Classical Electrodynamics - Jackson (Solucionario 1)

Solutions to Problems in Jackson,

Classical Electrodynamics, Third Edition

Homer Reid

December 8, 1999

Chapter 2

Problem 2.1

A point charge q is brought to a position a distance d away from aninfinite plane conductor held at zero potential. Using the methodof images, find:

(a) the surface-charge density induced on the plane, and plot it;

(b) the force between the plane and the charge by using Coulombslaw for the force between the charge and its image;

(c) the total force acting on the plane by integrating 2/20 overthe whole plane;

(d) the work necessary to remove the charge q from its position toinfinity;

(e) the potential energy between the charge q and its image (com-pare the answer to part d and discuss).

(f) Find the answer to part d in electron volts for an electronoriginally one angstrom from the surface.

(a) Well take d to be in the z direction, so the charge q is at (x, y, z) = (0, 0, d).The image charge is q at (0, 0,d). The potential at a point r is

(r) =q

4pi0

[1

|r dk| 1

|r + dk|]

The surface charge induced on the plane is found by differentiating this:

1

Homer Reids Solutions to Jackson Problems: Chapter 2 2

= 0 ddz

z=0

= q4pi

[(z d)|r + dk|3 +

(z + d)

|r + dk|3]

z=0

= qd2pi(x2 + y2 + d2)3/2

(1)

We can check this by integrating this over the entire xy plane and verifyingthat the total charge is just the value q of the image charge:

(x, y)dxdy = qd2pi

0

2pi0

rddr

(r2 + d2)3/2

= qd

0

rdr

(r2 + d2)3/2

= qd2

d2u3/2du

= qd2

2u1/2d2

= q

(b) The point of this problem is that, for points above the z axis, it doesntmatter whether there is a charge q at (0, 0, d) or an infinite grounded sheetat z = 0. Physics above the z axis is exactly the same whether we have thecharge or the sheet. In particular, the force on the original charge is the samewhether we have the charge or the sheet. That means that, if we assume thesheet is present instead of the charge, it will feel a reaction force equal to whatthe image charge would feel if it were present instead of the sheet. The forceon the image charge would be just F = q2/16pi0d

2, so this must be what thesheet feels.

(c) Total force on sheet

=1

20

0

2pi0

2dA

=q2d2

4pi0

0

rdr

(r2 + d2)3

=q2d2

8pi0

d2u3du

=q2d2

8pi0

12u2

d2

=q2d2

8pi0

[1

2d4

]


=q2

16pi0d2

in accordance with the discussion and result of part b.

(d) Work required to remove charge to infinity

=q2

4pi0

d

dz

(z + d)2

=q2

4pi0

2d

u2du

=q2

4pi0

1

2d

=q2

8pi0d

(e) Potential energy between charge and its image

=q2

8pi0d

equal to the result in part d.

(f)

q2

8pi0d=

(1.6 1019 coulombs )28pi(8.85 1012 coulombsV1m1)(1010 m )

= 7.2 (1.6 1019 coulombs 1 V )= 7.2 eV .

Problem 2.2

Using the method of images, discuss the problem of a point chargeq inside a hollow, grounded, conducting sphere of inner radius a.Find

(a) the potential inside the sphere;

(b) the induced surface-charge density;

(c) the magnitude and direction of the force acting on q.

(d) Is there any change in the solution if the sphere is kept at afixed potential V ? If the sphere has a total charge Q on itsinner and outer surfaces?


Problem 2.3

A straight-line charge with constant linear charge density is located perpendicularto the x y plane in the first quadrant at (x0, y0). The intersecting planes x =0, y 0 and y = 0, x 0 are conducting boundary surfaces held at zero potential.Consider the potential, fields, and surface charges in the first quadrant.

(a) The well-known potential for an isolated line charge at (x0, y0) is (x, y) =(/4pi0) ln(R

2/r2), where r2 = (x x0)2 + (y y0)2 and R is a constant.Determine the expression for the potential of the line charge in the presence ofthe intersecting planes. Verify explicitly that the potential and the tangentialelectric field vanish on the boundary surface.

