. The electric field is
where er is the radial unit vector, r/r. Sometimes you see Eqn. 2.18 rewritten as

If you encounter this version, do not get fooled into thinking the field is falling as r−3. It’s still r−2 because there’s an extra r at the top.
So here you have this formula. If you’re a person who likes to work with formulas this is all you need. You manipulate the stuff on paper, and you add different fields. But people like to visualize this. How do we visualize this? That’s the real question. Suppose someone asks you, what’s the height above sea level of a certain part of the United States? You’ve got some mountains. You’ve got some valleys. Somebody can give you a function that gives you the height at any point in the United States, but it’s more revealing for most of us to have some kind of a contour map. Each contour is a different height. If you go hiking, you want this map, not the corresponding function. Similarly, you want a pictorial representation of this electric field. Unlike the height function, which is a scalar, i.e., just a number at each point, the electric field is a vector E(r) at each point r.
Suppose I want to communicate to you pictorially the information contained in the function in Eqn. 2.18. I begin with the modest goal of describing E at just one point, 1, in Figure 2.2.
Like many of the figures I will show you, it is a two-dimensional cross section of a three-dimensional configuration.
I take that point and draw an arrow there to represent E(1). The length of the arrow gives the size of the field in some scale, so many centimeters of length for each newton/coulomb of field. That is the electric field at that point 1. Then I pick a few points, say eight in all, at the same radius. (The points lie on a circle in the plane of the paper, while real charges live in three dimensions. You should think of this as a cross section of what happens on a sphere of the same radius.) The points are also uniformly distributed to reflect the isotropy of the electric field. The figure is already telling you something: the field is radially outward and same in magnitude at points with the same r. Be very careful about what it is not telling you. An arrow is not telling you what is happening throughout the length of the arrow. It’s telling you what’s happening at the starting point, the tail. You understand the arrow is in your mind. It’s not really sticking out in space. It’s a property or a condition at that starting point, but we’ve got to draw it somehow, so we draw it that way. (If E were the velocity of a fluid in a river, the arrow starting at some point r would be the velocity at r only, even if the arrow is a foot long and passes over regions where the actual water velocity is totally different in magnitude and direction.)

Figure 2.2 The electric fields E, at a few representative points on circles of two different radii around a charge at the origin.
What happens when we go further out in r? If I put a test charge further away, it is still going to be repelled radially, but less. So I draw a few arrows at representative points 9–16 and make them shorter, to reflect the 1/r2 nature. I can draw a few more arrows and hope you get the picture from the few discrete sampling points.
Then someone had this clever idea: join all these arrows as in Figure 2.3. These are called field lines.
The actual charge and lines should be drawn in three dimensions but Figure 2.3 shows what happens in a representative plane, which I assume for illustrative purposes has 8 lines. What have I gained and what have I lost? Previously I knew the field direction only at the chosen points at some radii, but now I know it throughout each line. On the other hand, I have lost information on the magnitude of the field: the arrows, whose lengths encoded |E| are gone and replaced by the field lines that go on forever. They merely tell me in which direction E points, but not how big it is. They just tell me that the charge is pushing every (positive) test charge out radially and that the force is isotropic.

Figure 2.3 The electric field lines due to a charge at the origin. The actual charge and lines live in three dimensions, and the figure shows what happens in a representative plane, which I assume has 8 lines.
But, thanks to a miraculous property of the Coulomb force, namely that it falls like 1/r2, there is information even on the strength of the electric field. That information is contained in the density of electric field lines. By density of lines, I mean the number of lines crossing a surface perpendicular to the lines, divided by the area of that surface.
To grasp this, let us pick some convention, that for every coulomb of charge, we will draw a certain number of lines emanating from it, say 64. If we draw a sphere of some radius surrounding the coulomb, 64 lines will cross that sphere, everywhere perpendicular to the surface and of uniform density, reflecting the isotropy of the electric field of a point charge. If I draw a bigger sphere, the same 64 lines will cross that sphere also, but they will be less dense, with fewer lines per unit area. Since the area of the sphere grows as r2, the density of lines will fall as 1/r2. This is exactly how the field strength |E| ≡ E falls with distance.
This wonderful ability of field lines to encode the magnitude and direction of the field exists only because we are living in three dimensions (where the sphere surrounding the charge has an area that grows as r2) and dealing with a field that falls as 1/r2. For example, if a radial field that falls as 1/r3 is represented by such lines, their direction will faithfully represent the direction of the field, but their density will not represent the field strength E.

The lines help you visualize the field strength. Wherever the lines are dense, the field is strong. Wherever the lines are spread apart, the field is weak. It is a very precise statement. The only thing not precise is how many lines you want to draw per coulomb. That is really up to you, but you must be consistent. Once you choose 64 lines per coulomb, and you are dealing with a charge of two coulombs, you should draw 128 lines coming out of it uniformly spread out. As long as you do that, the number of lines crossing per unit area will be proportional to the field.
Now, no matter how many lines you pick per coulomb, there will be spaces between the lines. That does not mean the field is zero between the lines. The field is continuous in space and not concentrated literally on these lines. You must read between the lines. For example, at a point midway between two adjacent lines, the field is pointing midway as well, with an intensity given by the density of the lines at that radius.
Figure 2.4 The electric field due to a dipole. The two vectors shown at point D are the contributions E+ and E− to E from the two charges. Their vector sum will be horizontal.
Clearly, if we consider the field of a negative charge, the lines will point inward, reflecting the attraction felt by the test charge.
The notion of field lines extends beyond the field due to just one charge. Figure 2.4 shows the field due to a pair of charges ±q, called a dipole. The first thing I want you to notice is that very close to any one charge the lines

point uniformly and radially out or in depending on its sign, no matter how many other charges there are. This is because as we approach any charge, the field it produces diverges as 1/r2 and swamps the finite contributions from the others.
Next consider the field lines labeled A, B, C and D. Look at line A. It is clear that a (positive) test charge placed anywhere on A (which goes from the minus charge all the way to infinity) would be attracted to the minus end of the dipole, which attracts it more than the plus end repels it. The reverse argument explains the outward pointing line B, on which repulsion wins. On the line C, pointing from the plus to the minus, both charges apply a force to the right. Finally, look at the line labeled D. It contains the point D, which lies on the perpendicular bisector of the line joining the two charges. Notice the field line at D is horizontal. This directionality follows from a symmetry argument. The minus charge attracts the test charge on a line from D to itself; the plus charge repels it along a line joining it to D. Both forces have the same magnitude (since D is equidistant from them), canceling vertical components, and additive horizontal components.
The figure also makes it clear that any closed surface enclosing only the plus (minus) charge will intercept 10 lines going outward (inward). If we draw any surface enclosing both charges, the net flow in or out will be zero. This is your qualitative introduction to Gauss’s law, which relates the net (outgoing minus incoming) number of lines leaving a closed surface to the net enclosed charge. (If the surface is convoluted a field line may exit, reenter, and exit again for example. This will count as a net exit of one line.)
Figure 2.5 shows two identical positive charges. Far from both, it will look like the field of a point charge of double the strength. The number of lines crossing a closed surface enclosing both charges is the sum of the lines emanating from each. The closed surface I have shown is a nice ellipse, but the lines crossing it will not change if I distort it in any way that does not exclude either charge (once again an example of Gauss’s law).

