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Unformatted text preview: 4 Electrostatics 4—1 Statics We begin now our detailed study of the theory of electromagnetism. All of
electromagnetism is contained in the Maxwell equations. Maxwell’s equations: vE=£, (4.1)
60
an
vxE_—6—t, (4.2)
2 _§€ L
cVXB—at+60, (4.3)
VB=0. (4.4) The situations that are described by these equations can be very complicated.
We will consider ﬁrst relatively simple situations, and learn how to handle them
before we take up more complicated ones. The easiest circumstance to treat is one
in which nothing depends on the time—called the static case. All charges are
permanently ﬁxed in space, or if they do move, they move as a steady ﬂow in a
circuit (so p and j are constant in time). In these circumstances, all of the terms in
the Maxwell equations which are time derivatives of the ﬁeld are zero. In this
case, the Maxwell equations become: Electrostatics:
v E = l. (4.5)
6o
v x E = (4.6)
Magnetostatics:
 L ,
v x B — 60c, (4.7)
v B = 0. (4.8) You will notice an interesting thing about this set of four equations. It can
be separated into two pairs. The electric ﬁeld E appears only in the ﬁrst two, and
the magnetic ﬁeld B appears only in the second two. The two ﬁelds are not inter
connected. This means that electricity and magnetism are distinct phenomena so
long as charges and currents are static. The interdependence of E and B does not
appear until there are changes in charges or currents, as when a condenser is
charged, or a magnet moved. Only when there are sufﬁciently rapid changes, so
that the time derivatives in Maxwell’s equations become signiﬁcant, will E and B
depend on each other. Now if you look at the equations of statics you will see that the study of the
two subjects we call electrostatics and magnetostatics is ideal from the point of
view of learning about the mathematical properties of vector ﬁelds. Electrostatics
is a neat example of a vector ﬁeld with zero curl and a given divergence. Magnet
ostatics is a neat example of a ﬁeld with zero divergence and a given curl. The more
conventional—and you may be thinking, more satisfactory—way of presenting 4—1 4—1 Statics 42 Coulomb’s law; superposition
43 Electric potential 44 E = —V¢ 45 The ﬂux of E 46 Gauss’ law; the divergence of E
47 Field of a sphere of charge 48 Field lines; equipotential
surfaces Review: Chapters 13 and 14, Vol. I,
Work and Potential Energy 1 ~9X10’ 41! 60 [so] = coulornb’lnewtonmetera the theory of electromagnetism is to start ﬁrst with electrostatics and thus to learn
about the divergence. Magnetostatics and the curl are taken up later. Finally,
electricity and magnetism are put together. We have chosen to start with the
complete theory of vector calculus. Now we shall apply it to the special case of
electrostatics, the ﬁeld of E given by the ﬁrst pair of equations. We will begin with the simplest situations—ones in which the positions of all
charges are speciﬁed. If we had only to study electrostatics at this level (as We
shall do in the next two chapters), life would be very simple—in fact, almost
trivial. Everything can be obtained from Coulomb’s law and some integration,
as you will see. In many real electrostatic problems, however, we do not know,
initially, where the charges are. We know only that they have distributed them
selves in ways that depend on the properties of matter. The positions that the
charges take up depend on the E ﬁeld, which in turn depends on the positions of
the charges. Then things can get quite complicated. If, for instance, a charged
body is brought near a conductor or insulator, the electrons and protons in the
conductor or insulator will move around. The charge density p in Eq. (4.5) may
have one part that we know about, from the charge that we brought up; but there
will be other parts from charges that have moved around in the conductor. And
all of the charges must be taken into account. One can get into some rather subtle
and interesting problems. So although this chapter is to be on electrostatics, it will
not cover the more beautiful and subtle parts of the subject. It will treat only the
situation where we can assume that the positions of all the charges are known.
Naturally, you should be able to do that case before you try to handle the other
ones. 4—2 Coulomb’s law; superposition It would be logical to use Eqs. (4.5) and (4.6) as our starting points. It will
be easier, however, if we start somewhere else and come back to these equations.
The results will be equivalent. We will start with a law that we have talked about
before, called Coulomb’s law, which says that between two charges at rest there is
a force directly proportional to the product of the charges and inversely propor
tional to the square of the distance between. The force is along the straight line
from one charge to the other. Coulomb’s law: 1 q1q2
F1 = —— = — . 4.
