This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 6 The Electric Field in Various Circumstances 6—1 Equations of the electrostatic potential This chapter will describe the behavror of the electric ﬁeld in a number of
different circumstances. It Will prov1de some experience With the way the electric
ﬁeld behaves, and will describe some of the mathematical methods which are
used to ﬁnd this ﬁeld. We begin by pointing out that the Whole mathematical problem is the solution
of two equations, the Maxwell equations for electrostatics: v~E = L (6.1)
60
v x E = 0. (6.2) In fact, the two can be combined into a Single equation. From the second equation,
we know at once that we can describe the ﬁeld as the gradient of a scalar (see Section 3—7):
E = — W). (6.3) We may, if we wish, completely describe any particular electric ﬁeld in terms
of its potential <13. We obtain the differential equation that dz must obey by sub
stituting Eq. (6.3) into (6.1), to get v V¢— _ —i’—. (6.4) V'V¢=V¢=~i (6.5) so we write Eq. (6.4) as (6.6) The operator V2 is called the LaplaCian, and Eq (6 6) is called the Porsson equa
tion. The entire subject of electrostatics, from a mathematical pomt of view, is
merely a study of the solutions of the Single equation (6.6). Once 451s obtained by
solvmg Eq. (6.6) we can find E immediately from Eq. (6.3). We take up ﬁrst the speCial class of problems in which p is given as a function
of x, y, z. In that case the problem IS almost trivial, for we already know the
solution of Eq. (6.6) for the general case. We have shown that if p is known at
every point, the potential at pornt (l) is _ (2) (IV
M1) ‘ / 47mm: where p(2) is the charge den51ty, (1V2 is the volume element at pomt (2), and r12
is the distance between points (I) and (2). The solution of the diﬂerentiul equation
(6.6) is reduced to an integration over space. The solution (6.7) should be especially
noted, because there are many situations in physics that lead to equations like (6.7) V2 (something) = (something else), and Eq. (6.7) is a prototype of the solution for any of these problems. The solution of electrostatic ﬁeld problems is thus completely straightforward
when the p05itions of all the charges are known. Let’s see how it works in a few
examples. 6—1 6—1 6—2
6—3
6—4 6—5 66 6—7
6—8 6—9 Equations of the electrostatic
potential The electric dipole
Remarks on vector equations The dipole potential as a
gradient The dipole approximation for
an arbitrary distribution The ﬁelds of charged
conductors The method of images A point charge near a
conducting plane A point charge near a
conducting sphere 6—10 Condensers; parallel plates 6—11 Highvoltage breakdown 6—12 The ﬁeld emission microscope Revzew. Chapter 23, Vol. I, Resonance Fig. 6—1. A dipole: two charges
+q and —q the distance d apart. +0 ¢+ Fig. 6—2. The water molecule H20.
The hydrogen atoms have slightly less
than their share of the electron cloud; the
oxygen, slightly more. 6—2 The electric dipole First, take two point charges, +q and —q, separated by the distance d. Let
the zaxis go through the charges, and pick the origin halfway between, as shown
in Fig. 6—1. Then, using (4.24), the potential from the two charges is given by ¢(x.y. 2)
l q —q
_ 47reo lx/[z — (ti/2)]2 + x2 + y2 + [z + (d/2)]2 + x2 + y2]' (68) We are not going to write out the formula for the electric ﬁeld, but we can always
calculate it once we have the potential. So we have solved the problem of two
charges. There is an important special case in which the two charges are very close
together—which is to say that we are interested in the ﬁelds only at distances from
the charges large in comparison with their separation. We call such a close pair
of charges a dipole. Dipoles are very common. A “dipole” antenna can often be approximated by two charges separated by a
small distance—if we don’t ask about the ﬁeld too close to the antenna. (We are
usually interested in antennas with moving charges; then the equations of statics
do not really apply, but for some purposes they are an adequate approximation.) More important perhaps, are atomic dipoles. If there is an electric ﬁeld in
any material, the electrons and protons feel opposite forces and are displaced
relative to each other. In a conductor, you remember, some of the electrons
move to the surfaces, so that the ﬁeld inside becomes zero. In an insulator the
electrons cannot move very far; they are pulled back by the attraction of the nu
cleus. They do, however, shift a little bit. So although an atom, or molecule,
remains neutral in an external electric ﬁeld, there is a very tiny separation of its
positive and negative charges and it becomes a microscopic dipole. If we are
interested in the ﬁelds of these atomic dipoles in the neighborhood of ordinary
sized objects, we are normally dealing with distances large compared with the
separations of the pairs of charges. In some molecules the charges are somewhat separated even in the absence
of external ﬁelds, because of the form of the molecule. In a water molecule, for
example, there is a net negative charge on the oxygen atom and a net positive
charge on each of the two hydrogen atoms, which are not placed symmetrically
but as in Fig. 6—2. Although the charge of the whole molecule is zero, there is a
charge distribution with a little more negative charge on one side and a little
more positive charge on the other. This arrangement is certainly not as simple
as two point charges, but when seen from far away the system acts like a dipole.