(b) Determine the surface charge density on the plane y = 0, x 0. Plot /versus x for (x0 = 2, y0 = 1), (x0 = 1, y0 = 1), and (x0 = 1, y0 = 2).

(c) Show that the total charge (per unit length in z) on the plane y = 0, x 0 is

Qx = 2pi tan1

(x0y0

)

What is the total charge on the plane x = 0?

(d) Show that far from the origin [ 0, where =x2 + y2 and 0 =

x20 + y20 ] the leading term in the potential is

asym = 4pi0

(x0)(y0)(xy)

4.

Interpret.

(a) The potential can be made to vanish on the specified boundary surfacesby pretending that we have three image line charges. Two image charges havecharge density and exist at the locations obtained by reflecting the originalimage charge across the x and y axes, respectively. The third image charge hascharge density + and exists at the location obtained by reflecting the originalcharge through the origin. The resulting potential in the first quadrant is

(x, y) =

4pi0

(lnR2

r21 ln R

2

r22 ln R

2

r23+ ln

R2

r24

)

=

2pi0lnr2r3r1r4

(2)

where

r21 = [(x x0)2 + (y y0)2] r22 = [(x+ x0)2 + (y y0)2]


r23 = [(x x0)2 + (y + y0)2] r24 = [(x+ x0)2 + (y + y0)2].From this you can see that

when x = 0, r1 = r2 and r3 = r4 when y = 0, r1 = r3 and r2 = r4

and in both cases the argument of the logarithm in (2) is unity.

(b)

= 0 ddy

= 2pi

(1

r2

dr2dy

+1

r3

dr3dy

1r1

dr1dy

1r4

dr4dy

) y=0

We have dr1/dy = (y y0)/r1 and similarly for the other derivatives, so

= 2pi

(y y0r22

+y + y0r23

y y0r21

y + y0r24

) y=0

= y0pi

(1

(x x0)2 + y20 1

(x+ x0)2 + y20)

)

(c) Total charge per unit length in z

Qx =

0

dx

= y0pi

[

0

dx

(x x0)2 + y20

0

dx

(x+ x0)2 + y20

]

For the first integral the appropriate substitution is (x x0) = y0 tanu, dx =y0 sec

2 udu. A similar substitution works in the second integral.

= pi

[ pi/2tan1

x0

y0

du pi/2

tan1x0

y0

du

]

= pi

[pi

2 tan1 x0

y0 pi

2+ tan1

x0y0

]

= 2pi

tan1x0y0. (3)

The calculations are obviously symmetric with respect to x0 and y0. Thetotal charge on the plane x = 0 is (3) with x0 and y0 interchanged:

Qy = 2pi

tan1y0x0

Since tan1 x tan1(1/x) = pi/2 the total charge induced isQ =


which is, of course, also the sum of the charge per unit length of the three imagecharges.

(d) We have

=

4pi0lnr22r

23

r21r24

Far from the origin,

r21 = [(x x0)2 + (y y0)2]=

[x2(1 x0

x)2 + y2(1 y0

y)2

]

[x2(1 2x0

x) + y2(1 2y0

y

]=

[x2 2x0x+ y2 2y0y)

]= (x2 + y2)

[1 2xx0 + yy0

x2 + y2

]

Similarly,

r22 = (x2 + y2)

[1 2xx0 + yy0

x2 + y2

]

r23 = (x2 + y2)

[1 2xx0 yy0

x2 + y2

]

r24 = (x2 + y2)

[1 2xx0 yy0

x2 + y2

]

Next,

r21r24 = (x

2 + y2)2[1 4(xx0 + yy0)

2

(x2 + y2)2

]

r22r23 = (x

2 + y2)2[1 4(xx0 yy0)

2

(x2 + y2)2

]so

=

4pi0ln

1 4 (xx0yy0)2(x2+y2)2

1 4 (xx0+yy0)2(x2+y2)2

.