Figure 2.5 The electric field due to two positive charges. Far from both, it looks like the field of a point charge of double the strength. The number of lines crossing a closed surface enclosing both charges is the sum of the lines emanating from both.
What if we had two opposite but unequal charges, say 10 C and −5C, and associate 10 and 5 lines with each? You can draw the sketch yourself with the following features: near each charge you can forget the other, 5 lines will flow from the 10 C to the −5 C, and the rest will escape to infinity, becoming radially outward asymptotically, like those of a point charge 10 C −5 C = 5 C.
Consider finally a case of a continuous charge distribution. Two parallel metallic plates carry uniform charge densities ±σ (measured in C/m2). This is called a parallel plate capacitor and is depicted in the top half of Figure 2.6. At the left is the view looking down at an angle and at the right the view end-on, with the plates coming out of the paper. What do the field lines look like? We know they must start at the positive charges and end at the negative charges. The bottom half shows the deflection of a positively charged particle injected from the left with velocity v0.
If you imagine the plates to be very large in area, the figure shows the part far from the edges. (At the edges the lines bulge out a bit midway between the top to the bottom plate.) You should not simply accept even the qualitative aspects of the preceding picture. Look very near the positive

plate. In the absence of the negative plate, the field lines will be emanating perpendicularly away from it with equal density above and below by symmetry. The same goes for the negative plate, but with the lines flowing into it. If you superpose the two plates, you can see the two plates aid each other in the region between, with both producing downward pointing fields there, just like along the line joining the charges in a dipole. But, if you follow the dipole analogy, and consider points just above the top plate, you expect the fields from the two plates to oppose, but with the upper plate winning since it is closer. So some lines must point up just above the upper plate. Yet the figure shows no lines above the upper plate and has all the lines coming straight down, as if there is a perfect cancellation of the fields due to the two plates, despite the different separations. The same goes below the lower plate, where there are no field lines. The answer to this mystery will be revealed when we compute the field due to each plate later and find that the field due to an infinite plate of uniform density does not weaken at all as we move away from the plate! It is perpendicular to the plate, and it has the same magnitude no matter how far we go, even though the contribution from the individual charges on the plate falls as 1/r2. Consequently, the plates cancel each other completely outside the plates (above the top plate and below the bottom plate) and aid each other inside. So, the figure is correct only if it represents a finite section of an infinite parallel plate capacitor, or far from the edges of a very large capacitor. The real finite plate problem is far more complicated: doable in principle, but not easy.
Figure 2.6 The top half shows two views of a parallel plate capacitor and the field inside it. It is uniform except near the edges, where it bulges out (not shown). The bottom shows the trajectory of a positively charged particle shot into it from the left.

Given that the field is limited to the space between the plates, questions still persist. Why is the field uniform between the plates in the infinite capacitor, unchanging as we move up and down or side to side?
First of all, it must be clear that in the infinite capacitor, the field at a given plane parallel to the plates, say at a height y = 2 cm above the lower one, cannot vary as we move parallel to the plates, say in the x-direction. Every point at some y is like every other point: if we look to the left or right, from any of these points, we see the two plates running to ±∞.
Here is a more detailed argument, based on cause and effect. Suppose the field varies with the x coordinate, i.e., has a non-trivial profile with some features, some ups and downs in strength. If I slide the plates to the right by 2 cm, these features should follow. On the other hand, I can argue that they should not shift since the cause behind the field, namely the infinite, charged plates, looks exactly the same before and after I slide them. If the plates look the same after a horizontal shift, so must the field they produce.
Had the plates been finite, this would no longer have been true. There would have been a preferred midpoint and edges where the plates end. If you move this finite system horizontally, it will look different after the shift and so the field need not be x-independent. Indeed, it is not, with bulges at the end.
So the field is constant in x. Why is the field independent of the y coordinate as well? After all, the y dimension is finite and as a result not all y’s are equivalent. We can tell if we are moving toward or away from either plate. Well, suppose the field got weaker as we approached the middle. The lines must spread out, i.e., the spacing between them must increase. But this is impossible in the infinite case: if you move a line, say second from the left in Figure 2.6, away from its neighbor on the left, to weaken the field to the left, you move it closer to the neighbor on the right, increasing the field between them. Such variations with x are not allowed in the infinite capacitor, as we have seen. So the lines have no choice but to go straight down, preserving their density as y varies. Again, variation in x and y is allowed in a finite capacitor: the lines do get less dense as we move toward the center, and they bulge out at the two ends.
In any event, if the field is uniform, the force will be uniform, just like the force of gravity near the surface of the earth. Consequently the particle we shoot in from the left will follow a parabolic path, as depicted in the lower half of the figure. More on this later.

2.5 Field of a dipole
We will now buckle down and calculate the precise value of the electric field due to a dipole. We will write a formula that is good at all points, but evaluate it only at some select places where the calculation is easier. We will examine the field at distances large compared to the separation between the charges. In a later chapter we will find a more efficient way to find the field using the notion of a potential.
Figure 2.7 shows a charge q at (a,0) and a charge −q at (−a,0). Consider the field at a generic point (x, y). (Once we have the field in the xy-plane, we can simply rotate the figure around the x-axis to get the answer in three dimensions. In other words, the cross section on the xy-plane is identical to what we will find in any other planar slice through the x-axis. This point will be fortified soon with symmetry arguments.)
Figure 2.7 Dipole field: E± are due to ±q located at (±a,0).
Recall that the field at the point r due to a single charge q at the origin is
If the charge were not at the origin (as in the application that follows immediately), r would be the vector from where the charge is to where we want the field.
The field due to both ±q at a generic point (x, y) is the sum of the individual contributions E±. These in turn can be evaluated by setting r = r± in Eqn. 2.20 and adding them as follows:

This general formula may be a bit hard to digest. Here are some simpler special cases.
At a generic point on the x-axis (y = 0) both E± are horizontal and

(Remember that lim

and not .) For a point

like A with x > a, we can drop the absolute value sign and obtain is called

the dipole moment. The dipole moment is the product of q and the vector

2ai going from the negative to the positive charge.

For x≫a, the field becomes
because r, the radial distance from the center of the dipole to (x,y), equals x when y = 0.

For a point on the axis like E with x < −a you should go back to Eqn. 2.24 and verify that the field is invariant under x → − x and also points along the positive x-axis.
On the y-axis, at a point D with coordinates (0, y), I leave it to you to show that
For y≫a, the field becomes
These results with E ∝p, when x → ∞ or y → ∞ are to be expected. If we set a = 0 in Eqn. 2.23 for the sum of E±, we get E ≡0 as we must: the two charges sit on top of each other and fully neutralize each other. The total E as a function of a vanishes when a = 0. The net field is non-zero only because a ≠ 0 and the non-zero part will start out as the first power of a in a Taylor series (Chapter 16, Volume I). To keep the dimension of the field E the same, the extra a must really be , which is what we find in Eqns. 2.29 and 2.31 since p = 2aqi is proportional to a.
Recall that the field of a single charge, which looks like a hedgehog, is isotropic. If I rigidly rotate the distribution of field lines around any axis passing through the origin at any angle, they look the same. We may demand this on the basis of the following symmetry argument. You must agree the charge is the cause and the field is the effect. The effect cannot change if the cause does not. Rotating around the origin leaves the point charge alone: it stays where it is and, being a point, looks the same as well after the rotation. It follows that the resulting field distribution must be unaffected by rotation.
On the other hand, even if the dipole looks like a point as we go far away, E is not isotropic. The field knows that the dipole near the origin has chosen a direction in space, defined by p, unlike a single charge, which does not do that. A generic rotation around an arbitrary axis passing through the origin will change the orientation of the dipole (the cause), and the field (the effect) will change accordingly. On the other hand, a rotation around the

axis of the dipole will leave it alone and the E configuration it produces should be unaffected by such a rotation. This is why we were satisfied with finding E in the xy-plane. The answer in any other plane may be found by a rigid rotation around the x-axis.
2.5.1 Far field of dipole: general case Far from the dipole, the general formula Eqn. 2.23 simplifies, though it takes some more work to extract the part linear in a. Following the details will enhance your mathematical prowess if you suffer through them. Let us begin with the exact result
The answer is some function of a (and of course x and y), which vanishes at a = 0. Near this zero, the function will have a Taylor expansion in a. By dimensional analysis, the series has to be in a divided by a length and the only possible candidate is r, the distance from the center of the dipole. We are content to find just the first correction to zero. It will be proportional to a or, equally well, the dipole moment p = 2aqi.
Eqn. 2.32 has two parts, each with a numerator divided by the denominator, or the numerator times the inverse denominator. We can get the single power of a from either term and the a0 term from the other. If we get a1 from the numerator we may set a = 0 in the denominator and vice versa.
Consider the contribution from the positive charge
Here is some explanation. In the last line, the first term comes from keeping the a0 term, namely r, in the numerator and keeping up to linear terms in

the denominator (and hence dropping the a2 in the expansion of [x − a]2). The second term comes from keeping the a term in the numerator and setting a = 0 in the denominator. The terms kept are then The E− terms are obtained by changing q → −q, a → −a: to give a total of
E(to order a)
where I have invoked p = 2aqi, p · r = 2axq, and applied (1 + z)n = 1+nz + . . . , to obtain

