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A Student™s Guide to Maxwell™s Equations

Maxwell™s Equations are four of the most in¬‚uential equations in science: Gauss™s
law for electric ¬elds, Gauss™s law for magnetic ¬elds, Faraday™s law, and the
Ampere“Maxwell law. In this guide for students, each equation is the subject of
an entire chapter, with detailed, plain-language explanations of the physical
meaning of each symbol in the equation, for both the integral and differential
forms. The ¬nal chapter shows how Maxwell™s Equations may be combined to
produce the wave equation, the basis for the electromagnetic theory of light.
This book is a wonderful resource for undergraduate and graduate courses in
electromagnetism and electromagnetics. A website hosted by the author, and
available through www.cambridge.org/9780521877619, contains interactive
solutions to every problem in the text. Entire solutions can be viewed
immediately, or a series of hints can be given to guide the student to the ¬nal
answer. The website also contains audio podcasts which walk students through
each chapter, pointing out important details and explaining key concepts.
da n i e l fl eis ch is Associate Professor in the Department of Physics at
Wittenberg University, Ohio. His research interests include radar cross-section
measurement, radar system analysis, and ground-penetrating radar. He is a
member of the American Physical Society (APS), the American Association of
Physics Teachers (AAPT), and the Institute of Electrical and Electronics
Engineers (IEEE).
A Student™s Guide to
Maxwell™s Equations

DANIEL FLEISCH
Wittenberg University
CAMBRIDGE UNIVERSITY PRESS
Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, São Paulo

Cambridge University Press
The Edinburgh Building, Cambridge CB2 8RU, UK
Published in the United States of America by Cambridge University Press, New York
www.cambridge.org
Information on this title: www.cambridge.org/9780521877619

© D. Fleisch 2008


This publication is in copyright. Subject to statutory exception and to the provision of
relevant collective licensing agreements, no reproduction of any part may take place
without the written permission of Cambridge University Press.
First published in print format 2008


ISBN-13 978-0-511-39308-2 eBook (EBL)

ISBN-13 978-0-521-87761-9 hardback




Cambridge University Press has no responsibility for the persistence or accuracy of urls
for external or third-party internet websites referred to in this publication, and does not
guarantee that any content on such websites is, or will remain, accurate or appropriate.
Contents




page vii
Preface
ix
Acknowledgments

1 Gauss™s law for electric ¬elds 1
1.1 The integral form of Gauss™s law 1
The electric ¬eld 3
The dot product 6
The unit normal vector 7
˜
The component of E normal to a surface 8
The surface integral 9
The ¬‚ux of a vector ¬eld 10
The electric ¬‚ux through a closed surface 13
The enclosed charge 16
The permittivity of free space 18
Applying Gauss™s law (integral form) 20
1.2 The differential form of Gauss™s law 29
Nabla “ the del operator 31
Del dot “ the divergence 32
The divergence of the electric ¬eld 36
Applying Gauss™s law (differential form) 38
2 Gauss™s law for magnetic ¬elds 43
2.1 The integral form of Gauss™s law 43
The magnetic ¬eld 45
The magnetic ¬‚ux through a closed surface 48
Applying Gauss™s law (integral form) 50
2.2 The differential form of Gauss™s law 53
The divergence of the magnetic ¬eld 54
Applying Gauss™s law (differential form) 55


v
vi Contents


3 Faraday™s law 58
3.1 The integral form of Faraday™s law 58
The induced electric ¬eld 62
The line integral 64
The path integral of a vector ¬eld 65
The electric ¬eld circulation 68
The rate of change of ¬‚ux 69
Lenz™s law 71
Applying Faraday™s law (integral form) 72
3.2 The differential form of Faraday™s law 75
Del cross “ the curl 76
The curl of the electric ¬eld 79
Applying Faraday™s law (differential form) 80
4 The Ampere“Maxwell law 83
4.1 The integral form of the Ampere“Maxwell law 83
The magnetic ¬eld circulation 85
The permeability of free space 87
The enclosed electric current 89
The rate of change of ¬‚ux 91
Applying the Ampere“Maxwell law (integral form) 95
4.2 The differential form of the Ampere“Maxwell law 101
The curl of the magnetic ¬eld 102
The electric current density 105
The displacement current density 107
Applying the Ampere“Maxwell law (differential form) 108
5 From Maxwell™s Equations to the wave equation 112
The divergence theorem 114
Stokes™ theorem 116
The gradient 119
Some useful identities 120
The wave equation 122

125
Appendix: Maxwell™s Equations in matter
131
Further reading
132
Index
Preface




This book has one purpose: to help you understand four of the most
in¬‚uential equations in all of science. If you need a testament to the
power of Maxwell™s Equations, look around you “ radio, television,
radar, wireless Internet access, and Bluetooth technology are a few
examples of contemporary technology rooted in electromagnetic ¬eld
theory. Little wonder that the readers of Physics World selected Maxwell™s
Equations as “the most important equations of all time.”
How is this book different from the dozens of other texts on electricity
and magnetism? Most importantly, the focus is exclusively on Maxwell™s
Equations, which means you won™t have to wade through hundreds of
pages of related topics to get to the essential concepts. This leaves room
for in-depth explanations of the most relevant features, such as the dif-
ference between charge-based and induced electric ¬elds, the physical
meaning of divergence and curl, and the usefulness of both the integral
and differential forms of each equation.
You™ll also ¬nd the presentation to be very different from that of other
books. Each chapter begins with an “expanded view” of one of Maxwell™s
Equations, in which the meaning of each term is clearly called out. If
you™ve already studied Maxwell™s Equations and you™re just looking for a
quick review, these expanded views may be all you need. But if you™re a
bit unclear on any aspect of Maxwell™s Equations, you™ll ¬nd a detailed
explanation of every symbol (including the mathematical operators) in
the sections following each expanded view. So if you™re not sure of the
˜^
meaning of E  n in Gauss™s Law or why it is only the enclosed currents
that contribute to the circulation of the magnetic ¬eld, you™ll want to read
those sections.
As a student™s guide, this book comes with two additional resources
designed to help you understand and apply Maxwell™s Equations: an
interactive website and a series of audio podcasts. On the website, you™ll
¬nd the complete solution to every problem presented in the text in