41r€0 rfz e12 F2 ( F1 is the force on charge ql, cm is the unit vector in the direction to ql from (12,
and r1 2 is the distance between ql and q2. The force F 2 on q2 is equal and opposite
to F 1. The constant of proportionality, for historical reasons, is written as 1/41reo.
In the system of units which we use—the mks system—it is deﬁned as exactly
10‘7 times the speed of light squared. Now since the speed of light is approxi
mately 3 X 108 meters per second, the constant is approximately 9 X 109, and
the unit turns out to be newtonmeter2 per coulomb2 or voltmeter per coulomb. 1 _ _7 2 . .
4160 — 10 c (by deﬁnltlon) = 9.0 x 109 (by experiment). (4.10)
Unit: newton'meterz/coulombz, or voltmeter/coulomb. When there are more than two charges present—~the only really interesting
times—we must supplement Coulomb’s law with one other fact of nature: the
force on any charge is the vector sum of the Coulomb forces from each of the other
charges. This fact is called “the principle of superposition.” That’s all there is to
electrostatics. If we combine the Coulomb law and the principle of superposition,
there is nothing else. Equations (4.5) and (4.6)—the electrostatic equations—say
no more and no less. 4—2 When applying Coulomb's law, it is convenient to introduce the idea of an
electric ﬁeld. We say that the ﬁeld E(l) is the force per unit charge on ql (due to
all other charges). Dividing Eq. (4.9) by 91, we have, for one other charge besides ql: 1 ‘12 e 2
2 1'
471'60 r12 5(1) = (4.11) Also, we consider that 15(1) describes something about the point (1) even if 41
were not there—assuming that all other charges keep their same positions. We
say: EU) is the electric ﬁeld at the point (1). The electric ﬁeld E is a vector, so by Eq. (4.11) we really mean three equations
——one for each component. Writing out explicitly the xcomponent, Eq. (4.11)
means
:12. XI  x2 ,
471'60 [(xi — x2)2 + 01 — 1’2)2 + (21  Z2)2]3/2 and similarly for the other components. If there are many charges present, the ﬁeld E at any point (1) is a sum of the
contributions from each of the other charges. Each term of the sum will look like
(4.11) or (4.12). Letting q; be the magnitude of the jth charge, and r1, the dis
placement from q, to the point (1), we write Ez(x1,y1,21) = 1
5(1) = 2m ﬁle 1 (413)
7 0 11
Which means, of course,
1 q‘(x1  x)
E = ~ —————’——l’————— , 4. 4
3(xl’y1’zl) 2]: 41reo [(xl  xj)2 + (yi ‘ by + (21 “ 2:)213/2 ( 1 ) and so on. Often it is convenient to ignore the fact that charges come in packages like
electrons and protons, and think of them as being spread out in a continuous smear
——or in a “distribution,” as it is called. This is OK. so long as we are not interested
in what is happening on too small a scale. We describe a charge distribution by the “charge density,” p(x, y, 2). If the amount of charge in a small volume AVZ
located at the point (2) is Aqg, then p is deﬁned by 4‘12 = 9(2) AV2 (415) To use Coulomb’s law with such a description, we replace the sums of Eqs.
(4.13) or (4.14) by integrals over all volumes containing charges. Then we have 1 ] p(2)e12 de. 47750 r2
all 12
space E(l) = (4.16) Some people prefer to write where r12 is the vector displacement to (1) from (2), as shown in Fig. 4—1. The
integral for E is then written as 1 / p(2)r12 «in. 47760 r132 5(1) = (4.17) all space
When we want to calculate something with these integrals, we usually have to
write them out in explicit detail. For the xcomponent of either Eq. (4.16) or
(4.17), we would have (361 — x2)P(x2, y2, 22) d9‘2 dyz dzz WNW = f all
space (4.18) 4—3 1)s1X,Y.Z,l (2); (X2 Y: 22) Fig. 4—1. The electric field E at
point (l), from a charge distribution, is
obtained from an integral over the
distribution. Point (1) could also be inside
the distribution. Fig. 4—2. The work done in carrying
a charge from a to b is the negative of
the integral of F  d8 along the path
taken. We are not going to use this formula much. We write it here only to empha
size the fact that we have completely solved all the electrostatic problems in which
we know the locations of all of the charges. Given the charges, what are the ﬁelds ?