2.6 Response to a field
Having seen how to find the field in a variety of situations using Coulomb’s law, let us now consider the response of charges to the field using F = qE, starting with the parallel plate capacitor with a uniform field E = −jE0 in between the plates, as indicated in Figure 2.6. Suppose I shoot a particle of mass m and charge q from the left, with a velocity v0. What will be its position and velocity as it exits the plates?
The force on the charge is a constant, F = −qE0j, just like the force of gravity, which will produce an acceleration
The particle will follow a parabolic path given by
To compute its y coordinate when it exits the capacitor, we need to know for how long it “falls” at the rate above. That time is clearly t∗ = L/v0 where L is the width of the capacitor. (Even though the capacitor is of finite width, we use the constant E field of the infinite capacitor as a simplification.) As in the case of gravity, the time to go a certain distance horizontally is determined by the initial horizontal velocity and is unaffected by the acceleration in the vertical direction. So if you set t = t∗ in r(t), you will find out where it will end up.
Here is one way in principle to make pictures on television: shoot electrons from the left into the region between two pairs of plates, one as shown (perpendicular to the page) and another pair parallel to the page, with one member of the pair above and one below the page. This will cause

motion up and down and also in and out of the page. Place a fluorescent screen at the right, perpendicular to the beam. If you apply the right electric field, the electron will land on the screen and make a little glowing dot just where you want it. By scanning the screen many times a second, and by varying the field appropriately and modulating the intensity of the beam, you create the impression of a steady picture. (Actually, magnetic fields were used to deflect electrons in old cathode ray tubes.)
Figure 2.8 The forces and torque τ on a dipole p due to a uniform horizontal field E. The torque, computed with respect to the negative charge, has a magnitude τ = 2aqEsinθ and tends to align it with the applied field. The vector τ = p ×E vanishes only when p and E are parallel or anti-parallel.
2.6.1 Dipole in a uniform field What is the force of a uniform electric field on a dipole? Figure 2.8 shows a dipole made of charges ±q a distance 2a apart in a horizontal uniform electric field. It is assumed the charges are mounted at the ends of some rigid structure, like a rod. The force on the two charges is ±qE as shown. So the dipole as a whole will not feel any net force, because the two parts are getting pulled by opposite amounts. (If the electric field were not uniform, say it were stronger at the plus charge, the dipole would accelerate to the right.) The forces, which add up to nothing, collaborate in producing a torque. I hope you can see that the torque wants to align the dipole with the field. Recall that when the total force vanishes, the torque may be computed with respect to any point. Choosing it to be the location of − q, we find it has a magnitude (see Figure 2.8)
which turns clockwise. As a vector, the torque is given by the cross product

which points into the page. If you mount this dipole so it can swing in the plane of the paper, you could use it as an “electrical compass,” which will point along the local electric field. (We assume the rod supporting the charges at its ends has a non-zero moment of inertia I and the support has some friction, so that if it started out non-parallel to E, it will quickly align with E after some damped oscillations.)
The torque also vanishes when the dipole is anti-parallel to E. This is a state of unstable equilibrium: if disturbed, it will not return there but end up parallel to E. We can understand this in terms of energy.
Recall that a conservative force F(x) and the associated potential U(x) are related as follows:
Next recall the SAT analogy: “Torque is to force as angle is to displacement.” The torque here is τ = −pEsinθ, where the minus sign reflects its tendency to rotate the dipole clockwise, in the direction of decreasing θ. So we may now write
In going from Eqn. 2.52 to 2.53 we have dropped a possible additive constant in U(θ).
You see in Figure 2.9 that U(θ) is an inverted cosine with a minimum at θ = 0, which is a point of stable equilibrium, and a maximum at θ = π, which is a point of unstable equilibrium. The points ±π are one and the same. When perturbed about θ = 0, the dipole will execute simple harmonic motion. For small angles, κ, the restoring torque per angular displacement,

and ω, the frequency of oscillations, will be (in terms of the moment of inertia I)
Figure 2.9 The potential energy of a dipole, U = − pEcosθ, as a function of the angle θ it makes with a field.

CHAPTER 3
Gauss’s Law I
In the last chapter we learned that we should think in terms of electric fields and not direct action-at-a-distance between charges according to Coulomb’s law. In this parlance, we say charges produce fields as per Coulomb’s law, and the fields in turn act on charges as per F = qE.
The field E(r) is a condition at a point r, even if there is no charge at that point. This condition is revealed when we place a test charge q there and find a force qE(r) acting on it. The field due to many charges is the sum of the fields due to each.
Strictly speaking, Coulomb’s law is to be applied only in a static situation when the charges do not move, though we do apply it in some situations where they move slowly compared to c, as in circuits. In this chapter, we will assume a static distribution of charges and apply Coulomb’s law.
We saw how field lines can depict the state of the electric field: the lines point along the local field, and their areal density (lines per unit area perpendicular to the lines) is proportional to the field magnitude. We could use any number of lines per coulomb, but once we agreed on a convention, say 64 lines per coulomb, we had to stick to it. We looked at the field lines of a dipole as well as that of two equal charges.
We considered the dipole field quantitatively. The answer was expressed in terms of p = q(r+ − r −), the dipole moment of charges ±q located at r±.
We found a general expression for the field due to a dipole. We evaluated it exactly along the dipolar axis and perpendicular to the dipolar axis. As for a general direction, we considered the field only for distances r≫a, the distance between the charges. The main point was that the leading term for E fell as 1/r3.
We studied how charges responded to a field. We saw what happened to a charge shot into the space between plates of a capacitor, where the field was assumed to be uniform and perpendicular to the plates. Finally, we saw that a dipole moment in a field experiences a torque, p × E, which tries to line it

up with the field. With that torque one can associate a potential energy U = −p · E.
3.1 Field of an infinite line charge
Here is a standard problem. We have an infinite line of charge parallel to the x-axis, say a charged wire, of which a finite part is shown in Figure 3.1. Somebody has sprinkled it with a continuous density of λ coulombs per meter. If we cut out one meter of this wire, we will find λ coulombs there. We want to compute the electric field everywhere due to this charge distribution using Coulomb’s law. (Let us assume λ is positive; if it is negative, we just have to reverse the field everywhere.)
Consider a point P = (0,a) at a distance a from the wire. What can we say about the field there without doing the full calculation?
Figure 3.1 The field due to an infinite line charge with linear charge density λ. It is found by adding the contributions from tiny segments of width dx treated as point charges. The figure shows clearly that the fields dE1 and dE2 due to segments at x and −x have the same y-components and opposite xcomponents.
First of all, it must be intuitively clear that the field will be the same at all points at the distance a from the wire. Any x-dependence leads to the following contradiction. Suppose the field had a variation in the x-direction. If I slide the wire to the right by some amount, this pattern will shift by that amount and look different. On the other hand, since the wire looks the same before and after the shift, so must the field it produces. If the cause (the wire) looks the same after a shift in x, so must be the effect, the field it produces.
The field may, however, depend on y and it does. Next we may argue that the field has to point radially away from the wire; it cannot be tilted to one side or the other since the infinite wire does

not distinguish right from left. Here is another way to say it. Suppose the field were tilted to the right. Now rotate the wire by π around an axis perpendicular to it (the y-axis in Figure 3.1) and passing through P. The field lines will rotate as well and end up tilted to the left. But the rotated wire looks the same as the unrotated one and so must the field it produces. The only configuration that is unaffected by this rotation is a field that is everywhere perpendicular to the wire.
The argument fails if the wire is finite. A finite wire has some distinct features and special points like the midpoint and end points. It does not look the same if you slide it parallel to itself and so the field can vary with x. The field lines may tilt toward the left end if the point P is left of center and likewise to the right for points to the right of center. This distribution will still turn into itself under any operation that leaves the wire invariant, such as the above-mentioned rotation by π about its midpoint.
Returning to the infinite wire, let us find how the perpendicular field varies with a.
Look at Figure 3.1. Let us take a segment of wire centered at x and of length dx, which is so small that we can treat it as a point charge. Now the dx as drawn is not a point, but in the end, we’re going to make it arbitrarily small. The segment is like a point-charge q = λdx at a distance x from the origin. The infinitesimal electric field it produces at P has a magnitude
and points along the vector joining (x,0) and the point P. We need only keep its y component since the ultimate x component has to be zero, either by our earlier symmetry arguments or by the explicit consideration of the contribution from the similar segment at −x. Convince yourself by looking at the figure that when the two contributions are added, the horizontal parts will cancel and the vertical part will be double that due to either segment. Let us therefore double the vertical contribution from the segment on the right but remember to consider only x ≥ 0. Using
to project out the vertical part, we find the total vertical field by integration:

What next? The integral can be done by a clever substitution. What if that trick does not occur to us? It turns out we can go quite far by dimensional analysis. Let us express the coordinate x in terms of a, the only length in the problem, via the dimensionless variable w as

Then the limits for the integral over we have

are 0 and ∞. Since dx = adw

Thus we have the answer up to an overall multiplicative constant N, which is independent of λ and a. Even before we evaluate N we see a surprising thing: the field falls like 1/a and not 1/a2, even though each piece of the wire makes a contribution that falls like the inverse square of the distance. This is surprising but also inevitable for dimensional reasons. The answer now had to be proportional to λ, which is a charge per unit length and not charge. The presence of the prefactor λ robs the denominator of one power of the length, leaving behind one power of the only length in the problem, which is a.
This argument fails if the wire is finite: now we have another length L, the length of the wire, which can bring in factors like L/a (or any function of the dimensionless variable L/a) without messing up the dimensionality of

the answer. Indeed, in this case we expect, and find, that if we go to distances much greater than L, the field will be that of a point charge λL.
Let us now evaluate N by making the substitution
(I call the substitution variable θ rather than some other Greek letter, because in this case it is actually the θ in the figure: w = tanθ means x = atanθ.) Observe that the change of variable is such that all possible values of w can be obtained by some choice of θ because tanθ can go from 0 to ∞. Had we made the substitution w = cosθ, we could never obtain w > 1 (or x > a).
Continuing, we find
For future use remember that the field due to an infinite linear charge density at a distance a from the line and lying in the xy-plane has a magnitude
and points perpendicularly away from the wire. While Figure 3.1 is two-dimensional, the wire and the field pattern live in
three dimensions. What we have in the figure is a slice taken through the xy-plane. The full field configuration will be obtained by rigidly rotating the configuration shown about the x-axis. We can slice that three-dimensional configuration through any plane passing through the x-axis and we will get the same two-dimensional configuration. This is demanded by symmetry or

by the cause-effect relationship. If I rotate the wire around the x-axis by some angle, it looks the same. Therefore the field configuration it produces should also be invariant under that rotation. The field pattern in which the lines radiate uniformly and radially away from the line is the unique electrostatic configuration meeting this requirement. (The lines could also point radially inward, but that would correspond to negative λ < 0.) The end view, with the wire running perpendicular to the page, is shown in Figure 3.2.
Figure 3.2 The field due to an infinite line charge seen end-on, with the wire perpendicular to the page. The wire and the field distribution it produces are invariant under a rotation of the wire about itself.
In three dimensions, it is common to denote the distance measured perpendicular to the wire by ρ (and not a). So we should write
where eρ is a unit vector in the direction perpendicular to the wire. (In the xy-plane, eρ = ±j.)
3.2 Field of an infinite sheet of charge
Imagine an infinite plane with an areal charge density σ depicted in Figure 3.3. This means that if you cut out a tiny part of it, of area dA, it will have a charge σ dA. (By convention, λ stands for charge per unit length, σ for

charge per unit area, and ρ for charge per unit volume.) We want to stand at a point P a distance a from the plane and ask for the field there.
Figure 3.3 The field due to an infinite plane with charge density σ. Shown is one contributing annulus of radius r and thickness dr. It produces a field dE perpendicular to the plane, as shown by the long dark arrow. There is no parallel part due to cancellations between parts of the annulus that are diametrically opposite. Shown are two such contributions, dE1 and dE2, due to the two darkened parts of the annulus. The sum of such vectors due to all parts of the annulus is dE. The integral of dE over all annuli will give the final E due to the entire plane.
Once again, before jumping into the calculation, let us see what features follow from general considerations.
I think we can agree that the electric field at some point a meters in front of the plane will be independent of the other two coordinates parallel to the plane. Suppose the field varied as we moved parallel to the plane at fixed a, with some ups and downs in field strength. If I move the plane to the right by one inch, the pattern should follow. But the shifted plane looks exactly like the unshifted one. It has to produce exactly the same field, of the same magnitude and direction. This can happen only if the field does not vary under displacements parallel to the plane.
As for the direction, it has to be perpendicular to the plane, again for symmetry reasons. If you tilt it away from the perpendicular, which way would you tilt it? The infinite plane defines no unique direction except the one perpendicular to it. Suppose the field is tilted away from the perpendicular by an angle of 30 degrees, say in the direction of dE1. If I now rotate the plane around an axis perpendicular to it and passing through P, the direction of the tilt will rotate as well (ending up parallel to dE2 after

a rotation by π). But the rotated infinite plane looks the same as the unrotated one, and so must be the field it produces. The only field configuration that meets this demand is the one where the field is everywhere normal to the plane.
We can also argue from symmetry that the magnitudes of the field should be the same at two points that are on opposite sides of the plane and at the same distance from it. If the charged sheet lies in the xy-plane we require that E(z) = E(−z) = E(|z|). The charges on the sheet repel a test charge at a given distance from the plane with the same intensity whether the test charge be on one side of the plane or the other. The directions will of course be opposite, pointing away from the plane. Thus we may assert that
where k is a unit vector along the z-axis. While this is intuitively obvious, we could provide the cause-effect
argument by demanding that the field configuration should be unaffected if the charged plane is flipped over like a pancake by a rotation around the xaxis by π since the plane looks the same before and after. The configuration written above meets that requirement.
Armed with these anticipations based on general symmetry arguments, we turn to the calculation that will yield results in agreement with our expectations. Our strategy is as follows. We will draw a perpendicular to the plane passing through the point P where we want the field, as shown in Figure 3.3. We will divide the plane into concentric annuli or rings of radius r and width dr, find the contribution dE from each ring, and integrate them over all rings.
The contribution from a given ring may be readily inferred from Eqn. 1.12 for the force on charge q due to a ring carrying a linear density λ, at a point on its symmetry axis, z meters above it:

The first factor is the q1q2/(4π ε0) appropriate to the test charge and the loop, the second reflects the inverse square law, and the third is the cosine factor that projects out the component perpendicular to the plane of the loop, which alone survives when all contributions from the loop are added.
We may import this result after three modifications:
• Drop the test charge q to get the field from the force. • Set z = a. • Relate σ to λ, the charge per unit length of the annulus. A segment of length 1 along the annulus will have an area 1 · dr and contain 1 · σ dr coulombs. Thus the linear charge density in our problem is related to the areal charge density by
The resulting field, at a distance a from the plane, is
Since E in Eqn. 3.20 is the infinitesimal contribution from a ring of
⊥
infinitesimal width dr, we rewrite it explicitly as an infinitesimal
and obtain the total field by integration

Once again we may use scaling to figure out the a-dependence as follows. Setting
we obtain
The integral equals 1, as can be shown by the substitution z = w2. The preceding result is so important, I will repeat it and suggest you
memorize it: Field of infinite plane with charge density
The most striking aspect of the result is that the field does not decrease as we move away from the plane. It is independent of a, the perpendicular distance to the plane. Since each part of it makes a contribution that falls like 1/(a2 + r2), and we are increasing the distance to every segment of the plane as we increase a, the field should get weaker, right? And yet that does not happen.
We can understand why this had to be so on dimensional grounds. The field has dimensions of charge over distance squared. (Forget the ubiquitous 4π ε0, which is a constant.) For a single charge q the distance in question had to be r, the distance between the charge and the field location. For a

line charge, the answer had to be linear (by the superposition principle) in λ, which had units of charge over length, leaving room for just one length a, the distance to the wire, to appear in the denominator. For the plane the inevitable factor linear in σ, which has units of charge over distance squared, has used up all the inverse powers of length, leaving no room for a to appear either in the numerator or denominator. As I mentioned before in connection with the wire, the argument fails if the plane is of a finite extent, say a square of side L. In this case the answer is allowed to have factors like L2/a2 and indeed it will: for a≫L, the field will be that of a point charge q = σ L2.
To understand the a-independence of E in pictorial terms, consult Figure 3.3. Let us start at some a and reduce it to get closer to the plane. We find the contributions from individual segments of each of the rings do indeed go up since a2 + r2 decreases. However, the contributions, which point along the line joining the segment to the field location, become increasingly parallel to the plane as we approach it. (Look at dE1 and dE2 in the figure.) But we have seen that the parallel part gets canceled by symmetry (within each ring) and only the (tiny) perpendicular part survives. So there are two opposing factors as we get close to the plane: the contributions from individual segments of any given ring get bigger, but the useful component that survives the sum over segments, the perpendicular part, gets smaller. So you can give arguments why the field should get weaker and arguments why it should get stronger as a varies. To show that these two tendencies exactly cancel, you have to bite the bullet and do the calculation.
We can now find the field between the plates of a parallel plate capacitor (ignoring edge effects) with ±σ. In the region between the plates the fields due to the two plates add to a total of σ/ε0, pointing from the positive to the negative plate. In the region outside the plates the field vanishes because the two fields cancel, being of equal and opposite strength and independent of distance.
3.3 Spherical charge distribution: Gauss’s law
Now we turn to the more difficult case of a spherical charge distribution. Rather than attack it frontally, I will introduce you to a powerful idea called Gauss’s law, which will provide a shortcut.

