zotero-db/storage/N2NJHW4S/.zotero-ft-cache

Introduction to Electrodynamics
David J. Griffiths Reed College
-Prentice Hall
Prentice Hall Upper Saddle River, New Jersey 07458

Library of Congress Cataloging-in-Publication Data
Griffiths, David J. (David Jeffrey) Introduction to electrodynamics/ David J. Griffiths - 3rd ed.
p. cm. Includes biqliographical references and index. ISBN 0-13-805326-X I. Electrodynamics. I. Title. OC680.G74 1999 537.6--dc21

98-50525 CIP

Executive Editor: Alison Reeves Production Editor: Kim Deltas Manufacturing Manager: Trudy Pisciotti Art Director: Jayne Conte Cover Designer: Bruce Kenselaar Editorial Assistant: Gillian Keiff Composition: PreT£X, Inc.

-

© 1999, 1989, !981 by Prentice-Hall, Inc.
Upper Saddle River, New Jersey 07458

All rights reserved. No part of this book may be reproduced, in any form or by any means, without permission in writing from the publisher.

Reprinted with corrections September, I999

Printed in the United States of America
1098765
ISBN □ -13-805326-X
Prentice-Hall International (UK) Limited, London Prentice-Hall of Australia Pty. Limited, Sydney Prentice-Hall Canada Inc., Toronto Prentice-Hall Hispanoamericana, S.A., Mexico City Prentice-Hall of India Private Limited, New Delhi Prentice-Hall of Japan, Inc., Tokyo Prentice-Hall Asia Pte. Ltd., Singapore Editora Prentice-Hall do Brasil, Ltda., Rio de Janeiro

Contents

Preface

ix

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xi

1 Vector Analysis

1

1.1 Vector Algebra . . . . . . . . . . . . . . .

1. 1.1 Vector Operations . . . . . . . . . .

1.1.2 Vector Algebra: Component Form .

4

1.1.3 Triple Products . . . . . . . . . . .

7

1.1.4 Position, Displacement, and Separation Vectors

8

1.1.5 How Vectors Transform

10

1.2 Differential Calculus . . . . .

13

1.2.1 "Ordinary" Derivatives

13

1.2.2 Gradient . . . . .

13

1.2.3 The Operator V .

16

1.2.4 The Divergence .

17

1.2.5 The Curl . . . .

19

1.2.6 Product Rules . .

20

1.2.7 Second Derivatives

22

1.3 Integral Calculus . . . . .

24

1.3.1 Line, Surface, and Volume Integrals

24

1.3.2 The Fundamental Theorem of Calculus

28

1.3.3 The Fundamental Theorem for Gradients

29

1.3.4 The Fundamental Theorem for Divergences

31

1.3.5 The Fundamental Theorem for Curls .

34

1.3.6 Integration by Parts . . . . .

37

1.4 Curvilinear Coordinates . . . . . .

38

1.4.1 Spherical Polar Coordinates

38

1.4.2 Cylindrical Coordinates

43

1.5 The Dirac Delta Function . . . . . .

45

1.5.1 The Divergence ofr/r 2 ..

45

1.5.2 The One-Dimensional Dirac Delta Function .

46

iii

IV
1.5.3 The Three-Dimensional Delta Function 1.6 The Theory of Vector Fields . .
1.6.1 The Helmholtz Theorem 1.6.2 Potentials
2 Electrostatics 2.1 The Electric Field 2.1.1 Introduction . . . 2.1.2 Coulomb's Law . 2.1.3 The Electric Field . 2.1.4 Continuous Charge Distributions . 2.2 Divergence and Curl of Electrostatic Fields 2.2.1 Field Lines, Flux, and Gauss's Law 2.2.2 The Divergence of E . . . . 2.2.3 Applications of Gauss's Law 2.2.4 The Curl of E . . . . . . 2.3 Electric Potential . . . . . . . . 2.3 .1 Introduction to Potential 2.3.2 Comments on Potential . 2.3.3 Poisson's Equation and Laplace's Equation 2.3.4 The Potential of a Localized Charge Distribution 2.3.5 Summary; Electrostatic Boundary Conditions 2.4 Work and Energy in Electrostatics . . . . . . . . . 2.4.1 The Work Done to Move a Charge . . . . . . 2.4.2 The Energy of a Point Charge Distribution . . 2.4.3 The Energy of a Continuous Charge Distribution 2.4.4 Comments on Electrostatic Energy . 2.5 Conductors . . . . . . . 2.5.1 Basic Properties . . . . . . . . . . 2.5.2 Induced Charges . . . . . . . . . . 2.5.3 Surface Charge and the Force on a Conductor 2.5.4 Capacitors . . . . . . . . . . . . . . . . . .
3 Special Techniques 3.1 Laplace's Equation . . . . . . . . . . . . . . 3.1.1 Introduction . . . . . . . . . . . . . . 3.1.2 Laplace's Equation in One Dimension 3.1.3 Laplace's Equation in Two Dimensions 3.1.4 Laplace's Equation in Three Dimensions 3. I .5 Boundary Conditions and Uniqueness Theorems 3.1.6 Conductors and the Second Uniqueness Theorem 3.2 The Method of Images . . . . . . . 3.2.1 The Classic Image Problem 3.2.2 Induced Surface Charge ..

CONTENTS
50 52 52 53
58 58 58 59 60 61 65 65 69 70 76 77 77 79 83 83 87 90 90
91
93 95 96 96 98 102 103
110 110 110 111 112 114 116 118 121 121 123

CONTENTS

V

3.2.3 Force and Energy . . .

123

3.2.4 Other Image Problems

124

3.3 Separation of Variables . . . .

127

3.3.1 Cartesian Coordinates

127

3.3.2 Spherical Coordinates

137

3.4 Multipole Expansion . . . . .

146

3.4.1 Approximate Potentials at Large Distances

146

3.4.2 The Monopole and Dipole Terms . . . . .

149

3.4.3 Origin of Coordinates in Multipole Expansions

151

3.4.4 The Electric Field of a Dipole . . . . . . . . .

153

4 Electric Fields in Matter

160

4.1 Polarization . . . . .

160

4.1.1 Dielectrics .

160

4.1.2 Induced Dipoles

160

4.1.3 Alignment of Polar Molecules

163

4.1.4 Polarization . . . . . . .

166

4.2 The Field of a Polarized Object . . . .

166

4.2.1 Bound Charges . . . . . . . .

166

4.2.2 Physical Interpretation of Bound Charges

170

4.2.3 The Field Inside a Dielectric . . . . . . .

173

4.3 The Electric Displacement . . . . . . . . . . . .

175

4.3.1 Gauss's Law in the Presence of Dielectrics

175

4.3.2 A Deceptive Parallel .

178

4.3.3 Boundary Conditions . . . . . . . . . . . .

178

4.4 Linear Dielectrics . . . . . . . . . . . . . . . . . .

179

4.4.1 Susceptibility, Permittivity, Dielectric Constant

179

4.4.2 Boundary Value Problems with Linear Dielectrics .

186

4.4.3 Energy in Dielectric Systems .

191

4.4.4 Forces on Dielectrics . . . . . . . . . . . . . . . .

193

5 Magnetostatics

202

5.1 The Lorentz Force Law

202

5.1.1 Magnetic Fields .

202

5.1.2 Magnetic Forces

204

5.1.3 Currents . . . . .

208

5.2 The Biot-Savart Law ..

215

5.2.1 Steady Currents .

215

5.2.2 The Magnetic Field of a Steady Current

215

5.3 The Divergence and Curl of B . . . .

221

5.3.1 Straight-Line Currents . . . . .

221

5.3.2 The Divergence and Curl of B .

222

5.3.3 Applications of Ampere's Law .

225

5.3.4 Comparison of Magnetostatics and Electrostatics

232

vi
5.4 Magnetic Vector Potential . . . . . . . . . . . . . . . . 5.4.1 The Vector Potential . . . . . . . . . . . . . . 5.4.2 Summary; Magnetostatic Boundary Conditions 5.4.3 Multipole Expansion of the Vector Potential .
6 Magnetic Fields in Matter 6.1 Magnetization . . . . . . . . . . . . . . . . . . . 6.1.1 Diamagnets, Paramagnets, Ferromagnets 6.1.2 Torques and Forces on Magnetic Dipoles 6.1.3 Effect of a Magnetic Field on Atomic Orbits 6.1.4 Magnetization . . . . . . 6.2 The Field of a Magnetized Object . . . . . . . . . 6.2.1 Bound Currents . . . . . . . . . . . . . . . 6.2.2 Physical Interpretation of Bound Currents . 6.2.3 The Magnetic Field Inside Matter . . . 6.3 The Auxiliary Field H . . . . . . . . . . . . . . 6.3.1 Ampere's law in Magnetized Materials 6.3.2 A Deceptive Parallel . 6.3.3 Boundary Conditions . . . . . . . . . . 6.4 Linear and Nonlinear Media . . . . . . . . . . 6.4.1 Magnetic Susceptibility and Permeability 6.4.2 Ferromagnetism
7 Electrodynamics 7.1 Electromotive Force . 7.1.1 Ohm's Law . 7.1.2 Electromotive Force 7.1.3 Motional emf . . . 7.2 Electromagnetic Induction . 7.2.1 Faraday's Law ... 7.2.2 The Induced Electric Field 7.2.3 Inductance . . . . . . . . 7.2.4 Energy in Magnetic Fields 7.3 Maxwell's Equations . . . . . . . 7.3.1 Electrodynamics Before Maxwell 7.3.2 How Maxwell Fixed Ampere's Law 7.3 .3 Maxwell's Equations . . . . . . 7.3.4 Magnetic Charge . . . . . . . . 7.3.5 Maxwell's Equations in Matter . 7.3.6 Boundary Conditions . . . . . .

CONTENTS
234 234 240 242
255 255 255 255 260 262 263 263 266 268 269 269 273 273 274 274 278
285 285 285 292 294 301 301 305 310 317 321 321 323 326 327 328 331

CONTENTS

vii

8 Conservation Laws

34S

8.1 Charge and Energy . . . . . . .

345

8.1.1 The Continuity Equation

345

8.1.2 Poynting's Theorem . .

346

8.2 Momentum . . . . . . . . . . .

349

8.2.1 Newton's Third Law in Electrodynamics

349

8.2.2 Maxwell's Stress Tensor ..

351

8.2.3 Conservation of Momentum

355

8.2.4 Angular Momentum

358

9 Electromagnetic Waves

364

9.1 Waves in One Dimension

364

9.1.1 The Wave Equation .

364

9.1.2 Sinusoidal Waves ..

367

9.1.3 Boundary Conditions: Reflection and Transmission .

370

9.1.4 Polarization . . . . . . . . . . .

373

9.2 Electromagnetic Waves in Vacuum ...

375

9.2.1 The Wave Equation for E and B

375

9.2.2 Monochromatic Plane Waves ..

376

9.2.3 Energy and Momentum in Electromagnetic Waves

380

9.3 Electromagnetic Waves in Matter . . . . . . . . . . . . . .

382

9.3.1 Propagation in Linear Media . . . . . . . . . . . .

382

9.3.2 Reflection and Transmission at Normal Incidence .

384

9.3.3 Reflection and Transmission at Oblique Incidence .

386

9.4 Absorption and Dispersion . . . . . . . . . . .

392

9.4.1 Electromagnetic Waves in Conductors , ..

392

9.4.2 Reflection at a Conducting Surface . . . . .

396

9.4.3 The Frequency Dependence of Permittivity

398

9.5 Guided Waves . . . . . . . . . . . . . . . . . . .

405

9.5.1 Wave Guides . . . . . . . . . . . . . . .

405

9.5.2 TE Waves in a Rectangular Wave Guide .

408

9.5.3 The Coaxial Transmission Line . . . . .

411

10 Potentials and Fields

416

10.1 The Potential Formulation . . . . .

416

10.1.1 Scalar and Vector Potentials

416

10.1.2 Gauge Transformations . . .

419

10.1.3 Coulomb Gauge and Lorentz* Gauge

421

10.2 Continuous Distributions . . .

422

10.2.1 Retarded Potentials . .

422

10.2.2 Jefimenko's Equations

427

10.3 Point Charges . . . . . . . . .

429

10.3.1 Lienard-Wiechert Potentials

429

10.3.2 The Fields of a Moving Point Charge

435

viii
11 Radiation 11.1 Dipole Radiation . . . . . . . . . 11.1.1 What is Radiation? . . . . 11.1.2 Electric Dipole Radiation . 11.1.3 Magnetic Dipole Radiation . 11.1.4 Radiation from an Arbitrary Source 11.2 Point Charges . . . . . . . . . . . . . . . 11.2.1 Power Radiated by a Point Charge . 11.2.2 Radiation Reaction . . . . . . . . . 11.2.3 The Physical Basis of the Radiation Reaction
12 Electrodynamics and Relativity 12.1 The Special Theory of Relativity 12.1.1 Einstein's Postulates .. 12.1.2 The Geometry of Relativity 12.1.3 The Lorentz Transformations . 12.1.4 The Structure of Spacetime .. 12.2 Relativistic Mechanics . . . . . . . . 12.2.1 Proper Time and Proper Velocity . 12.2.2 Relativistic Energy and Momentum 12.2.3 Relativistic Kinematics . 12.2.4 Relativistic Dynamics . . . . . . . 12.3 Relativistic Electrodynamics . . . . . . . . 12.3.1 Magnetism as a Relativistic Phenomenon 12.3.2 How the Fields Transform . . . . . 12.3.3 The Field Tensor . . . . . . . . . . 12.3.4 Electrodynamics in Tensor Notation 12.3.5 Relativistic Potentials . . . . . . .
A Vector Calculus in Curvilinear Coordinates A.1 Introduction A.2 Notation . A.3 Gradient . A.4 Divergence A.5 Curl A.6 Laplacian
B The Helmholtz Theorem
C Units
Index

CONTENTS
443 443 443 444 451 454 460 460 465 469
477 477 477 483 493 500 507 507 509 511 516 522 522 525 535 537 541
547 547 547 548 549 552 554
555
558
562

Preface
This is a textbook on electricity and magnetism, designed for an undergraduate course at the junior or senior level. It can be covered comfortably in two semesters, maybe even with room to spare for special topics (AC circuits, numerical methods, plasma physics, transmission lines, antenna theory, etc.) A one-semester course could reasonably stop after Chapter 7. Unlike quantum mechanics or thermal physics (for example), there is a fairly general consensus with respect to the teaching of electrodynamics; the subjects to be included, and even their order of presentation, are not particularly controversial, and textbooks differ mainly in style and tone. My approach is perhaps less formal than most; I think this makes difficult ideas more interesting and accessible.
For the third edition I have made a large number of small changes, in the interests of clarity and grace. I have also modified some notation to avoid inconsistencies or ambiguities.
z, x, z, Thus the Cartesian unit vectors J, and k have been replaced with y, and so that all
vectors are bold, and all unit vectors inherit the letter of the corresponding coordinate. (This also frees up k to be the propagation vector for electromagnetic waves.) It has always bothered me to use the same letter r for the spherical coordinate (distance from the origin) and the cylindrical coordinate (distance from the z axis). A common alternative for the latter is p, but that has more important business in electrodynamics, and after an exhaustive search I settled on the underemployed letter s; I hope this unorthodox usage will not be confusing.
Some readers have urged me to abandon the script letter -t (the vector from a source point r' to the field point r) in favor of the more explicit r - r'. But this makes many equations
distractingly cumbersome, especially when the unit vector -t is involved. I know from my
own teaching experience that unwary students are tempted to read -t as r-it certainly makes the integrals easier! I have inserted a section in Chapter 1 explaining this notation, and I
hope that will help. If you are a student, please take note: -t = r - r', which is not the same
as r. If you're a teacher, please warn your students to pay close attention to the meaning of -t. I think it's good notation, but it does have to be handled with care.
The main structural change is that I have removed the conservation laws and potentials from Chapter 7, creating two new short chapters (8 and 10). This should more smoothly accommodate one-semester courses, and it gives a tighter focus to Chapter 7.
I have added some problems and examples (and removed a few that were not effective). And I have included more references to the accessible literature (particularly the American Journal of Physics). I realize, of course, that most readers will not have the time or incli-
lX

X

PREFACE

nation to consult these resources, but I think it is worthwhile anyway, if only to emphasize that electrodynamics, notwithstanding its venerable age, is very much alive, and intriguing new discoveries are being made all the time. I hope that occasionally a problem will pique your curiosity, and you will be inspired to look up the reference--some of them are real gems.
As in the previous editi,ons, I distinguish two kinds of problems. Some have a specific pedagogical purpose, and should be worked immediately after reading the section to which they pertain; these I have placed at the pertinent point within the chapter. (In a few cases the solution to a problem is used later in the text; these are indicated by a bullet (•) in the left margin.) Longer problems, or those of a more general nature, will be found at the end
of each chapter. When I teach the subject i assign some of these, and work a few of them
in class. Unusually challenging problems are flagged by an exclamation point (!) in the margin. Many readers have asked that the answers to problems be provided at the back of the book; unfortunately, just as many are strenuously opposed. I have compromised, supplying answers when this seems particularly appropriate. A complete solution manual is available (to instructors) from the publisher.
I have benefitted from the comments of many colleagues-I cannot list them all here. But I would like to thank the following people for suggestions that contributed specifically to the third edition: Burton Brody (Bard), Steven Grimes (Ohio), Mark Heald (Swarthmore), Jim McTavish (Liverpool), Matthew Moelter (Puget Sound), Paul Nachman (New Mexico State), Gigi Quartapelle (Milan), Carl A. Roner (West Virginia), Daniel Schroeder (Weber State), Juri Silmberg (Ryerson Polytechnic), Walther N. Spjeldvik (Weber State), Larry Tankersley (Naval Academy), and Dudley Towne (Amherst). Practically everything I know about electrodynamics-certainly about teaching electrodynamics-I owe to Edward Purcell.
David J. Griffiths

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What is electrodynamics, and how does it fit into the general scheme of physics?

Four Realms of Mechanics In the diagram below I have sketched out the four great realms of mechanics:

Classical Mechanics (Newton)
Special Relativity (Einstein)

Quantum Mechanics (Bohr, Heisenberg, Schrodinger, et al.)
Quantum Field Theory (Dirac, Pauli, Feynman,
Schwinger, et al.)