Answer: Do this integral. So there is nothing to the subject; it is just a case of
doing complicated integrals over three dimensions—strictly a job for a computing
machine! With our integrals we can ﬁnd the ﬁelds produced by a sheet of charge, from
a line of charge, from a spherical shell of charge, or from any speciﬁed distribution.
It is important to realize, as we go on to draw ﬁeld lines, to talk about potentials,
or to calculate divergences, that we already have the answer here. It is merely a
matter of it being sometimes easier to do an integral by some clever guesswork
than by actually carrying it out. The guesswork requires learning all kinds of
strange things. In practice, it might be easier to forget trying to be clever and al
ways to do the integral directly instead of being so smart. We are, however, going
to try to be smart about it. We shall go on to discuss some other features of the
electric ﬁeld. 43 Electric potential First we take up the idea of electric potential, which is related to the work done
in carrying a charge from one point to anOIher. There is some distribution of
charge, which produces an electric ﬁeld. We ask about how much work it would
take to carry a small charge from one place to another. The work done against
the electrical forces in carrying a charge along some path is the negative of the com
ponent of the electrical force in the direction of the motion, integrated along the
path. If we carry a charge from point a to point b, b
W=—/ F'ds, where F is the electrical force on the charge at each point, and ds is the differential
vector displacement along the path. (See Fig. 4—2.) It is more interesting for our purposes to consider the work that would be
done in carrying one unit of charge. Then the force on the charge is numerically
the same as the electric ﬁeld. Calling the work done against electrical forces in this
case W(unit), we write b
W(unit) = —/ E  ds. (4.19)
Now, in general, what we get with this kind of an integral depends on the path we
take. But if the integral of (4.19) depended on the path from a to b, we could get
work out of the ﬁeld by carrying the charge to b along one path and then back to a
on the other. We would go to b along the path for which W is smaller and back
along the other, getting out more work than we put in. There is nothing impossible, in principle, about getting energy out of a ﬁeld.
We shall, in fact, encounter ﬁelds where it is possible. It could be that as you move
a charge you produce forces on the other part of the “machinery.” If the “ma
chinery” moved against the force it would lose energy, thereby keeping the total
energy in the world constant. For electrostatics, however, there is no such “ma
chinery.” We know What the forces back on the sources of the ﬁeld are. They are
the Coulomb forces on the charges responsible for the ﬁeld. If the other charges
are ﬁxed in position—as we assume in electrostatics only—these back forces can
do no work on them. There is no way to get energy from them—provided, of
course, that the principle of energy conservation works for electrostatic situations.
We believe that it will work, but let’s just show that it must follow from Coulomb’s
law of force. We consider ﬁrst what happens in the ﬁeld due to a single charge 4. Let
point a be at the distance r1 from q, and point b at r2. Now we carry a different
charge, which we will call the “test” charge, and whose magnitude we choose to 44 be one unit, from a to b. Let’s start with the easiest possible path to calculate. We
carry our test charge ﬁrst along the arc of a circle, then along a radius, as shown in
part (a) of Fig. 4—3. Now on that particular path it is child’s play to ﬁnd the work
done (otherwise we wouldn’t have picked it). First, there is no work done at all
on the path from a to a'. The ﬁeld is radial (from Coulomb’s law), so it is at right
angles to the direction of motion. Next, on the path from a’ to b, the ﬁeld is in the
direction of motion and varies as l/r2. Thus the work done on the test charge
in carrying it from a to b would be b b
' _ q dr _ _q_ l _ 1
*ﬁEdS——4T€o a’ﬁ_—47r€0(i Now let’s take another easy path. For instance, the one shown in part (b) of
Fig. 4—3. It goes for awhile along an arc of a circle, then radially for awhile, then
along an are again, then radially, and so on. Every time we go along the circular
parts, we do no work. Every time we go along the radial parts, we must just
integrate l/r2. Along the ﬁrst radial stretch, we integrate from r, to run then
along the next radial stretch from rav to run, and so on. The sum of all these in
tegrals is the same as a single integral directly from r, to rb. We get the same answer
for this path that we did for the ﬁrst path we tried. It is clear that we would get
the same answer for any path which is made up of an arbitrary number of the same
kinds of pieces.