vii
viii Preface


interactive format “ which means that you™ll be able to view the entire
solution at once, or ask for a series of helpful hints that will guide you to
the ¬nal answer. And if you™re the kind of learner who bene¬ts from
hearing spoken words rather than just reading text, the audio podcasts are
for you. These MP3 ¬les walk you through each chapter of the book,
pointing out important details and providing further explanations of key
concepts.
Is this book right for you? It is if you™re a science or engineering
student who has encountered Maxwell™s Equations in one of your text-
books, but you™re unsure of exactly what they mean or how to use them.
In that case, you should read the book, listen to the accompanying
podcasts, and work through the examples and problems before taking a
standardized test such as the Graduate Record Exam. Alternatively, if
you™re a graduate student reviewing for your comprehensive exams, this
book and the supplemental materials will help you prepare.
And if you™re neither an undergraduate nor a graduate science student,
but a curious young person or a lifelong learner who wants to know more
about electric and magnetic ¬elds, this book will introduce you to the
four equations that are the basis for much of the technology you use
every day.
The explanations in this book are written in an informal style in which
mathematical rigor is maintained only insofar as it doesn™t get in the way
of understanding the physics behind Maxwell™s Equations. You™ll ¬nd
plenty of physical analogies “ for example, comparison of the ¬‚ux of
electric and magnetic ¬elds to the ¬‚ow of a physical ¬‚uid. James Clerk
Maxwell was especially keen on this way of thinking, and he was careful
to point out that analogies are useful not because the quantities are alike
but because of the corresponding relationships between quantities. So
although nothing is actually ¬‚owing in a static electric ¬eld, you™re likely
to ¬nd the analogy between a faucet (as a source of ¬‚uid ¬‚ow) and
positive electric charge (as the source of electric ¬eld lines) very helpful in
understanding the nature of the electrostatic ¬eld.
One ¬nal note about the four Maxwell™s Equations presented in this
book: it may surprise you to learn that when Maxwell worked out his theory
of electromagnetism, he ended up with not four but twenty equations that
describe the behavior of electric and magnetic ¬elds. It was Oliver Heaviside
in Great Britain and Heinrich Hertz in Germany who combined and sim-
pli¬ed Maxwell™s Equations into four equations in the two decades after
Maxwell™s death. Today we call these four equations Gauss™s law for electric
¬elds, Gauss™s law for magnetic ¬elds, Faraday™s law, and the Ampere“
Maxwell law. Since these four laws are now widely de¬ned as Maxwell™s
Equations, they are the ones you™ll ¬nd explained in the book.
Acknowledgments




This book is the result of a conversation with the great Ohio State radio
astronomer John Kraus, who taught me the value of plain explanations.
Professor Bill Dollhopf of Wittenberg University provided helpful sug-
gestions on the Ampere“Maxwell law, and postdoc Casey Miller of the
University of Texas did the same for Gauss™s law. The entire manuscript
was reviewed by UC Berkeley graduate student Julia Kregenow and
Wittenberg undergraduate Carissa Reynolds, both of whom made sig-
ni¬cant contributions to the content as well as the style of this work.
Daniel Gianola of Johns Hopkins University and Wittenberg graduate
Melanie Runkel helped with the artwork. The Maxwell Foundation of
Edinburgh gave me a place to work in the early stages of this project, and
Cambridge University made available their extensive collection of James
Clerk Maxwell™s papers. Throughout the development process, Dr. John
Fowler of Cambridge University Press has provided deft guidance and
patient support.




ix
1
Gauss™s law for electric ¬elds




In Maxwell™s Equations, you™ll encounter two kinds of electric ¬eld: the
electrostatic ¬eld produced by electric charge and the induced electric ¬eld
produced by a changing magnetic ¬eld. Gauss™s law for electric ¬elds
deals with the electrostatic ¬eld, and you™ll ¬nd this law to be a powerful
tool because it relates the spatial behavior of the electrostatic ¬eld to the
charge distribution that produces it.


1.1 The integral form of Gauss™s law
There are many ways to express Gauss™s law, and although notation
differs among textbooks, the integral form is generally written like this:
I
˜  ^ da ¼ qenc Gauss™s law for electric fields (integral form).
En
e0
S

The left side of this equation is no more than a mathematical description
of the electric ¬‚ux “ the number of electric ¬eld lines “ passing through a
closed surface S, whereas the right side is the total amount of charge
contained within that surface divided by a constant called the permittivity
of free space.
If you™re not sure of the exact meaning of ˜˜¬eld line™™ or ˜˜electric ¬‚ux,™™
don™t worry “ you can read about these concepts in detail later in this
chapter. You™ll also ¬nd several examples showing you how to use
Gauss™s law to solve problems involving the electrostatic ¬eld. For
starters, make sure you grasp the main idea of Gauss™s law:

Electric charge produces an electric ¬eld, and the ¬‚ux of that ¬eld
passing through any closed surface is proportional to the total charge
contained within that surface.


1
2 A student™s guide to Maxwell™s Equations


In other words, if you have a real or imaginary closed surface of any size
and shape and there is no charge inside the surface, the electric ¬‚ux
through the surface must be zero. If you were to place some positive
charge anywhere inside the surface, the electric ¬‚ux through the surface
would be positive. If you then added an equal amount of negative charge
inside the surface (making the total enclosed charge zero), the ¬‚ux would
again be zero. Remember that it is the net charge enclosed by the surface
that matters in Gauss™s law.
To help you understand the meaning of each symbol in the integral
form of Gauss™s law for electric ¬elds, here™s an expanded view:


Dot product tells you to find
Reminder that the The amount of
ˆ
the part of E parallel to n
electric field is a charge in coulombs
(perpendicular to the surface)
vector
Reminder that this The unit vector normal Reminder that only
integral is over a to the surface the enclosed charge
qenc
«
closed surface contributes

E n da =
ˆ
0
S
The electric An increment of The electric
surface area in m2
field in N/C permittivity
of the free space
Tells you to sum up the
contributions from each Reminder that this is a surface
portion of the surface integral (not a volume or a line integral)



How is Gauss™s law useful? There are two basic types of problems that
you can solve using this equation:
(1) Given information about a distribution of electric charge, you can
¬nd the electric ¬‚ux through a surface enclosing that charge.
(2) Given information about the electric ¬‚ux through a closed surface,
you can ¬nd the total electric charge enclosed by that surface.
The best thing about Gauss™s law is that for certain highly symmetric
distributions of charges, you can use it to ¬nd the electric ¬eld itself,
rather than just the electric ¬‚ux over a surface.
Although the integral form of Gauss™s law may look complicated, it is
completely understandable if you consider the terms one at a time. That™s
exactly what you™ll ¬nd in the following sections, starting with ˜ the
E,
electric ¬eld.
3
Gauss™s law for electric ¬elds