Newtonian mechanics was found to be inadequate in the early years of this century-it's all right in "everyday life," but for objects moving at high speeds (near the speed of light) it is incorrect, and must be replaced by special relativity (introduced by Einstein in 1905); for objects that are extremely small (near the size of atoms) it fails for different reasons, and is superseded by quantum mechanics (developed by Bohr, Schrodinger, Heisenberg, and many others, in the twenties, mostly). For objects that are both very fast and very small (as is common in modem particle physics), a mechanics that combines relativity and quantum principles is in order: this relativistic quantum mechanics is known as quantum field theory-it was worked out in the thirties and forties, but even today it cannot claim to be a completely satisfactory system. In this book, save for the last chapter, we shall work exclusively in the domain of classical mechanics, although electrodynamics extends with unique simplicity to the other three realms. (In fact, the theory is in most respects automatically consistent with special relativity, for which it was, historically, the main stimulus.)
xi

xii Four Kinds of Forces

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Mechanics tells us how a system will behave when subjected to a given force. There are just four basic forces known (presently) to physics: I list them in the order of decreasing strength:

I. Strong 2. Electromagnetic 3. Weak 4. Gravitational

The brevity of this list may surprise you. Where is friction? Where is the "normal" force that keeps you from falling through the floor? Where are the chemical forces that bind molecules together? Where is the force of impact between two colliding billiard balls? The answer is that all these forces are electromagnetic. Indeed, it is scarcely an exaggeration to say that we live in an electromagnetic world-for virtually every force we experience in everyday life, with the exception of gravity, is electromagnetic in origin.
The strong forces, which hold protons and neutrons together in the atomic nucleus, have extremely short range, so we do not "feel" them, in spite of the fact that they are a hundred times more powerful than electrical forces. The weak forces, which account for certain kinds of radioactive decay, are not only of short range; they are far weaker than electromagnetic ones to begin with. As for gravity, it is so pitifully feeble (compared to all of the others) that it is only by virtue ofhuge mass concentrations (like the earth and the sun) that we ever notice it at all. The electrical repulsion between two electrons is 1042 times as large as their gravitational attraction, and if atoms were held together by gravitational (instead of electrical) forces, a single hydrogen atom would be much larger than the known universe.
Not only are electromagnetic forces overwhelmingly the dominant ones in everyday life, they are also, at present, the only ones that are completely understood. There is, of course, a classical theory of gravity (Newton's law of universal gravitation) and a relativistic one (Einstein's general relativity), but no entirely satisfactory quantum mechanical theory of gravity has been constructed (though many people are working on it). At the present time there is a very successful (if cumbersome) theory for the weak interactions, and a strikingly attractive candidate (called chromodynamics) for the strong interactions. All these theories draw their inspiration from electrodynamics; none can claim conclusive experimental verification at this stage. So electrodynamics, a beautifully complete and successful theory, has become a kind of paradigm for physicists: an ideal model that other theories strive to emulate.
The laws of classical electrodynamics were discovered in bits and pieces by Franklin, Coulomb, Ampere, Faraday, and others, but the person who completed the job, and packaged it all in the compact and consistent form it has today, was James Clerk Maxwell. The theory is now a little over a hundred years old.

Xlll
The Unification of Physical Theories
In the beginning, electricity and magnetism were entirely separate subjects. The one dealt with glass rods and cat's fur, pith balls, batteries, currents, electrolysis, and lightning; the other with bar magnets, iron filings, compass needles, and the North Pole. But in 1820 Oersted noticed that an electric current could deflect a magnetic compass needle. Soon afterward, Ampere correctly postulated that all magnetic phenomena are due to electric charges in motion. Then, in 1831, Faraday discovered that a moving magnet generates an electric current. By the time Maxwell and Lorentz put the finishing touches on the theory, electricity and magnetism were inextricably intertwined. They could no longer be regarded as separate subjects, but rather as two aspects of a single subject: electromagnetism.
Faraday had speculated that light, too, is electrical in nature. Maxwell's theory provided spectacular justification for this hypothesis, and soon optics-the study of lenses, mirrors, prisms, interference, and diffraction-was incorporated into electromagnetism. Hertz, who presented the decisive experimental confirmation for Maxwell's theory in 1888, put it this way: "The connection between light and electricity is now established . . . In every flame, in every luminous particle, we see an electrical process . . . Thus, the domain of electricity extends over the whole of nature. It even affects ourselves intimately: we perceive that we possess ... an electrical organ-the eye." By 1900, then, three great branches of physics, electricity, magnetism, and optics, had merged into a single unified theory. (And it was soon apparent that visible light represents only a tiny "window" in the vast spectrum of electromagnetic radiation, from radio though microwaves, infrared and ultraviolet, to xrays and gamma rays.)
Einstein dreamed of a further unification, which would combine gravity and electrodynamics, in much the same way as electricity and magnetism had been combined a century earlier. His unified field theory was not particularly successful, but in recent years the same impulse has spawned a hierarchy of increasingly ambitious (and speculative) unification schemes, beginning in the 1960s with the electroweak theory of Glashow, Weinberg, and Salam (which joins the weak and electromagnetic forces), and culminating in the 1980s with the superstring theory (which, according to its proponents, incorporates all four forces in a single "theory of everything"). At each step in this hierarchy the mathematical difficulties mount, and the gap between inspired conjecture and experimental test widens; nevertheless, it is clear that the unification of forces initiated by electrodynamics has become a major theme in the progress of physics.
The Field Formulation of Electrodynamics
The fundamental problem a theory of electromagnetism hopes to solve is this: I hold up a bunch of electric charges here (and maybe shake them around)-what happens to some other charge, over there? The classical solution takes the form of a field theory: We say that the space around an electric charge is permeated by electric and magnetic fields (the electromagnetic "odor," as it were, of the charge). A second charge, in the presence of these fields, experiences a force; the fields, then, transmit the influence from one charge to the other-they mediate the interaction.

xiv

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When a charge undergoes acceleration, a portion of the field "detaches" itself, in a sense, and travels off at the speed of light, carrying with it energy, momentum, and angular momentum. We call this electromagnetic radiation. Its existence invites (if not compels) us to regard the fields as independent dynamical entities in their own right, every bit as "real" as atoms or baseballs. Our interest accordingly shifts from the study of forces between charges to the theory of the fields themselves. But it takes a charge to produce an electromagnetic field, and it takes another charge to detect one, so we had best begin by reviewing the essential properties of electric charge.

Electric Charge
l. Charge comes in two varieties, which we call "plus" and "minus," because their effects tend to cancel (if you have +q and - q at the same point, electrically it is the same as having no charge there at all). This may seem too obvious to warrant comment, but I encourage you to contemplate other possibilities: what if there were 8 or IO different species of charge? (In chromodynamics there are, in fact, three quantities analogous to electric charge, each of which may be positive or negative.) Or what if the two kinds did not tend to cancel? The extraordinary fact is that plus and minus charges occur in exactly equal amounts, to fantastic precision, in bulk matter, so that their effects are almost completely neutralized. Were it not for this, we would be subjected to enormous forces: a potato would explode violently if the cancellation were imperfect by as little as one part in 1010 .
2. Charge is conserved: it cannot be created or destroyed-what there is now has always been. (A plus charge can "annihilate" an equal minus charge, but a plus charge cannot simply disappear by itself-something must account for that electric charge.) So the total charge of the universe is fixed for all time. This is called global conservation of charge. Actually, I can say something much stronger: Global conservation would allow for a charge to disappear in New York and instantly reappear in San Francisco (that wouldn't affect the tota[), and yet we know this doesn't happen. If the charge was in New York and it went to San Francisco, then it must have passed along some continuous path from one to the other. This is called local conservation of charge. Later on we'll see how to formulate a precise mathematical law expressing local conservation of charge-it's called the continuity equation.
3. Charge is quantized. Although nothing in classical electrodynamics requires that it be so, the fact is that electric charge comes only in discrete lumps-integer multiples of the basic unit of charge. If we call the charge on the proton +e, then the electron carries charge -e, the neutron charge zero, the pi mesons +e, 0, and -e, the carbon nucleus +6e, and so on (never 7.392e, or even 1/2e). 1 This fundamental unit of charge is extremely small, so for practical purposes it is usually appropriate to ignore quantization altogether. Water, too, "really" consists of discrete lumps (molecules); yet, if we are dealing with reasonably large large quantities of it we can treat it as a continuous fluid. This is in fact much closer to Maxwell's own view; he knew nothing of electrons and protons-he must have pictured
j j 1Actually, protons and neutrons are composed of three quarks, which carry fractional charges(± e and± e).
However, free quarks do not appear to exist in nature, and in any event this does not alter the fact that charge is quantized; it merely reduces the size of the basic unit.

xv
charge as a kind of "jelly" that could be divided up into portions of any size and smeared out at will.
These, then, are the basic properties of charge. Before we discuss the forces between charges, some mathematical tools are necessary; their introduction will occupy us in Chapter 1.
Units
The subject of electrodynamics is plagued by competing systems of units, which sometimes render it difficult for physicists to communicate with one another. The problem is far worse than in mechanics, where Neanderthals still speak of pounds and feet; for in mechanics at least all equations look the same, regardless of the units used to measure quantities.
Newton's second law remains F = ma, whether it is feet-pounds-seconds, kilograms-
meters-seconds, or whatever. But this is not so in electromagnetism, where Coulomb's law may appear variously as
-q1,z,q2-24~ (Gaussi.an),
Of the systems in common use, the two most popular are Gaussian (cgs) and SI (mks). Ele-
mentary particle theorists favor yet a third system: Heaviside-Lorentz. Although Gaussian units offer distinct theoretical advantages, most undergraduate instructors seem to prefer SI, I suppose because they incorporate the familiar household units (volts, amperes, and watts). In this book, therefore, I have used SI units. Appendix C provides a "dictionary" for converting the main results into Gaussian units.

Chapter 1
Vector Analysis

I. 1 Vector Algebra
1.1.1 Vector Operations
If you walk 4 miles due north and then 3 miles due east (Fig. 1.1 ), you will have gone a total of 7 miles, but you're not 7 miles from where you set out-you're only 5. We need an arithmetic to describe quantities like this, which evidently do not add in the ordinary way. The reason they don't, of course, is that displacements (straight line segments going from one point to another) have direction as well as magnitude (length), and it is essential to take both into account when you combine them. Such objects are called vectors: velocity, acceleration, force and momentum are other examples. By contrast, quantities that have magnitude but no direction are called scalars: examples include mass, charge, density, and temperature. I shall use boldface (A, B, and so on) for vectors and ordinary type
for scalars. The magnitude of a vector A is written IAI or, more simply, A. In diagrams,
vectors are denoted by arrows: the length of the arrow is proportional to the magnitude of the vector, and the arrowhead indicates its direction. Minus A (-A) is a vector with the
3mi

4

A

mi

5mi

Figure I.I

Figure 1.2

2

CHAPTER I. VECTOR ANALYSIS

same magnitude as A but of opposite direction (Fig. 1.2). Note that vectors have magnitude and direction but not location: a displacement of 4 miles due north from Washington is represented by the same vector as a displacement 4 miles north from Baltimore (neglecting,
of course, the curvature of the earth). On a tliagram, therefore, you can slide the arrow
around at will, as long as you don't change its length or direction. We define four vector operations: addition and three kinds of multiplication.
(i) Addition of two vectors. Place the tail of B at the head of A; the sum, A + B, is
the vector from the tail of A to the head of B (Fig. 1.3). (This rule generalizes the obvious procedure for combining two displacements.) Addition is commutative:
A+B=B+A;
3 miles east followed by 4 miles north gets you to the same place as 4 miles north followed by 3 miles east. Addition is also associative:
(A+ B) + C =A+ (B + C).
To subtract a vector (Fig. 1.4), add its opposite:
A - B =A+ (-B).

B ...

(A+B)

(B+A)

B Figure 1.3

(A-BQ Figure 1.4

(ii) Multiplication by a scalar: Multiplication of a vector by a positive scalar a multiplies the magnitude but leaves the direction unchanged (Fig. 1.5). (If a is negative, the direction is reversed.) Scalar multiplication is distributive:

a(A+B) =aA+aB.

(iii} Dot product of two vectors. The dot product of two vectors is defined by

A· B = AB cos 0,

(1.1)

where 0 is the angle they form when placed tail-to-tail (Fig. 1.6). Note that A •B is itself a scalar (hence the alternative natne scalar product). Tlie dot product is commutative,

A-B =B·A,

1.1. VECTOR ALGEBRA

3

Figure 1.5

Figure 1.6

and distributive,

A · (B + C) = A · B + A · C.

(1.2)

GeometricaJly, A • B is the product of A times the projection of B along A (or the product
of B times the projection of A along B). If the two vectors are parallel, then A • B = AB.
In particular, for any vector A,
( 1.3)

If A and B are perpendicular, then A • B = 0.

Example 1.1

Let C = A - B (Fig. 1.7), and calculate the dot product of C with itself.

Solution:

C · C = (A - B) · (A - B) = A · A - A · B - B ·A+ B · B,

or
c2 = A2 + B2 - 2ABcos0.

This is the law of cosines.

(iv) Cross product of two vectors. The cross product of two vectors is defined by

Ax B = ABsin0n,

(1.4)

where nis a unit vector (vector of length I) pointing perpendicular to the plane of A and
B. (I shall use a hat () to designate unit vectors.) Of course, there are two directions perpendicular to any plane: "in" and "out." The ambiguity is resolved by the right-hand
rule: let your fingers point in the direction of the first vector-and curl around (via the smaJler
angle) toward the second; then your thumb indicates the direction of n. (In Fig. 1.8 A x B
points into the page; B x A points out of the page.) Note that A x Bis itself a vector (hence the alternative name vector product). The cross product is distributive,

AX (B + C) =(AX B) +(AX C),

(1.5)

4 C
Figure 1.7

CHAPTER]. VECTORANALYSIS
--------------,
I I I I I I I I I
L.......,_ _ _ _ _ _ _ _ _ _ ,
B
Figure 1.8

but not commutative. In fact,

(B X A)= -(AX B).

(1.6)

Geometrically, IA x BI is the area of the parallelogram generated by A and B (Fig. 1.8). If two vectors are parallel, their cross product is zero. In particular,

AxA=O

for any vector A.

Problem 1.1 Using the definitions in Eqs. 1.1 and 1.4, and appropriate diagrams, show that the dot product and cross product are distributive, a) when the three vectors are coplanar; b) in the general case.
Problem 1.2 ls the cross product associative?
(A X B) X C =?= A X (BX C).
If so, prove it; if not, provide a counterexample.

1.1.2 Vector Algebra: Component Form
In the previous section I defined the four vector operations (addition, scalar multiplication, dot product, and cross product) in "abstract" form-that is, without reference to any partic-
ular coordinate system. In practice, it is often easier to set up Cartesian coordinates x, y, z
x, and work with vector "components." Let y, and zbe unit vectors parallel to the x, y, and
z axes, respectively (Fig. l.9(a)). An arbitrary vector A can be expanded in terms of these basis vectors (Fig. I.9(b)):

1.1. VECTOR ALGEBRA

5

z

z

z xy y

X

X

(a)

(b)

Figure 1.9

The numbers Ax, Ay, and A2 , are called components of A; geometrically, they are the projections of A along the three coordinate axes. We can now reformulate each of the four vector operations as a rule for manipulating components:

A+ B = (Axx + Ayy + A2 z) + (Bxx + Byy + B2 z)

= (Ax+ Bx)x + (Ay + By)Y + (A 2 + B2 )z.

(1.7)

(i) Rule: To add vectors, add like components.

aA = (aAx)x + (aAy)Y + (aA2 )z.

(1.8)

(ii) Rule: To multiply by a scalar, multiply each component.

x, z Because y, and are mutually perpendicular unit vectors,

Accordingly,

A· B

(Axi + Ayy + A2z) • (Bxi + Byy + B7 z) AxBx + AyBy + AzBz.

(iii) Rule: To calculate the dot product, multiply like components, and add. In particular,

(1.9) (1.10)

so (1.11)
(This is, if you like, the three-dimensional generalization of the Pythagorean theorem.) Note that the dot product of A with any unit vector is the component of A along that direction
x z (thus A· = Ax, A· y = Ay, and A· = A2 ).

6 Similarly, 1
Therefore,

CHAPTER I. VECTOR ANALYSIS

y x y X X X =

= Z X Z = 0,

y -y XX =

XX = z,

y X Z = -z X y = x,

ZXX= -X X Z = y.

(1.12) (1.13)

This cumbersome expression can be written more neatly as a determinant:

xy z
AxB= Ax Ay Az Bx By Bz

(1.14)

(iv) Rule: To calculate the cross product,form the determinant whose.first row is x, y, z,
whose second row is A (in component form), and whose third row is B.

Example 1.2

Find the angle between the face diagonals of a cube.
Solution: We might as well use a cube of side 1, and place it as shown in Fig. 1.10, with one corner at the origin. The face diagonals A and B are
A = I x+ 0 y + 1 z; B = 0 x + I y + 1 z.

X (I, 0, 0)

(0, 1, 0)
y

Figure 1.10

1These signs pertain to a right-handed coordinate system (x-axis out of the page, y-axis to the right, z-axis up,
= or aoy rotated version thereof). In a lefi-handed system (z-axis down) the signs are reversed: x x y -z, aod so
on. We shall use right-handed systems exclusively.

1.1. VECTOR ALGEBRA

7

So, in component form,

A-B=l•0+0-1+1·1=1.

On the other hand, in "abstract" form,

A• B = ABcos0 = ..fi...fi.cos0 = 2cos0.

Therefore,

cos0 = 1/2, or 0 = 60°.

Of course, you can get the answer more easily by drawing in a diagonal across the top of the cube, completing the equilateral triangle. But in cases where the geometry is not so simple,

this device of comparing the abstract and compohent forms of the dot product can be a very

efficient means of finding angles.

Problem 1.3 Find the angle between the body diagonals of a cube.
Problem 1.4 Use the cross product to find the cdmponents of the unit vector nperpendicular
to the plane shown in Fig. 1.1 I.

1.1.3 Triple Products
Since the cross product of two vectors is itself a vector, it can be dotted or crossed with a third vector to form a triple product.
(i) Scalar triple product: A • (B x C). Geometrically, IA• (B x C)I is the volume
of the parallelepiped generated by A, B, and C, since IB x Cl is the area of the base, and IA cos 0 I is the altitude (Fig. 1.12). Evidently,
A. (BX C) = B. (C X A)= C. (AX B),
for they all correspond to the same figure. Note that "alphabetical" order is preserved-in view of Eq. 1.6, the "nohalphabetical" triple products,
= A. (C X B) = B. (A X C) C . (B X A),

z

y
X
Figure 1.11

Figure 1.12

8

CHAJYJ'ER 1. VECTOR ANALYSIS

have the opposite sign. In component form,

Ax Ay Az
A · (B x C) = Bx By B2
Cx Cy C2
Note that the dot and cross can be interchanged:
A . (B X C) = (A X B) . C

(1.16)

(this follows immediately from Eq. 1.15); however, the placement of the parentheses is critical: (A• B) x C is a meaningless expression-you can't make a cross product from a scalar and a vector.
(ii) Vector triple product: A x (B x C). The vector triple product can be simplified by the so-called BAC-CAB rule:

A x (B x C) = B(A · C) - C(A · B).

( 1.17)

Notice that

(AX B) X C = -C X (AX B) = -A(B. C) + B(A. C)

is an entirely different vector. Incidentally, all higher vector products can be similarly reduced, often by repeated application of Eq. 1.17, so it is never necessary for an expression to contain more than one cross product in any term. For instance,

= (Ax B) · (C x D)

(A· C)(B · D) - (A· D)(B · C);

AX (B X (C X D)) = B(A. (C X D)) - (A. B)(C X D).