What about smooth paths? Would we get the same answer? We discussed
this point previously in Chapter 13 of Vol. 1. Applying the same arguments used
there, we can conclude that work done in carrying a unit charge from a to b is independent of the path.
. b
W(un1t)} = _ f E. a," a—+b (4.20) 3:31. ‘ Since the work done depends only on the endpoints, it can be represented as
the difference between two numbers. We can see this in the following way. Let’s
choose a reference point P0 and agree to evaluate our integral by using a path that
always goes by way of point P0. Let ¢(a) stand for the work done against the ﬁeld
in going from P0 to point a, and let ¢(b) be the work done in going from P0 to
point b (Fig. 4—4). The work in going to P0 from a (on the way to b) is the negative
of ¢(a), so we have that b
—/ Eds = ¢(b) — ¢(a). (4.21) Since only the difference in the function :1; at two points is ever involved, we
do not really have to specify the location of P0. Once we have chosen some
reference point, however, a number 45 is determined for any point in space; ¢ is
then a scalar ﬁeld. It is a function of x, y, 2. We call this scalar function the elec
trostatic potential at any point. Electrostatic potential: P
¢(P) = —/ Eds. (4.22) P o
For convenience, we will often take the reference point at inﬁnity. Then, for a single charge at the origin, the potential ¢ is given for any point (x, y, 2)—
using Eq. (4.20): 1
we. y. z) = 4—736—0 ; (4.23) The electric ﬁeld from several charges can be written as the sum of the electric
ﬁeld from the ﬁrst, from the second, from the third, etc. When we integrate the
sum to ﬁnd the potential we get a sum of integrals. Each of the integrals is the 4~S Fig. 43.
from a to b the same work is done along
either path. In carrying a test charge "(a  b)  Mb)  ﬂu) b vac ~ b) . Mb) "(2,  a)  to) P, Fig. 4—4. The work done in going
along any path from a to b is the negative
of the work from some point P0 to a plus
the work from P0 to b. potential from one of the charges. We conclude that the potential 45 from a lot of
charges is the sum of the potentials from all the individual charges. There is a
superposition principle also for potentials. Using the same kind of arguments by
which we found the electric ﬁeld from a group of charges and for a distribution of
charges, we can get the complete formulas for the potential ¢ at a point we call (1): _ 1 qr ¢(1)_ ‘ 4m ’3, (4.24)
_ I pawl/2, d>(l) — 41‘10/ ’12 (4.25) Remember that the potential 91. has a physical signiﬁcance: it is the potential
energy which a unit charge would have if brought to the speciﬁed point in space
from some reference point. 4—4 E =—V¢ Who cares about ¢? Forces on charges are given by E, the electric ﬁeld. The
point is that E can be obtained easily from ¢—it is as easy, in fact, as taking a
derivative. Consider two points, one at x and one at (x + dx), but both at the
same y and z, and ask how much work is done in carrying a unit charge from one
point to the other. The path is along the horizontal line from x to x + dx. The
work done is the diﬁ‘erence in the potential at the two points: AW = ¢(x + AX,J’,Z) "' ¢(x:ysz) = 3—:'Ax‘ But the work done against the ﬁeld for the same path is AW: —fEds = —E,Ax. We see that
 _ it.
E, — ax (4.26)
Similarly, Eu = —6¢/6y, E, = —a¢/az, or, summarizing with the notation of
vector analysis,
E = —V¢. (4.27) This equation is the differential form of Eq. (4.22). Any problem with speciﬁed
charges can be solved by computing the potential from (4.24) or (4.25) and using
(4.27) to get the ﬁeld. Equation (4.27) also agrees with what we found from vector
calculus: that for any scalar ﬁeld q} b
[1 WI: ' d: = Mb)  ¢(a) (428) According to Eq. (4.25) the scalar potential ()5 is given by a threedimensional
integral similar to the one we had for E. Is there any advantage to computing ¢
rather than E ? Yes. There is only one integral for 4:, while there are three integrals
for E—because it is a vector. Furthermore, l/r is usually a little easier to integrate
than x/ra. It turns out in many practical cases that it is easier to calculate 4: and
then take the gradient to ﬁnd the electric ﬁeld, than it is to evaluate the three
integrals for E. It is merely a practical matter. There is also a deeper physical signiﬁcance to the potential ¢. We have shown
that E of Coulomb’s law is obtained from E = — grad 4., when «t is given by
(4.22). But if E is equal to the gradient of a scalar ﬁeld, then we know from the
vector calculus that the curl of E must vanish: V X E = 0. (4.29) But that is just our second fundamental equation of electrostatics, Eq. (4.6). We
have shown that Coulomb’s law gives an E ﬁeld that staisﬁes that condition. So
far, everything is all right. We had really proved that V X E was zero before we deﬁned the potential.