˜ The electric ¬eld
E
To understand Gauss™s law, you ¬rst have to understand the concept of
the electric ¬eld. In some physics and engineering books, no direct def-
inition of the electric ¬eld is given; instead you™ll ¬nd a statement that an
electric ¬eld is ˜˜said to exist™™ in any region in which electrical forces act.
But what exactly is an electric ¬eld?
This question has deep philosophical signi¬cance, but it is not easy to
answer. It was Michael Faraday who ¬rst referred to an electric ˜˜¬eld of
force,™™ and James Clerk Maxwell identi¬ed that ¬eld as the space around
an electri¬ed object “ a space in which electric forces act.
The common thread running through most attempts to de¬ne the
electric ¬eld is that ¬elds and forces are closely related. So here™s a very
pragmatic de¬nition: an electric ¬eld is the electrical force per unit charge
exerted on a charged object. Although philosophers debate the true
meaning of the electric ¬eld, you can solve many practical problems by
thinking of the electric ¬eld at any location as the number of newtons of
electrical force exerted on each coulomb of charge at that location. Thus,
the electric ¬eld may be de¬ned by the relation
˜
˜  Fe ; °1:1Þ
E
q0
where ˜e is the electrical force on a small1 charge q0 . This de¬nition
F
makes clear two important characteristics of the electric ¬eld:
(1) ˜ is a vector quantity with magnitude directly proportional to force
E
and with direction given by the direction of the force on a positive
test charge.
(2) ˜ has units of newtons per coulomb (N/C), which are the same as
E
volts per meter (V/m), since volts ¼ newtons · meters/coulombs.
In applying Gauss™s law, it is often helpful to be able to visualize the
electric ¬eld in the vicinity of a charged object. The most common
approaches to constructing a visual representation of an electric ¬eld are
to use a either arrows or ˜˜¬eld lines™™ that point in the direction of
the ¬eld at each point in space. In the arrow approach, the strength of the
¬eld is indicated by the length of the arrow, whereas in the ¬eld line


1
Why do physicists and engineers always talk about small test charges? Because the job of
this charge is to test the electric ¬eld at a location, not to add another electric ¬eld into the
mix (although you can™t stop it from doing so). Making the test charge in¬nitesimally
small minimizes the effect of the test charge™s own ¬eld.
4 A student™s guide to Maxwell™s Equations




-
+




Positive point charge Negative point charge Infinite line of
positive charge




Infinite plane of Positively charged Electric dipole with
negative charge conducting sphere positive charge on left

Figure 1.1 Examples of electric ¬elds. Remember that these ¬elds exist
inthree dimensions; full three-dimensional (3-D) visualizations are available
on the book™s website.


approach, it is the spacing of the lines that tells you the ¬eld strength
(with closer lines signifying a stronger ¬eld). When you look at a drawing
of electric ¬eld lines or arrows, be sure to remember that the ¬eld exists
between the lines as well.
Examples of several electric ¬elds relevant to the application of Gauss™s
law are shown in Figure 1.1.
Here are a few rules of thumb that will help you visualize and sketch
the electric ¬elds produced by charges2:

 Electric ¬eld lines must originate on positive charge and terminate on
negative charge.
 The net electric ¬eld at any point is the vector sum of all electric ¬elds
present at that point.
 Electric ¬eld lines can never cross, since that would indicate that the
¬eld points in two different directions at the same location (if two or
more different sources contribute electric ¬elds pointing in different
directions at the same location, the total electric ¬eld is the vector sum

2
In Chapter 3, you can read about electric ¬elds produced not by charges but by changing
magnetic ¬elds. That type of ¬eld circulates back on itself and does not obey the same
rules as electric ¬elds produced by charge.
5
Gauss™s law for electric ¬elds


Table 1.1. Electric ¬eld equations for simple objects

Point charge (charge ¼ q) ˜ ¼ 1 q ^ (at distance r from q)
E r
4pe0 r 2

Conducting sphere (charge ¼ Q) ˜ ¼ 1 Q ^ (outside, distance r from
E r
4pe0 r 2 center)
˜ ¼ 0 (inside)
E

˜ ¼ 1 Q ^ (outside, distance r from
Uniformly charged insulating
E r
sphere (charge ¼ Q, radius ¼ r0) 4pe0 r 2 center)

˜ ¼ 1 Qr ^ (inside, distance r from
E r
3
4pe0 r0 center)

˜ ¼ 1 k ^ (distance r from line)
In¬nite line charge (linear
E r
charge density ¼ k) 2pe0 r

˜¼ r ^
In¬nite ¬‚at plane (surface E n
charge density ¼ r) 2e0



of the individual ¬elds, and the electric ¬eld lines always point in the
single direction of the total ¬eld).
 Electric ¬eld lines are always perpendicular to the surface of a
conductor in equilibrium.
Equations for the electric ¬eld in the vicinity of some simple objects
may be found in Table 1.1.
So exactly what does the ˜ in Gauss™s law represent? It represents the
E
total electric ¬eld at each point on the surface under consideration. The sur-
face may be real or imaginary, as you™ll see when you read about the
meaning of the surface integral in Gauss™s law. But ¬rst you should consider
the dot product and unit normal that appear inside the integral.
6 A student™s guide to Maxwell™s Equations


 The dot product
When you™re dealing with an equation that contains a multiplication
symbol (a circle or a cross), it is a good idea to examine the terms on
both sides of that symbol. If they™re printed in bold font or are wearing
vector hats (as are ˜ and ^ in Gauss™s law), the equation involves vector
E n
multiplication, and there are several different ways to multiply vectors
(quantities that have both magnitude and direction).
In Gauss™s law, the circle between ˜ and ^ represents the dot product
E n
(or ˜˜scalar product™™) between the electric ¬eld vector ˜ and the unit
E
normal vector ^ (discussed in the next section). If you know the Cartesian
n
components of each vector, you can compute this as
˜  ˜ ¼ A x Bx þ Ay By þ Az Bz : °1:2Þ
AB

Or, if you know the angle h between the vectors, you can use
˜  ˜ ¼ j˜ ˜ cos h; °1:3Þ
AB AjjBj

where j˜ and j˜ represent the magnitude (length) of the vectors. Notice
Aj Bj
that the dot product between two vectors gives a scalar result.
To grasp the physical signi¬cance of the dot product, consider vectors
˜ and ˜ that differ in direction by angle h, as shown in Figure 1.2(a).
A B
For these vectors, the projection of ˜ onto ˜ is j˜ cos h, as shown
A B Aj
in Figure 1.2(b). Multiplying this projection by the length of ˜ gives
B
j˜ ˜ cos h. Thus, the dot product ˜  ˜ represents the projection of ˜
AjjBj AB A
onto the direction of ˜ multiplied by the length of ˜ 3 The usefulness of
B B.
this operation in Gauss™s law will become clear once you understand the
meaning of the vector ^.
n

(a) (b)
A A
B
B
u u
The projection of A onto B: |A| cos u
3|B|
multiplied by the length of B:
gives the dot product A B: |A||B|cos u

Figure 1.2 The meaning of the dot product.