Problem 1.5 Prove the BAC-CAB rule by writing out both sides in component form. Problem 1.6 Prove that
[A x (B x C)] + [B x (C x A)]+ [C x (A x B)] = 0. Under what conditions does A x (B x C) = (A x B) x C?
1.1.4 Position, Displacement, and Separation Vectors
The location of a point in three dimensions can be described by listing its Cartesian coordinates {x, y, z). The vector to that point from the origin (Fig. 1.13) is called the position vector:
(1.19)

1.1. VECTOR ALGEBRA

z
X

r

(x, y, z)

z: r

I

y

Figure 1.13

9
source point
j
'field pom• t
Figure 1.14

I will reserve the letter r for this purpose, throughout the book. Its magnitude,

(1.20)

is the distance from the origin, and

A r xx+yy+zz r = - = ---,=:;:===~
r Jx2+y2+z2

(1.21)

is a unit vector pointing radially outward. The infinitesimal displacement vector, from
(x, y, z) to (x + dx, y + dy, z + dz), is

dl = dx x+ dy y+ dz z.

(1.22)

(We could call this dr, since that's what it is, but it is useful to reserve a special letter for infinitesimal displacements.)
In electrodynamics one frequently encounters problems involving two points-typically, a source point, r', where an electric charge is located, and a field point, r, at which you are calculating the electric or magnetic field (Fig. 1.14). It pays to adopt right from the start some short-hand notation for the separation vector from the source point to the field point. I shall use for this purpose the script letter .Ii:

Ji= r - r'.

(1.23)

Its magnitude is

I/,= lr-r'I,

and a unit vector in the direction from r' to r is

4 = -4
1-

=

r - r' -Ir -- -r'I.

(1.24) (1.25)

IO

CHAJYI'ER 1. VECTOR ANALYSIS

In Cartesian coordinates,
.. = (x - x')x + (y - y')y + (z - z')z,

(1.26)

4 = )<x - x') 2 + (y - y')2 + (z - z')2 , A (x - x')x + (y - y')y + (z - z')z
4=~============
J(x - x')2 + (y - y')2 + (z - z')2
(from which you can begjn to appreciate the advantage of the script-'/, notation).

(1.27) ( 1.28)

Problem 1.7 Find the separatiop vector4 from the source point (2,8,7) to the field point (4,6,8). Determine its magnitude (4), and construct the unit vector .i.

1.1.5 How Vectors Transfo11D
The definition of a vector as "a quantity with a magnitude and direction" is not altogether s;itisfactory: What precisely does "directjpn" mean?2 This may seem a pedantic question, but we shall shortly encounter a sm:cies of derivative that looks rather like a vector, and we'll want to know for sure whether it is one. You might be inclined to say that a vector is anything that has three components that combil}e properly under addition. Well, how about this: We have a barrel of fruit that contains Nx pears, Ny apples, and Nz bananas.
= ls N Nxi + Nyy + NzZ a vector? It has three components, and when you add another
barrel with Mx pears, My apples, and Mz bananas the result is (Nx +Mx) pears, (Ny +My)
qpples, (Nz + Mz) bananas. So it does add like a vector. Yet it's obviously not a vector, in
the physicist's sense of the word, because it doesn't really have a direction. What exactly is wrong with it?
The answer is that N d9es not transforrp properly when you change coordinates. The coordinate frqme we use to describe positions in space is of course entirely arbitrary, but there is a sp~cific geometrical transformation law for converting vector components from
one frame to (mother. Suppose, for instance, the x, y, zsystem is rotated by angle¢, relative to x, y, z, about the common x = x axes. Frqµi Fig. 1.15,
Ay = Acos0, Az = A sin 0,
while
= = Ay = A cos0 A cos(0 - ¢) A(cos0 cos¢+ sin0 sin¢) = cos</)Ay + sin¢Az,
= = Az = A sin0 A sin(0 - ¢) A(sin0 cos¢ - cos0 sin¢) = -sin</)Ay +cos¢Az.
2This section can be skipped without Joss of continuity.

1.1. VECTOR ALGEBRA

11

z
z

y
y
Figure 1.15

We might express this conclusion in matrix notation:

sin</>) ( Ay) cos</> Az •

(1.29)

More generally, for rotation about an arbitrary axis in three dimensions, the transformation law takes the form

Rxx ( Ryx
Rv:

(1.30)

or, more compactly,

L 3
A;= R;jAj,
}=I

(1.31)

where the index 1 stands for x, 2 for y, and 3 for z. The elements of the matrix R can be ascertained, for a given rotation, by the same sort of geometrical arguments as we used for a rotation about the x axis.
Now: Do the components of N transform in this way? Of course not-it doesn't matter what coordinates you use to represent positions in space, there is still the same number of apples in the barrel. You can't convert a pear into a banana by choosing a different set of axes, but you can turn Ax into Ay. Formally, then, a vector is any set of three components that transforms in the same manner as a displacement when you change coordinates. As always, displacement is the model for th~ behavior of all vectors.
By the way, a (second-rank) tensor is a quantity with nine components, T.u, Txy, Txz, Tyx, ... , Tzz, which transforms with two factors of R:

T xx = Rxx (Rxx Tu + Rxy Txy + Rxz Ttz) + Rxy (Rxx Tyx + Rxv Tvy + Rxz Tyz)
+Rx2 (R.nTzx + RxyTzy + R.uTzz), ...

12

CHAPTER 1. VECTOR ANALYSIS

or, more compactly,

=LL 3 3

Tij

RikRJtTtt.

k=I l=I

(1.32)

In general, an nth-rank tensor has n indices and 3n components, and transforms with n factors of R. In this hierarchy, a vector is a tensor of rank 1, and a scalar is a tensor of rank zero.

Problem 1.8
(a) Prove that the two-dimensional rotation matrix (1.29) preserves dot products. (That is,
show that AyBy + AzBz = AyBy + AzBz,)
(b) What constraints must the elements (Rij) of the three-dimensional rotation matrix (1.30) satisfy in order to preserve the length of A (for all vectors A)?
Problem 1.9 Find the transformation matrix R that describes a rotation by 120° about an axis from the origin through the point (1, 1, 1). The rotation is clockwise as you look down the axis toward the origin.
Problem 1.10
(a) How do the components of a vector transform under a translation of coordinates (x = x, y = y - a, z = z, Fig. l.16a)?
(b) How do the components of a vector transform under an inversion of coordinates (x = - x, y = -y, z = -z, Fig. l.16b)?
(c) How does the cross product (1.13) of two vectors transform under inversion? [The crossproduct of two vectors is properly called a pseudovector because of this "anomalous" behavior.] Is the cross product of two pseudovectors a vector, or a pseudovector? Name two pseudovector quantities in classical mechanics.
(d) How does the scalar triple product of three vectors transform under inversions? (Such an object is called a pseudoscalar.)

z z
,...a......,
X x

z x

y y
(a)

y
X

Figure 1.16

y
z (b)

1.2. DIFFERENTIAL CALCULUS

13

1.2 Differential Calculus

1.2.1 "Ordinary" Derivatives
Question: Suppose we have a function of one variable: f (x). What does the derivative, df/dx, do for us? Answer: It tells us how rapidly the function f (x) varies when we change the argument x by a tiny amount, dx:

(1.33)

In words: If we change x by an amount dx, then f changes by an amount df; the derivative
is the proportionality factor. For example, in Fig. l.l 7(a), the function varies slowly with x, and the derivative is correspondingly small. In Fig. l.l 7(b), f increases rapidly with x,
and the derivative is large, as you move away from x = 0.
Geometrical Interpretation: The derivative df / dx is the slope of the graph off versus x.

f

f

(a)

X

Figure 1.17

(b)

X

1.2.2 Gradient
Suppose, now, that we have a function of three variables-say, the temperature T(x, y, z) in a room. (Start out in one comer, and set up a system of axes; then for each point (x, y, z) in the room, T gives the temperature at that spot.) We want to generalize the notion of "derivative" to functions like T, which depend not on one but on three variables.
Now a derivative is supposed to tell us how fast the function varies, if we move a little distance. But this time the situation is more complicated, because it depends on what direction we move: If we go straight up, then the temperature will probably increase fairly rapidly, but if we move horizontally, it may not change much at all. In fact, the question "How fast does T vary?" has an infinite number of answers, one for each direction we might choose to explore.
Fortunately, the problem is not as bad as it looks. A theorem on partial derivatives states that
(1.34)

14

CHAPTER 1. VECTOR ANALYSIS

This tells us how T changes when we alter all three variables by the infinitesimal amounts dx, dy, dz. Notice that we do not require an infinite number of derivatives-three will suffice: the partial derivatives along each of the three coordinate directions.
Equation 1.34 is reminiscent of a dot product:

dT

( -axT,+ -aTy, +a-Tz-) • (dxx, + dyy, + d zz,)

ax ay az

(VT) . (di),

(1.35)

where

aT, aT, ar, VT= -x+-y+-z
ax ay az

( 1.36)

is the gradient of T. VT is a vector quantity, with three components; it is the generalized

derivative we have been looking for. Equation 1.35 is the three-dimensional version of

Eq. 1.33.

Geometrical Interpretation ofthe Gradient: Like any vector, the gradient has magnitude

and direction. To determine its geometrical meaning, let's rewrite the dot product (I .35) in

abstract form:

dT =VT· di= IVTlldll cos 0,

( 1.37)

where 0 is the angle between VT and di. Now, if we.fix the magnitude ldll and search around in various directions (that is, vary 0), the maximum change in T evidentally occurs when 0 = 0 (for then cos 0 = 1). That is, for a fixed distance ldll, dT is greatest when I move in the same direction as VT. Thus:

The gradient VT points in the direction of maximum increase of the function T.

Moreover:

The magnitude IVTI gives the slope (rate of increase) along this maximal
direction.

Imagine you are standing on a hillside. Look all around you, and find the direction of steepest ascent. That is the direction of the gradient. Now measure the slope in that direction (rise over run). That is the magnitude of the gradient. (Here the function we're talking about is the height of the hill, and the coordinates it depends on are positionslatitude and longitude, say. This function depends on only two variables, not three, but the geometrical meaning of the gradient is easier to grasp in two dimensions.) Notice from Eq. 1.37 that the direction of maximum descent is opposite to the direction of maximum ascent, while at right angles (0 = 90°) the slope is zero (the gradient is perpendicular to the contour lines). You can conceive of surfaces that do not have these properties, but they always have "kinks" in them and correspond to nondifferentiable functions.
What would it mean for the gradient to vanish? If VT = 0 at (x, y, z), then dT = 0
for small displacements about the point (x, y, z). This is, then, a stationary point of the function T(x, y, z). It could be a maximum (a summit), a minimum (a valley), a saddle

1.2. DIFFERENTIAL CALCULUS

15

point (a P¥S), or a "shoulder." This is analogous to the situation for functions of one variable, where a vanishing derivative signals a maximum, a minimum, or an inflection. In particular, ~f you want to locate the extrema of a function of three variables, set its gradient equal to zero.
Example 1.3
Find the gradient of r = Jx 2 + y2 + z2 (the magnitude of the position vector).
SolutioTT;

Vr

=

-arxA +-aryA +-arzA
ax ay az

I

2x

. 1

2y

. l

2z

.

-
2

--;:
J x2

:+=y2=+=z=2 x

+

---
2 J

;:
x2

:+=y=2

+ =

=
z2

Y+-2-✓-x;2::+=y=2 +=z=2 z

=

xx+yy+zz
-J-x;2::+=y=2 +=z=2

=

r r

=

A
r.

Does this make sense? Well, it says that the distance from the origin increases most rapidly in the radial direction, and that its rate of increase in that direction is 1... just what you'd expect.

Problem 1.11 Find the gradients of the following functions:

= + {a) /(x, y, z) x2 + y3 z4_

(b) f(x, Y, z) = x 2y 3z4.

= r (c) f(x, y, z)

sin(y)ln(z).

Problem 1.12 The height of a certain hill (in feet) is given by
+ h(x, y) = 10(2xy - 3x2 - 4y2 - l8x 28y + 12),

where y is the distance (in miles) nortp, x the distance east of South Hadley. (a) Where is the top of the hill located? (b) How high is the hill? (c) How steep is the slope (in feet per mile) at a point I mile north and one mile east of South Hadley? In what direction is the slope steepest, at that point?

•

Problem 1.13 Leu. be the separation vector from a fixed point (x', y', z') to the point (x, y, z),

and let 1/,, be its length. Show that

(a) V(l/,,2) = 24.

= (b) V(l/1/,,) -4/1/,,2.

(c) What is the general formula for V(1/,,n )?

16

CHAPTER 1. VECTOR ANALYSIS

Problem 1.14 Suppose that f is a function of two variables (y and z) only. Show that the
gradient Vf = (8f/8y)y + (8f/8z)z transforms as a vector under rotations. Eq. 1.29. [Hint:
(8fj'Jy) = (8j/8y)(8y/8y) + (8f/8z)(8z/8y), and the analogous formula for aJ;az. We know that y = y cos cp + z sin c/J and z = -y sin c/J + z cos cp; "solve" these equations for y and
z (as functions of y and z), and compute the needed derivatives ay/ oy, oz/8y, etc.]

1.2.3 The Operator V
The gradient has the formal appearance of a vector, V, "multiplying" a scalar T:
(1.38)
(For once I write the unit vectors to the left, just so no one will think this means ai/ax, and
so on-which would be zero, since xis constant.) The term in parentheses is called "del'':
(1.39)
Of course, del is not a vector, in the usual sense. Indeed. it is without specific meaning until we provide it with a function to act upon. Furthermore, it does not "multiply" T; rather, it is an instruction to differentiate what follows. To be precise, then, we should say that V is a vector operator that acts upon T, not a vector that multiplies T.
With this qualification, though, V mimics the behavior of an ordinary vector in virtually every way; almost anything that can be done with other vectors can also be done with V, if we merely translate "multiply" by "act upon." So by all means take the vector appearance of V seriously: it is a marvelous piece of notational simplification, as you will appreciate if you ever consult Maxwell's original work on electromagnetism, written without the benefit ofV.
Now an ordinary vector A can multiply in three ways:
1. Multiply a scalar a : Aa; 2. Multiply another vector B, via the dot product: A • B; 3. Multiply another vector via the cross product: A x B.
Correspondingly, there are three ways the operator V can act:
1. On a scalar function T : VT (the gradient); 2. On a vector function v, via the dot product: V • v (the divergence); 3. On a vector function v, via the cross product: V xv (the curl).
We have already discussed the gradient. In the following sections we examine the other two vector derivatives: divergence and curl.

1.2. DIFFERENTIAL CALCULUS

17

1.2.4 The Divergence
From the definition of V we construct the divergence:

V-v =

(1.40)

Observe that the divergence of a vector function vis itself a scalar V •v. (You can't have the divergence of a scalar: that's meaningless.)
Geometrical Interpretation: The name divergence is well chosen, for V•Vis a measure of how much the vector v spreads out (diverges) from the point in question. For example, the vector function in Fig. 1.18a has a large (positive) divergence (if the arrows pointed in, it would be a large negative divergence), the function in Fig. 1.1 Sb has zero divergence, and the function in Fig. 1.1 Sc again has a positive divergence. (Please understand that v here is afunction-there's a different vector associated with every point in space. In the diagrams,

l l

(a)

(b)

l Ill Ill
(c)
Figure 1.18

CHAPTER I. VECTOR ANALYSIS

of course, I can only draw the arrows at a few representative locations.) Imagine standing
at the edge of a pond. Sprinkle some sawdust or pine needles on the surface. If the material
spreads out, then you dropped it at a point of positive divergence; if it collects together, you dropped it at a point of negative divergence. (The vector function v in this model is the velocity of the water-this is a two-dimensional example, but it helps give one a "feel" for what the divergence means. A point of positive divergence is a source, or "faucet"; a point of negative divergence is a sink, or "drain.")
Example 1.4
z, z, z. Suppose the functions in Fig. 1.18 are Va = r = xx+ y y + z Vb = and Ve = z
Calculate their divergences.
Solution:

a

a

a

V · Va = -;--(x) + -(y) + -(z) = l + 1 + 1 = 3.

ax

ay

oz

As anticipated, this function has a positive divergence.

as expected.

a

a

a

V ·Vb= -a(x0) + -ay(0) + -oz(1) = 0 + 0 + 0 = 0,

a

a

a

V ·Ve= -(0) + -(0) + -(z) = 0 + 0 + l = 1.

ax

ay

oz

Problem 1.15 Calculate the divergence of the following vector functions:
x z. + (a) Va = x2 3xz2 y - 2xz
{b) Vb= xyx + 2yzy + 3zxz.
x (c) Ve= y2 + (2xy + z2)y + 2yzz.

•

Problem 1.16 Sketch the vector function

and compute its divergence. The answer may surprise you ... can you explain it?
Problem 1.17 In two dimensions, show that the divergence transforms as a scalar under rotations. [Hint: Use Eq. 1.29 to determine Vy and Vz, and the method of Prob. 1.14 to calculate
the derivatives. Your aim is to show that avy/oy + ovz!oz = OVy/oy + ovzfoz.]

1.2. DIFFERENTIAL CALCULUS

19

1.2.5 The Curl

From the definition of V we construct the curl:

X

y

Z

V XV

a;ax a;ay a;az

Notice that the curl of a vector function vis, like any cross product, a vector. (You cannot have the curl of a scalar; that's meaningless.)
Geometrical Interpretation: The name curl is also well chosen, for V x vis a measure of how much the vector v "curls around" the point in question. Thus the three functions in Fig. 1.18 all have zero curl (as you can easily check for yourself), whereas the functions in Fig. 1.19 have a substantial curl, pointing in the z-direction, as the natural right-hand rule would suggest. Imagine (again) you are standing at the edge of a pond. Float a small paddlewheel (a cork with toothpicks pointing out radially would do); if it starts to rotate, then you placed it at a point of nonzero curl. A whirlpool would be a region of large curl.
z

.

y

_,,._.._-...-- y

X

(a)

Figure 1.19

Example 1.5

Suppose the function sketched in Fig. 1.19a is Va = -yx + xy, and that in Fig. 1.19b is vb = xy. Calculate their curls.

Solution:

x y z

V x Va= a;ax a;ay a;az = 2z,

-y

X

Q

and

x y z

V X Vb= a;ax a;ay a;az =z.

0

X

0

20

CHAPTER 1. VECTOR ANALYSIS

As expected, these curls point in the+zdirection. (Incidentally, they both have zero divergence, as you might guess from the pictures: nothing is "spreading out"... it just "curls around.")

Problem 1.18 Calculate the curls of the vector functions in Prob. 1.15.
Problem 1.19 Construct a vector function that has zero divergence and zero curl everywhere.. (A constant will do the job, of course, but make it something a little more interesting than that!)

1.2.6 Product Rules

The calculation of ordinary derivatives is facilitated by a number of general rules, such as

the sum rule:

d

df dg

dx (f + g) = dx + dx '

the rule for multiplying by a constant:

the product rule:

d

df

-(kf) =k-,

dx

dx

d

dg df

dx (Jg)= f dx + g dx'

and the quotient rule:

df dg

d dx

(1g)

=

gb

-fb
g2

•

Similar relations hold for the vector derivatives. Thus,

V(f+g)=V/+Vg, V • (A+ B) = (V • A) + (V •B),

V X (A + B) = (V X A) + (V X B)'
and
V(k/) = kV f, V · (kA) = k(V · A), V X (kA) = k(V X A),
as you can check for yourself. The product rules are not quite so simple. There are two ways to construct a scalar as the product of two functions:

f g (product of two scalar functions), A• B (dot product of two vector functions),

and two ways to make a vector:

/A (scalar times vector), AxB (cross product of two vectors).