Imagine a solid ball of charge density ρ (measured in C/m3). We want to find the field due to this ball.
Now, when we did a similar problem in gravitation, we assumed that when you’re outside the sphere, the whole sphere acts like a point mass with the entire mass sitting at the center, and that when you are inside (as in our analysis of dark matter), the mass inside the chosen radius acts like a point mass at the center and the mass outside does not contribute.
Since the electrostatic force also obeys the inverse square law, it should not be surprising that we may replace the word “mass” by the word “charge” in the preceding paragraph. But now we want to prove all this, rather than assume it.
This is what took Newton a long time. He knew it was true but he couldn’t prove it, because for that, he had to first develop integral calculus. Even today, to find the field due to a sphere using integration is quite difficult. Think about what you have to do. Look at Figure 3.4. You want the field at point P at location r. You have to divide the sphere into tiny little cubes centered at r′, each carrying charge equal to the density ρ(r′) (which happens to be constant in this case) times the volume of the cube, d3r′. A typical cube will create a field dE(r) as shown. You have to integrate the dE(r)’s from every tiny cube in the sphere. But the contribution from each cube will have a different magnitude and direction. Adding all these vectors is a tough problem that we are going to finesse by invoking a very powerful notion called Gauss’s law. As a prelude, we need to cover some mathematical ideas involving areas and surface integrals.
Figure 3.4 The field due to a spherical charge distribution. Each tiny cube d3r′ located at r′ makes its contribution dE(r) to E(r) as per Coulomb’s law. These contributions have to be vectorially added to obtain E(r).

3.4 Digression on the area vector dA
Imagine I am holding up a tiny little planar area, like a postage stamp, in three dimensions at some location r. I want you to be able to visualize this area. What can I do to specify it besides telling you it is located at r? The first thing I can tell you is how big it is. I say it is dA square meters in size. I then have to tell you in which plane it lies. How do I do that?
Suppose it lies in the xy-plane. Rather than say “lies in xy-plane,” I could just as well say it lies perpendicular to the z-axis. I could then associate a vector dA with this area, of magnitude dA and direction along the z-axis. But there are two ways to draw the perpendicular to the xy-plane: up or down the z-axis. To further specify the area, to make it an oriented or signed area, I will draw arrows that run around its perimeter in one of two possible directions. The area vector dA will point along the thumb of our right hand if we curl the fingers around the loop in the sense of the arrows. This is called the right-hand rule. It is illustrated in Figure 3.5 by the two areas in the xy-plane, given by dA1 = −kdA1 and dA2 = −kdA2. (An area without the arrows on its perimeter is like a vector without its head and tail marked.)
Figure 3.5 The figure shows a generic (shaded) area floating in three dimensions. The area vector dA is given by the right-hand rule applied to the arrows running around the edges. Also shown are two areas lying in the xy-plane with a common edge. Their sum is an area with the common edge (shown by a dotted line) deleted. If we use that dotted line as a hinge and rotate the second area out of the xy-plane (as indicated), their sum is a non-planar area, bounded by the uncanceled edges.
The upper part of the figure shows a generic (shaded) area vector dA, floating in three dimensions. Its direction is determined by the sense of the arrows running around the edges as per the right-hand rule.

Only a planar area can be represented as a vector. All infinitesimal areas can be treated as planar. Finite areas that are non-planar, like a hemisphere or magic carpet, cannot be represented by a single vector: we cannot reconstruct an entire macroscopic surface, with all its undulations, given just a magnitude and a direction.
The use of the right-hand rule in defining areas might remind you of the cross product and indeed there is such an interpretation of areas. Consider an area shaped like a parallelogram, whose adjacent edges are defined by two vectors B and C with angle θ between them. Then A = B × C is the area of the parallelogram, with magnitude |BC sinθ| and direction given by the right-hand rule. Infinitesimal areas are bounded by infinitesimal vectors.
Using vectors to describe areas or combining two vectors to get a third by the cross product is possible only in three dimensions where every plane has a unique normal, up to a sign. In four dimensions you cannot have a cross product of two vectors that yields a vector. If you pick two non-planar vectors, the plane they define will have two orthogonal directions perpendicular to it.
3.4.1 Composition of areas
Even though infinitesimal areas are given by vectors, the natural rule for combining them is different from vector addition, unless all the areas lie in one plane. I introduce the rule through an analogy, with one fewer dimension; see Figure 3.6.
Suppose we want to construct a curve in two or three dimensions, given any number of tiny vectors. Each vector has two boundary points: its tip and its tail, which are assigned opposite signs. To form the curve, we string these vectors along: the tail of the second vector touches the tip of the first, the tail of the third the tip of the second, and so on to the last one. The resulting curve has only two boundary points: the tail of the first and the tip of the last. All other boundary points have canceled in pairs when we joined them head-to-tail. Of course, the perfectly smooth curve is realized only in the limit of an infinite number of infinitesimal vectors.

Figure 3.6 A curve C joining points 1 and 2 in the plane, composed of little vectors added tip-to-tail. The tip and tail are the boundaries of each arrow. When two arrows are glued, the touching tip and tail are erased. At the end only the tail of the first vector and the tip of the last vector survive. These are the boundaries of C. The formation of the curve by gluing arrows is not to be confused with vector addition, which would give V12. If the points 1 and 2 were also glued, we would have a closed loop, while the vector sum V12 would vanish.
Do not confuse this composition of the curve with the vector sum, which would be a straight line going from the tail of the first vector to the tip of the last. Whereas the vector sum remembers only the bottom line, the curve remembers every vector that went into its composition. For example, if the curve is closed, say a circle, the vector sum would simply vanish.
There is a similar rule for combining areas to form two-dimensional surfaces. Consider the two areas dA1 and dA2 in Figure 3.5. To combine them, we superpose the right edge of dA1 and the left edge of dA2 with their opposing arrows. (This is analogous to placing the tail of one vector on the tip of the previous in forming a curve.) We delete the overlapping parts that carry opposite arrows. The “sum” of the areas is bounded by the remaining edges.
Look at the deleted portion shown by a dotted line. If we use that dotted line as a hinge and rotate the second area out of the xy-plane, their sum, bounded by the uncanceled edges, is now a non-planar area. In this manner, a generic surface in three dimensions may be formed by gluing together little areas or plaquettes and deleting the common edges, as illustrated in Figure 3.7. The arrows that used to run around the interior plaquettes have been canceled by the neighboring plaquettes with counterpropagating arrows. What remains are arrows around the perimeter, which run along the boundary of their union or sum.