The (x2 + y2) term in the denominator grows much more quickly than the(xx0 + yy0) term, so in the asymptotic limit we can use ln(1 + ) to find

=

4pi0

[4(xx0 yy0)

2

(x2 + y2)2+ 4

(xx0 + yy0)2

(x2 + y2)2

]

=

4pi0

[4(x2x20 + y2y20 2xyx0y0) + 4(x2x20 + y2y20 + 2xyx0y0)(x2 + y2)2

]


=

4pi0

[16xyx0y0(x2 + y2)2

]

=4

pi0

(xy)(x0y0)

(x2 + y2)2.

Problem 2.4

A point charge is placed a distance d > R from the center of anequally charged, isolated, conducting sphere of radius R.

(a) Inside of what distance from the surface of the sphere is thepoint charge attracted rather than repelled by the chargedsphere?

(b) What is the limiting value of the force of attraction when thepoint charge is located a distance a(= dR) from the surfaceof the sphere, if a R?

(c) What are the results for parts a and b if the charge on thesphere is twice (half) as large as the point charge, but stillthe same sign?

Lets call the point charge q. The charged, isolated sphere may be replacedby two image charges. One image charge, of charge q1 = (R/d)q at radiusr1 = R

2/d, is needed to make the potential equal at all points on the sphere.The second image charge, of charge q2 = q q1 at the center of the sphere,is necessary to recreate the effect of the additional charge on the sphere (theadditional charge is the extra charge on the sphere left over after you subtractthe surface charge density induced by the point charge q).

The force on the point charge is the sum of the forces from the two imagecharges:

F =1

4pi0

[qq1[

d R2d]2 + qq2d2

](4)

=q2

4pi0

[ dR[d2 R2]2 +

d2 + dR

d4

](5)

As d R the denominator of the first term vanishes, so that term wins,and the overall force is attractive. As d , the denominator of both termslooks like d4, so the dR terms in the numerator cancel and the overall force isrepulsive.

(a) The crossover distance is found by equating the two bracketed terms in (5):


dR

[d2 R2]2 =d2 + dR

d4

d4R = (d+R)[d2 R2]20 = d5 2d3R2 2d2R3 + dR4 +R5

I used GnuPlot to solve this one graphically. The root is d/R=1.6178.

(b) The idea here is to set d = R+ a = R(1 + a/R) and find the limit of (4) asa 0.

F =q2

4pi0

[R2(1 + aR )[

R2(1 + aR )2 R2]2 +

R2[(1 + aR )

2 + (1 + aR )]

R4(1 + aR )4

]

q2

4pi0

[R2 aR4a2R2

+(2R+ 3a)(R 4a)

R4

]

The second term in brackets approaches the constant 2/R2 as a 0. The firstterm becomes 1/4a2. So we have

F q2

16pi0a2.

Note that only the first image charge (the one required to make the sphere anequipotential) contributes to the force as d a. The second image charge,the one which represents the difference between the actual charge on the sphereand the charge induced by the first image, makes no contribution in this limit.That means that the limiting value of the force will be as above regardless ofthe charge on the sphere.

(c) If the charge on the sphere is twice the point charge, then q2 = 2q q1 =q(2 +R/d). Then (5) becomes

F =q2

4pi0

[ dR

[d2 R2]2 +2d2 + dR

d4

]and the relevant equation becomes

0 = 2d5 4d3R2 2d2R3 + 2dR4 +R5.Again I solved graphically to find d/R = 1.43. If the charge on the sphere ishalf the point charge, then

F =q2

4pi0

[ dR

[d2 R2]2 +d2 + 2dR

2d4

]and the equation is

0 = d5 2d3R2 4d2R3 + dR4 + 2R5.The root of this one is d/R=1.88.


Problem 2.5

(a) Show that the work done to remove the charge q from a dis-tance r > a to infinity against the force, Eq. (2.6), of agrounded conducting sphere is

W =q2a

8pi0(r2 a2) .

Relate this result to the electrostatic potential, Eq. (2.3),and the energy discussion of Section 1.11.