1.2. DIFFERENTIAL CALCULUS

21

Accordingly, there are six product rules, two for gradients:

(i)

V(fg) = JV g + gV f,

(ii)

V (A . B) = A X (V X B) + B X (V X A) + (A . V)B + (B . V)A,

two for divergences:

(iii)

V • (fA)= f(V ·A)+ A· (Vf),

(iv) and two for curls: (v)

V . (A X B) = B. (V X A) - A. (V X B), V x (f A) = f (V x A) - Ax (V f),

(vi)

V x (Ax B) = (B · V)A - (A· V)B + A(V · B) - B(V -A).

You will be using these product rules so frequently that I have put them on the inside front cover for easy reference. The proofs come straight from the product rule for ordinary derivatives. For instance,

V • (JA) =

It is also possible to formulate three quotient rules:
gV f-fVg
g2
g(V -A) -A· (Vg)
g2
g(V X A) +AX (Vg)
g2
However, since these can be obtained quickly from the corresponding product rules, I haven't bothered to put them on the inside front cover.

22

CHAPTER 1. VECTOR ANALYSIS

Problem 1.20 Prove product rules (i), (iv), and (v).
Problem 1.21 (a) If A and B are two vector functions, what does the expression (A • V)B mean? (That is,
what are its x, y, and z components in terms of the Cartesian components of A, B, and V?)
(b) Compute (r · V)r, where r is the unit vector defined in Eq. 1.21. (c) For the functions in Prob. 1.15, evaluate (va • V)vb.
Problem 1.22 (For masochists only.) Prove product rules (ii) and (vi). Refer to Prob. 1.21 for the definition of (A • V)B.
Problem 1.23 Derive the three quotient rules.
Problem 1.24 (a) Check product rule (iv) (by calculating each term separately) for the functions
A =xx + 2y y + 3z z; B = 3yx- 2xy.
(b) Do the same for product rule (ii). (c) The same for rule (vi).

1.2.7 Second Derivatives

The gradient, the divergence, and the curl are the only first derivatives we can make with V; by applying V twice we can construct five species of second derivatives. The gradient VT is a vector, so we can take the divergence and curl of it:
(1) Divergence of gradient: V • (VT).
(2) Curl of gradient: V x (VT).

The divergence V • v is a scalar-all we can do is take its gradient:

(3) Gradient of divergence: V (V • v).
The curl V x v is a vector, so we can take its divergence and curl:

(4) Divergence of curl: V • (V x v). (5) Curl of curl: V x (V xv).

This exhausts the possibilities, and in fact not all of them give anything new. Let's consider them one at a time:

(l) V · (VT)

( xA aa-x +yA aa-y +zA aa-z ) • (a-aTxxA + -aayryA +a-azrzA) -aa2xT2+aa-2yr2+aa2-zr2 •

(1.42)

1.2. DIFFERENTIAL CALCULUS

23

This object, which we write V2 T for short, is called the Laplacian of T; we shall be studying it in great detail later on. Notice that the Laplacian of a scalar T is a scalar. Occasionally, we shall speak of the Laplacian of a vector, V2v. By this we mean a vector quantity whose x-component is the Laplacian of Vx, and so on:3

(1.43)

This is nothing more than a convenient extension of tqe meaning of V2. (2) The curl of a gradient is always zero:

V x (VT)= 0.

(1.44)

This is an important fact, which we shall use repeatedly; you can easily prove it from the definition of V, Eq. 1.39. Beware: You might think Eq. 1.44 is "obviously" true-isn't it just (V x V)T, and isn't the cross product of any vector (in this case, V) with t~self always zero? This reasoning is suggestive but not quite conclusive, since V is an operator and does not "multiply" in the usual way. The proof of Eq. 1.44, in fact, hinges on the equality of cross derivatives:

aax (~:) = :Y (::) •

(1.45)

If you think I'm being fussy, test your intuition on this one:

(VT) X (VS).

Is that always zero? (It would be, of course, if you replaced the V's by an ordinary vector.)
(3) V(V •v) for some reason seldom occurs in physical applications, and it has not been
given any special name of its own-it's just the gradient of the divergence. Notice that
V(V • v) is not the same as the Laplacian of a vector: V2v = (V • V)v I- V(V •v).
(4) The divergence of a curl, like the curl of a gradient, is always zero:

V. (V xv)= 0.

(1.46)

You can prove this for yourself. (Again, there is a fraudulent short-cut proof, using the
= vector identity A• (B x C) (A x B) • C.)
(5) As you can check from the definition of V:

V x (V x v) = V(V • v) - V2v.

(1.47)

So curl-of-curl gives nothing new; the first term is just number (3) and the second is the Laplacian (of a vector). (In fact, Eq. 1.47 is often used to define the Laplacian of a vector, in preference to Eq. 1.43, which makes specific reference to Cartesian coordinates.)

Really, then, there are just two kinds of second derivatives: the Laplacian (which is of fundamental importance) and the gradient-of-divergence (which we seldom encounter).
3In curvilinear coordinates, where the unit vectors themselves qepeqd on position, they too must be differentiated (see Sect. 1.4.1).

24

CHAPTER 1. VECTOR ANALYSIS

We could go through a similar ritual to work out third derivatives, but fortunately second derivatives suffice for practically all physical applications.
A final word on vector differential calculus: It all flows from the operator V, and from taking seriously its vector character. Even if you remembered only the definition of V, you should be able, in principle, to reconstruct all the rest.

Problem 1.25 Calculate the Laplacian of the following functions:
(a) Ta = x 2 + 2xy + 3z +4.
(b) Tb= sinx siny sinz. (c) Tc= e-Sx sin4y cos 3z.
(d) v = x 2 x+ 3xz2 y - 2xz z.
Problem 1.26 Prove that the divergence of a curl is always zero. Check it for function Va in Prob. 1.15.
Problem 1.27 Prove that the curl of a gradient is always zero. Check it for function (b) in Prob. 1.11.

L.3 Integral Calculus
1.3.1 Line, Surface, and Volume Integrals
In electrodynamics we encounter several different kinds of integrals, among which the most important are line (or path) integrals, surface integrals (or flux), and volume integrals.
(a) Line Integrals. A line integral is an expression of the form

fb V · di, laP

( 1.48)

where vis a vector function, di is the infinitesimal displacement vector (Eq. 1.22), and the

integral is to be carried out along a prescribed path P from point a to point b (Fig. 1.20). If

the path in question forms a closed loop (that is, if b = a), I shall put a circle on the integral

sign:

f v • di.

(1.49)

At each point on the path we take the dot product of v (evaluated at that point) with the displacement di to the next point on the path. To a physicist, the most familiar example of
a line integral is the work done by a force F: W = f F • di.
Ordinarily, the value of a line integral depends critically on the particular path taken from a to b, but there is an important special class of vector functions for which the line inteiJral is independent of the path, and is determined entirely by the end points. It will be our business in due course to characterize this special class of vectors. (A force that has this property is called conservative.)

1.3. INTEGRAL CALCULUS
z b
y
X
Figure 1.20

25
y 2

2

X

Figure 1.21

Example 1.6
Calculate the line integral of the function v = y2 x+ 2x(y + 1) yfrom the point a = (1, 1, 0)
f to the point b = (2, 2, 0), along the paths (1) and (2) in Fig. 1.21. What is v •di for the loop
that goes from a to b along (1) and returns to a along (2)?
Solution: As always, di = dx x+ dy y + dz z. Path (1) consists of two parts. Along the ''horizontal" segment dy = dz = 0, so (i) di = dx x, y = I, v · di = y2 dx = dx, so f v · di = f? dx = 1. On the "vertical" stretch dx = dz = 0, so
(ii)dl=dyy, x=2, v-dl=2x(y+l)dy=4(y+1)dy,so fv-dl=4f?(y+l)dy= 10.
By path (1 ), then,
Meanwhile, on path (2) x = y, dx = dy, and dz= 0, so di = dx x+ dx y, v •di = x 2 dx + 2x (x + I) dx = (3x2 + 2x) dx,
so
I~= Lb v •di= fi\3x2 + 2x) dx = (x3 + x2) 10.
(The strategy here is to get everything in terms of one variable; I could just as well have eliminated x in favor of y.)
For the loop that goes out (l) and back (2), then,
fv-dl=ll-10=1.

26

CHAPTER 1. VECTOR ANALYSIS

(b) Surface Integrals. A surface integral is an expression of the form
Lv-da,

(1.50)

where vis again some vector function, and dais an infinitesimal patch of area, with direction perpendicular to the surface (Fig. 1.22). There are, of course, two directions perpendicular to any surface, so the sign of a surface integral is intrinsically ambiguous. If the surface is closed (forming a "balloon"), in which case I shall again put a circle on the integral sign
f v-da,

then tradition dictates that "outward" is positive, but for open surfaces it's arbitrary. If v
J describes the flow of a fluid (mass per unit area per unit time), then v •da represents the
total mass per unit time passing through the surface-hence the alternative name, "flux." Ordinarily, the value of a surface integral depends on the particular surface chosen, but
there is a special class of vector functions for which it is independent of the surface, and is determined entirely by the boundary line. We shall soon be in a position to characterize this special class.

z

(v)

(ii)

y
X
Figure 1.22

X 2

(iii)

2

y

Figure 1.23

Example 1.7
Calculate the surface integral ofv = 2xz x+ (x + 2) y+ y(z2 - 3) zover five sides (excluding
the bottom) of the cubical box (side 2) in Fig. 1.23. Let "upward and outward" be the positive direction, as indicated by the arrows. Solution: Taking the sides one at a time:
(i) x = 2, da = dy dz x, v •da = 2xz dy dz= 4z dy dz, so
j fo fo v •da = 4 2 dy 2 z dz = 16.
(ii) x = 0, da = -dydzx, v · da = -2xzdydz = 0, so
j v•da=0.

1.3. INTEGRAL CALCULUS

27

(iii) y = 2, da = dx dz y, v •da = (x + 2) dx dz, so

fo fo / v •da = 2(x + 2) dx 2 dz = 12.

(iv) y = 0, da = -dxdzy, v •da = -(x + 2)dxdz, so
f v-da=-fo\x+2)dx fo2 dz=-12.

(v) z = 2, da = dx dy z, v · da = y(z2 - 3) dx dy = y dx dy. so

la [2

2

/ v •da = lo dx y dy = 4.

Evidently the total flux is
f v · da = 16 + 0 + 12 - 12 + 4 = 20.
lsurface

(c) Volume Integrals. A volume integral is an expression of the form
Iv T dr,

(1.51)

where T is a scalar function and dr is an infinitesimal volume element. In Cartesian

coordinates,

dr=dxdydz.

( 1.52)

For example, if Tis the density of a substance (which might vary from point to point), then the volume integral would give the total mass. Occasionally we shall encounter volume integrals of vector functions:
I I I I vdr = (vx X+ Vy y+ Vz z)dr =XI VxdT + y VydT + z VzdT; ( 1.53)

because the unit vectors are constants, they come outside the integral.
Example 1.8
Calculate the volume integral of T = xyz2 over the prism in Fig. 1.24.
Solution: You can do the three integrals in any order. Let's do x first: it runs from Oto (1- y); then y (it goes from Oto 1); and finally z (0 to 3):

lo ~I lo3 z2 dz I (I - y) 2ydy = 2I (9)( T1I)= s3·

2 0

0

28

CHAPTER 1. VECTOR ANALYSIS

y
X
Figure 1.24

x z Problem 1.28 Calculate the line integral of the function v = x 2 + 2yz y + y2 from the
origin to the point (1,1,1) by three different routes:
(a) (0,0,0) ➔ (1,0,0) ➔ (1, 1,0) ➔ (I, I, 1);
(b) (0, 0, 0) ➔ (0, 0, I) ➔ (0, 1, l) ➔ (l, I, I);
(c) The direct straight line. (d) What is the line integral around the closed loop that goes out along path (a) and back along path (b)?
Problem 1.29 Calculate the surface integral of the function in Ex. 1.7, over the bottom of the box. For consistency, let "upward" be the positive direction. Does the surface integral depend only on the boundary line for this function? What is the total flux over the closed surface of the box (including the bottom)? [Note: For the closed surface the positive direction is "outward," and hence "down," for the bottom face.]
Problem 1.30 Calculate the volume integral of the function T = z2 over the tetrahedron with
comers at (0,0,0), (l,0,0), (0,1,0), and (0,0,1).

1.3.2 The Fundamental Theorem of Calculus

Suppose f (x) is a function of one variable. The fundamental theorem of calculus states:

1b

df -dx

=

f(b)

-

f(a).

a dx

(1.54)

In case this doesn't look familiar, let's write it another way:

1b F(x) dx = f (b) - f(a),

where df/dx = F(x). The fundamental theorem tells you how to integrate F(x): you
think up a function f (x) whose derivative is equal to F.

1.3. INTEGRAL CALCULUS

29

Geometrical Interpretation: According to Eq. 1.33, df = (df/dx)dx is the infinitesi-
mal change inf when you go from (x) to (x + dx). The fundamental theorem (1.54) says
that if you chop the interval from a to b (Fig. 1.25) into many tiny pieces, dx, and add up the increments df from each little piece, the result is (not surprisingly) equal to the total change in f: f (b) - f (a). In other words, there are two ways to determine the total change in the function: either subtract the values at the ends or go step-by-step, adding up all the tiny increments as you go. You'll get the same answer either way.
Notice the basic format of the fundamental theorem: the integral of a derivative over an interval is given by the value of the function at the end points (boundaries). In vector calculus there are three species of derivative (gradient, divergence, and curl), and each has its own "fundamental theorem," with essentially the same format. I don't plan to prove these theorems here; rather, I shall explain what they mean, and try to make them plausible. Proofs are given in Appendix A.

f(x) f(b) f(a)

z b
y

a dx

XX

X

Figure 1.25

Figure 1.26

1.3.3 The Fundamental Theorem for Gradients
Suppose we have a scalar function of three variables T(x, y, z). Starting at point a, we move a small distance di, (Fig. 1.26). According to Eq. 1.37, the function Twill change by an amount
dT =(VT)· dli.
Now we move a little further, by an additional small displacement dlz; the incremental
change in Twill be (VT)• dh. In this manner, proceeding by infinitesimal steps, we make the journey to point b. At each step we compute the gradient of T (at that point) and dot it into the displacement di. .. this gives us the change in T. Evidently the total change in T in going from a to b along the path selected is

lb (VT) . di= T(b) - T(a). p

(1.55)

30

CHAPTER 1. VECTOR ANALYSIS

This is called the fundamental theorem for gradients; like the "ordinary" fundamental theorem, it says that the integral (here a line integral) of a derivative (here the gradient) is given by the value of the function at the boundaries (a and b).
Geometrical Interpretation: Suppose you wanted to determine the height of the Eiffel Tower. You could climb the stairs, using a ruler to measure the rise at each step, and adding them all up (that's the left side of Eq. 1.55), or you could place altimeters at the top and the bottom, and subtract the two readings (that's the right side); you should get the same answer either way (that's the fundamental theorem).
Incidentally, as we found in Ex. 1.6, line integrals ordinarily depend on the path taken from a to b. But the right side of Eq. 1.55 makes no reference to the path-only to the end points. Evidently, gradients have the special property that their line integrals are path independent:
J: Corollary 1: (VT) • di is independent of path taken from a to b.
Corollary 2: :f (VT) - di = 0, since the beginning and end points
are identical, and hence T(b) - T(a) = 0.

Example 1.9
Let T = xy2, and take point a to be the origin (0, 0. 0) and b the point (2, l, 0). Check the
fundamental theorem for gradients.
Solution: Although the integral is independent of path, we must pick a specific path in order to evaluate it. Let's go out along the x axis (step i) and then up (step ii) (Fig. 1.27). As always,
x di= dxx +dyy +dzz; VT= y2 + 2xyy.
(i)y=0; dl=dxx, VT-dl=y2dx=0,so
1VT• di= 0.

(ii) x = 2; di= dyy, VT· di= 2xydy = 4ydy, so

k (VT· di= f14ydy = 2y21 1 = 2.

l1

o

y

b ii)

2

X

Figure 1.27

1.3. INTEGRAL CALCULUS

31

Evidently the total line integral is 2. Is this consistent with the fundamental theorem? Yes:
T(b) - T(a) = 2 - 0 = 2.
Now, just to convince you that the answer is independent of path, let me calculate the same integral along path iii (the straight line from a to b):
(iii) y = ½x, dy = ½dx. VT· dl = _v2dx + 2xydy = ¾x2 dx, so

f [
Jiii

VT·

dl

=

2
lo

¾x 2

dx

=

¼x

3

2
a 1

=

2.

Problem 1.31 Check the fundamental theorem for gradients, using T = x2 + 4xy + 2yz3 , the points a = (0, 0, 0), b = (I, 1, 1), and the three paths in Fig. 1.28:
(a) (0,0.0) ➔ (L0,0) ➔ (1.1,0) ➔ (1, 1.1); (b) (0,0.0) ➔ (0,0, l) ➔ (0, 1, l) ➔ (1, I, 1);
(c) the parabolic path z = x 2; y = x.

z
•
y (a)

z

y

X

(b)

Figure 1.28

z

• I
I
y

X

(c)

1.3.4 The Fundamental Theorem for Divergences
The fundamental theorem for divergences states that:

J / (V •v) dr = v • da.

V

S

(1.56)

In honor, I suppose of its great importance, this theorem has at least three special names: Gauss's theorem, Green's theorem, or, simply, the divergence theorem. Like the other "fundamental theorems," it says that the integral of a derivative (in this case the divergence) over a region (in this case a volume) is equal to the value of the function at the boundary

32

CHAPTER 1. VECTOR ANALYSIS

(in this case the surface that bounds the volume). Notice that the boundary term is itself an integral (specifically, a surface integral). This is reasonable: the "boundary" of a line is just two end points, but the boundary of a volume is a (closed) surface.
Geometrical Interpretation: If v represents the flow of an incompressible fluid, then the flux of v (the right side of Eq. 1.56) is the total amount of fluid passing out through the surface, per unit time. Now, the divergence measures the "spreading out" of the vectors from a point-a place of high divergence is like a "faucet," pouring out liquid. If we have lots of faucets in a region filled with incompressible fluid, an equal amount pf liquid will be forced out through the boundaries of the region. In fact, there are two ways we could determine how much is being produced: (a) we could count up all the faucets, recording how much each puts out, or (b) we could go around the boundary, measuring the flow at each point, and add it all up. You get the same answer either way:
f / (faucets within the volume)= (flow out through the surface).

This, in essence, is what the divergence theorem says.

Example 1.10

Check the divergence theorem using the function
V =ix+ (2xy + z2)y + (2yz)z

and the unit cube situated at the origin (Fig. 1.29).