As we shall see a little later, the ﬁeld at large distances is not sensitive to the
ﬁne details. Let’s look, then, at the ﬁeld of two opposite charges with a small separation
d. If d becomes zero, the two charges are on top of each other, the two potentials
cancel, and there is no ﬁeld. But if they are not exactly on top of each other, we
can get a good approximation to the potential by expanding the terms of (6.8) in
a power series in the small quantity d (using the binomial expansion). Keeping
terms only to ﬁrst order in d, we can write (—‘2—1)2~22—zd. x2+y2+zz=r2. 2
(2—521)+x2+y2zr2—zd=r2(l—E(—1)s It is convenient to write Then and 1 z 1 z 1 (1 — 2:1)—1I2.
[2 — (av/2)? + x2 + y2 r2i1 — (zd/r2)i r '2 6—2 Using the binomial expansion again for [1 — (zd/r2)]_“2—and throwing away
terms with higher powers than the square of d—~we get 1 lzd
70%;?) ____1___~1(1_ 12d).
[2+ (d/2)]2 + x2 + y2 r 272 The difference of these two terms gives for the potential Similarly, —— qd. (6.9) ¢>(x, y, 2) = The potential, and hence the ﬁeld, which is its derivative, is proportional to qd,
the product of the charge and the separation. This product is deﬁned as the
dipole moment of the two charges, for which we W111 use the symbol p (do not
confuse with momentum!): p = qd. (6.10)
Equation (6.9) can also be written as l pcosO
47reo r2 d>(x, y, Z) = ’ (611)
since z/r = cos 0, where 0 is the angle between the axis of the d1pole and the
radius vector to the point (x, y, z}——see F ig. 6—1. The potential of a dipole decreases
as l/r2 for a given direction from the axis (whereas for a point charge it goes as
l/r). The electric ﬁeld E of the dipole wrll then decrease as 1/r3. We can put our formula into a vector form if we deﬁne p as a vector whose
magnitude is p and whose direction is along the axis of the dipole, pointing from
q_ toward q+. Then cos 0 = per, (6.12) where e, rs the unit radial vector (Fig. 6—3). We can also represent the point
(x, y, z) by r. Then D' o] t t' l:  ‘
1p epoenla (W) = 1 p er: “Lu. (6.13) This formula 1s valld for a dipole with any orientation and position if r represents
the vector from the dipole to the point of interest. If we want the electric ﬁeld of the dipole we can get it by taking the gradient
of ¢. For example, the zcomponent of the ﬁeld 1s —6¢/az. For a dipole oriented
along the zaxis we can use (6.9): _§2__P33__P 122
62 ' 41reo 62 r3 — 41reo r3 r5 5
or 3 2o 1
_ p cos — _
E2 — 4760 ————————r3 (6.14) The x and ycomponents are p 32x p 3zy_ __ , E = _,_ ._
471'60 r5 1’ 41reo r5 Ex: These two can be combined to give one component directed perpendicular to the
zaxis, which we will call the transverse component E1: E1=VE3+E2= p 3~2«/x2+y2 1/ 47r60 r5 or 3 o ‘ 0
E; p cos s1n . — 411'60 r3 (6.15) 6—3 p Fig. 6—3. Vector
dipole. notation for 0 Fig. 6—4.
dipole. The electric ﬁeld of a g. The transverse component E i is in the xy plane and points directly away from
the axis of the dipole. The total ﬁeld, of course, is E: x/E§+Ei. The dipole ﬁeld varies inversely as the cube of the distance from the dipole.