You could have obtained the same result by ¬nding the projection of ˜ onto the direction
3
B
˜ and then multiplying by the length of ˜
of A A.
7
Gauss™s law for electric ¬elds


^
n The unit normal vector
The concept of the unit normal vector is straightforward; at any point on a
surface, imagine a vector with length of one pointing in the direction per-
pendicular to the surface. Such a vector, labeled ^, is called a ˜˜unit™™ vector
n
because its length is unity and ˜˜normal™™ because it is perpendicular to the
surface. The unit normal for a planar surface is shown in Figure 1.3(a).
Certainly, you could have chosen the unit vector for the plane in
Figure 1.3(a) to point in the opposite direction “ there™s no fundamental
difference between one side of an open surface and the other (recall that
an open surface is any surface for which it is possible to get from one side
to the other without going through the surface).
For a closed surface (de¬ned as a surface that divides space into an
˜˜inside™™ and an ˜˜outside™™), the ambiguity in the direction of the unit
normal has been resolved. By convention, the unit normal vector for a
closed surface is taken to point outward “ away from the volume enclosed
by the surface. Some of the unit vectors for a sphere are shown in Figure
1.3(b); notice that the unit normal vectors at the Earth™s North and South
Pole would point in opposite directions if the Earth were a perfect sphere.
You should be aware that some authors use the notation d˜ rather a
than ^ da. In that notation, the unit normal is incorporated into the
n
vector area element d˜ which has magnitude equal to the area da and
a,
direction along the surface normal ^. Thus d˜ and ^ da serve the same
n a n
purpose.




Figure 1.3 Unit normal vectors for planar and spherical surfaces.
8 A student™s guide to Maxwell™s Equations


˜ ^ The component of ˜ normal to a surface
En E
If you understand the dot product and unit normal vector, the meaning of
˜  ^ should be clear; this expression represents the component of the
En
electric ¬eld vector that is perpendicular to the surface under consideration.
If the reasoning behind this statement isn™t apparent to you, recall that
the dot product between two vectors such as ˜ and ^ is simply the pro-
E n
jection of the ¬rst onto the second multiplied by the length of the second.
Recall also that by de¬nition the length of the unit normal is one °j^j ¼ 1),
n
so that
˜  ^ ¼ j˜ nj cos h ¼ j˜ cos h; °1:4Þ
En Ejj^ Ej

where h is the angle between the unit normal n and ˜ This is the com-
^ E.
ponent of the electric ¬eld vector perpendicular to the surface, as illus-
trated in Figure 1.4.
Thus, if h ¼ 90 , ˜ is perpendicular to ^, which means that the electric
E n
¬eld is parallel to the surface, and ˜  ^ ¼ j˜ cos°90 Þ ¼ 0. So in this case
En Ej
the component of ˜ perpendicular to the surface is zero.
E
Conversely, if h ¼ 0 , ˜ is parallel to ^, meaning the electric ¬eld is
E n
perpendicular to the surface, and ˜  ^ ¼ j˜ cos°0 Þ ¼ j˜ In this case,
En Ej Ej.
the component of ˜ perpendicular to the surface is the entire length of ˜
E E.
The importance of the electric ¬eld component normal to the surface
will become clear when you consider electric ¬‚ux. To do that, you
should make sure you understand the meaning of the surface integral
in Gauss™s law.



Component of E normal
^
to surface is E n
^ E
n




Surface




Figure 1.4 Projection of ˜ onto direction of ^.
E n
9
Gauss™s law for electric ¬elds

R
°Þda The surface integral
S
Many equations in physics and engineering “ Gauss™s law among them “
involve the area integral of a scalar function or vector ¬eld over a spe-
ci¬ed surface (this type of integral is also called the ˜˜surface integral™™).
The time you spend understanding this important mathematical oper-
ation will be repaid many times over when you work problems in
mechanics, ¬‚uid dynamics, and electricity and magnetism (E&M).
The meaning of the surface integral can be understood by considering a
thin surface such as that shown in Figure 1.5. Imagine that the area
density (the mass per unit area) of this surface varies with x and y, and
you want to determine the total mass of the surface. You can do this by
dividing the surface into two-dimensional segments over each of which
the area density is approximately constant.
For individual segments with area density ri and area dAi, the mass of
each segment is ri dAi, and the mass of the entire surface of N segments is
PN
i¼1 ri dAi . As you can imagine, the smaller you make the area
given by
segments, the closer this gets to the true mass, since your approximation
of constant r is more accurate for smaller segments. If you let the seg-
ment area dA approach zero and N approach in¬nity, the summation
becomes integration, and you have
Z
Mass ¼ r°x; yÞ dA:
S

This is the area integral of the scalar function r(x, y) over the surface S. It
is simply a way of adding up the contributions of little pieces of a
function (the density in this case) to ¬nd a total quantity. To understand
the integral form of Gauss™s law, it is necessary to extend the concept of
the surface integral to vector ¬elds, and that™s the subject of the next
section.


Density approximately constant over
each of these areas (dA1, dA2, . . . , dAN)
Area density (s)
varies across surface

s1 s2 s3


y
x
Mass = s1 dA1+ s2 dA2+ . . . + sN dAN.
Density = s(x,y)

Figure 1.5 Finding the mass of a variable-density surface.
10 A student™s guide to Maxwell™s Equations

R
˜n
s A  ^ da The ¬‚ux of a vector ¬eld
In Gauss™s law, the surface integral is applied not to a scalar function
(such as the density of a surface) but to a vector ¬eld. What™s a vector
¬eld? As the name suggests, a vector ¬eld is a distribution of quantities in
space “ a ¬eld “ and these quantities have both magnitude and direction,
meaning that they are vectors. So whereas the distribution of temperature
in a room is an example of a scalar ¬eld, the speed and direction of the
¬‚ow of a ¬‚uid at each point in a stream is an example of a vector ¬eld.
The analogy of ¬‚uid ¬‚ow is very helpful in understanding the meaning
of the ˜˜¬‚ux™™ of a vector ¬eld, even when the vector ¬eld is static and
nothing is actually ¬‚owing. You can think of the ¬‚ux of a vector ¬eld
over a surface as the ˜˜amount™™ of that ¬eld that ˜˜¬‚ows™™ through that
surface, as illustrated in Figure 1.6.
In the simplest case of a uniform vector ¬eld ˜ and a surface S per-
A
pendicular to the direction of the ¬eld, the ¬‚ux U is de¬ned as the product
of the ¬eld magnitude and the area of the surface:

U ¼ j˜ · surface area: °1:5Þ
Aj

This case is shown in Figure 1.6(a). Note that if ˜ is perpendicular to the
A
surface, it is parallel to the unit normal ^:
n
If the vector ¬eld is uniform but is not perpendicular to the surface, as
in Figure 1.6(b), the ¬‚ux may be determined simply by ¬nding the
component of ˜ perpendicular to the surface and then multiplying that
A
value by the surface area:

U ¼ ˜  ^ · °surface areaÞ: °1:6Þ
An
While uniform ¬elds and ¬‚at surfaces are helpful in understanding the
concept of ¬‚ux, many E&M problems involve nonuniform ¬elds and
curved surfaces. To work those kinds of problems, you™ll need to
understand how to extend the concept of the surface integral to vector
¬elds.

(a) (b)


A n A
n


Figure 1.6 Flux of a vector ¬eld through a surface.
11
Gauss™s law for electric ¬elds


(a) (b)
ni
Component of A perpendicular
to this surface element is A ° ni
u
A
A

Surface
S
A



Figure 1.7 Component of ˜ perpendicular to surface.
A



Consider the curved surface and vector ¬eld ˜ shown in Figure 1.7(a).
A
Imagine that ˜ represents the ¬‚ow of a real ¬‚uid and S a porous mem-
A
brane; later you™ll see how this applies to the ¬‚ux of an electric ¬eld
through a surface that may be real or purely imaginary.
Before proceeding, you should think for a moment about how you
might go about ¬nding the rate of ¬‚ow of material through surface S.
You can de¬ne ˜˜rate of ¬‚ow™™ in a few different ways, but it will help to
frame the question as ˜˜How many particles pass through the membrane
each second?™™
To answer this question, de¬ne ˜ as the number density of the ¬‚uid
A
(particles per cubic meter) times the velocity of the ¬‚ow (meters per
second). As the product of the number density (a scalar) and the velocity
(a vector), ˜ must be a vector in the same direction as the velocity, with
A
units of particles per square meter per second. Since you™re trying to
¬nd the number of particles per second passing through the surface,
dimensional analysis suggests that you multiply ˜ by the area of the
A
surface.
But look again at Figure 1.7(a). The different lengths of the arrows are
meant to suggest that the ¬‚ow of material is not spatially uniform,
meaning that the speed may be higher or lower at various locations
within the ¬‚ow. This fact alone would mean that material ¬‚ows through
some portions of the surface at a higher rate than other portions, but you
must also consider the angle of the surface to the direction of ¬‚ow. Any
portion of the surface lying precisely along the direction of ¬‚ow will
necessarily have zero particles per second passing through it, since the
¬‚ow lines must penetrate the surface to carry particles from one side to
12 A student™s guide to Maxwell™s Equations


the other. Thus, you must be concerned not only with the speed of ¬‚ow
and the area of each portion of the membrane, but also with the com-
ponent of the ¬‚ow perpendicular to the surface.
Of course, you know how to ¬nd the component of ˜ perpendicular
A
to the surface; simply form the dot product of ˜ and ^, the unit normal to
A n
the surface. But since the surface is curved, the direction of ^ depends on
n
which part of the surface you™re considering. To deal with the different ^
n
(and ˜ at each location, divide the surface into small segments, as shown
A)
in Figure 1.7(b). If you make these segments suf¬ciently small, you can
assume that both ^ and ˜ are constant over each segment.
n A
Let ^i represent the unit normal for the ith segment (of area dai); the
n
¬‚ow through segment i is (˜i  ^i ) dai, and the total is
An
P
˜i  ^i dai :
¬‚ow through entire surface ¼ An
i

It should come as no surprise that if you now let the size of each
segment shrink to zero, the summation becomes integration.
Z
Flow through entire surface ¼ ˜  ^ da: °1:7Þ
An
S

For a closed surface, the integral sign includes a circle:
I
˜  ^ da: °1:8Þ
An
S

This ¬‚ow is the particle ¬‚ux through a closed surface S, and the similarity
to the left side of Gauss™s law is striking. You have only to replace the
vector ¬eld ˜ with the electric ¬eld ˜ to make the expressions identical.
A E
13
Gauss™s law for electric ¬elds

H
˜  ^ da
SE n The electric ¬‚ux through a closed surface
On the basis of the results of the previous section, you should understand
that the ¬‚ux UE of vector ¬eld ˜ through surface S can be determined
E
using the following equations:
UE ¼ j˜ · °surface areaÞ ˜ is uniform and perpendicular to S; °1:9Þ
Ej E


UE ¼ ˜  ^ · °surface areaÞ ˜ is uniform and at an angle to S; °1:10Þ
En E

Z
˜  ^ da ˜ is non-uniform and at a variable angle to S: °1:11Þ
UE ¼ En E
S

These relations indicate that electric ¬‚ux is a scalar quantity and has units
of electric ¬eld times area, or Vm. But does the analogy used in the
previous section mean that the electric ¬‚ux should be thought of as a ¬‚ow
of particles, and that the electric ¬eld is the product of a density and a
velocity?
The answer to this question is ˜˜absolutely not.™™ Remember that when
you employ a physical analogy, you™re hoping to learn something about
the relationships between quantities, not about the quantities themselves.
So, you can ¬nd the electric ¬‚ux by integrating the normal component of
the electric ¬eld over a surface, but you should not think of the electric
¬‚ux as the physical movement of particles.
How should you think of electric ¬‚ux? One helpful approach follows
directly from the use of ¬eld lines to represent the electric ¬eld. Recall
that in such representations the strength of the electric ¬eld at any point is
indicated by the spacing of the ¬eld lines at that location. More specif-
ically, the electric ¬eld strength can be considered to be proportional to
the density of ¬eld lines (the number of ¬eld lines per square meter) in a
plane perpendicular to the ¬eld at the point under consideration. Inte-
grating that density over the entire surface gives the number of ¬eld lines
penetrating the surface, and that is exactly what the expression for
electric ¬‚ux gives. Thus, another way to de¬ne electric ¬‚ux is
electric flux °UE Þ  number of field lines penetrating surface.
There are two caveats you should keep in mind when you think of electric
¬‚ux as the number of electric ¬eld lines penetrating a surface. The ¬rst is
that ¬eld lines are only a convenient representation of the electric ¬eld,
which is actually continuous in space. The number of ¬eld lines you
14 A student™s guide to Maxwell™s Equations