Solution: In this case

= V • v 2(x + y),

and
fv 2(x + y)d-r = 2 lof 1lof 1lof\x + y)dxdydz,

fo fo\x+y)dx=½+Y, {c½+y)dy=1, 1 1dz=1.

(v)
z 1

(ii)

y
l(vi)
Figure 1.29

1.3. INTEGRAL CALCULUS

33

Evidently,

/ V -vd-r: = 2.
V

So much for the left side of the divergence theorem. To evaluate the surface integral we must consider separately the six sides of the cube:

(i)

(ii)

(iii)

la la 1· / v •da =

1

1 (2x

+

z2)

dx

dz

=

(iv)

(v)

j fo fo v. da = 1 12ydxdy = 1.

(vi)

j lo lo v •da = - 1 10 dx dy = 0.

So the total flux is: as expected.

f v •da = 1- 1+ ~ - 1+ 1 + 0 = 2,
s

x z. Problem 1.32 Test the divergence theorem for the function v = (xy) + (2yz) y + (3zx)
Take as your volume the cube shown in Fig. 1.30, with sides of length 2.

34

CHAPTER 1. VECTOR ANALYSIS

z
2,1-------,,

2 y
X
Figure 1.30

1.3.5 The Fundamental Theorem for Curls

The fundamental theorem for curls, which goes by the special name of Stokes' theorem,

states that

J f= (V x v) - da v • di.

s

p

(1.57)

As always, the integral of a derivative (here, the curl) over a region (here, a patch of suiface) is equal to the value of the function at the boundary (here, the perimeter of the patch). As in the case of the divergence theorein, the boundary term is itself an integral-specifically, a closed line integral.
Geometrical Interpretation: Recall that the curl measures the "twist" of the vectors v; a region of high curl is a whirlpool-if you put a tiny paddle wheel there, it will rotate. Now, the integral of the curl over some surface (or, more precisely, the flux of the curl through that surface) represents the "total amount of swirl," and we can determine that swirl just as well by going around the edge and finding how much the flow is following the boundary (Fig. 1.31). You may find this a rather forced interpretation of Stokes' theorem, but it's a helpful mnemonic, if nothing else.
You might have noticed an apparent ambiguity in Stokes' theorem: concerning the boundary line integral, which way are we supposed to go around (clockwise or counterclockwise)? If we go the "wrorig" way we'll pick up an overall sign error. The answer is that it doesn't matter which way you go as long as you are consistent, for there is a compensating sign ambiguity in the surface integtal: Which way does da point? For a closed surface (as in the divergence theorem) da points in the direction of the outward normal; but for an open surface, which way is "out?" Consistency in Stokes' theorem (as in all such matters) is given by the right-hand rule: If your fingers point in the direction of the line integral, then your thumb fixes the direction of da (Fig. 1.32).
Now, there are plenty of surfaces (infinitely many) that share any given boundary line. Twist a paper clip into a loop and dip it in soapy water. The soap film constitutes a surface, with the wire loop as its boundary. If you blow on it, the soap film will expand, making a larger surface, with the same boundary. Ordinarily, a flux integral depends critically on what surface you integrate over, but evidently this is not the case with curls. For Stokes'

1.3. INTEGRAL CALCULUS Figure 1.31

35
da
~
di ------Figure 1.32

theorem says that j(V xv)• dais equal to the line integral ofv around the boundary, and the latter makes no reference to the specific surface you choose.

Corollary 1: f (V x v) • da depends only on the boundary line, not
on the particular surface used.

Corollary 2:

f (V x v) · da = 0 for any closed surface, since the
boundary line, like the mouth of a balloon, shrinks down to a point, and hence the right side of Eq. 1.57 vanishes.

These corollaries are analogous to those for the gradient theorem. We shall develop the parallel further in due course.

Example 1.11

Suppose v = (2xz + 3y2)y + (4yz 2)z. Check Stokes' theorem for the square surface shown
in Fig. 1.33.

Solution: Here

V x v = (4z2 - 2x) i + 2zz and da = dy dzi.

z (iii) I

(iv) .

(ii)

X

(i)

y

Figure 1.33

36

CHAPTER I. VECTOR ANALYSIS

(In saying that da points in the x direction, we are committing ourselves to a counterclockwise
x, line integral. We could as well write da = -dy dz but then we would be obliged to go
clockwise.) Since x = 0 for this surface,

Now, what about the line integral? We must break this up into four segments:

f Jd (i) X =0, z = 0, v-dl=3y2 dy, v ·di= 3y 2 dy = I,

Jd 1· f (ii) X = 0, y = I. v • di = 4z2 dz,

v • di = 4z2 dz =

f (iii) X =0, z = 1, v-dl=3y 2 dy, v ·di=!? 3y2 dy = -1,

(iv) X = 0, y = 0, v ·di= 0, So

Jv • di = ff Odz = 0.

It checks.
A point of strategy: notice how I handled step (iii). There is a temptation to write di = -dy y
here, since the path goes to the left. You can get away with this, if you insist, by running the
x z integral from 0 - I. Personally, I prefer to say di = dx + dy y + dz always (never any
minus signs) and let the limits of the integral take care of the direction.

Problem 1.33 Test Stokes' theorem for the function v = (xy) x+ (2yz) y+ (3zx) z, using the
triangular shaded area of Fig. 1.34.
Problem 1.34 Check Corollary I by using the same function and boundary line as in Ex. 1. 11, but integrating over the five sides of the cube in Fig. 1.35. The back of the cube is open.

z 2

X

2 y

Figure 1.34

z

(v)

-(iv)

X

i(ii)

(iii) y

Figure 1.35

1.3. INTEGRAL CALCULUS

37

1.3.6 Integration by Parts

The technique known (awkwardly) as integration by parts exploits the product rule for

derivatives:

(:!) :x (jg)= f

+ g (~~)-

Integrating both sides, and invoking the fundamental theorem:

l

b a

-d ( dx

f g)

dx

=

f g Ib = lb f
a a

(d- g) dx dx

+ lb g (d- f)
a dx

dx,

or

[b (dg)

[b (df)

b

la f dx dx=-Ja g dx dx+fgla·

( 1.58)

That's integration by parts. It pertains to the situation in which you are called upon to integrate the product of one function (f) and the derivative of another (g); it says you can transfer the derivative from g to f, at the cost of a minus sign and a boundary term.

Example 1.12

Evaluate the integral

Solution: The exponential can be expressed as a derivative:

in this case, then, f(x) = x, g(x) = -e-x, and df/dx = l, so

We can exploit the product rules of vector calculus, together with the appropriate fundamental theorems, in exactly the same way. For example, integrating
V • (f A) = f (V ·A)+ A· (V f)
over a volume, and invoking the divergence theorem, yields
f f f V • (fA) dr = f(V •A) dr + A· (V f) dr =ffA· da,

or

Iv Iv f(V •A) dr = - A· (V f) dr +ifA· da.

(1.59)

38

CHAPTER 1. VECTOR ANALYSIS

Here again the integrand is the product of one function (f) and the derivative (in this case the divergence) of another (A), and integration by parts licenses us to transfer the derivative from A to / (where it becomes a gradient), at the cost of a minus sign and a boundary term (in this case a surface integral).
You might wonder how often one is likely to encounter an integral involving the product of one function and the derivative of another; the answer is surprisingly often, and integration by parts turns out to be one of the most powerful tools in vector calculus.

Problem 1.35 (a) Show that

Is f(V x A) -da = fs[A x (V f)] -da + { f A• di.

(b) Show that

iv B • (V x A) dr = iv A· (V x B) dr +£(A x B) •da.

(1.60) (1.61)

1.4 Curvilinear Coordinates

1.4.1 Spherical Polar Coordinates

The spherical polar coordinates (r, 0, </)) of a point P are defined in Fig. 1.36; r is the distance from the origin (the magnitude of the position vector), 0 (the angle down from the z axis) is called the polar angle, and¢ (the angle around from the x axis) is the azimuthal angle. Their relation to Cartesian coordinates (x, y, z) can be read from the figure:

= = x rsin0cos¢, y r sin0 sin¢, z = rcos0.

(l.62)

z

X
Figure 1.36

1.4. CURVILINEAR COORDINATES

39

J, Figure 1.36 also shows three unit vectors, r, 0, pointing in the direction of increase
of the corresponding coordinates. They constitute an orthogonal (mutually perpendicular)
basis set Gust like x, y, z), and any vector A can be expressed in terms of them in the usual
way:
(l.63)

l A,, Ae, and A¢ are the radial, polar, and azimuthal components of A. In terms of the
Cartesian unit vectors,

r

sin 0 cos¢ x+ sin 0 sin¢ y+ cos 0 z,

0

cos 0 cos¢ x+ cos 0 sin¢ y - sin 0 z,

(1.64)

J

- sin ¢ x+ cos¢ y,

as you can easily check for yourself (Prob. 1.37). I have put these formulas inside the back cover, for easy reference.
J But there is a poisonous snake lurking here that I'd better warn you about: r, 0, and
are associated with a particular point P, and they change dfrection as P moves around. For
example, r always points radially outward, but "radially outward" can be the x direction,
they direction, or any other direction, depending on where you are. In Fig. 1.37, A= yand
r B = -y, and yet both of them would be written as in spherical cootdinates. One could Jee, take account of this by explicitly indicating the point ofreference: r(0, ¢ ), 0(0, </J), </J),
but this would be cumbersome, and as long as you are alert to the problem I don't think it
will cause difficulties.4 In particular, do not nai"vely combihe the spherical components of
vectors associated with different points (in Fig. 1.37, A+ B = 0, not 2r, and A • B = -1,
not+ 1). Beware of differentiating a vector that is expressed in spherical coordinates, since
the unit vectors themselves are functions of position (Br/B0 = 0, for example). And do
J not taker, 0, and outside an integral, as we did with x, y, and zin Eq. 1.53. In general,
if you're uncertain about the validity of an operation, reexpress the problem in Cartesian coordinates, where this difficulty does not arise.

z

B

A

-1

y

X

Figure 1.37

41 claimed on the very first page ihat vectors have no location, and I'll stand by that. The vectors themselves live "out there," completely independent of o~r choice of coordinates. But the notation we use to represent them does depend on the point in question, in curvilinear coordinates.

40

CHAPTER 1. VECTOR ANALYSIS

An infinitesimal displacement in the r direction is simply dr (Fig. 1.38a), just as an
infinitesimal element of length in the x direction is dx:

dl, = dr.

( 1.65)

On the other hand, an infinitesimal element of length in the 0 direction (Fig. 1.38b) is not
just d0 (that's an angle-it doesn't even have the right units for a length), but rather r d0:

die =rd0.

(l.66)

Similarly, an infinitesimal element of length in the~ direction (Fig. 1.38c) is r sin 0 d</J:

di¢ = r sin 0 d</J.

(1.67)

Thus, the general infinitesimal displacement di is

J. di = dr i- + r d0 0+ r sin 0 d</J

(1.68)

x z This plays the role (in line integrals, for example) that di = dx + dy y+ dz played in
Cartesian coordinates.

dr

r

l a ~rde

~

rsin0

(a)

(b)

(c)

Figure 1.38

The infinitesimal volume element dr, in spherical coordinates, is the product of the three infinitesimal displacements:

dr = dl, die di¢ = r 2 sin 0 dr d0 d</J.

(1.69)

I cannot give you a general expression for surface elements da, since these depend on the orientation of the surface. You simply have to analyze the geometry for any given case (this goes for Cartesian and curvilinear coordinates alike). If you are integrating over the surface of a sphere, for instance, then r is constant, whereas 0 and¢ change (Fig. 1.39), so

da1 = die dl</> i- = r2 sin0d0d¢i-.

On the other hand, if the surface lies in the xy plane, say, so that 0 is constant (to wit: n /2)

while r and ¢ vary, then

da2 = dl, dl</> 0 = r dr d</J 0.

1.4. CURVILINEAR COORDINATES

41

z

Figure 1.39

Notice, finally, that r ranges from O to oo, </J from O to 2n, and 0 from O to n (not 2n-that would count every point twice).5

Example 1.13

Find the volume of a sphere of radius R.

Solution:

V

= r fd•=1R

f rr 2 r 2 sin0drd0d<P

r=O }0=0 jc/>=0

(Not a big surprise.)
So far we have talked only about the geometry of spherical coordinates. Now I would like to "translate" the vector derivatives (gradient, divergence, curl, and Laplacian) into r, 0, </J notation. In principle this is entirely straightforward: in the case of the gradient,
aTA aT. aT.
VT= -axx + -ayy + -azz' for instance, we would first use the chain rule to reexpress the partials:
5Alternatively, you could run rp from Oto rr (the "eastern hemisphere") and cover the "western hemisphere" by
extending 0 from rr up to 2rr. But this is very bad notation, since, among other things, sin 0 will then run negative, and you'll have to put absolute value signs around that term in volume and surface elements (area and volume being intrinsically positive quantities).

42

CHAPTER 1. VECTOR ANALYSIS

The terms in parenthe~es could~ worked out from Eq. 1.62--or rather, the inverse of those equations (Prob. 1.36). Theq we'd do the same for aT;ay and aT;az. Finally, we'd
z r, i substitute in the formulas for i, y, and in terms of 8, and (Prob. 1.37). It would take
an hour to figure out tqe gradient in SP,herical coordinates by this brute-force method. I suppose this is how it was first done, bu~ there is a much more efficient indirect approach, explained in Appendix A, which has the extra advantage of treating all coordinate systems at once. I described the "straightforw~rd" method only to show you that there is nothing
subtle or mysterious abqµt transforming to spherical coordinates: you're expressing the same quantity (gradient, divergence, or whatever) in different notation, that's all.
. Here, then, are the vector deriv&tives in spqerical coordinates: -

Gradient: Divergence:

ar A 1 ar

1 ar A

VT =;cc - r + - - 8 + - - - t f , .

ar r a0 r sin0 a4>

(1.70)

V

· v

=

1 a -r 2-a(r

r

2

vr)

1 a. +r-s-i-n (0sam 0 0ve)

+

1 av<I> - r si- n 0-a4>.

(l.71)

Curl:

Vxv

- .1- [ -a(sin0v¢) - -ave] rA + -1 [ -.1- -av-r - -a(rvc/>)] (A)

r sm 0 a0

a4>

r sm 0 a4> ar

+

-1 [ -a(rv0)- -avr] tA,t.

r ar

a0

(l.72)

Laplacian:

V2T =r12-a-ar

( r 2-ar) ar

+r 2-s1in-0

-a
a0

(

s•m0aT- ) ae

+r-2 s~i1n2-0-aa42->T2.

For reference, these formulas are listed inside the front cover.

(1.73)

Prstblem 1.36 Find formulas for r, 0, rp in terms of x, y, z (the inverse, in other words, of
Eq. l.~2).

•

Problem 1.37 Express the unit vectors r, 8, ~ in terms of x, y, z (that is, derive Eq. l.64).

,..,..?
Check your answers several ways (r •r ==

I,

l:A l •A t/> ? == 0,

A
r

x

0A=?=

A
t/>,

. . . ).

Also work out the

inverse formulas, giving x, y, zin terms of r, 8, $ (and 0, ¢).

•

Problem 1.38

r, (a) Check the divergence theorem for the function v1 = r 2 using as your volume the sphere
of radius R, centered at the origin.

b) Do the same for v2 = (l/r2)r. (If the answer surprises you, look back at Prob. l.l6.)

1.4. CURVILINEAR COORDINATES

43

z z

y R
X
Figure 1.40

2 y
X
Figure 1.41

Problem 1.39 Compute the divergence of the function
v = (rcos0) r + (r sin0)0 + (r sine cos¢)¢.
Check the divergence theorem for this function, using as your volume the inverted hemispherical bowl of radius R, resting on the xy plane and centered at the origin (Fig. 1.40).
Problem 1.40 Compute the gradient and Laplacian of the function T = r (cos 0 + sin 0 cos cp).
Check the Laplacian by converting T to Cartesian coordinates and using Eq. 1.42. Test the gradient theorem for this function, using the path shown in Fig. 1.41, from (0, 0, 0) to (0, 0, 2).

1.4.2 Cylindrical Coordinates
The cylindrical coordinates (s, </J, z) of a point P are defined in Fig. 1.42. Notice that ¢ has the same meaning as in spherical coordinates, and z is the same as Cartesian; s is the distance to P from the z axis, whereas the spherical coordinate r is the distance from the origin. The relation to Cartesian coordinates is

x = s cos¢, y = s sin¢, z = z.

(1.74)

The unit vectors (Prob. 1.41) are
s
J
z

l x cos¢ + sin¢ y,
x - sin¢ +cos¢ y,
z.

(l.75)

The infinitesimal displacements are

d[.,. =ds, di¢= sd</J, dl2 = dz,

(l.76)

44

CHAPTER 1. VECTOR ANALYSIS

z

X
Figure 1.42

so
dl = ds s+ s d</J ~ + dz z,

and the volume element is

dr=sdsd</Jdz.

The range of sis O-+ oo, ¢ goes from O-+ 2n, and z from -oo to oo. The vector derivatives in cylindrical coordinates are:

Gradient: Divergence: Curl:

ar A 1 aT A aT A

VT= -s+--<J,+-z.

as s a</)

az

V

• v

=

l a
--(svs)

+

-l av-</>

+

-avz.

s as

s a</) az

Vxv= ( - 1 av-z -a- vq,) sA + (a- vs - a-vz) <A J,+1- [ -a(svq,)a-v-s] zA .

s a</) az

az as

s as

a</)

Laplacian:

a v' 2T =s1-a-s

( aT) sa-s

+1-a2-r +a2-r
s2 a¢ 2 az2 •

These formulas are also listed inside the front cover.

(1.77) (l.78) (1.79) (l.80)
(1.82)

Problem 1.41 Express the cylindrical unit vectors s, ¢, zin terms of x, y, z(that is, derive Eq. 1.75). "Invert" your formulas to get x, y, zin terms of s, ¢, z(and¢).

1.5. THE DIRAC DELTA FUNCTION

45

z

2
X

2 y

Figure 1.43

Figure 1.44

Problem 1.42 (a) Find the divergence of the function
V = s(2 + sin2 <p)S + S sinrpcosrp 4, + 3z Z.
(b) Test the divergence theorem for this function, using the quarter-cylinder (radius 2, height 5) shown in Fig. 1.43. (c) Find the curl of v.

1.5 The Dirac Delta Function

1.5.1 The Divergence of r/ r2

Consider the vector function

V=

1 2r

r-.

(1.83)

At every location, vis directed radially outward (Fig. 1.44); if ever there was a function that ought to have a large positive divergence, this is it. And yet, when you actually calculate

the divergence (using Eq. 1.71), you get precisely zero:

(1.84)

(You will have encountered this paradox already, if you worked Prob. 1.16.) The plot thickens if you apply the divergence theorem to this function. Suppose we integrate over a sphere of radius R, centered at the origin (Prob. 1.38b); the surface integral is

/(;i)·(R fv-da =

2 sin0d0d¢r)

(fo (!arr sin 0 d0) 2rr d<j)) = 4n.