On the axis, at 6 = 0, it is tWice as strong as at 0 = 90°. At both of these special
angles the electric ﬁeld has only a z—component, but of opposite sign at the two
places (Fig. 6—4). 6—3 Remarks on vector equations This is a good place to make a general remark about vector analysis. The
fundamental proofs can be expressed by elegant equations in a general form, but
in making various calculations and analyses it is always a good idea to choose
the axes in some convenient way. Notice that when we were ﬁnding the potential
of a dipole we chose the z—axis along the direction of the dipole, rather than at some
arbitrary angle. This made the work much eaSier. But then we wrote the equations
in vector form so that they would no longer depend on any particular coordinate
system. After that, we are allowed to choose any coordinate system we wish,
knowing that the relation is, in general, true. It clearly doesn’t make any sense to
bother with an arbitrary coordinate system at some complicated angle when you
can choose a neat system for the particular problem—provided that the result can
ﬁnally be expressed as a vector equation. So by all means take advantage of the
fact that vector equations are independent of any coordinate system. On the other hand, if you are trying to calculate the divergence of a vector,
instead of just looking at V  E and wondering what it is, don’t forget that it can always be spread out as
6E, 6E, 6E,
6x W + a: ‘ If you can then work out the x, y, and zcomponents of the electric ﬁeld and
differentiate them, you will have the divergence. There often seems to be a feeling
that there is something inelegant—some kind of defeat involved—in writing out
the components; that somehow there ought always to be a way to do everything
with the vector operators. There is often no advantage to it. The ﬁrst time we
encounter a particular kind of problem, it usually helps to write out the components
to be sure we understand what is going on. There is nothing inelegant about put
ting numbers into equations, and nothing inelegant about substituting the deriva
tives for the fancy symbols. In fact, there is often a certain cleverness in doing
just that. Of course when you publish a paper in a professional journal it will look
better—and be more easily understood—if you can write everything in vector form.
Besides, it saves print. 6—4 The dipole potential as a gradient We would like to point out a rather amusing thing about the dipole formula,
Eq. (6.13). The potential can also be written as (,2 1 1,.V<l). (6.16) 47reo r If you calculate the gradient of l/r, you get and Eq. (6.16) is the same as Eq. (6.13). How did we think of that? We Just remembered that e./r2 appeared in the
formula for the ﬁeld of a point charge, and that the ﬁeld was the gradient of a
potential which has a 1/r dependence. 6—4 There is a physical reason for being able to write the dipole potential in the
form of Eq. (6.16). Suppose we have a point charge q at the origin. The potential
at the point P at (x, y, z) is _ 9..
¢0 — r
(Let’s leave off the l/41reo while we make these arguments; we can stick it in at
the end.) Now if we move the charge +q up a distance Az, the potential at P will
change a little, by, say, A¢+. How much is A44? Well, it is just the amount that the potential would change if we were to leave the charge at the origin and move
P downward by the same distance Az (Fig. 6—5). That is, Aqﬁ = —%A2, where by A2 we mean the same as (1/2. So, using ¢ = q/r, we have that the po—
tential from the positive charge is a d
45+ = g ‘ age) 5' {617)
Applying the same reasoning for the potential from the negative charge, we can write
_ :1 i :1 é . ¢—— r+az<r>2 (6.18) The total potential is the sum of (6.17) and (6.18): 6 ¢ = ¢+ + ¢_ = ‘a (13).; (6.19) ll 6 1
—6—2 (7) 4d For other orientation of the dipole, we could represent the displacement of
the positive charge by the vector Ar+. We should then write Eq. (6.17) as A¢+ = _V¢0 'Ar+, where Ar is then to be replaced by 11/2. Completing the derivation as before,
Eq. (6.19) would then become
1 . This is the same as Eq. (6.16), if we replace qd = p, and put back the 1/471'60.
Looking at it another way, we see that the dipole potential, Eq. (6.13), can be
interpreted as 4; = —pVcI>0, (6.20) where (£0 = l/47reor is the potential of a unit point charge. Although we can always ﬁnd the potential of a known charge distribution by
an integration, it is sometimes possible to save time by getting the answer with a
clever trick. For example, one can often make use of the superposition principle.
If we are given a charge distribution that can be made up of the sum of two dis
tributions for which the potentials are already known, it is easy to ﬁnd the de
sired potential by just adding the two known ones. One example of this is our
derivation of (6.20), another is the following. Suppose we have a spherical surface with a distribution of surface charge
that varies as the cosine of the polar angle. The integration for this distribution is
fairly messy. But, surprisrngly, such a distribution can be analyzed by super
position. For imagine a sphere with a uniform volume density of positive charge,
and another sphere with an equal uniform volume density of negative charge, 6—5 Fig. 6—5. The potential at P from a
point charge at A2 above the origin is the
same as the potential at P’(Az below P)
from the same charge at the origin. Fig. 6—6. Two uniformly charged spheres, superposed w ment, are equivalent to a nonuniform — — —— _. _. _. _ _. distribution of surface Fig. 6—7. Computation of the po
tential at a point P at a large distance from a set of charges. ith a slight displace charge. (0) 'l' (b) = (c) originally superposed to make a neutral—that is, uncharged—sphere. If the
positive sphere is then displaced slightly with respect to the negative sphere, the
body of the uncharged sphere would remain neutral, but a little positive charge will
appear on one side, and some negative charg. Wlll appear on the opposite side,
as illustrated in Fig. 6—6. If the relative displacement of the two spheres is small,
the net charge is equivalent to a surface charge (on a spherical surface), and the
surface charge density will be proportional to the cosme of the polar angle. Now if we want the potential from this distribution, we do not need to do an
integral. We know that the potential from each of the spheres of charge is—for
points outside the sphere—the same as from a point charge. The two displaced
spheres are like two point charges; the potential is just that of a dipole. In this way you can show that a charge distribution on a sphere of radius a
with a surface charge density a = 0'0 cos 9
produces a ﬁeld outside the sphere which is just that of a dipole whose moment is 47r00a3
p = 3 u It can also be shown that inside the sphere the ﬁeld is constant, with the value 0'0
E = _ .