Figure 3.7 A generic surface in three dimensions obtained by gluing together tiny areas or plaquettes. The arrows that used to run around the interior plaquettes have been canceled by the neighboring plaquettes with counter-propagating arrows. What remains are arrows around the perimeter, which define the boundary of the sum. Also shown for later use is one highlighted interior area dS and the electric field vector E at that point. The orientation of this area is indicated by the arrow on one edge.
3.4.2 An application of the area vector
Let us put the concept of the area vector to work. Imagine a tube with a rectangular cross section of height h and width w carrying some fluid moving with velocity v parallel to its length, as shown in Figure 3.8. What is the flux Φ, the volume of fluid flowing past any cross section per second?
To find Φ, we pick as a checkpoint the leftmost area A in the figure and ask how much fluid goes past it in one second. To this end, at some time t = 0 we introduce some tiny beads into the fluid at A. After 1 second, the beads would have moved a distance v · 1 and will be resident on the middle area in the figure, which is a shifted duplicate of A. The fluid that has crossed the checkpoint in one second is contained between these two areas. It is a parallelepiped of base A = wh and height v · 1 as shown in the figure. Thus

Figure 3.8 A tube of cross-sectional area A = wh, carrying a fluid with a velocity v parallel to A. To monitor the flux (volume flow per second) past the area A shown at the left, we sprinkle some beads into the fluid at t = 0. One second later the beads end up at the middle area. The volume between these two fronts is the flow per second, Φ = Av = A· v. The rightmost area A′ is bigger than A by a factor 1/cosθ but intercepts the same amount of flux or flow per second. As shown in the text, Φ′ = A′ · v = A′vcosθ = Av = A· v = Φ . The inset shows the volume contained between two tilted areas A′ at times t = 0 and t = 1, separated by v . 1 meters.
Because v and A are the magnitudes of the parallel vectors v and A, we may rewrite the Φ above as their dot product:
Remember that the area, A, if considered as a planar object, lies perpendicular to the flow but the area vector A, as defined above, is parallel to v. So the cosθ factor that enters the dot product is simply cos0 = 1.
Invoking dot product in the present case, when it is just the product of the magnitudes of the parallel vectors A and v, seems like overkill. But it is introduced to cover a more general case depicted in Figure 3.8. Look at the rightmost area A′, which also goes from the ceiling to the floor but with its plane tilted by an angle θ from the vertical. Now

has the same base w as A but a longer side (h/cosθ). Let us compute the flux through A′. If we wait one second, the points in A′ will move a distance v · 1 downstream and create a replica of A′ there. The flux Φ′ is the volume trapped between these two tilted areas. This volume (shown in the inset) is the product of the width w and the area of the parallelogram of base v, side h′, and height h. Recall that the area of a parallelogram is base times height. Thus
which is the same as Φ = v · A = vwh. Thus even though A′ is bigger than A by a factor 1/cosθ, it intercepts the same flux because it is tilted by θ relative to v.
A given area can intercept the greatest flux (say of a fluid) by orienting its plane perpendicular to the flow, or its area vector parallel to the flow. Likewise, it intercepts no fluid at all if it lies in a plane parallel to the flow, or if its area vector is perpendicular to the flow. Most importantly, for intermediate angles, the correct multiplicative factor to use with vA is cosθ. This appears naturally in the dot product, which therefore seems tailormade for computing fluxes.
We shall use the term “flux” to denote the dot product of an area vector with any other vector V, even if V is not a velocity. In what follows the vector in question will be E, the electric field.

Figure 3.9 The figure shows the two-dimensional cross section of field lines emanating from a charge q. Thus the circle S represents a sphere. It is evident that these lines cross any surface enclosing the charge. Two surfaces, a sphere S and a generic one S′, are shown. Since the number of lines crossing a surface is proportional to the surface integral of E, it means the latter has the same value on any surface surrounding q. The side views of tiny areas dS and dS′ on the two surfaces are shown, along with the local value of the field E and E′. Whereas dS and E are parallel on the sphere S, dS′ and E′ corresponding to the general case are at an angle θ. The third surface S′′ on the lower left-hand corner encloses no charge and has no net lines flowing in or out.

3.5 Gauss’s law through pictures

Consider a charge q and the field lines coming out of it. Let us assign k lines per coulomb, where k is an arbitrary constant. The following statements should be obvious from Figure 3.9.

• The number of lines passing through a sphere S centered on the charge is independent of its radius r and equals kq, the number emanating from q.

• The same number of lines pass through any closed surface such as S′ that surrounds the charge.

• If there are several charges qi, i = 1 . . . n inside the closed surface S′, the number of lines

crossing is simply the sum

This may not be so obvious, since when many charges are

present, the lines assume complicated shapes instead of going straight to infinity. So look at the field lines due to two positive charges in Figure 2.5. The charges emit kq1 and kq2 lines respectively (which happen to be equal in this example). None of these lines can terminate on the other charge, since they are both positive. So all the kq1 + kq2 lines have to go out of S′ and terminate on negative charges outside or escape to infinity. (Again if S′ is very convoluted, a line may go in and out of it an odd number of times before finally escaping.)

Suppose next q2 is negative, i.e., q2 = −|q2| with |q2| < q1. Now k|q2| lines will terminate on q2 and the rest, kq1 − k|q2| = k(q1 + q2), will terminate on negative charges outside S or escape to infinity, after possibly going in and out of S a few times. The argument is readily generalized to any number of charges, of any sign and magnitude. We may assert that if qi are the charges inside a generic surface S′,

where qenc is the total charge enclosed in S′. If you understand Eqn. 3.32 based on the pictures, you understand Gauss’s law, for this is what it essentially is, once we express the number of lines leaving S′ in terms of the electric field. We will do that in stages. First consider the special case when S′ is a sphere S centered on a single charge q and consider the areal density of flux lines. These lines cross the sphere perpendicularly, or, if you like, are parallel to the normal to the surface.
But since E(r), the magnitude of the electric field on S, is given by
we may write
Therefore the electric field is proportional to the lines per unit area, where the area lies in a plane perpendicular to the lines of E or, equivalently, the area vector is parallel to the lines and to E.

Consider now a tiny area dS sitting on the surface of this sphere. I use dS instead of dA to signify that this little area is part of a surface S, and I will follow this notation from now on. Letting dS stand for its magnitude,
Let us now re-express product E(r)dS in terms of the corresponding vectors E and dS, which are both radial. Thus
This now allows us to reach a very important relationship:
Therefore E · dS, the electric flux coming out of the area dS, is proportional to the lines crossing the surface. The proportionality constant is ε0k, where ε0 is a fixed number and k is up to us to choose (once and for all).
If we cover the surface of the sphere with tiny little patches dS and add the contributions from all of them to the two sides of Eqn. 3.40, make the patches smaller and smaller, and turn the sum into an integral we obtain:
The integral on the right is called the surface integral of E over S. The symbol ∮ means the surface is closed.

Since the lines crossing the sphere are independent of the radius, we may now assert that the surface integral of E over the sphere S is also independent of its radius r.
Next consider an arbitrary surrounding surface S′ surrounding the charge q as shown in Figure 3.9. We know the total number of lines crossing it are again the same, namely kq. How do we express this result in terms of E′? If we cover this surface with patches, the area vectors dS′ will not generally be radial. The number of lines these patches intercept will not be the product kε0E′(r)dS′, but rather kε0E′(r)dS′ cosθ, where θ is the angle between E′ and dS′. If you think of the lines as the flow of something, from the fluid flow analogy it is evident that a given area will intercept the most lines if its area vector is parallel to E′, and that as it rotates off this direction, their number will diminish by the geometrical factor cosθ. We have therefore the result that
Since the number of lines crossing a generic surface is independent of its shape as long as it surrounds the charge, we deduce the corresponding fact about the surface integral of the electric field over any generic surface:
where we have dropped the prime on E and S, where the latter, from now on, will refer to the general surface, spherical or not. Eqn. 3.45 is Gauss’s law for a single charge.
There is no arbitrary constant k in this relation and there should not be. Whereas the lines we draw to aid our imagination have a density that does

depend on k, the electric field at a point is uniquely defined by the charges that produce it or the force it exerts on a test charge. Therefore its integral on a closed surface better not depend on k. The result above is simply a property of the electric field as given by Coulomb’s law and does not rely on the notion of field lines. The field lines helped us anticipate the final answer, which can, however, be derived by explicit computation.
As an illustration, consider the field of a point charge and spherical surface S centered on it. By direct computation
The steps leading to Eqn. 3.48 need some explanation. There a surface integral is evaluated by inspection and the answer is simply written down. What happened to the integration? The answer is that the integrand,
, is a constant on the sphere. So E(r) may be pulled out of the integral, like a number 19 can be pulled out. The integral of E(r)dS over the sphere then reduces to the product of this constant E(r) and the area of the sphere. (Here is an analogy. If f(x) = f0, a constant, the definite integral over an interval of length L is the area of a rectangle of height f0 and base L. More formally, f0 may be pulled out of the integral and the remaining integral of dx is just L.)
If the surface S is not spherical, it takes more work to show that Eqn. 3.48 still holds by invoking the notion of a solid angle. The pictorial argument in terms of lines spared us that effort.
We now want to extend Gauss’s law to many charges qi, i = 1 . . . n. Now we forget all about lines of force, which can be very complicated. Instead we use superposition of the fields to these charges, each of which obeys Gauss’s law. Each charge qi produces its own Ei that obeys

for any closed surface S containing the charge. By summing both sides over i we obtain Gauss’s law in all its generality,

where

is the total electric field and qenc is the total charge enclosed

by S.