( 1.85)

46

CHAPTER 1. VECTOR ANALYSIS

But the volume integral, JV • v dr, is zero, if we are really to believe Eq. 1.84. Does this
mean that the divergence theorem is false? What's going on here?
The source of the problem is the point r = 0, where v blows up (and where, in Eq. 1.84,
we have unwittingly divided by zero). It is quite true that V • v = 0 everywhere except
the origin, but right at the origin the situation is more complicated. Notice that the surface integral (1.85) is independent of R; if the divergence theorem is right (and it is), we should
J get (V •v) dr = 4n for any sphere centered at the origin, no matter how small. Evidently
the entire contribution must be coming from the point r = O! Thus, V · v has the bizarre
property that it vanishes everywhere except at one point, and yet its integral (over any
volume containing that point) is 4n. No ordinary function behaves like that. (On the other
hand, a physical example does come to mind: the density (mass per unit volume) of a point
particle. It's zero except at the exact location of the particle, and yet its integral is finitenamely, the mass of the particle.) What we have stumbled on is a mathematical object known to physicists as the Dirac delta function. It arises in many branches of theoretical
physics. Moreover, the specific problem at hand (the divergence of the function r/ r 2 ) is not
just some arcane curiosity-it is, in fact, central to the whole theory of electrodynamics.
So it is worthwhile to pause here and study the Dirac delta function with some care.

1.5.2 The One-Dimensional Dirac Delta Function
The one dimensional Dirac delta function, 8(x ), can be pictured as an infinitely high, infinitesimally narrow "spike," with area 1 (Fig. 1.45). That is to say:

8(x) = { O, 00,

if X =/=- 0 }
if X = 0

( 1.86)

and

1_: 8(x) dx = I.

( 1.87)

Technically, 8(x) is not a function at all, since its value is not finite at x = 0. In the
mathematical literature it is known as a generalized function, or distribution. It is, if you

o(x)

X
Figure 1.45

1.5. THE DIRAC DELTA FUNCTION

47

-1/2-1/4 1/4 1/2 X (a)

-I -1/2 1/2 I x (b)

Figure 1.46

like, the limit of a sequence of functions, such as rectangles Rn (x), of height n and width
1/ n, or isosceles triangles Tn (x ), of height n and base 2/ n (Fig. 1.46). If f (x) is some "ordinary" function (that is, not another delta function-in fact, just
to be on the safe side let's say that f (x) is continuous), then the product f(x)o(x) is zero
everywhere except at x = 0. It follows that

f(x)o(x) = f(0)o(x).

( 1.88)

(This is the most important fact about the delta function, so make sure you understand why

it is true: since the product is zero anyway except at x = o,' we may as well replace f (x)

by the value it assumes at the origin.) In particular

L: L: = J(x)o(x) dx f(0)

8(x) dx = f(0).

(1.89)

Under an integral, then, the delta function "picks out" the value of f(x) at x = 0. (Here

and below, the integral need not run from -oo to +oo; it is sufficient that the domain extend

across the delta function, and -E to +E would do as well.)

f Of course, we can shift the spike from x = 0 to some other point, x = a (Fig. 1.47):

8(x-a)={ O,

i'ff x =fa a } wi•th 00 8(x - a) dx = I.

( 1.90)

00,

1x=a

-oo

Equation 1.88 becomes

f(x)o(x - a)= f(a)o(x - a),

(1.91)

and Eq. 1.89 generalizes to
L: f(x)o(x - a) dx = f(a).

(1.92)

Example 1.14
Evaluate the integral

{3
fo x 38(x-2)dx.

48

CHAPTER 1. VECTOR ANALYSIS

O(x-a)

a

X

Figure 1.47

Solution: The delta function picks out the value of x 3 at the point x = 2, so the integral is 23 = 8. Notice, however, that if the upper limit had been 1 (instead of 3) the answer would be
0, because the spike would then be outside the domain of integration.

Although 8 itself is not a legitimate function, integrals over 8 are perfectly acceptable.
In fact, it's best to think of the delta function as something that is always intended for use
under an integral sign. In particular, two expressions involving delta functions (say, D1 (x) and D2 (x)) are considered equal if 6

= 1_: f(x)Di(x)dx 1_: f(x)D2(x)dx,

( 1.93)

for all ("ordinary") functions f(x).

Example 1.15

Show that

8(kx)

=

1 -8(x),

lkl

where k is any (nonzero) constant. (In particular, /i(-x) = 8(x).)

(1.94)

Solution:

For an arbitrary test function f(x), consider the integral
1-: f(x)8(kx)dx.

Changing variables, we let y = kx, so that x = y / k, and dx = I/ k dy. If k is positive, the
integration still runs from -oo to +oo, but if k is negative, then x = oo implies y = -oo, and
6This is not as arbitrary as it may sound. The crucial point is that the integrals must be equal for any f(x).
Suppose D1 (x) and D2(x) actually differed, say, in the neighborhood of the point x = 17. Then we could pick a
= function f (x) that was sharply peaked about x 17, and the integrals would not be equal.

1.5. THE DIRAC DELTA FUNCTION

49

vice versa, so the order of the limits is reversed. Restoring the "proper" order costs a minus

1 1 sign. Thus 00 J(x)S(kx)dx = ± 00 f(y/k)S(y)d-ky = ±-l f(0) = -I f(0).

-oo

-oo

k

lkl

(The lower signs apply when k is negative, and we account for this neatly by putting absolute

value bars around the final k, as indicated.) Under the integral sign, then, 8(kx) serves the

same purpose as (1/lkl).S(x):

1 1 00 f (x)8(kx) dx = 00 f(x) [-1k ll(x)] dx.

-oo

-oo

I I

According to criterion 1.93, therefore, S(kx) and (1/lkl)o(x) are equal.

Problem 1.43 Evaluate the following integrals:

Jg (b) cosx S(x - :rc)dx.
(c) f~ x 38(x + l)dx.
(d) f~ ln(x + 3) /l(x + 2) dx.
Problem 1.44 Evaluate the following integrals:

(b) JJ(x 3 + 3x + 2) /l(l - x) dx.

(d) Jl1_00 8(x - b) dx.

Problem 1.45

(a) Show that [Hint: Use integration by parts.]

d x-(8(x))

=

-/l(x).

dx

I (b) Let 0(x) be the step function: I, 0(x) =
0,

ifx>O ]· ifx :::0

Show thatd0/dx = /l(x).

(1.95)

50

CHAPTER 1. VECTOR ANALYSIS

1.5.3 The Three-Dimensional Delta Function

It is an easy matter tp generalize the delta function to three dimensions:

83 (r) = 8(x) 8(y) 8(z).

(l .96)

(As always, r = xx+ y y+z zis the position vector, extending from the origin to the point
(x, y, z)). This three-dimensional delta function is zero everywhere except at (0, 0, 0),
where it blows up. Its volume integral is 1:

1-oo 1-oo 1-oo [

= f f o3(r) dr rX! 00 00 8(x) 8(y) 8(z) dx dy dz= 1.

Jllll space

(1.97)

And, generalizing Eq. 1.92,

1

f (r)83(r - a) dr = f (a).

all space

(l.98)

As in the one-dimensional case, integration with 8 picks out the value of the function f at the location of the spike.
We are now in a position to resolve the paradox introduced in Sect. 1.5.1. As you will
recall, we found that the divergence of r/r 2 is zero everywhere except at the origin, and
yet its integral over any volume containing the origin is a constant (to wit: 4n ). These are precisely the defining conditions for the Dirac delta function; evidently

(1.99)

More generally,

(l.100)

where, as always, -'l- is the separation vector: -'l- = r - r'. Note that differentiation here is
with respect tor, while r' is held constant. Incidentally, since

(l.101)

(Prob. 1.13), it follows that

(l.102)

1.5. THE DIRAC DELTA FUNCTION

51

Example 1.16

Evaluate the integral

fv (:Z) J = (r2 + 2) V.

dr,

where Vis a sphere of radius R centered at the origin.

Solution 1: Use Eq. 1.99 to rewrite the divergence, and Eq. 1.98 to do the integral:

This one-line solution demonstrates something of the power and beauty of the delta function, but I would like to show you a second method, which is much more cumbersome but serves to illustrate the method of integration by parts, Sect. 1.3 .6.
r/ Solution 2: Using Eq. 1.59, we transfer the derivative from r 2 to (r 2 + 2):

l=-f :2 -[V(r2 +2)Jdr+f(r2 +2)~ • da.

The gradient is

V(r2 + 2) = 2ri-,

so the volume integral becomes

f ~ f dr = ~r2 sin0drd0d<j) = Sn foR rdr = 4nR2 .

Meanwhile, on the boundary of the sphere (where r = R), da = R2 sin 0 d0 d¢ i-,

so the surface integral becomes
f(R2 + 2) sin 0 d0 d<j) = 4n(R2 + 2).

Putting it all together, then,
J = -4n R2 + 4n(R2 + 2) = Sn,
as before.

52

CHAPTER 1. VECTOR ANALYSIS

Problem 1.46 (a) Write an expression for the electric charge density p(r) of a point charge q at r'. Make sure that the volume integral of p equals q. (b) What is the charge density of an electric dipole, consisting of a point charge -q at the origin and a point charge +q at a? (c) What is the charge density of a uniform, infinitesimally thin spherical shell of radius Rand total charge Q, centered at the origin? [Beware: the integral over all space must equal Q.]
Problem 1.47 Evaluate the following integrals:
J (a) all space (r2 + r ·a+ a 2 )1i 3 (r - a) dr, where a is a fixed vector and a is its magnitude.
z. (b) fv lr-bi 283 (5r) dr, where Vis a cube of side 2, centered on the origin, and b = 4 y+ 3
(c) fv(r 4 + r 2(r · c) + c4)83(r - c) dr, where Vis a sphere of radius 6 about the origin,
c = 5 x+ 3 y+ 2 z, and c is its magnitude.
(d) fv r • (d - r)/i 3 (e - r) dr, where d = (I. 2, 3), e = (3, 2, 1), and Vis a sphere of radius
1.5 centered at (2, 2, 2).
Problem 1.48 Evaluate the integral

(where Vis a sphere of radius R, centered at the origin) by two different methods, as in Ex. 1.16.

1.6 The Theory of Vector Fields
1.6.1 The Helmholtz Theorem
Ever since Faraday, the laws of electricity and magnetism have been expressed in terms of electric and magnetic fields, E and B. Like many physical laws, these are most compactly expressed as differential equations. Since E and B are vectors, the differential equations naturally involve vector derivatives: divergence and curl. Indeed, Maxwell reduced the entire theory to four equations, specifying respectively the divergence and the curl of E and B.7
7Strictly speaking, this is only true in the static case; in general, the divergence and curl are given in terms of time derivatives of the fields themselves.

1.6. THE THEORY OF VECTOR FIELDS

53

Maxwell's formulation raises an important mathematical question: To what extent is a vector function determined by its divergence and curl? In other words, if I tell you that the divergence ofF (which stands for E or B, as the case may be) is a specified (scalar) function D,
V-F=D,
and the curl of Fis a specified (vector) function C,
V X F =C,
(for consistency, C must be divergenceless,
V-C=O,
because the divergence of a curl is always zero), can you then determine the function F? Well. .. not quite. For example, as you may have discovered in Prob. 1.19, there are
many functions whose divergence and curl are both zero everywhere-the trivial case F = 0,
of course, but also F = y z x+ zx y+ xy z, F = sin x cosh y x- cos x sinh y y, etc. To solve
a differential equation you must also be supplied with appropriate boundary conditions. In electrodynamics we typically require that the fields go to zero "at infinity" (far away from all charges).8 With that extra information the Helmholtz theorem guarantees that the field is uniquely determined by its divergence and curl. (A proof of the Helmholtz theorem is given in Appendix B.)

1.6.2 Potentials

If the curl of a vector field (F) vanishes (everywhere), then F can be written as the gradient

of a scalar potential (V):

V x F = 0 {=:=} F = -VV.

(l.103)

(The minus sign is purely conventional.) That's the essential burden of the following theorem:

Theorem 1:

Curl-less (or "irrotational") fields. The following conditions are equivalent (that is, F satisfies one if and only
if it satisfies all the others):
J: (a) V x F = 0 everywhere.
(b) F • di is independent of path, for any given end
points.
(c) j F •di = 0 for any closed loop. (d) F is the gradient of some scalar, F = -V V.

8In some textbook problems the charge itself extends to infinity (we speak, for instance, of the electric field of an infinite plane, or the magnetic field of an infinite wire). In such cases the normal boundary conditions do not apply, and one must invoke symmetry arguments to determine the fields uniquely.

54

CHAPTER I. VECTOR ANALYSIS

The scalar potential is not unique-any constant can be added to V with impunity, since this will not affect its gradient.
If the divergence of a vector field (F) vanishes (everywhere), then F can be expressed as the curl of a vector potential (A):

V · F = 0 ~ F = V x A.

(1.104)

That's the main conclusion of the following theorem:

Theorem 2:

Divergence-less (or "solenoidal") fields. The following
conditions are equivalent:
(a) V • F = 0 everywhere.
J (b) F-da is independent of surface, for any given bound-
ary line.
(c) j F •da = 0 for any closed surface.
(d) Fis the curl of some vector, F = V x A.

The vector potential is not unique-the gradient of any scalar function can be added to A without affecting the curl, since the curl of a gradient is zero.
You should by now be able to prove all the connections in these theorems, save for the ones that say (a), (b), or (c) implies (d). Those are more subtle, and will come later.
Incidentally, in all cases (whatever its curl and divergence may be) a vector field F can be
written as the gradient of a scalar plus the curl of a vector:

F=-VV+VxA (always).

(I.I 05)

Problem 1.49
(a) Let F1 = x 2 zand F2 = xx+ y y+ z z. Calculate the divergence and curl of F1 and F2.
Which one can be written as the gradient of a scalar? Find a scalar potential that does the job. Which one can be written as the curl of a vector? Find a suitable vector potential.
(b) Show that F3 = yz x+ zx y+ xy zcan be written both as the gradient of a scalar and as
the curl of a vector. Find scalar and vector potentials for this function.
Problem 1.50 For Theorem I show that (d) =} (a), (a) =} (c), (c) =} (b), (b) =} (c), and
(c) =} (a).
Problem 1.51 For Theorem 2 show that (d) =} (a), (a) =} (c), (c) =} (b), (b) =} (c), and (c) =} (a).

1.6. THE THEORY OF VECTOR FIELDS

55

Problem 1.52 (a) Which of the vectors in Problem 1. 15 can be expressed as the gradient of a scalar? Find a scalar function that does the job.
(b) Which can be expressed as the curl of a vector? Find such a vector.

More Problems on Chapter 1
Problem 1.53 Check the divergence theorem for the function
r 0 - = V r 2 COS 0 + r 2 COS <p r 2 COS 0 sin <p ~,
using as your volume one octant of the sphere of radius R (Fig. I.48). Make sure you include the entire surface. [Answer: T{ R4/4]
Problem 1.54Check Stokes' theorem using the function v = ay x+bx y(a andb are constants)
and the circular path of radius R, centered at the origin in the xy plane. [Answer: T{ R2(b-a)]
Problem 1.55 Compute the line integral of

along the triangular path shown in Fig. 1.49. Check your answer using Stokes' theorem. [Answer: 8/3]

Problem 1.56 Compute the line integral of
V = (r cos2 0) r - 0 (r COS 0 sin 0) + 3r ~
around the path shown in Fig. 1.50 (the points are labeled by their Cartesian coordinates). Do it either in cylindrical or in spherical coordinates. Check your answer, using Stokes' theorem. [Answer: 3T{/2]

z R
R y
X
Figure 1.48

z 2
y
X
Figure 1.49

z

(0, 1,2)

(0,1,0) y
Figure 1.50

56
z

CHAPTER 1. VECTOR ANALYSIS

X (a,Q,Q)
Figure 1.51

X

y

Figure 1.52

z, Problem 1.57 Check Stokes' theorem for the function v = y using the triangular surface
shown in Fig. 1.51. [Answer: a2]
Problem 1.58 Check the divergence theorem for the function

using the volume of the "ice-cream cone" shown in Fig. 1.52 (the top surface is spherical, with
radius Rand centered at the origin). [Answer: (:,r R4 /12)(2:,r + 3v'3)]

Problem 1.59 Here are two cute checks of the fundamental theorems:
= (a) Combine Corollary 2 to the gradient theorem with Stokes' theorem (v VT, in this case).
Show that the result is consistent with what you already knew about second derivatives.
(b) Combine Corollary 2 to Stokes' theorem with the divergence theorem. Show that the result is consistent with what you already knew.

•

Problem 1.60 Although the gradient, divergence, and curl theorems are the fundamental in-

tegral theorems of vector calculus, it is possible to derive a number of corollaries from them.

Show that:

(a)fv(VT) dr = :fs T da. [Hint: Let v = cT, where c is a constant, in the divergence theorem; use the product rules.]
(b) fv(V xv) dr = - §5 v x da. [Hint: Replace v by (v x c) in the divergence theorem.]
= = + (c) fv[T'v 2U (VT) - (VU)] dr :fs(TVU) • da. [Hint: Let v TVU in the divergence
theorem.]

(d) fv(TV 2U -UV2T) dr = :fs(TVU -UVT) · da. [~omment: this is known as Green's
theorem; it follows from (c), which is sometimes called Green's identity.]
(e) fs VT x da = - Pp T di. [Hint: Let v = cT in Stokes' theorem.]

1.6. THE THEORY OF VECTOR FIELDS

57

•

Problem 1.61 The integral

(l.106)

is sometimes called the vector area of the surface S. If S happens to be flat, then lal is the
ordinary (scalar) area, obviously.

(a) Find the vector area of a hemispherical bowl of radius R.

(b) Show that a= 0 for any closed surface. [Hirit: Use Prob. l.60a.]

(c) Show that a is the same for all surfaces sharing the same boundary.

(d) Show that

a=½ fr x di,

(1.107)

where the integral is around the boundary line. [Hint: One way to do it is to draw the cone subtended by the loop at the origin. Divide the conical surface up into infinitesimal triangular wedges, each with vertex at the origin and opposite side di, and exploit the geometrical interpretation of the cross product (Fig. 1.8).]

(e) Show that

fcc•r)dl=axc,

(1.108)

for any constant vector c. [Hint: let T = c • r in Prob. l .60e.]