(b) Repeat the calculation of the work done to remove the charge qagainst the force, Eq. (2.9), of an isolated charged conductingsphere. Show that the work done is

W =1

4pi0

[q2a

2(r2 a2) q2a

2r2 qQ

r

].

Relate the work to the electrostatic potential, Eq. (2.8), andthe energy discussion of Section 1.11.

(a) The force is

|F | = q2a

4pi0

1

y3(1 a2/y2)2directed radially inward. The work is

W =

r

Fdy (6)

=q2a

4pi0

r

dy

y3(1 a2/y2)2

=q2a

4pi0

r

ydy

(y2 a2)2

=q2a

4pi0

r2a2

du

2u2

=q2a

4pi0

12u

r2a2

=q2a

8pi0(r2 a2) (7)

To relate this to earlier results, note that the image charge q = (a/r)q islocated at radius r = a2/r. The potential energy between the point charge and


its image is

PE =1

4pi0

(qq

|r r|)

=1

4pi0

( q2ar(r a2/r)

)

=1

4pi0

( q2ar2 a2

)(8)

Result (7) is only half of (8). This would seem to violate energy conservation. Itwould seem that we could start with the point charge at infinity and allow it tofall in to a distance r from the sphere, liberating a quantity of energy (8), whichwe could store in a battery or something. Then we could expend an energyequal to (7) to remove the charge back to infinity, at which point we wouldbe back where we started, but we would still have half of the energy saved inthe battery. It would seem that we could keep doing this over and over again,storing up as much energy in the battery as we pleased.

I think the problem is with equation (8). The traditional expression q1q2/4pi0rfor the potential energy of two charges comes from calculating the work neededto bring one charge from infinity to a distance r from the other charge, and itis assumed that the other charge does not move and keeps a constant chargeduring the process. But in this case one of the charges is a fictitious imagecharge, and as the point charge q is brought in from infinity the image chargemoves out from the center of the sphere, and its charge increases. So the simpleexpression doesnt work to calculate the potential energy of the configuration,and we should take (7) to be the correct result.

(b) In this case there are two image charges: one of the same charge and locationas in part a, and another of charge Q q at the origin. The work needed toremove the point charge q to infinity is the work needed to remove the pointcharge from its image charge, plus the work needed to remove the point chargefrom the extra charge at the origin. We calculated the first contribution above.The second contribution is

r

q(Q q)dy4pi0y2

= 14pi0

r

[qQ

y2+q2a

y3

]dy

= 14pi0

qQy q2a

2y2

r

= 14pi0

[qQ

r+q2a

2r2

]

so the total work done is

W =1

4pi0

[q2a

2(r2 a2) q2a

2r2 qQ

r

].


Review of Greens Functions

Some problems in this and other chapters use the Greens function technique.Its useful to review this technique, and also to establish my conventions sinceI define the Greens function a little differently than Jackson.

The whole technique is based on the divergence theorem. Suppose A(x) isa vector valued function defined at each point x within a volume V . Then

V

( A(x)) dV =

S

A(x) dA (9)

where S is the (closed) surface bounding the volume V . If we take A(x) =(x)(x) where and are scalar functions, (9) becomes

V

[((x)) ((x)) + (x)2(x)] dV =

S

(x)

n

x

dA

where /n is the dot product of ~ with the outward normal to the surfacearea element. If we write down this equation with and switched and subtractthe two, we come up with

V

[2 2] dV =

S

[

n

n

]dA. (10)

This statement doesnt appear to be very useful, since it seems to require thatwe know over the whole volume to compute the left side, and both and/n on the boundary to compute the right side. However, suppose we couldchoose (x) in a clever way such that 2 = (x x0) for some point x0within the volume. (Since this is a function of x which also depends on x0as a parameter, we might write it as x0(x).) Then we could use the siftingproperty of the delta function to find

(x0) =

V

[x0(x

)2(x)] dV + S

[(x)

x0n

x

x0(x)

n

x

]dA.

If is the scalar potential of electrostatics, we know that 2(x) = (x)/0,so we have

(x0) = 10

V

x0(x)(x)dV +

S

[(x)

x0n

x

x0(x)

n

x

]dA.