The charges qi have to be inside S to contribute to the surface integral of

E, or, equivalently, the lines flowing out of S. Consider for example an

empty surface S′′ in Figure 3.9 with the charge q lying outside. Any field

line emanating from the charge that enters the surface will necessarily also

exit since there is no charge inside for it to terminate. Lines coming in are

counted as negative and those coming out are described as positive, and the

positive and negative contributions will cancel precisely. In terms of the

electric field, E · dS will be negative where lines enter and positive where

they leave, and the integral over S′′ will be zero.

For future use I repeat the algorithm for computing the surface integral of

E over any surface S, closed or not. Consult Figure 3.7.

• Tile the surface with tiny areas or patches dS(ri) located at ri. For a closed surface the area vectors are defined to point outward. • On each tiny area compute the flux dΦ(ri) = E(ri) · dS(ri). • Do the sum
• Repeat with smaller and smaller patches till the sum converges to some limit. That defines ʃSE · dS.

In some special cases like the field of a point charge the integral can be done analytically, but in all cases it has a well-defined numerical value, which can be determined as above.

3.5.1 Continuous charge density

Suppose S contains a continuous blob of charge, with charge density ρ(r) instead of discrete charges qi. To write down Gauss’s law we need the total charge enclosed. A tiny cube of size d3r = dxdydz at r will enclose ρ(r)d3r coulombs, and the enclosed charge will be this quantity integrated over the volume V within the closed surface S.
So the form of Gauss’s law we will find most useful is as follows:
where
In future I will also use
C = ∂S, which means the closed loop S is the boundary of the surface S. (3.53)

CHAPTER 4
Gauss’s Law II: Applications
In the last chapter we encountered Gauss’s law:
On the left-hand side we have the surface integral of the electric field over a closed surface S, which is the boundary of a volume V. On the righthand side is the total charge enclosed by S divided by ε0. The charge enclosed is the volume integral of the charge density ρ(r) if continuous, and the sum over point charges qi if discrete.
The surface S, called the Gaussian surface, is a theoretical construct to help our calculations. It may be chosen at will, and for every choice of S, there is a corresponding equality. The Gaussian surface could sometimes coincide with a real surface (say of a conductor).
As for the left-hand side, recall the algorithm for computing the surface integral: divide S into little areas or patches dS, add the contributions E · dS from each area (where E is the electric field on that tiny area), and take the limit of an infinite number of patches of infinitesimal size. Often the only way to do this integral is by numerical means, though occasionally an analytic evaluation may be possible, such as when E is due to a point charge and S is a surrounding concentric sphere.
As for the right-hand side, the integrals of ρ can be done by inspection in all the cases we will discuss in this chapter. In general you will have to do a multiple integral of the charge density ρ over the volume V.
4.1 Applications of Gauss’s law
As a first application, consider the field due to a uniform spherical ball of charge Q and radius R centered on the origin. Since the sphere of charge

looks the same if we rotate it around any axis passing through the origin, the field distribution must have this property. The only solution to this requirement is the hedgehog field, with lines fanning out equally in all directions, with the same density at all points of a given r. In terms of the field, the allowed configuration is of the form
We just need to find E(r) and will do so using Gauss’s law. To find the field outside the charged sphere, we choose as the Gaussian surface S, a sphere of radius r > R, as depicted in the top left half of Figure 4.1. The calculation proceeds as with a point charge:

Figure 4.1 The use of Gauss’s law to find the field due to a uniform solid ball of charge outside (top left) and inside (top right) its radius R by using a spherical Gaussian surface S of appropriate radius. The graph at the bottom shows E(r), the radial field as a function of r. It rises linearly for r≤R and thereafter falls as 1/r2.
which is the field of a point charge Q at the origin. Note that Gauss’s law gives just one piece of information about E(r): its
integral over a surface S. One cannot infer from that a whole function E(r). For example, if I say
that
what can you say about f (x)? It could be f(x) = 7, f(x) = 7 + sinx, etc. But if I said f(x) is a constant f0 over the region of integration, you could deduce f0 = 7 as follows:
The moral is that if a function is a constant over a region of integration, its integral equals that constant times the length or area or volume of the

integration region. This is what happened in Eqn. 4.3: the surface integral was E(r) × 4π r2.
Equating this to Q/ε0, we obtained Eqn. 4.7. Gauss’s law can be used to deduce the entire field E(r) only when
symmetry arguments can be used to reduce the unknown to just one number, E(r) on the Gaussian surface. Had S been a sphere, but the charge a non-sphere with bumps and lumps here and there, the surface integral of E would still be known to be Q/ε0, but one could not use this to find E(r) anywhere because it would vary over S. Similarly, had the charge been spherical but S not spherical, we would again have a result that was true, but not helpful in finding E(r) anywhere on S.
Next we want to find the field inside the sphere of charge. So we take for the Gaussian surface a sphere of radius r < R, as shown in the top right half of Figure 4.1. The calculation proceeds as for r > R but with one change: the charge enclosed is not all of Q but only qenc, the amount enclosed by the sphere of radius r. Since the density is uniform, the ratio of the enclosed charge to the total charge is the ratio of their volumes:
If you do not like this argument, let me rewrite this result as follows:
where the first factor is the charge density and the second factor is the volume in question.
The surface integral of the field is the same as before and Gauss’s law takes the form

Thus the field actually grows from zero as we move out, and it reaches a maximum of Q/(4π ε0R2) at the surface. Thereafter, it falls like 1/r2. The field, radial in all cases, is as follows for all values of r:
and is depicted in the bottom part of Figure 4.1. The two expressions agree on the surface of the ball r = R.
Why does E(r) grow (linearly) with r when r < R? Because, as r increases, the enclosed charge grows as r3 and the field it produces, acting as a point charge at the origin, falls as 1/r2. Once we go outside the sphere, for r > R, the field falls like 1/r2, since we do not pick up any extra charge as we increase r, the radius of the Gaussian surface.
These results may be taken over verbatim for gravity, with the understanding that the force is always attractive. Consider in particular the linear force, which points toward the center inside a spherical mass. This linear (restoring) force implies simple harmonic motion. If the spherical mass in question is the earth, this has the following interesting consequence. If you drill a very narrow hole passing through the center of the earth (so narrow that the mass you scooped out does not affect the preceding answer for the field) and drop an object into it, it will oscillate back and forth between where you are and the diametrically opposite point on the globe. I

invite you to show that gravity.)

(First write down Gauss’s law for

4.2 Field inside a shell
Consider a uniformly charged solid sphere of radius R2 from which a concentric sphere of radius R1 < R2 has been scooped out. We want the field due to this hollow shell. By Gauss’s law, for r > R2 this hollow sphere will act like a point charge centered at the origin. How about inside the hollow region, for r < R1? By applying Gauss’s law to a Gaussian surface of radius r < R1, we see that the field inside is zero because the charge enclosed is zero. This result is equally true for the gravitational force.
Let us try to understand the absence of the field inside a hollow shell directly in terms of Coulomb’s law. This discussion is optional.
I will only show that the field inside a hollow shell of radius R and infinitesimal thickness is zero. I am done, because the original shell of finite thickness R2 − R1 can be built out of concentric, infinitesimally thin shells of radius ranging from R1 to R2, each of which contributes a zero to the total.
Consider then a point P inside such a shell of radius R and infinitesimal thickness, as depicted in Figure 4.2. Assume the shell has a surface charge density σ. (I invite you to show that σ = ρdr, where ρ is the uniform density of the charged sphere and dr is the thickness of the shell.)