(b) (c)
(a)




Zero net flux Positive flux Negative flux

Figure 1.8 Flux lines penetrating closed surfaces.


choose to draw for a given ¬eld is up to you, so long as you maintain
consistency between ¬elds of different strengths “ which means that ¬elds
that are twice as strong must be represented by twice as many ¬eld lines
per unit area.
The second caveat is that surface penetration is a two-way street; once
the direction of a surface normal ^ has been established, ¬eld line com-
n
ponents parallel to that direction give a positive ¬‚ux, while components
in the opposite direction (antiparallel to ^) give a negative ¬‚ux. Thus, a
n
surface penetrated by ¬ve ¬eld lines in one direction (say from the top
side to the bottom side) and ¬ve ¬eld lines in the opposite direction (from
bottom to top) has zero ¬‚ux, because the contributions from the two
groups of ¬eld lines cancel. So, you should think of electric ¬‚ux as the net
number of ¬eld lines penetrating the surface, with direction of penetra-
tion taken into account.
If you give some thought to this last point, you may come to an
important conclusion about closed surfaces. Consider the three boxes
shown in Figure 1.8. The box in Figure 1.8(a) is penetrated only by
electric ¬eld lines that originate and terminate outside the box. Thus,
every ¬eld line that enters must leave, and the ¬‚ux through the box must
be zero.
Remembering that the unit normal for closed surfaces points away
from the enclosed volume, you can see that the inward ¬‚ux (lines entering
the box) is negative, since ˜  ^ must be negative when the angle between
En
˜ and ^ is greater than 90 . This is precisely cancelled by the outward ¬‚ux
E n
(lines exiting the box), which is positive, since ˜  ^ is positive when the
En
angle between ˜ and ^ is less than 90 .
E n
Now consider the box in Figure 1.8(b). The surfaces of this box are
penetrated not only by the ¬eld lines originating outside the box, but also
by a group of ¬eld lines that originate within the box. In this case, the net
number of ¬eld lines is clearly not zero, since the positive ¬‚ux of the lines
15
Gauss™s law for electric ¬elds


that originate in the box is not compensated by any incoming (negative)
¬‚ux. Thus, you can say with certainty that if the ¬‚ux through any closed
surface is positive, that surface must contain a source of ¬eld lines.
Finally, consider the box in Figure 1.8(c). In this case, some of the ¬eld
lines terminate within the box. These lines provide a negative ¬‚ux at the
surface through which they enter, and since they don™t exit the box, their
contribution to the net ¬‚ux is not compensated by any positive ¬‚ux.
Clearly, if the ¬‚ux through a closed surface is negative, that surface must
contain a sink of ¬eld lines (sometimes referred to as a drain).
Now recall the ¬rst rule of thumb for drawing charge-induced electric
¬eld lines; they must originate on positive charge and terminate on
negative charge. So, the point from which the ¬eld lines diverge in Figure
1.8(b) marks the location of some amount of positive charge, and the
point to which the ¬eld lines converge in Figure 1.8(c) indicates the
existence of negative charge at that location.
If the amount of charge at these locations were greater, there would be
more ¬eld lines beginning or ending on these points, and the ¬‚ux through
the surface would be greater. And if there were equal amounts of positive
and negative charge within one of these boxes, the positive (outward) ¬‚ux
produced by the positive charge would exactly cancel the negative
(inward) ¬‚ux produced by the negative charge. So, in this case the ¬‚ux
would be zero, just as the net charge contained within the box would be
zero.
You should now see the physical reasoning behind Gauss™s law: the
electric ¬‚ux passing through any closed surface “ that is, the number of
electric ¬eld lines penetrating that surface “ must be proportional to the
total charge contained within that surface. Before putting this concept to
use, you should take a look at the right side of Gauss™s law.
16 A student™s guide to Maxwell™s Equations


qenc The enclosed charge
If you understand the concept of ¬‚ux as described in the previous section,
it should be clear why the right side of Gauss™s law involves only the
enclosed charge “ that is, the charge within the closed surface over which
the ¬‚ux is determined. Simply put, it is because any charge located out-
side the surface produces an equal amount of inward (negative) ¬‚ux and
outward (positive) ¬‚ux, so the net contribution to the ¬‚ux through the
surface must be zero.
How can you determine the charge enclosed by a surface? In some
problems, you™re free to choose a surface that surrounds a known
amount of charge, as in the situations shown in Figure 1.9. In each of
these cases, the total charge within the selected surface can be easily
determined from geometric considerations.
For problems involving groups of discrete charges enclosed by surfaces
of any shape, ¬nding the total charge is simply a matter of adding the
individual charges.
X
Total enclosed charge ¼ qi :
i


While small numbers of discrete charges may appear in physics and
engineering problems, in the real world you™re far more likely to encounter
charged objects containing billions of charge carriers lined along a wire,
slathered over a surface, or arrayed throughout a volume. In such cases,
counting the individual charges is not practical “ but you can determine
the total charge if you know the charge density. Charge density may be
speci¬ed in one, two, or three dimensions (1-, 2-, or 3-D).



Point Charged
Multiple
charge Charged
line
point charges
plane




Enclosing
pillbox
enclosing
Enclosing Enclosing
cube
sphere cylinder

Figure 1.9 Surface enclosing known charges.
17
Gauss™s law for electric ¬elds




Dimensions Terminology Symbol Units

k
1 Linear charge C/m
density
r C/m2
2 Area charge density
q C/m3
3 Volume charge
density

If these quantities are constant over the length, area, or volume under
consideration, ¬nding the enclosed charge requires only a single multi-
plication:
1-D : qenc ¼ k L °L ¼ enclosed length of charged lineÞ; °1:12Þ


2-D : qenc ¼ rA °A ¼ enclosed area of charged surfaceÞ; °1:13Þ


3-D : qenc ¼ qV °V ¼ enclosed portion of charged volumeÞ: °1:14Þ
You are also likely to encounter situations in which the charge density
is not constant over the line, surface, or volume of interest. In such cases,
the integration techniques described in the ˜˜Surface Integral™™ section of
this chapter must be used. Thus,
Z
1-D : qenc ¼ k dl where k varies along a line; °1:15Þ
L