We had shown that the work done around a closed path is zero. That is, that fEds=o for any path. We saw in Chapter 3 that for any such ﬁeld V X E must be zero
everywhere. The electric ﬁeld in electrostatics is an example of a curlfree ﬁeld. You can practice your vector calculus by proving that V X E is zero in a dif
ferent way—by computing the components of V X E for the ﬁeld of a point charge,
as given by Eq. (4.11). If you get zero, the superposition principle says you would
get zero for the ﬁeld of any charge distribution. We should point out an important fact. For any radial force the work done is
independent of the path, and there exists a potential. If you think about it, the
entire argument we made above to show that the work integral was independent
of the path depended only on the fact that the force from a single charge was
radial and spherically symmetric. It did not depend on the fact that the dependence
on distance was as l/r2—there could have been any r dependence. The existence
of a potential, and the fact that the curl of E is zero, comes really only from the
symmetry and direction of the electrostatic forces. Because of this, Eq. (4—28)——
or (4.29)—can contain only part of the laws of electricity. 4—5 The ﬂux of E We will now derive a ﬁeld equation that depends speciﬁcally and directly on
the fact that the force law is inverse square. That the ﬁeld varies inversely as the
square of the distance seems, for some people, to be “only natural,” because “that’s
the way things spread out.” Take a light source with light streaming out: the
amount of light that passes through a surface cut out by a cone with its apex at
the source is the same no matter at what radius the surface is placed. It must be so
if there is to be conservation of light energy. The amount of light per unit area—
the intensity—must vary inversely as the area cut by the cone, i.e., inversely as the
square of the distance from the source. Certainly the electric ﬁeld should vary
inversely as the square of the distance for the same reason! But there is no such
thing as the “same reason” here. Nobody can say that the electric ﬁeld measures
the ﬂow of something like light which must be conserved. If we had a “model”
of the electric ﬁeld in which the electric ﬁeld vector represented the direction and
speed—say the current—of some kind of little “bullets” which were ﬂying out,
and if our model required that these bullets were conserved, that none could ever
disappear once it was shot out of a charge, then we might say that we can “see”
that the inverse square law is necessary. On the other hand, there would necessarily
be some mathematical way to express this physical idea. If the electric ﬁeld were
like conserved bullets going out, then it would vary inversely as the square of the
distance and we would be able to describe that behavior by an equation—which
is purely mathematical. Now there is no harm in thinking this way, so long as we
do not say that the electric ﬁeld is made out of bullets, but realize that we are
using a model to help us ﬁnd the right mathematics. Suppose, indeed, that we imagine for a moment that the electric ﬁeld did
represent the ﬂow of something that was conserved—everywhere, that is, except
at charges. (It has to start somewhere!) We imagine that whatever it is ﬂows out
of a charge into the space around. If E were the vector of such a ﬂow (as h is for
heat ﬂow), it would have a l /r2 dependence near a point source. Now we wish to
use this model to ﬁnd out how to state the inverse square law in a deeper or more
abstract way, rather than simply saying “inverse square.” (You may wonder
why we should want to avoid the direct statement of such a simple law, and want
instead to imply the same thing sneakily in a different way. Patience! It will turn
out to be useful.) 4—7 Point Charge surface S is zero. Fig. 4—7. Any volume can be thought
of as completely made up of inﬁnitesimal
truncated cones. The flux of E from one
end of each conical segment is eqqu and
opposite to the flux from the other end.
The total flux from the surface S is
therefore zero. b
q
5‘" a
Fig. 48. If a charge is inside a surface, the ﬂux out is not zero. Fig. 4—5. The ﬂux of E out of the 3/ //’ Fig. 4—6. The ﬂux of 5 out of the Point Charge surface S is zero. We ask: What is the “ﬂow” of E out of an arbitrary closed surface in the
neighborhood of a point charge? First let’s take an easy surface—the one shown
in Fig. 4—5. If the E ﬁeld is like a ﬂow, the net ﬂow out of this box should be zero.