(11)Equation (11) allows us to find the potential at an arbitrary point x0 as

long as we know within the volume and both and /n on the boundary.boundary. Usually we do know within the volume, but we only know either or /n on the boundary. This lack of knowledge can be accommodated bychoosing such that either its value or its normal derivative vanishes on theboundary surface, so that the term which we cant evaluate drops out of thesurface integral. More specifically,


if we know but not /n on the boundary (Dirichlet boundary con-ditions), we choose such that = 0 on the boundary. Then

(x0) = 10

V

x0(x)(x)dV +

S

(x)x0n

x

dA. (12)

if we know /n but not on the boundary (Neumann boundaryconditions), we choose such that /n = 0 on the boundary. Then

(x0) = 10

V

x0(x)(x)dV +

S

x0(x)

n

x

dA. (13)

Again, in both cases the function x0(x) has the property that

2x0(x) = (x x0).



Homer Reid

December 8, 1999

Chapter 2: Problems 11-20

Problem 2.11

A line charge with linear charge density is placed parallel to, and adistance R away from, the axis of a conducting cylinder of radius b heldat fixed voltage such that the potential vanishes at infinity. Find

(a) the magnitude and position of the image charge(s);

(b) the potential at any point (expressed in polar coordinates with theorigin at the axis of the cylinder and the direction from the originto the line charge as the x axis), including the asymptotic form farfrom the cylinder;

(c) the induced surface-charge density, and plot it as a function of anglefor R/b=2,4 in units of /2pib;

(d) the force on the charge.

(a) Drawing an analogy to the similar problem of the point charge outsidethe conducting sphere, we might expect that the potential on the cylinder canbe made constant by placing an image charge within the cylinder on the lineconducting the line charge with the center of the cylinder, i.e. on the x axis.Suppose we put the image charge a distance R < b from the center of thecylinder and give it a charge density . Using the expression quoted in Problem2.3 for the potential of a line charge, the potential at a point x due to the linecharge and its image is

(x) =

4pi0ln

R2

|x Ri|2

4pi0ln

R2

|xRi|2

1


=

4pi0ln|xRi|2|x Ri|2 .

We want to choose R such that the potential is constant when x is on thecylinder surface. This requires that the argument of the logarithm be equal tosome constant at those points:

|xRi|2|x Ri|2 =

orb2 + R2 2Rb cos = b2 + R2 2Rb cos.

For this to be true everywhere on the cylinder, the term must drop out, whichrequires R = R. We can then rearrange the remaining terms to find

R =b2

R.

This is also analogous to the point-charge-and-sphere problem, but there aredifferences: in this case the image charge has the same magnitude as the originalline charge, and the potential on the cylinder is constant but not zero.

(b) At a point (, ), we have

=

4pi0ln

2 + R2 2R cos2 + R2 2R cos .

For large , this becomes

4pi0

ln1 2R cos1 2R cos

.

Using ln(1 x) = (x + x2/2 + ), we have

4pi0

2(RR) cos

=

2pi0

R(1 b2/R2) cos

(c)

= 0

r=b

= 4pi

[2b 2R cos

b2 + R2 2bR cos 2b 2R cos

b2 + R2 2bR cos]

= 2pi

[b b2R cos

b2 + b4

R2 2 b3

R cos bR cos

b2 + R2 2bR cos

]


Multiplying the first term by R2/b2 on top and bottom yields

= 2pi

[R2

b bR2 + b2 2bR cos

]

= 2pib

[R2 b2

R2 + b2 2bR cos]

(d) To find the force on the charge, we note that the potential of the imagecharge is

(x) = 4pi0

lnC2

|xR i|2 .

with C some constant. We can differentiate this to find the electric field due tothe image charge:

E(x) = (x) = 4pi0

ln |xR i|2

= 4pi0

2(xR i)|xRi|2 .

The original line charge is at x = R, y = 0, and the field there is

E = 2pi0

1

RR i =

2pi0

R

R2 b2 i.