•

Problem 1.62

(a) Find the divergence of the function

r
V= -.
r

First compute it directly, as in Eq. 1.84. Test your result using the divergence theorem, as in
Eq. 1.85. Is there a delta function at the origin, as there was for r/ r 2? What is the general = formula for the divergence of rnr? [Answer: V • (rnr) (n + 2)rn-I, unless n = -2, in

which case it is 4rr83 (r)]

(b) Find the curl of rnr. Test your conclusion using Prob. 1.60b. [Answer: V x (rnr) = 0]

Chapter 2
Electrostatics
2.1 The Electric Field
2.1.1 Introduction
The fundamental problem electromagnetic theory hopes to solve is this (Fig. 2.1 ): We have some electric charges, qi, q2, q3, ... (call them source charges); what force do they exert on another charge, Q (call it the test charge)? The positions of the source charges are given (as functions of time); the trajectory of the test particle is to be calculated. In general, both the source charges and the test charge are in motion.
The solution to this problem is facilitated by the principle of superposition, which states that the interaction between any two charges is completely unaffected by the presence of others. This means that to determine the force on Q, we can first compute the force F1, due to q1 alone (ignoring all the others); then we compute the force F2, due to q2 alone; and so
on. Finally, we take the vector sum of all these individual forces: F = F1 + F2 + F3 + ...
Thus, if we can find the force on Q due to a single source charge q, we are, in principle, done (the rest is just a question of repeating the same operation over and over, and adding it all up). 1
Well, at first sight this sounds very easy: Why don't I just write down the formula for the force on Q due to q, and be done with it? I could, and in Chapter IO I shall, but you would be shocked to see it at this stage, for not only does the force on Q depend on the separation distance,z, between the charges (Fig. 2.2), it also depends on both their velocities and on the acceleration of q. Moreover, it is not the position, velocity, and acceleration of q right now that matter: Electromagnetic "news" travels at the speed of light, so what concerns Q is the position, velocity, and acceleration q had at some earlier time, when the message left.
1The principle of superposition may seem "obvious" to you, but it did not have to be so simple: if the electromagnetic force were proportional to the square of the total source charge, for instance, the principle of superposition
would not hold, since (q1 + q2) 2 i= qf + qi (there would be "cross terms" to consider). Superposition is not a
logical necessity, but an experimental fact.
58

2.1. THE ELECTRIC FIELD

59

•

•Q

• •

"Source" charges

"Test" charge

Figure 2.1

Figure 2.2

Therefore, in spite of the fact that the basic question ("What is the force on Q due to q?") is easy to state, it does not pay to confront it head on; rather, we shall go at it by stages. In the meantime, the theory we develop will permit the solution of more subtle electromagnetic problems that do not present themselves in quite this simple format. To begin with, we shall consider the special case of electrostatics in which all the source charges are stationary (though the test charge may be moving).

2.1.2 Coulomb's Law
What is the force on a test charge Q due to a single point charge q which is at rest a distance !z, away? The answer (based on experiments) is given by Coulomb's law:

F= _I_qQli_
4rrEo 1),2

(2.1)

The constant Eo is called the permitivity of free space. In SI units, where force is in Newtons (N), distance in meters (m), and charge in coulombs (C),

= -12 c2
EQ 8.85 X 10 - N-m-2 .
In words, the force is proportional to the product of the charges and inversely proportional to the square of the separation distance. As always (Sect. 1.1.4), "- is the separation vector from r' (the location of q) tor (the location of Q):

4=r-r';

(2.2)

I), is its magnitude, and li is its direction. The force points along the line from q to Q; it is repulsive if q and Q have the same sign, and attractive if their signs are opposite.
Coulomb's law and the principle of superposition constitute the physical input for electrostatics-the rest, except for some special properties of matter, is mathematical elaboration of these fundamental rules.

60

CHAPTER 2. ELECTROSTATICS

Problem2.1 (a) Twelve equal charges, q, are situated at the comers of a regular 12-sided polygon (for instance, one on each numeral of a clock face). What is the net force on a test charge Q at the center? (b) Suppose one of the 12 q's is removed (the one at "6 o'clock"). What is the force on Q? Explain your reasoning carefully. (c) Now 13 equal charges, q, are placed at the comers of a regular 13-sided polygon. What is the force on a test charge Q at the center? (d) If one of the 13 q's is removed, what is the force on Q? Explain your reasoning.
2.1.3 The Electric Field
If we have several point charges q1, qz, ... , qn, at distances 1J,1, 1),2, ... , !z,n from Q, the total force on Q is evidently

or (2.3)

where

= - - E(r)

1 4rrEo

~ 2 ~

qi A 4i.

.
l=

1

I),. l

(2.4)

Eis called the electric field of the source charges. Notice that it is a function of position (r), because the separation vectors 4i depend on the location of the field point P (Fig. 2.3). But it makes no reference to the test charge Q. The electric field is a vector quantity that varies

X
z
Figure 2.3

2.1. THE ELECTRIC FIELD

61

from point to point and is determined by the configuration of source charges; physically, E(r) is the force per unit charge that would be exerted on a test charge, if you were to place one at P.
What exactly is an electric field? I have deliberately begun with what you might call the "minimal" interpretation of E, as an intermediate step in the calculation of electric forces. But I encourage you to think of the field as a "real" physical entity, filling the space in the neighborhood of any electric charge. Maxwell himself came to believe that electric and magnetic fields represented actual stresses and strains in an invisible primordial jellylike "ether." Special relativity has forced us to abandon the notion of ether, and with it Maxwell's mechanical interpretation of electromagnetic fields. (It is even possible, though cumbersome, to formulate classical electrodynamics as an "action-at-a-distance" theory, and dispense with the field concept altogether.) I can't tell you, then, what a field is-only how to calculate it and what it can do for you once you've got it.

Problem2.2
(a) Find the electric field (magnitude and direction) a distance z above the midpoint between two equal charges, q, a distanced apart (Fig. 2.4). Check that your result is consistent with
» what you'd expect when z d.
(b) Repeat part (a), only this time make the right-hand charge -q instead of +q.

p
z
q d/2 d/2 q Figure 2.4

(a) Continuous distiibution "- •P

~ di'
(b) Line charge, 11.
dr:'

(c) Surface charge, a

(d) Volume charge, p

Figure 2.5

2.1.4 Continuous Charge Distributions
Our definition of the electric field (Eq. 2.4), assumes that the source of the field is a collection of discrete point charges qi. If, instead, the charge is distributed continuously over some region, the sum becomes an integral (Fig. 2.5a):

E(r) = -1- / 2I 4, dq. 47l'EO 1-

(2.5)

62

CHAPTER 2. ELECTROSTATICS

If the charge is spread out along a line (Fig. 2.5b), with charge-per-unit-length A, then
dq = Adi' (where di' is an element of length along the line); if the charge is smeared
out over a surface (Fig. 2.5c), with charge-per-unit-area r,, then dq = r, da' (where da'
is an element of area on the surface); and if the charge fills a volume (Fig. 2.5d), with
charge-per-unit-volume p, then dq = p dr:' (where dr:' is an element of volume):
~ dq ~ Adi'~ r, da' p dr:'.

Thus the electric field of a line charge is

E(r) = -1- /
4m,0

-A(-r2-')lidl'; !z,

(2.6)

p

for a surface charge, and for a volume charge,

E(r) =

1 / --

-rr(-r2-')1Az.da,;

4rrEo

!z,

s

E(r) = -1- / -p(-r2-')1zA.dr:1•

47TEO

!z,

V

(2.7) (2.8)

Equation 2.8 itself is often referred to as "Coulomb's law," because it is such a short step from the original (2.1), and because a volume charge is in a sense the most general and realistic case. Please note carefully the meaning of"- in these formulas. Originally, in Eq. 2.4, lz.; stood for the vector from the source charge q; to the field point r. Correspondingly, in Eqs. 2.5-2.8, "- is the vector from dq (therefore from di', da', or dr:') to the field point
r.2
Example 2.1

Find the electric field a distance z above the midpoint of a straight line segment of length 2L, which carries a uniform line charge A (Fig. 2.6).
Solution: It is advantageous to chop the line up into symmetrically placed pairs (at ±x), for then the horizontal components of the two fields cancel, and the net field of the pair is

dE =2-1- (A--d2-x) cos0 zA .
4nEo 1J,

2 Warning: The unit vector4 is not constant; its direction depends on the source point r', and hence it cannot be
taken outside the integrals 2.5-2.8. In practice, you must work with Cartesian components (x, y, z are constant,
and do come out), even if you use curvilinear coordinates to perform the integration.

2.1. THE ELECTRIC FIELD

63

z

dx

-L

~+L x

Figure 2.6

Here cos0 = z/JJ.,,JJ., = Jz 2 + x 2, and x runs from Oto L:

= E

-I- loL 2).z dx

4nEo o (z2 + x2)3/2

[ ]I = 2).z

L
X

4rrEo z2Jz2+x2 0

= 1

2>.L

4rrEo zJz2 + L2'

and it aims in the z-direction.
» For points far from the line (z L), this result simplifies:

I 2>.L E~----
- 4rrEo z2 '
which makes sense: From far away the line "looks" like a point charge q = 2>.L, so the field
reduces to that of point charge q/(4rrEoz2). In the limit L ➔ oo, on the other hand, we obtain the field of an infinite straight wire:

I 2>. E=---·
4rrEo z '

or, more generally,

1 2). E=---
4rrEo s '

(2.9)

where s is the distance from the wire.

Problem 2.3 Find the electric field a distance z above one end of a straight line segment of length L (Fig. 2. 7), which carries a uniform line charge>.. Check that your formula is consistent
» with what you would expect for the case z L.

64
TP
I I 1Z I I
L
Figure 2.7

CHAPTER 2. ELECTROSTATICS

TP
I I 1Z I
/ •I 7
I-a-I
Figure 2.8

TP
I I IZ I
G
Figure 2.9

Problem 2A Find the electric field a distance z above the center of a square loop (side a)
carrying uniform line charge)._ (Fig. 2.8). [Hint: Use the result of Ex. 2.1.]
Problem 2.5 Find the electric field a distance z above the center of a circular loop of radius r (Fig. 2.9), which carries a uniform line charge A.
Problem 2.6 Find the electric field a distance z above the center of a flat circular disk of radius R (Fig. 2.10), which carries a uniform surface charge a. What does your formula give in the
» limit R ➔ oo? Also check the case z R.
Problem 2.7 Find the electric field a distance z from the center of a spherical surface of radius R (Fig. 2.11 ), which carries a uniform charge density a. Treat the case z < R (inside) as well
as z > R (outside). Express your answers in terms of the total charge q on the sphere. [Hint: Use the law of cosines to write JJ, in terms of R and 0. Be sure to take the positive square root:
JR2 + z2 - 2Rz = (R - z) if R > z, but it's (z - R) if R < z.]
Problem 2.8 Use your result in Prob. 2.7 to find the field inside and outside a sphere of radius R, which carries a uniform volume charge density p. Express your answers in terms of the
total charge of the sphere, q. Draw a graph of IEI as a function of the distance from the center.

TP
I I IZ I
~
Figure2.1O

Figure 2.11

2.2. DIVERGENCE AND CURL OF ELECTROSTATIC FIELDS

65

2.2 Divergence and Curl of Electrostatic Fields

2.2.1 Field Lines, Flux, and Gauss's Law

In principle, we are done with the subject of electrostatics. Equation 2.8 tells us how to

compute the field of a charge distribution, and Eq. 2.3 tells us what the force on a charge Q

placed in this field will be. Unfortunately, as you may have discovered in working Prob. 2.7,

the integrals involved in computing E can be formidable, even for reasonably simple charge

distributions. Much of the rest of electrostatics is devoted to assembling a bag of tools and

tricks for avoiding these integrals. It all begins with the divergence and curl of E. I shall

calculate the divergence of E directly from Eq. 2.8, in Sect. 2.2.2, but first I want to show

you a more qualitative, and perhaps more illuminating, intuitive approach.

Let's begin with the simplest possible case: a single point charge q, situated at the

origin:

E (r)

=

1 qA
---r.

4nEo r2

(2.10)

To get a "feel" for this field, I might sketch a few representative vectors, as in Fig. 2.12a. Because the field falls off like 1/ r2, the vectors get shorter as you go farther away from the origin; they always point radially outward. But there is a nicer way to represent this field, and that's to connect up the arrows, to form field lines (Fig. 2.12b). You might think that I have thereby thrown away information about the strength of the field, which was contained in the length of the arrows. But actually I have not. The magnitude of the field is indicated by the density of the field lines: it's strong near the center where the field lines are close together, and weak farther out, where they are relatively far apart.
In truth, the field-line diagram is deceptive, when I draw it on a two-dimensional surface, for the density of lines passing through a circle of radius r is the total number divided by the circumference (n/2nr), which goes like (1/r), not (l/r 2 ). But if you imagine the model in three dimensions (a pincushion with needles sticking out in all directions), then the density of lines is the total number divided by the area of the sphere (n /4n r 2 ), which does go like (l/r2).

♦

' t ,,

·-7'~11/~/E--

E

" t 1'.

♦ (a)

(b)

Figure 2.12

66

CHAPTER 2. ELECTROSTATICS

Equal but opposite charges
Figure 2.13
Such diagrams are also convenient for representing more complicated fields. Of course, the number of lines you draw depends on how energetic you are (and how sharp your pencil
is), though you ought to inciude enough to get an accurate sense <?f the field, and you must
be consistent: If charge q gets 8 lines, then 2q deserves 16. And you must space them fairly-they emanate from a point charge symmetrically in all directions. Field lines begin on positive charges and end on negative ones; they cannot simply terminate in midair, though they may extend out to infinity. Moreover, field lines can never cross-at the intersection, the field would have two different ditections at once! With all this in mind, it is easy to sketch the field of any simple configuration of point charges: Begin by drawing the lines in the neighborhood of each charge, and then connect them up or extend them to infinity (Figs. 2.13 and 2.14).

Equal charges Figure 2.14

2.2. DIVERGENCE AND CURL OF ELECTROSTATIC FIELDS

67

/
/ Figure 2.15

In this model the flux of E through a surface S,
= <t> E LE ·da,

(2.11)

is a measure of the "number of field lines" passing through S. I put this in quotes because of course we can only draw a representative sample of the field lines-the total number would be infinite. But for a given sampling rate the flux is proportional to the number of lines drawn, because the field strength, remember, is proportional to the density of field lines (the number per unit area), and hence E •da is proportional to the number of lines passing through the infinitesimal area da. (The dot product picks out the component of da along the direction ofE, as indicated in Fig. 2.15. It is only the area in the plane perpendicular to E that we have in mind when we say that the density of field lines is the number per unit area.)
This suggests that the flux through any closed surface is a measure of the total charge inside. For the field lines that originate on a positive charge must either pass out through the surface or else terminate on a negative charge inside (Fig. 2.16a). On the other hand, a charge outside the surface will contribute nothing to the total flux, since its field lines pass in one side and out the other (Fig. 2.16b). This is the essence of Gauss's law. Now let's make it quantitative.

q

(a)

(b)

Figure 2.16

68

CHAPTER 2. ELECTROSTATICS

In the case of a point charge q at the origin, the flux of E through a sphere of radius r is

jJ.E. da =

/ -14nEo

( rq2 r)

• (r2 sin0d0d¢r) =

_!_q. Eo

(2.12)

Notice that the radius of the sphere cancels out, for while the surface area goes up as r 2 , the field goes down as 1/ r 2 , and so the product is constant. In terms of the field-line picture, this makes good sense, since the same number of field lines passes through any sphere centered at the origin, regardless of its size. In fact, it didn't have to be a sphere-any closed surface, whatever its shape, would trap the same number of field lines. Evidently the .fiux through any surface enclosing the charge is q /Eo.
Now suppose that instead of a single charge at the origin, we have a bunch of charges scattered about. According to the principle of superposition, the total field is the (vector) sum of all the individual fields:
n

The flux through a surface that encloses them all, then, is

For any closed surface, then,

(2.13)

where Qenc is the total charge enclosed within the surface. This is the quantitative statement of Gauss's law. Although it contains no information that was not already present in Coulomb's law and the principle of superposition, it is of almost magical power, as you will see in Sect. 2.2.3. Notice that it all hinges on the 1/ r 2 character of Coulomb's law; without that the crucial cancellation of the r's in Eq. 2.12 would not take place, and the total flux of E would depend on the surface chosen, not merely on the total charge enclosed. Other 1/ r 2 forces (I am thinking particularly of Newton's law of universal gravitation) will obey "Gauss's laws" of their own, and the applications we develop here carry over directly.
As it stands, Gauss's law is an integral equation, but we can readily turn it into a differential one, by applying the divergence theorem:
f f E •da = (V •E)d't'.

S

V

Rewriting Qenc in terms of the charge density p, we have
f Qenc = pdr.
V

2.2. DIVERGENCE AND CURL OF ELECTROSTATIC FIELDS

69

So Gauss's law becomes

f /(V-E)dr= (~)dr.

V

V

And since this holds for any volume, the integrands must be equal:

(2.14)

Equation 2.14 carries the same message as Eq. 2.13; it is Gauss's law in differential form. The differential version is tidier, but the integral form has the advantage in that it accommodates point, line, and surface charges more naturally.

Problem 2.9 Suppose the electric field in some region is found to be E = kr 3i:, in spherical
coordinates (k is some constant). (a) Find the charge density p. (b) Find the total charge contained in a sphere of radius R, centered at the origin. (Do it two different ways.)
Problem 2.10 A charge q sits at the back comer of a cube, as shown in Fig. 2.17. What is the flux of E through the shaded side?

q
Figure 2.17

2.2.2 The Divergence of E

Let's go back, now, and calculate the divergence of E directly from Eq. 2.8:

f E(r) = _1_ 4m,0

4
2

p(rI)

dr

I

.

I/,

all space

(2.15)

(Originally the integration was over the volume occupied by the charge, but I may as
well extend it to all space, since p = 0 in the exterior region anyway.) Noting that the

70

CHAPTER 2. ELECTROSTATICS

r-dependence is contained in -t = r - r', we have

V · E = -1- / V · (4- ) p(r/) dr /.

4JTEQ

1/,.2

This is precisely the divergence we calculated in Eq. 1.100:

(!) V •

= 4rr8\-t).

Thus

v. E = - 1- / 4rr8\r - r')p(r') dr' = _!_p(r),

4JTEo

Eo

(2.16)

which is Gauss's law in differential form (2.14). To recover the integral form (2.13), we

run the previous argument in reverse-integrate over a volume and apply the divergence

theorem:

J. f = = ~ = / V • E dr

E · da
j

EQ

p dr

_!__ Qenc•
EQ

V

S

V

2.2.3 Applications of Gauss's Law
I must interrupt the theoretical development at this point to show you the extraordinary power of Gauss's law, in integral form. When symmetry permits, it affords by far the quickest and easiest way of computing electric fields. I'll illustrate the method with a series of examples.

Example 2.2
Find the field outside a uniformly charged solid sphere of radius R and total charge q.
Solution: Draw a spherical surface at radius r > R (Fig. 2.18); this is called a "Gaussian surface" in the trade. Gauss's law says that for this surface (as for any other)

sJ.j

E · da =

_!___ Qenc,
EQ

and Qenc = q. At first glance this doesn't seem to get us very far, because the quantity we

want (E) is buried inside the surface integral. Luckily, symmetry allows us to extract E from

under the integral sign: E certainly points radially outward,3 as does da, so we can drop the

dot product,

f f E •da = IEI da,

s

s

3Ifyou doubt that Eis radial, consider the alternative. Suppose, say, that it points due east, at the '·equator." But the orientation of the equator is perfectly arbitrary-nothing is spinning here, so there is no natural "north-south" axis-any argument purporting to show that E points east could just as well be used to show it points west, or north, or any other direction. The only unique direction on a sphere is radial.