Figure 4.2 The aim is to show that the field at a generic point P inside a hollow shell is zero. The figure shows two oppositely pointing cones of identical opening angle that meet at P and intersect the sphere in two caps, shown as dark ellipses. The same number of field lines emitted by the test charge at P pierce the two caps. This is shown in the text to imply that the charges on these two caps exert equal and opposite forces on a test charge at P. It is possible to cover the entire shell using canceling pairs of cones.
If P is the center of the shell, we can argue by symmetry that the field there has to vanish: a non-zero E at the center necessarily has to point in some arbitrary direction, violating the rotational symmetry of the problem. But we can see more directly that the field has to be zero because for every tiny patch of charge on the shell pushing a test charge one way, there is a diametrically opposite patch that exerts an equal and opposite force.
But the result is stronger; it says E(r) = 0 even for a point off-center, like P in the figure. We would like to show that this too follows from the cancellation of forces exerted by charges in different segments of the shell. To this end consider two cones of the same opening angle pointing away from P in opposite directions and intersecting the shell on two caps. The opening angle of the cones is infinitesimal, as are the planar areas they pierce through, denoted by dS1 and dS2.
Instead of showing that (the charges on) the caps exert equal and opposite forces on a unit test charge at P, we will show the unit test charge exerts equal and opposite forces on the (charges on) the two caps. We are then done because action and reaction are equal and opposite in Coulomb’s law: if the forces the test charge at P exerts on the caps are equal and opposite, so are the forces the caps exert on the test charge. Since it is

possible to surround the point P with such canceling pairs of cones, we know the net force of the shell on the charge at P will be zero.
So imagine a unit test charge placed at P and the lines emanating isotropically from it. Since the cones have the same opening angle, they contain the same number of field lines and thus the number of lines crossing the two caps is equal. Now the number of lines crossing the caps is, by Eqn. 3.42, kε0E1 · dS1 and kε0E2 · dS2 where E1 and E2 are the fields produced at the caps by the test charge at P.
Next we collect some relevant facts.
• The area vectors dS1 and dS2 are radial, being parts of a sphere. • The electric fields E1 and E2 point outward along the symmetry axis of the two head-to-head cones. • The angles between the area vectors and the corresponding field vectors are the same in both patches, and are denoted by θ. This equality follows from the fact that the angles opposite to the indicated θ’s lie at the base of an isosceles triangle (whose two equal sides are the radius R and whose base is the chord connecting dS1 and dS2).
We put all this together and reason as follows:
where σ dSi = dqi is the charge on cap i, i = 1 or 2. The caps will behave as point charges σ dS1 and σ dS2 when we take the opening angles of the cones to zero.
Look at Eqn. 4.21. It says dq1 E1, the magnitude of the force the unit test charge at P exerts on the charges residing in dS1 through the field E1 it creates there, is equal to dq2 E2, the force the unit test charge at P exerts on the charges residing in dS2 through the field E2 it creates there. The two forces of course have opposite directions, pointing away from the test

charge. But if the test charge exerts equal and opposite forces on the caps, they in turn must exert equal and opposite forces on the test charge, because in Coulomb’s law action and reaction are equal and opposite. (Recall F12 = −F21.)
The argument relating the flux lines intercepted by the two caps to the fields E1 and E2 relies on the inverse square law of the electric force. Conversely one of the earliest tests of the inverse square law was the absence of field inside a hollow sphere.
4.3 Field of an infinite charged wire, redux
We have already seen how symmetry demands that the field of an infinite wire with linear charge density λ is constant if we move parallel to the wire at a fixed distance ρ, and points radially away from it:
We found that E(ρ) = λ/(2π ε0ρ) by doing an integral along the wire. Now we will rederive E(ρ) by using Gauss’s law. The trick is to find a
Gaussian surface on which there is a single unknown, E(ρ). A natural choice is a cylinder of radius ρ coaxial with the wire, as shown in Figure 4.3, since the field is constant in magnitude all over it. However, it is not enough to take just the curvy sides of the cylinder; we need the two flat sides at both ends, since the Gaussian surface has to be closed in order for the law to work, for it to enclose a definite amount of charge.
The radius of the cylinder is clearly ρ since we want E(ρ), but what should be its length L? Since the Gaussian surface is a figment of our imagination and not really wrapped around the wire, we can choose any length we want and then desperately hope that the answer will not depend on this arbitrary L.
Look at Figure 4.3. The charge enclosed within the cylinder is λ · L, from the very definition of λ as the charge per unit length. So we begin with

Figure 4.3 By symmetry, the field due to an infinite wire is radial and of constant magnitude at a fixed distance ρ from the wire. The Gaussian surface is a coaxial cylinder of radius ρ and has an arbitrary length L. The charge enclosed is simply λL. The two flat faces make no contribution to the flux since E and dS are perpendicular. The curved face, on which the flux density is constant, makes a contribution E(ρ) · 2πρL.
The surface breaks up into three parts: the two flat ends and the curved face parallel to the wire.
We seem to have a problem with the flat faces, since E(ρ) is not a constant on the entire face because different parts of it are at different distances from the wire. On the other hand, we have seen that Gauss’s law is useful only when there is just one constant E on the entire surface. Luckily we are saved by the fact that the area vectors dS and field E are perpendicular on these two faces: dS is parallel to the wire while E is perpendicular to it so the flux through the flat faces is zero. Or if you like, the field lines run parallel to the flat faces and so none cross it.
We are then left with the curved face on which the area vectors dS are radial and E(ρ) is a constant and radial. (Remember that for a closed surface, the area vector is defined as positive if it points outward.) So Gauss’s law tells us

The arbitrary length L has canceled out, as it must. We get the answer so easily only because of the high symmetry of the
problem. For example, if the wire had been non-uniformly charged, with λ = λ(x), we could still equate the flux over the cylinder to the charge enclosed (the integral of λ(x) over the length L). However, since E varies in magnitude and direction (not always radial) this will only tell us something about the integral of E over the surface and not about its value at any one place. On the other hand, if the line charge is replaced by a uniform cylindrical distribution, E may be found everywhere using Gauss’s law and symmetry.
4.4 Field of an infinite plane, redux
Consider an infinite plane, which we take to be the xy-plane. It has a uniform charge density σ.
Recall what the symmetry arguments tell us. The field is independent of x or y (but could depend on z) and must point perpendicularly away from the plane with the same magnitude at z and −z. That is, E must have the form
To find E(|z|) we need a Gaussian surface on whose various parts E is either constant or perpendicular to the area vector. Such a surface is shown in Figure 4.4. It is a cylinder of cross section A, with its symmetry axis parallel to the z-axis and its flat faces at ±z.
The area A is arbitrary and hopefully will drop out of the answer. The charge enclosed is clearly σ A, where A is the area of the circle the cylinder encloses as it pierces the plane. In contrast to the infinite wire, this time the curved side of the cylinder makes no contribution to the surface integral

since the field is parallel to the curved side and the area vector is normal to it. (The field lines cross the two flat faces but not the curved face.) As for the flat faces, on the upper face we have kA· kE(|z|) = A · E(|z|). The same contribution comes from the lower face where both E and the area have flipped their orientation to yield (−kA) · (−kE(|z|)) = A · E(|z|). Gauss’s law then tells us
Figure 4.4 Shown is an infinite plane with charge density σ. Symmetry tells us the field is everywhere normal to the plane and constant in magnitude as we move parallel to the plane. The Gaussian surface is a cylinder of area A and height 2z, symmetrically located with respect to the plane. The charge enclosed is σ A. As for the flux, or surface integral of E, the curved side makes no contribution because the area vector and field are perpendicular, while the two flat faces make equal contributions of E(|z|)A each, where E(|z|) is the constant value of the field strength a distance |z| from the charged sheet.
The area has dropped out as it must, and, remarkably, there is no dependence on z, the coordinate perpendicular to the plane.
4.5 Conductors
Consider a chunk of copper, which is a good conductor. In a good conductor not all of the electrons in the atoms are tied to the nuclei, but shared

communally. They are free to move around the material but not to leave it. The conductor is like a swimming pool for the electrons: they can swim freely inside but cannot scale the walls at the boundary. If they try that, all the nuclei will exert a force to pull them back. The energy needed to rip an electron out of the material is called the work function. There are good and bad conductors, and we will discuss a perfect conductor in which the charges can move freely in response to the smallest field.
We will now make many predictions about conductors, mainly using Gauss’s law.
4.5.1 Field inside a perfect conductor is zero
The first property that follows by definition is that in a static situation, the electric field inside a perfect conductor is zero. Had there been a field, the charges would have been moving but we have been assured it is an electrostatic situation. So there can be no field. The no-field rule does not hold in the non-static case.
For example, it fails if there is a field E in space and I suddenly insert a chunk of conductor shaped like a rectangular slab into that region, as shown in the top half of Figure 4.5. Initially there will be a field inside the conductor. It will start moving the electrons (whose charge is negative) in the opposite direction. These will pile up on the left face as shown, leaving behind positive charges on the other face, due to nuclei whose electrons have drifted away. The pileup will continue until the internally generated field due to the two layers cancels the applied field E. For an infinite slab, we know the two faces would produce a field σ/ε0 in the region between them, opposing the external E. For a finite slab there will be some complications near the ends but the field inside the conductor will still end up vanishing.