That is what we get if by the “ﬂow” from this surface we mean the surface integral
of the normal component of E—that is, the ﬂux of E. On the radial faces, the nor
mal component is zero. On the spherical faces, the normal component En is just
the magnitude of E—minus for the smaller face and plus for the larger face. The
magnitude of E decreases as Mr”, but the surface area is proponional to r2, so
the product is independent of r. The ﬂux of E into face a is just cancelled by the
ﬂux out of face b. The total ﬂow out of S is zero, which is to say that for this
surface [S E. da = o. (4.30) Next we show that the two end surfaces may be tilted with respect to the
radial line without changing the integral (4.30). Although it is true in general, for
our purposes it is only necessary to show that this is true when the end surfaces are
small, so that they subtend a small angle from the source—in fact, an inﬁnitesimal
angle. In Fig. 4—6 we show a surface S whose “sides” are radial, but whose “ends”
are tilted. The end surfaces are not small in the ﬁgure, but you are to imagine the
situation for very small end surfaces. Then the ﬁeld E will be sufﬁciently uniform
over the surface that we can use just its value at the center. When we tilt the sur
face by an angle 0, the area is increased by the factor l/cos 0. But E... the compo
nent of E normal to the surface, is decreased by the factor cos 9. The product
En M is unchanged. The ﬂux out of the whole surface S is still zero. Now it is easy to see that the ﬂux out of a volume enclosed by any surface S
must be zero. Any volume can be thought of as made up of pieces, like that in
Fig. 4—6. The surface will be subdivided completely into pairs of end surfaces,
and since the ﬂuxes in and out of these end surfaces cancel by pairs, the total ﬂux
out of the surface will be zero. The idea is illustrated in Fig. 4P7. We have the
completely general result that the total ﬂux of E out of any surface S in the ﬁeld
of a point charge is zero. But notice! Our proof works only if the surface S does not surround the charge.
What would happen if the point charge were inside the surface? We could still
divide our surface into pairs of areas that are matched by radial lines through the
charge, as shown in Fig. 4—8. The ﬂuxes through the two surfaces are still equal—
by the same arguments as before—only now they have the same sign. The ﬂux
out of a surface that surrounds a charge is not zero. Then what is it? We can ﬁnd
out by a little trick. Suppose we “remove” the charge from the “inside” by sur
rounding the charge by a little surface S’ totally inside the original surface S, as
shown in Fig. 4—9. Now the volume enclosed between the two surfaces S and S’
has no charge in it. The total ﬂux out of this volume (including that through S’)
is zero, by the arguments we have given above. The arguments tell us, in fact, that
the ﬂux into the volume through S’ is the same as the ﬂux outward through S. 48 We can choose any shape we wish for S’, so let’s make it a sphere centered on
the charge, as in Fig. 410. Then we can easily calculate the ﬂux through it. If the
radius of the little sphere is r, the value of E everywhere on its surface is 1 .4.
47reo r2 ’ and is directed always normal to the surface. We ﬁnd the total ﬂux through 5’ if
we multiply this normal component of E by the surface area: 1 q 2 q
’ _ ———— 1r — )
Flux through the suface S < 60 r2) (4 r ) 60 (4.31) a number independent of the radius of the sphere! We know then that the ﬂux
outward through S is also q/eo——a value independent of the shape of S so long as
the charge q is inside. We can write our conclusions as follows: 0; q outside S
f E,, da = q (4.32) —— ; q inside S any surface S 60 Let’s return to our “bullet” analogy and see if it makes sense. Our theorem
says that the net ﬂow of bullets through a surface is zero if the surface does not
enclose the gun that shoots the bullets. If the gun is enclosed in a surface, whatever
size and shape it is, the number of bullets passing through is the same—it is given
by the rate at which bullets are generated at the gun. It all seems quite reasonable
for conserved bullets. But does the model tell us anything more than we get
simply by writing Eq. (4.32)? No one has succeeded in making these “bullets” do
anything else but produce this one law. After that, they produce nothing but
errors. That is why today we prefer to represent the electromagnetic ﬁeld purely
abstractly. 4—6 Gauss’ law; the divergence of E Our nice result, Eq. (4.32), was proved for a single point charge. Now suppose
that there are two charges, a charge q] at one point and a charge qg at another.