2.2. DIVERGENCE AND CURL OF ELECTROSTATIC FIELDS

71

Gaussian ...-surface
Figure 2.18

and the magnitude of E is constant over the Gaussian surface, so it comes outside the integral:

j j !Elda= IEI da = !El 4nr 2.

s

s

Thus or

IEl4nr 2 = -1q,
EQ
1 qA
E=---r. 4rrEo r 2

Notice a remarkable feature of this result: The field outside the sphere is exactly the same as it would have been if all the charge had been concentrated at the center.

Gauss's law is always true, but it is not always useful. If p had not been uniform (or, at any rate, not spherically symmetrical), or if I had chosen some other shape for my Gaussian
surface, it would still have been true that the flux of Eis (1/Eo)q, but I would not have been certain that E was in the saine direction as da and constant in magnitude over the
surface, and without that I could not pull IEI out of the integral. Symmetry is crucial to this
application of Gauss's law. As far as I know, there are only three kinds of symmetry that work:

l. Spherical symmetry. Make your Gaussian surface a concentric sphere. 2. Cylindrical symmetry. Make your Gaussian surface a coaxial cylinder
(Fig. 2.19). 3. Plane symmetry. Use a Gaussian "pillbox," which straddles the surface
(Fig. 2.20).
Although (2) and (3) technically require infinitely long cylinders, and planes extending to infinity in all directions, we shall often use them to get approximate answers for "long" cylinders or "large" plane surfaces, at points far from the edges.

72

CHAPTER 2. ELECTROSTATICS

\ Gaussian surface Figure 2.19

Figure 2.20

Example2.3

A long cylinder (Fig. 2.21) carries a charge density that is proportional to the distance from
= the axis: p ks, for some constant k. Find the electric field inside this cylinder.

Solution: Draw a Gaussian cylinder of length l and radius s. For this surface, Gauss's law states:

s = jJ. E · da

_!_Qenc-
EQ

The enclosed charge is

J J Qenc = p dr = (ks') (s' ds' dq, dz) = 2n kl fos s'2 ds' = in kls 3.

(I used the volume element appropriate to cylindrical coordinates, Eq. 1.78, and integrated¢ from Oto 211', dz from Oto l. I put a prime on the integration variable s1, to distinguish it from the radius s of the Gaussian surface.)

Figure 2.21

2.2. DIVERGENCE AND CURL OF ELECTROSTATIC FIELDS

73

Now, symmetry dictates that E must point radially outward, so for the curved portion of the Gaussian cylinder we have:
JE-da= jlElda=IElj da=IE12rrsl,

while the two ends contribute nothing (here Eis perpendicular to da). Thus,

IE12rrsl = -I-r2rk!s3.
EQ 3

or, finally,

E= -1k s 2As.
3EQ

Example 2.4

An infinite plane carries a uniform surface charge c,. Find its electric field.
Solution: Draw a "Gaussian pillbox," extending equal distances above and below the plane (Fig. 2.22). Apply Gauss's law to this surface:

J.j

E · da = _!__ QencEQ

In this case, Qenc = c, A, where A is the area of the lid of the pillbox. By symmetry, E points

away from the plane (upward for points above, downward for points below). Thus, the top and

bottom surfaces yield

JE •da = 2AiEI,

E

E

Figure 2.22

74

CHAPTER 2. ELECTROSTATICS

whereas the sides contribute nothing. Thus

2A

IEI

=

I -u A,

EQ

or (2.17)

where nis a unit vector pointing away from the surface. In Prob. 2.6, you obtained this same
result by a much more laborio!1s method.
It seems surprising, at first, thaf the field of an infinite plane is independent ofhow far away you are. What about the I/r 2 iq Coulomb's law? Well, the point is that as you move farther and farther away from the plane, more and more charge comes into your "field of view" (a cone shape extending out from your eye), and this compensates for the diminishing influence of any particular piece. The electric field of a sphere falls off like 1/ r2; the electric field of an infinite line falls off Iilce 1/ r; and the electric field of an infinite plane does not fall off at all.

Although the direct u&e of Gauss's law to compute electric fields is limited to cases of spherical, cylindrical, and planar symmetry, we can put together combinations of objects possessing such symmetry, even though the arrangement as a whole is not symmetrical. For example, invoking the principle of superposition, we could find the field in the vicinity of two uniformly charged parallel cylinders, or a sphere near an infinite charged plane.
Example2.5
Two infinite parallel planes carry equal Qµt opposite uniform charge densities ±u (Fig. 2.23).
Find the field in each of the three regions: (i), to the left of both, (ii) between them, (iii) to the right of both.
Solution: The left plate produces a field (l/2Eo)u which points away from it (Fig. 2.24}-to
the left in region (i) and to th~ right in regions (ii) and (iii). The right plate, being negatively
charged, produces a field (l/2Eo)P-, which points toward it-to the right in regions (i) and (ii) and to the left in region (iii). The two fields cancel in regions (i) and (iii); they conspire in region (ii). Conclusion: The field is (1/Eo)u, and points to the right, between the planes; elsewhere it is zero.

(i)

(ii)

(iii)

+CT

-CT

Figure 2.23

E+

E+

E+

E

E

E

(i)

(ii)

(iii)

+CT

-CJ

Figure 2.24

2.2. DIVERGENCE AND CURL OF ELECTROSTATIC FIELDS

75

Problem 2.11 Use Gauss's law to find the electric field inside and outside a spherical shell of radius R, which carries a uniform surface charge density a. Compare your answer to Prob. 2.7.
Problem 2.12 Use Gauss's law to find the electric field inside a uniformly charged sphere (charge density p). Compare your answer to Prob. 2.8.
Problem 2.13 Find the electric field a distance s from an infinitely long straight wire, which carries a uniform line charge),__ Compare Eq. 2.9.
Problem 2.14 Find the electric field inside a sphere which carries a charge density proportional
to the distance from the origin, p = kr, for some constant k. [Hint: This charge density is not
uniform, and you must integrate to get the enclosed charge.]
Problem 2.15 A hollow spherical shell carries charge density
k
P = r2
in the region a .::: r .::: b (Fig. 2.25). Find the electric field in the three regions: {i) r < a, {ii)
a < r < b, (iii) r > b. Plot IEI as a function of r.
Problem 2.16 A long coaxial cable (Fig. 2.26) carries a uniform volume charge density p on the inner cylinder (radius a), and a uniform surface charge density on the outer cylindrical shell (radius b). This surface charge is negative and of just the right magnitude so that the cable as a whole is electrically neutral. Find the electric field in each of the three regions: (i) inside the inner cylinder (s < a), (ii) between the cylinders (a < s < b), (iii) outside the cable
(s > b). Plot IEI as a function of s.
Problem 2.17 An infinite plane slab, of thickness 2d, carries a uniform volume charge density
p (Fig. 2.27). Find the electric field, as a function of y, where y = 0 at the center. Plot E
versus y, calling E positive when it points in the +y direction and negative when it points in the -y direction.

•

Problem 2.18 Two spheres, each of radius R and carrying uniform charge densities +P and

-p, respectively, are placed so that they partially overlap (Fig. 2.28). Call the vector from the

positive center to the negative center d. Show that the field in the region of overlap is constant,

and find its value. [Hint: Use the answer to Prob. 2.12.]

Figure 2.25

Figure 2.26

76

CHAPTER 2. ELECTROSTATICS

z

y
X

Figure 2.27

Figure 2.28

2.2.4 The Curl of E
I'll calculate the curl ofE, as I did the divergence in Sect. 2.2.1, by studying first the simplest possible configuration: a point charge at the origin. In this case
I q~ E=---r.
4m=o r 2 Now, a glance at Fig. 2.12 should convince you that the curl of this field has to be zero, but I suppose we ought to come up with something a little more rigorous than that. What if we calculate the line integral of this field from some point a to some other point b (Fig. 2.29):

lb E. di.

i, In spherical coordinates, di= dr r + r d0 0+ r sin 0 d</> so

Therefore,

E ·di= - 1-!Ldr. 4rrEo r2

f f b E. di - _1_ b !Ldr - --=-!._ Cj_ lrb - _1_ (!]_ - !]_)
la - la 4rrEo r2 - 4rrEo r ra - 4rrEo ra rb '

(2.18)

where ra is the distance from the origin to the point a and rb is the distance to b. The
integral around a closed path is evidently zero (for then ra = rb):

(2.19)

2.3. ELECTRIC POIBNTIAL

77

z b

y
X
Figure 2.29

and hence, applying Stokes' theorem,
(2.20)
Now, I proved Eqs. 2.19 and 2.20 only for the field of a single point charge at the origin, but these results make no reference to what is, after all, a perfectly arbitrary choice of coordinates; they also hold no matter where the charge is located. Moreover, if we have many charges, the principle of superposition states that the total field is a vector sum of their individual fields:
so
Thus, Eqs. 2.19 and 2.20 hold for any static charge distribution whatever.
Problem 2.19 Calculate V x E directly from Eq. 2.8, by the method of Sect. 2.2.2. Refer to Prob. 1.62 if you get stuck.

2.3 Electric Potential
2.3.1 Introduction to Potential
The electric field E is not just any old vector function; it is a very special kind of vector
function, one whose curl is always zero. E = yx, for example, could not possibly be
an electrostatic field; no set of charges, regardless of their sizes and positions, could ever produce such a field. In this section we're going to exploit this special property of electric fields to reduce a vector problem (finding E) down to a much simpler scalar problem. The first theorem in Sect. 1.6.2 asserts that any vector whose curl is zero is equal to the gradient of some scalar. What I'm going to do now amounts to a proof of that claim, in the context of electrostatics.

78

CHAPTER 2. ELECTROSTATICS

.tJ' (ii)

Figure 2.30

Because V x E = 0, the line integral of E around any closed loop is zero (that follows from Stokes' theorem). Because :f E • di = 0, the line integral of E from point a to point
bis the same for all paths (otherwise you could go out along path (i) and return along path
(ii)-Fig. 2.30-and obtain :f E •di of. 0). Because the line integral is independent of path,
we can define a function4

=- /~ I V(r)

E ·di.I

(2.21)

Here O is some standard reference point on which we have agreed beforehand; V then depends only on the point r. It is called the electric potential.
Evidently, the potential difference between two points a and b is

V(b) - V(a)

- 1:E-dl+ 1:E-dl

_J; 1 -1 ° = E . di _

E • di

b E • di.

(2.22)

Now, the fundamental theorem for gradients states that

V(b) - V(a) = i\vv). di,
so
1\v V) •dl = - lb E . di.
Since, finally, this is true for any points a and b, the integrands must be equal:
(2.23)
Equation 2.23 is the differential version of Eq. 2.21; it says that the electric field is the gradient of a scalar potential, which is what we set out to prove.
4To avoid any possible ambiguity I should perhaps put a prime on the integration variable:
1; V(r) = - E(r'). di'.
But this makes for cumbersome notation, and I prefer whenever possible to reserve the primes for source points. However, when (as in Ex. 2.6) we calculate such integrals explicitly, I shall put in the primes.

80

CHAPTER 2. ELECTROSTATICS

(iii) The reference point 0. There is an essential ambiguity in the definition of potential, since the choice of reference point O was arbitrary. Changing reference points amounts to adding a constant K to the potential:

V'(r)=- la(E , -dl=- lfo,0 E-dl- l{orE-dl=K+V(r),
where K is the line integral of E from the old reference point O to the new one O'. Of course, adding a constant to V will not affect the potential difference between two points:

V'(b) - V'(a) = V(b) - V(a),

since the K's cancel out. (Actually, it was already clear from Eq. 2.22 that the potential difference is independent of O, because it can be written as the line integral of E from a to b, with no reference to 0.) Nor does the ambiguity affect the gradient of V:

vv' = vv,

since the derivative of a constant is zero. That's why all such V's, differing only in their

choice of reference point, correspond to the same field E.

Evidently potential as such carries no real physical significance, for at any given point

we can adjust its value at will by a suitable relocation of O. In this sense it is rather like

altitude: If I ask you how high Denver is, you will probably tell me its height above sea level,

because that is a convenient and traditional reference point. But we could as well agree

to measure altitude above Washington D.C., or Greenwich, or wherever. That would add

(or, rather, subtract) a fixed amount from all our sea-level readings, but it wouldn't change

anything about the real world. The only quantity of intrinsic interest is the difference in

altitude between two points, and that is the same whatever your reference level.

Having said this, however, there is a "natural" spot to use for O in electrostatics-

analogous to sea level for altitude-and that is a point infinitely far from the charge. Or-

dinarily, then, we "set the zero of potential at infinity." (Since V (0) = 0, choosing a

reference point is equivalent to selecting a place where V is to be zero.) But I must warn

you that there is one special circumstance in which this convention fails: when the charge

distribution itself extends to infinity. The symptom of trouble, in such cases, is that the

potential blows up. For instance, the field of a uniformly charged plane is (er/2E0 )fi., as we found in Ex. 2.4; if we naively put O = oo, then the potential at height z above the plane

becomes

lz I

1

V(z) = - - a dz= --a(z - oo).

00 2Eo

2E0

The remedy is simply to choose some other reference point (in this problem you might use the origin). Notice that the difficulty occurs only in textbook problems; in "real life" there is no such thing as a charge distribution that goes on forever, and we can always use infinity as our reference point.

2.3. ELECTRIC POTENTIAL

81

(iv) Potential obeys the superposition principle. The original superposition principle of electrodynamics pertains to the force on a test charge Q. It says that the total force on Q is the vector sum of the forces attributable to the source charges individually:

Dividing through by Q, we find that the electric field, too, obeys the superposition principle:

Integrating from the common reference point to r, it follows that the potential also satisfies such a principle:
That is, the potential at any given point is the sum of the potentials due to all the source charges separately. Only this time it is an ordinary sum, not a vector sum, which makes it a lot easier to work with.
(v) Units of Potential. In our units, force is measured in newtons and charge in coulombs, so electric fields are in newtons per coulomb. Accordingly, potential is measured in newton-meters per coulomb or joules per coulomb. A joule per coulomb is called a volt.
Example2.6
Find the potential inside and outside a spherical shell of radius R (Fig. 2.31), which carries a uniform surface charge. Set the reference point at infinity.

Figure 2.31

82

CHAPTER 2. ELECTROSTATICS

Solution: From Gauss's law, the field outside is

where q is the total charge on the sphere. The field inside is zero. For points outside the sphere (r > R),

J V(r)=- r E-dl=--1- 1r -qdr1 = -1- -q Ir =

q

o

4nEo 00 r 12

4JTEo r' 00 4nEo r

To find the potential inside the sphere (r < R), we must break the integral into two sections, using in each region the field that prevails there:

V(r)=--1- 1R -qd r, - 1r (O),dr=1--q- IR +0=-I --q.

4nEo 00 r12

R

4nEo r' 00

4nEo R

Notice that the potential is not zero inside the shell, even though the field is. V is a constant
in this region, to be sure, so that V V = 0-that's what matters. In problems of this type you
must always work your way in from the reference point; that's where the potential is "nailed down." It is tempting to suppose that you could figure out the potential inside the sphere on the basis of the field there alone, but this is false: The potential inside the sphere is sensitive to what's going on outside the sphere as well. Ifl placed a second uniformly charged shell out at radius R' > R, the potential inside R would change, even though the field would still be zero. Gauss's law guarantees that charge exterior to a given point (that is, at larger r) produces no
net.field at that point, provided it is spherically or'cylindrically symmetric; but there is no such rule for potential, when infinity is used as the reference point.

Problem 2.21 Find the potential inside and outside a uniformly charged solid sphere whose radius is R and whose total charge is q. Use infinity as your reference point. Compute the gradient of Vin each region, and check that it yields the correct field. Sketch V(r).
Problem 2.22 Find the potential a distance s from an infinitely long straight wire that carries a uniform line charge A. Compute the gradient of your potential, and check that it yields the correct field.
Problem 2.23 For the charge configuration of Prob. 2.15, find the potential at the center, using infinity as your reference point.
Problem 2.24 For the configuration of Prob. 2.16, find the potential difference between a point on the axis and a point on the outer cylinder. Note that it is not necessary to commit yourself to a particular reference point if you use Eq. 2.22.

2.3. ELECTRIC POTENTIAL

83

2.3.3 Poisson's Equation and Laplace's Equation
We found in Sect. 2.3.1 that the electric field can be written as the gradient of a scalar potential.
E = -VV.
The question arises: What do the fundamental equations for E,

and V x E = 0,

look like, in terms of V? Well, V •E = V •(-V V) = -V2 V, so, apart from that persisting
minus sign, the divergence of Eis the Laplacian of V. Gauss's law then says that

(2.24)

This is known as Poisson's equation. In regions where there is no charge, so that p = 0,
Poisson's equation reduces to Laplace's equation,
(2.25)
We'll explore these equations more fully in Chapter 3. So much for Gauss's law. What about the curl law? This says that
V x E = V x (-V V)
must equal zero. But that's no condition on V-curl of gradient is always zero. Of course, we used the curl law to show that E could be expressed as the gradient of a scalar, so it's not
really surprising that this works out: V x E = 0 permits E = - V V; in return, E = - V V guarantees V x E = 0. It takes only one differential equation (Poisson's) to determine V,
because V is a scalar; for E we needed two, the divergence and the curl.

2.3.4 The Potential of a Localized Charge Distribution
I defined Vin terms ofE (Eq. 2.21). Ordinarily, though, it's E that we're looking for (ifwe already knew E there wouldn't be much point in calculating V). The idea is that it might be easier to get V first, and then calculate E by taking the gradient. Typically, then, we know where the charge is (that is, we know p ), and we want to find V. Now, Poisson's equation relates V and p, but unfortunately it's "the wrong way around": it would give us p, if we knew V, whereas we want V, knowing p. What we must do, then, is "invert" Poisson's equation. That's the program for this section, although I shall do it by roundabout means, beginning, as always, with a point charge at the origin.

84

CHAPTER 2. ELECTROSTATICS

p

Figure 2.32

Setting the reference point at infinity, the potential of a point charge q at the origin is

1r Ir V(r) = ---I -qdrI = -l- -q

41rEo 00 r'2

41rEo r' 00

I q 41rEo r

(You see here the special virtue of using infinity for the reference point: it kills the lower limit on the integral.) Notice the sign of V; presumably the conventional minus sign in the definition of V (Eq. 2.21) was chosen precisely in order to make the potential of a positive charge come out positive. It is useful to remember that regions of positive charge are potential "hills," regions of negative charge are potential "valleys," and the electric field points "downhill," from plus toward minus.
In general, the potential of a point charge q is

V(r) = -1-q, 41l'EO 1-

(2.26)

where 1-, as always, is the distance from the charge to r (Fig. 2.32). Invoking the superposition principle, then, the potential of a collection of charges is

I n q; V(r)=-I:-,
41l'EQ i=J 1/,i

(2.27)

or, for a continuous distribution,
I V(r) = - I -Idq. 41l'EO I/,
In particular, for a volume charge, it's

(2.28)

= - V(r)

1- / p(r') dr'.

41l'EQ

I/,

(2.29)

This is the equation we were looking for, telling us how to compute V when we know p; it is, if you like, the "solution" to Poisson's equation, for a localized charge distribution.5 I
5Equation 2.29 is an example of the Helmholtz theorem (Appendix B), in the context of electrostatics, where the curl ofE is zero and its divergence is p/€0.