The force per unit width on the line charge is

F = E = 2

2pi0

R

R2 b2

tending to pull the original charge in toward the cylinder.

Problem 2.12

Starting with the series solution (2.71) for the two-dimensional potentialproblem with the potential specified on the surface of a cylinder of radiusb, evaluate the coefficients formally, substitute them into the series, andsum it to obtain the potential inside the cylinder in the form of Poissonsintegral:

(, ) =1

2pi

2pi0

(b, )b2 2

b2 + 2 2b cos( )d

What modification is necessary if the potential is desired in the region ofspace bounded by the cylinder and infinity?


Referring to equation (2.71), we know the bn are all zero, because the lnterm and the negative powers of are singular at the origin. We are left with

(, ) = a0 +

n=1

n {an sin(n) + bn cos(n)} . (1)

Multiplying both sides successively by 1, sin n, and cosn and integratingat = b gives

a0 =1

2pi

2pi0

(b, )d (2)

an =1

pibn

2pi0

(b, ) sin(n)d (3)

bn =1

pibn

2pi0

(b, ) cos(n)d. (4)

Plugging back into (1), we find

(, ) =1

pi

2pi0

(b, )

{1

2+

n=1

(b

)n[sin(n) sin(n) + cos(n) cos(n)]

}d

=1

pi

2pi0

(b, )

{1

2+

n=1

(b

)ncosn( )

}. (5)

The bracketed term can be expressed in closed form. For simplicity definex = (/b) and = ( ). Then

1

2+

n=1

xn cos(n) =1

2+

1

2

n=1

[xnein + xnein

]

=1

2+

1

2

[1

1 xei +1

1 xei 2]

=1

2+

1

2

[1 xei xei + 11 xei xei + x2 2

]

=1

2+

[1 x cos

1 + x2 2x cos 1]

=1

2+

x cos x21 + x2 2x cos

=1

2

[1 x2

1 + x2 2x cos]

.

Plugging this back into (5) gives the advertised result.


Problem 2.13

(a) Two halves of a long hollow conducting cylinder of inner radius b areseparated by small lengthwise gaps on each side, and are kept at differentpotentials V1 and V2. Show that the potential inside is given by

(, ) =V1 + V2

2+

V1 V2pi

tan1(

2b

b2 2 cos)

where is measured from a plane perpendicular to the plane through thegap.(b) Calculate the surface-charge density on each half of the cylinder.

This problem is just like the previous one. Since we are looking for anexpression for the potential within the cylinder, the correct expansion is (1)with expansion coefficients given by (2), (3) and (4):

a0 =1

2pi

2pi0

(b, )d

=1

2pi

[V1

pi0

d + V2

2pipi

d

]

=V1 + V2

2

an =1

pibn

[V1

pi0

sin(n)d + V2

2pipi

sin(n)d

]

= 1npibn

[V1 |cosn|pi0 + V2 |cosn|2pipi

]= 1

npibn[V1(cosnpi 1) + V2(1 cosnpi)]

=

{0 , n even2(V1 V2)/(npibn) , n odd

}

bn =1

pibn

[V1

pi0

cos(n)d + V2

2pipi

cos(n)d

]

=1

npibn

[V1 |sinn|pi0 + V2 |sin n|2pipi

]= 0.

With these coefficients, the potential expansion becomes

(, ) =V1 + V2

2+

2(V1 V2)pi

n odd

1

n

(b

)nsinn. (6)


Here we need an auxiliary result:

n odd

1

nxn sin n =

1

2i

n odd

1

n(iy)n[einpi ein] (x = iy)

=1

2

n=0

(1)n2n + 1

[(yei)2n+1 (yei)2n+1]

=1

2

[tan1(yei) tan1(yei)] (7)

where in the last line we just identified the Taylor series for the inverse tangentfunction. Next we need an identity:

tan1 1 tan1 2 = tan1(

1 21 + 12

).