The problem looks more diﬂicult. The electric ﬁeld whose normal component we
integrate for the ﬂux is the ﬁeld due to both charges. That is, if E1 represents the
electric ﬁeld that would have been produced by ql alone, and E2 represents the
electric ﬁeld produced by q2 alone, the total electric ﬁeld is E = E1 + £2. The
ﬂux through any closed surface S is [S (Em + E2.)da = [S Elnda + [S 32mg. (4.33) The ﬂux with both charges present is the ﬂux due to a single charge plus the ﬂux
due to the other charge. If both charges are outside S, the ﬂux through S is zero.
If ql is inside S but q 2 is outside, then the ﬁrst integral gives ql/eo and the second
integral gives zero. If the surface encloses both charges, each will give its contribu
tion and we have that the ﬂux is (q 1 + q 2)/e0. The general rule is clearly that the
total ﬂux out of a closed surface is equal to the total charge inside, divided by 60. Our result is an important general law of the electrostatic ﬁeld, called Gauss’
law. G ’1 . ‘ '
wuss 0W / En da = W, (4.34)
0
23¥r§i35°sd
01'
/ Enda = QED" : (4'35)
0
h “$5132”?
W ere
Qua = 2 qt (4'36) inside 3 4—9 Surface
8’ Fig. 49. The ﬂux through 5 is the
same as the ﬂux through 5’. Fig. 4—10. The ﬂux through a spheri
cal surface containing a point charge q is q/Eo. Chem
Distrlpbution\ Surface 5
\ \ /
Fig. 4—11. Using Gouss' law to ﬁnd the ﬁeld of a uniform sphere of charge. If we describe the location of charges in terms of a charge density p, we can con
sider that each inﬁnitesimal volume dV contains a “point” charge p dV. The sum
over all charges is then the integral Qm = (437) / pdV. volume
inside 5
From our derivation you see that Gauss’ law follows from the fact that the
exponent in Coulomb’s law is exactly two. A l/r3 ﬁeld, or any l/r" ﬁeld with
n ¢ 2, would not give Gauss’ law. So Gauss’ law is just an expression, in a dif
ferent form, of the Coulomb law of forces between two charges. In fact, working
back from Gauss’ law, you can derive Coulomb’s law. The two are quite equiva
lent so long as we keep in mind the rule that the forces between charges is radial.
We would now like to write Gauss’ law in terms of derivatives. To do this,
we apply Gauss’ law to an inﬁnitesimal cubical surface. We showed in Chapter 3
that the ﬂux of E out of such a cube is V  E times the volume (IV of the cube. The
charge inside of dV, by the deﬁnition of p, is equal to p dV, so Gauss’ law gives v EdV = “1V,
60
Of
vE = 3—. (4.38)
50 The differential form of Gauss’ law is the ﬁrst of our fundamental ﬁeld equations of
electrostatics, Eq. (4.5). We have now shown that the two equations of electro
statics, Eqs. (4.5) and (4.6), are equivalent to Coulomb’s law of force. We will
now consider one example of the use of Gauss’ law. (We will come later to many
more examples.) 4—7 Field of a sphere of charge One of the difﬁcult problems we had when we studied the theory of gravita
tional attractions was to prove that the force produced by a solid sphere of matter
was the same at the surface of the sphere as it would be if all the matter were
concentrated at the center. For many years Newton didn’t make public his
theory of gravitation, because he couldn’t be sure this theorem was true. We
proved the theorem in Chapter 13 of Vol. I by doing the integral for the
potential and then ﬁnding the gravitational force by using the gradient. Now we
can prove the theorem in a most simple fashion. Only this time we will prove the
corresponding theorem for a uniform sphere of electrical charge. (Since the laws
of electrostatics are the same as those of gravitation, the same proof could be
done for the gravitational ﬁeld.) We ask: What is the electric ﬁeld E at a point P anywhere outside the surface
of a sphere ﬁlled with a uniform distribution of charge? Since there is no “special”
direction, we can assume that E is everywhere directed away from the center of the
sphere. We consider an imaginary surface that is spherical and concentric with
the sphere of charge, and that passes through the point P (Fig. 4—11). For this
surface, the flux outward is [En da = E41rR2. Gauss’ law tells us that this ﬂux is equal to the total charge Q of the sphere (over so): E47rR2 = 9,
60
01‘
_ 1 Q
E _ 4m 5, (4.39) 4—10 I Fig. 442. Field lines and equipotential surfaces for a positive point charge. which is the same formula we would have for a point charge Q. We have proved
Newton’s problem more easily than by doing the integral. It is, of course, a false
kind of easiness—it has taken you some time to be able to understand Gauss’ law,
so you may think that no time has'really been saved. But after you have used the
theorem more and more, it begins to pay. It is a question of eﬁiciency. 4—8 Field lines; equipotential surfaces We would like now to give a geometrical description of the electrostatic ﬁeld.