Z
2-D : qenc ¼ r da where r varies over a surface; °1:16Þ
S


Z
3-D : qenc ¼ q dV where q varies over a volume: °1:17Þ
V

You should note that the enclosed charge in Gauss™s law for electric ¬elds
is the total charge, including both free and bound charge. You can read
about bound charge in the next section, and you™ll ¬nd a version of
Gauss™s law that depends only on free charge in the Appendix.
Once you™ve determined the charge enclosed by a surface of any size
and shape, it is very easy to ¬nd the ¬‚ux through that surface; simply
divide the enclosed charge by e0, the permittivity of free space. The
physical meaning of that parameter is described in the next section.
18 A student™s guide to Maxwell™s Equations


e0 The permittivity of free space
The constant of proportionality between the electric ¬‚ux on the left side
of Gauss™s law and the enclosed charge on the right side is e0, the
permittivity of free space. The permittivity of a material determines
its response to an applied electric ¬eld “ in nonconducting materials
(called ˜˜insulators™™ or ˜˜dielectrics™™), charges do not move freely, but
may be slightly displaced from their equilibrium positions. The relevant
permittivity in Gauss™s law for electric ¬elds is the permittivity of free
space (or ˜˜vacuum permittivity™™), which is why it carries the subscript
zero.
The value of the vacuum permittivity in SI units is approximately
8.85 · 10À12 coulombs per volt-meter (C/Vm); you will sometimes see the
units of permittivity given as farads per meter (F/m), or, more funda-
mentally, (C2s2/kg m3). A more precise value for the permittivity of free
space is

e0 ¼ 8.8541878176 · 10À12 C/Vm.

Does the presence of this quantity mean that this form of Gauss™s law is
only valid in a vacuum? No, Gauss™s law as written in this chapter is
general, and applies to electric ¬elds within dielectrics as well as those in
free space, provided that you account for all of the enclosed charge,
including charges that are bound to the atoms of the material.
The effect of bound charges can be understood by considering what
happens when a dielectric is placed in an external electric ¬eld. Inside the
dielectric material, the amplitude of the total electric ¬eld is generally less
than the amplitude of the applied ¬eld.
The reason for this is that dielectrics become ˜˜polarized™™ when placed
in an electric ¬eld, which means that positive and negative charges are
displaced from their original positions. And since positive charges are
displaced in one direction (parallel to the applied electric ¬eld) and
negative charges are displaced in the opposite direction (antiparallel to
the applied ¬eld), these displaced charges give rise to their own electric
¬eld that opposes the external ¬eld, as shown in Figure 1.10. This makes
the net ¬eld within the dielectric less than the external ¬eld.
It is the ability of dielectric materials to reduce the amplitude of an
electric ¬eld that leads to their most common application: increasing the
capacitance and maximum operating voltage of capacitors. As you
may recall, the capacitance (ability to store charge) of a parallel-plate
capacitor is
19
Gauss™s law for electric ¬elds


Induced
Dielectric
No dielectric present field
“ +
External +

electric
“ +
field
“ +

Displaced
charges

Figure 1.10 Electric ¬eld induced in a dielectric.



eA
C¼ ;
d
where A is the plate area, d is the plate separation, and e is the permittivity
of the material between the plates. High-permittivity materials can
provide increased capacitance without requiring larger plate area or
decreased plate spacing.
The permittivity of a dielectric is often expressed as the relative per-
mittivity, which is the factor by which the material™s permittivity exceeds
that of free space:
relative permittivity er ¼ e=e0 :
Some texts refer to relative permittivity as ˜˜dielectric constant,™™ although
the variation in permittivity with frequency suggests that the word ˜˜con-
stant™™ is better used elsewhere. The relative permittivity of ice, for example,
changes from approximately 81 at frequencies below 1 kHz to less than 5 at
frequencies above 1 MHz. Most often, it is the low-frequency value of
permittivity that is called the dielectric constant.
One more note about permittivity; as you™ll see in Chapter 5, the
permittivity of a medium is a fundamental parameter in determining the
speed with which an electromagnetic wave propagates through that
medium.
20 A student™s guide to Maxwell™s Equations

H
˜  ^ da ¼ qenc =e0 Applying Gauss™s law (integral form)
sE n
A good test of your understanding of an equation like Gauss™s law is
whether you™re able to solve problems by applying it to relevant situ-
ations. At this point, you should be convinced that Gauss™s law relates
the electric ¬‚ux through a closed surface to the charge enclosed by that
surface. Here are some examples of what can you actually do with that
information.

Example 1.1: Given a charge distribution, ¬nd the ¬‚ux through a closed
surface surrounding that charge.
Problem: Five point charges are enclosed in a cylindrical surface S. If the
values of the charges are q1 ¼ þ3 nC, q2 ¼ À2 nC, q3 ¼ þ2 nC, q4 ¼ þ4 nC,
and q5 ¼ À1 nC, ¬nd the total ¬‚ux through S.



S
q3 q5
q1
q4
q2


Solution: From Gauss™s law,
I
˜  ^ da ¼ qenc :
UE ¼ En
e0
S

For discrete charges, you know that the total charge is just the sum of the
individual charges. Thus,
X
qenc ¼ Total enclosed charge ¼ qi
i
¼ °3 À 2 þ 2 þ 4 À 1Þ · 10À9 C
¼ 6 · 10À9 C

and

6 · 10À9 C
qenc
UE ¼ ¼ ¼ 678 Vm:
8:85 · 10À12 C=Vm
e0

This is the total ¬‚ux through any closed surface surrounding this group of
charges.
21
Gauss™s law for electric ¬elds


Example 1.2: Given the ¬‚ux through a closed surface, ¬nd the enclosed
charge.
Problem: A line charge with linear charge density k ¼ 10À12 C/m passes
through the center of a sphere. If the ¬‚ux through the surface of the
sphere is 1.13 · 10À3 Vm, what is the radius R of the sphere?

Charged
line




L
Sphere
encloses
portion of
line



Solution: The charge on a line charge of length L is given by q ¼ kL. Thus,
qenc kL
UE ¼ ¼ ;
e0 e0
and
U E e0
L¼ :
k
Since L is twice the radius of the sphere, this means

U E e0 UE e0
2R ¼ or R ¼ :
k 2k

Inserting the values for UE, e0 and k, you will ¬nd that R ¼ 5 · 10À3 m.