(I derived this one by drawing some triangles and doing some algebra.) Withthis, (7) becomes

n odd

1

nxn sin n =

1

2tan1

(2iy sin

1 + y2

)

=1

2tan1

(2x sin

1 x2)

.

Using this in (6) with x = /b gives

(, b) =V1 + V2

pi+

V1 V2pi

tan1(

2b sin

b2 2)

.

(Evidently, Jackson and I defined the angle differently).


Problem 2.15

(a) Show that the Green function G(x, y; x, y) appropriate for Dirichletboundary conditions for a square two-dimensional region, 0 x 1, 0 y 1, has an expansion

G(x, y; x, y) = 2

n=1

gn(y, y) sin(npix) sin(npix)

where gn(y, y) satisfies(

2

y2 n2pi2

)gn(y, y

) = (y y) and gn(y, 0) = gn(y, 1) = 0.

(b) Taking for gn(y, y) appropriate linear combinations of sinh(npiy)

and cosh(npiy) in the two regions y < y and y > y, in accord with theboundary conditions and the discontinuity in slope required by the sourcedelta function, show that the explicit form of G is

G(x, y; x, y) =

2

n=1

1

npi sinh(npi)sin(npix) sin(npix) sinh(npiy)]

where y< (y>) is the smaller (larger) of y and y.

(I have taken out a factor 4pi from the expressions for gn and G, in accordancewith my convention for Greens functions; see the Greens functions reviewabove.)

(a) To use as a Greens function in a Dirichlet boundary value problem G mustsatisfy two conditions. The first is that G vanish on the boundary of the regionof interest. The suggested expansion of G clearly satisfies this. First, sin(npix)is 0 when x is 0 or 1. Second, g(y, y) vanishes when y is 0 or 1. So G(x, y; x, y)vanishes for points (x, y) on the boundary.

The second condition on G is

2G =(

2

x2+

2

y2

)G = (x x) (y y). (8)

With the suggested expansion, we have

2

x2G = 2

n=1

gn(y, y) sin(npix)

[n2pi2 sin(npix)]2

y2G = 2

n=1

2

y2gn(y, y

) sin(npix) sin(npix)


We can add these together and use the differential equation satisfied by gn tofind

2G = (y y) 2

n=1

sin(npix) sin(npix)

= (y y) (x x)

since the infinite sum is just a well-known representation of the function.

(b) The suggestion is to take

gn(y, y) =

{An1 sinh(npiy

) + Bn1 cosh(npiy), y < y;

An2 sinh(npiy) + Bn2 cosh(npiy

), y > y.(9)

The idea to use hyperbolic sines and cosines comes from the fact that sinh(npiy)and cosh(npiy) satisfy a homogeneous version of the differential equation for gn(i.e. satisfy that differential equation with the function replaced by zero).Thus gn as defined in (9) satisfies its differential equation (at all points excepty = y) for any choice of the As and Bs. This leaves us free to choose thesecoefficients as required to satisfy the boundary conditions and the differentialequation at y = y.

First lets consider the boundary conditions. Since y is somewhere between0 and 1, the condition that gn vanish for y

= 0 is only relevant to the top lineof (9), where it requires taking Bn1 = 0 but leaves An1 undetermined for now.The condition that gn vanish for y

= 1 only affects the lower line of (9), whereit requires that

0 = An2 sinh(npi) + Bn2 cosh(npi)

= (An2 + Bn2)enpi + (An2 + Bn2)enpi (10)

One way to make this work is to take

An2 + Bn2 = enpi and An2 + Bn2 = enpi.

Then

Bn2 = enpi + An2 2An2 = enpi enpi

so An2 = cosh(npi) and Bn2 = sinh(npi).

With this choice of coefficients, the lower line in (9) becomes

gn(y, y) = cosh(npi) sinh(npiy)+sinh(npi) cosh(npiy) = sinh[npi(1y)] (11)

for (y > y). Actually, we havent completely determined An2 and Bn2; we couldmultiply (11) by an arbitrary constant n and (10) would still be satisfied.