The two laws of electrostatics, one that the ﬂux is proportional to the charge inside
and the other that the electric ﬁeld is the gradient of a potential, can also be repre
sented geometrically. We illustrate this with two examples. First, we take the ﬁeld of a point charge. We draw lines in the direction of the
ﬁeld—lines which are always tangent to the ﬁeld, as in Fig. 4—12. These are called
ﬁeld lines. The lines show everywhere the direction of the electric vector. But we
also wish to represent the magnitude of the vector. We can make the rule that the
strength of the electric ﬁeld will be represented by the “density” of the lines. By
the density of the lines we mean the number of lines per unit area through a sur
face perpendicular to the lines. With these two rules we can have a picture of the
electric ﬁeld. For a point charge, the density of the lines must decrease as 1/r2.
But the area of a spherical surface perpendicular to the lines at any radius r increases
as r2, so if we always keep the same number of lines for all distances from the
charge, the density will remain in proportion to the magnitude of the ﬁeld. We can
guarantee that there are the same number of lines at every distance if we insist
that the lines be continuous—that once a line is started from the charge, it never
stops. In terms of the ﬁeld lines, Gauss’ law says that lines should start only at
plus charges and stop at minus charges. The number which leave a charge q must
be equal to q/eo. Now, we can ﬁnd a similar geometrical picture for the potential 115. The easiest
way to represent the potential is to draw surfaces on which :1: is a constant. We call
them equipotential surfaces—surfaces of equal potential. Now what is the geometri 4—11 A Note about Units Quantity Unit
F newton
Q coulomb
L meter
W joule
p ~ Q/L3 coulomb/meter3
l/eo ~ FL2/Q2 newtonmeter2/coulomb2
E ~ F/ Q newton/coulomb
4» ~ W/ Q joule/coulomb = volt
E ~ ¢/L volt/ meter l/eo ~ EL2/ Q voltmeter/coulomb Fig. 413. Field lines and equipotentials for two equal and opposite point charges. cal relationship of the equipotential surfaces to the ﬁeld lines? The electric ﬁeld is
the gradient of the potential. The gradient is in the direction of the most rapid
change of the potential, and is therefore perpendicular to an equipotential surface.
If E were not perpendicular to the surface, it would have a component in the
surface. The potential would be changing in the surface, but then it wouldn’t be
an equipotential. The equipotential surfaces must then be everywhere at right
angles to the electric ﬁeld lines. For a point charge all by itself, the equipotential surfaces are spheres centered
at the charge. We have shown in Fig. 4—12 the intersection of these spheres with a
plane through the charge. As a second example, we consider the ﬁeld near two equal charges, a positive
one and a negative one. To get the ﬁeld is easy. The ﬁeld is the superposition of
the ﬁelds from each of the two charges. So, we can take two pictures like Fig. 4—12
and superimpose them—impossible! Then we would have ﬁeld lines crossing each
other, and that’s not possible, because E can’t have two directions at the same point.
The disadvantage of the ﬁeldline picture is now evident. By geometrical argu
ments it is impossible to analyze in a very simple way where the new lines go.
From the two independent pictures, we can’t get the combined picture. The
principle of superposition, a simple and deep principle about electric ﬁelds, does
not have, in the ﬁeldline picture, an easy representation. The ﬁeldline picture has its uses, however, so we might still like to draw the
picture for a pair of equal (and opposite) charges. If we calculate the ﬁelds from
Eq. (4.13) and the potentials from (4.23), we can draw the ﬁeld lines and equi
potentials. Figure 4—13 shows the result. But we ﬁrst had to solve the problem
mathematically! 4—12 ...
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 Spring '09
 LeeKinohara
 Physics, Electrostatics

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