360
If 0 is the angle from the positive zaxis, the electric ﬁeld inside the sphere is in the
negative z—direction. The example we have just conSIdered is not as artiﬁcial as it may appear; we will encounter it again in the theory of dielectrics. 6—5 The dipole approximation for an arbitrary distribution The dipole ﬁeld appears in another circumstance both interesting and im
portant. Suppose that we have an ObjCCt that has a complicated distribution of
charge—like the water molecule (Fig. 6—2)—and we are interested only in the
ﬁelds far away. We will show that it is possible to ﬁnd a relatively simple expression
for the ﬁelds which is appropriate for distances large compared with the size of
the object. We can think of our ObJCCt as an assembly of point charges qz in a certain limited
region, as shown in Fig. 6—7. (We can, later, replace q, by p dV if we wish.) Let
each charge q, be located at the displacement d, from an origin chosen somewhere 6—6 in the middle of the group of charges. What is the potential at the point P, located
at R, where R is much larger than the maximum 11,? The potential from the
whole collection is given by _ 1 ﬂ
¢ _ 4m 1 n, (6.21) where r, is the distance from P to the charge q, (the length of the vector R ~ d,).
Now if the distance from the charges to P, the point of observation, is enormous,
each of the r,’s can be approximated by R. Each term becomes q,/R, and we
can take l/R out as a factor in front of the summation. This gives us the simple
result _ 1 l _ Q
¢ — 47TEO R 2 qt — 47r€0R5 (622) where Q is just the total charge of the whole object. Thus we ﬁnd that for points
far enough from any lump of charge, the lump looks like a point charge. The
result is not too surprising. But what if there are equal numbers of positive and negative charges? Then
the total charge Q of the object is zero. This is not an unusual case; in fact, as we
know, objects are usually neutral. The water molecule is neutral, but the charges
are not all at one point, so if we are close enough we should be able to see some
eﬁects of the separate charges. We need a better approximation than (6.22) for
the potential from an arbitrary distribution of charge in a neutral object. Equation
(6.21) is still precrse, but we can no longer just set r, = R. We need a more accu
rate expression for r,. If the point P is at a large distance, r, will differ from R to
an excellent approximation by the projection of d on R, as can be seen from
Fig. 67. (You should imagine that P is really farther away than is shown in the
ﬁgure.) In other words, if e, is the unit vector in the direction of R, then our next
approximation to r, is r, z R ~ 11, e,. (6.23)
What we really want is l /r,, which, since d, << R, can be written to our approxima
tion as 1 1 d
~ _ ‘L ' er .
71 ~ R (1 + R ) (6.24) Substituting this in (6.21), we get that the potential is _ 1 Q dye, )
¢‘4ﬁ(§+;q‘ R2 +.... (6.25) The three dots indicate the terms of higher order in d/R that we have neglected.
These, as well as the ones we have already obtained, are successive terms in a Taylor
expansnon of Ur, about UK in powers of d1/R. The ﬁrst term in (6.25) is what we got before; it drops out if the object is
neutral. The second term depends on l/RZ, just as for a dipole. In fact, if we deﬁne P = 2 qt d1 (626)
as a property of the charge distribution, the second term of the potential (6.25) is __ 1 per
(bu—47760 R2 , (6.27) precisely a dipole potential. The quantity p is called the dipole moment of the
distribution. It is a generalization of our earlier deﬁnition, and reduces to it for
the special case of two point charges. Our result is that, far enough away from any mess of charges that 15 as a
whole neutral, the potential is a dipole potential. It decreases as l/R2 and varies
as cos B—and its strength depends on the dipole moment of the distribution of
charge. It is for these reasons that dipole ﬁelds are important, since the Simple
case of a pair of pomt charges is quite rare. 6—7 Fig. 6—8. The field lines and equipo
tentials for two point charges. CON DUCTOR / Fig. 6—9. The ﬁeld outside a con
ductor shaped like the equipotential A
of Fig. 6—8. The water molecule, for example, has a rather strong dipole moment. The
electric ﬁelds that result from this moment are responsible for some of the im
portant properties of water. For many molecules, for example C02, the dipole
moment vanishes because of the symmetry of the molecule. For them we should
expand still more accurately, obtai...
View
Full Document
 Spring '09
 LeeKinohara
 Physics

Click to edit the document details