Example 1.3: Find the ¬‚ux through a section of a closed surface.
Problem: A point source of charge q is placed at the center of curvature of
a spherical section that extends from spherical angle h1 to h2 and from u1
to u2. Find the electric ¬‚ux through the spherical section.

Solution: Since the surface of interest in this problem is open, you™ll have
to ¬nd the electric ¬‚ux by integrating the normal component of the
electric ¬eld over the surface. You can then check your answer using
Gauss™s law by allowing the spherical section to form a complete sphere
that encloses the point charge.
22 A student™s guide to Maxwell™s Equations

R
The electric ¬‚ux UE is S ˜  ^ da, where S is the spherical section of
En
interest and ˜ is the electric ¬eld on the surface due to the point charge
E
at the center of curvature, a distance r from the section of interest.
From Table 1.1, you know that the electric ¬eld at a distance r from a
point charge is

˜ ¼ 1 q ^:
E r
4pe0 r 2

Before you can integrate this over the surface of interest, you have to
consider ˜  ^ (that is, you must ¬nd the component of the electric ¬eld
En
perpendicular to the surface). That is trivial in this case, because the unit
normal ^ for a spherical section points in the outward radial direction
n
(the ^ direction), as may be seen in Figure 1.11. This means that ˜ and ^
r E n
are parallel, and the ¬‚ux is given by
Z Z Z Z
1q
˜  ^ da ¼ j˜ nj cos°0 Þ da ¼ j˜ da ¼
UE ¼ En Ejj^ Ej da:
4pe0 r 2
S S S S


Since you are integrating over a spherical section in this case, the logical
choice for coordinate system is spherical. This makes the area element r2
sin h dh dU, and the surface integral becomes




da
rs
n
in
u
r du




u u df
r sin
du
da = (rdu)(r sin udf)



df

f




Figure 1.11 Geomentry of sperical section.
23
Gauss™s law for electric ¬elds

ZZ Z Z
1 q2 q
UE ¼ r sin h dh df ¼ sin h dh df;
4pe0 r 2 4pe0
h f h f

which is easily integrated to give

q
UE ¼ °cos h1 À cos h2 Þ°f2 À f1 Þ:
4pe0
As a check on this result, take the entire sphere as the section (h1 ¼ 0,
h2 ¼ p, u1 ¼ 0, and u2 ¼ 2p). This gives

q q
UE ¼ °1 À °À1ÞÞ °2p À 0Þ ¼ ;
e0
4pe0
exactly as predicted by Gauss™s law.

Example 1.4: Given ˜ over a surface, ¬nd the ¬‚ux through the surface
E
and the charge enclosed by the surface.
Problem: The electric ¬eld at distance r from an in¬nite line charge with
linear charge density k is given in Table 1.1 as

˜ ¼ 1 k ^:
E r
2pe0 r



r




h




Use this expression to ¬nd the electric ¬‚ux through a cylinder of radius r
and height h surrounding a portion of an in¬nite line charge, and then use
Gauss™s law to verify that the enclosed charge is kh.
24 A student™s guide to Maxwell™s Equations


Solution: Problems like this are best approached by considering the ¬‚ux
through each of three surfaces that comprise the cylinder: the top, bot-
tom, and curved side surfaces. The most general expression for the
electric ¬‚ux through any surface is
Z
UE ¼ ˜  ^ da;
En
S
which in this case gives
Z
1k
^  ^ da:
UE ¼ rn
2pe0 r
S


n




n
n
n n
n




n


Consider now the unit normal vectors of each of the three surfaces: since
the electric ¬eld points radially outward from the axis of the cylinder, ˜ is
E
perpendicular to the normal vectors of the top and bottom surfaces and
parallel to the normal vectors for the curved side of the cylinder. You
may therefore write
Z
1k
^  ^top da ¼ 0;
UE; top ¼ rn
S 2pe0 r


Z
1k
^  ^bottom da ¼ 0;
UE; bottom ¼ rn
2pe0 r
S


Z Z
1k 1k
^  ^side da ¼
UE; side ¼ rn da;
2pe0 r 2pe0 r
S S
25
Gauss™s law for electric ¬elds


and, since the area of the curved side of the cylinder is 2prh, this gives
1k kh
UE;side ¼ °2prhÞ ¼ :
e0
2pe0 r

Gauss™s law tells you that this must equal qenc /e0, which veri¬es that the
enclosed charge qenc ¼ kh in this case.

Example 1.5: Given a symmetric charge distribution, ¬nd ˜
E:
Finding the electric ¬eld using Gauss™s law may seem to be a hopeless
task. After all, while the electric ¬eld does appear in the equation, it is only
the normal component that emerges from the dot product, and it is only
the integral of that normal component over the entire surface that is propo-
rtional to the enclosed charge. Do realistic situations exist in which it is
possible to dig the electric ¬eld out of its interior position in Gauss™s law?
Happily, the answer is yes; you may indeed ¬nd the electric ¬eld
using Gauss™s law, albeit only in situations characterized by high sym-
metry. Speci¬cally, you can determine the electric ¬eld whenever you™re
able to design a real or imaginary ˜˜special Gaussian surface™™ that
encloses a known amount of charge. A special Gaussian surface is one on
which
(1) the electric ¬eld is either parallel or perpendicular to the surface
normal (which allows you to convert the dot product into an
algebraic multiplication), and
(2) the electric ¬eld is constant or zero over sections of the surface (which
allows you to remove the electric ¬eld from the integral).

Of course, the electric ¬eld on any surface that you can imagine around
arbitrarily shaped charge distributions will not satisfy either of these
requirements. But there are situations in which the distribution of charge
is suf¬ciently symmetric that a special Gaussian surface may be imagined.
Speci¬cally, the electric ¬eld in the vicinity of spherical charge distribu-
tions, in¬nite lines of charge, and in¬nite planes of charge may be
determined by direct application of the integral form of Gauss™s law.
Geometries that approximate these ideal conditions, or can be approxi-
mated by combinations of them, may also be attacked using Gauss™s law.
The following problem shows how to use Gauss™s law to ¬nd the
electric ¬eld around a spherical distribution of charge; the other cases are
covered in the problem set, for which solutions are available on the
website.
26 A student™s guide to Maxwell™s Equations


Problem: Use Gauss™s law to ¬nd the electric ¬eld at a distance r from the
center of a sphere with uniform volume charge density q and radius a.
Solution: Consider ¬rst the electric ¬eld outside the sphere. Since the
distribution of charge is spherically symmetric, it is reasonable to expect
the electric ¬eld to be entirely radial (that is, pointed toward or away

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