Next we need to make sure that the two halves of (9) match up at y = y:

An1 sinh(npiy) = n sinh[npi(1 y)]. (12)


0

10000

20000

30000

40000

50000

60000

70000

0 0.2 0.4 0.6 0.8 1

g(ypri

me)

yprime

Figure 1: gn(y, y) from Problem 2.15 with n=5, y=.41

This obviously happens when

An1 = n sinh[npi(1 y)] and n = n sinh(npiy)where n is any constant. In other words, we have

gn(y, y) =

{n sinh[npi(1 y)] sinh(npiy), y < y;n sinh[npi(1 y)] sinh(npiy), y > y.

= n sinh[npi(1 y>)] sinh(npiy defined as in the problem. Figure 1 shows a graph of thisfunction n = 5, y = .41.

The final step is to choose the normalization constant n such that gn sat-isfies its differential equation:(

2

2y2 n2pi2

)gn(y, y

) = (y y). (14)

To say that the left-hand side equals the delta function requires two things:

that the left-hand side vanish at all points y 6= y, and that its integral over any interval (y1, y2) equal 1 if the interval contains

the point y = y, and vanish otherwise.

The first condition is clearly satisfied regardless of the choice of n. The secondcondition may be satisfied by making gn continuous, which we have alreadydone, but giving its first derivative a finite jump of unit magnitude at y = y:


ygn(y, y

)

y=y+

y=y= 1.

Differentiating (13), we find this condition to require

npin [ cosh[npi(1 y)] sinh(npiy) sinh[npi(1 y)] cosh(npiy)] = npin sinh(npi) = 1

so (14) is satisfied if

n = 1npi sinh(npi)

.

Then (13) is

gn(y, y) = sinh[npi(1 y>)] sinh(npiy)] sinh(npiy


(x)dV = dxdy throughout the entire volume. Then we can plug in (15) tofind

(x0) =2

pi0

n=1

1

n sinh(npi)

10

10

sinh[npi(1y>)] sinh(npiy


Problem 2.17

(a) Construct the free-space Green function G(x, y; x, y) for two-dimensional electrostatics by integrating 1/R with respect to z zbetween the limits Z, where Z is taken to be very large. Show thatapart from an inessential constant, the Green function can be writtenalternately as

G(x, y; x, y) = ln[(x x)2 + (y y)2]= ln[2 + 2 2 cos( )].

(b) Show explicitly by separation of variables in polar coordinates that theGreen function can be expressed as a Fourier series in the azimuthalcoordinate,

G =1

2pi

eim()gm(,

)

where the radial Green functions satisfy

1

(

gm

) m

2

2gm =

( )

.

Note that gm(, ) for fixed is a different linear combination of the

solutions of the homogeneous radial equation (2.68) for < andfor > , with a discontinuity of slope at = determined by thesource delta function.

(c) Complete the solution and show that the free-space Green function hasthe expansion

G(, ; , ) =1

4piln(2>)

1

2pi

m=1

1

m

(

)m cos[m( )]

where ) is the smaller (larger) of and .

(As in Problem 2.15, I modified the text of the problem to match with myconvention for Greens functions.)(a)

R = [(x x)2 + (y y)2 + (z z)2]1/2 [a2 + u2]1/2 , a = [(x x)2 + (y y)2]1/2 , u = (z z).

Integrating, ZZ

du

[a2 + u2]1/2=

ln [(a2 + u2)1/2 + u] +ZZ


= ln(Z2 + a2)1/2 + Z

(Z2 + a2)1/2 Z

= ln(1 + (a2/Z2))1/2 + 1

(1 + (a2/Z2))1/2 1

ln 2 +a2

2Z2

a2

2Z2

= ln4Z2 + a2

a2

= ln[4Z2 + a2] ln a2.

Since Z is much bigger than a, the first term is essentially independent of a andis the nonessential constant Jackson is talking about. The remaining term isthe 2D Greens function:

G = lna2 = ln[(x x)2 + (y y)2] in rectangular coordinates= ln[2 + 2 2 cos( )] in cylindrical coordinates.

(b) The 2d Greens function is defined by2G(, ; , )dd = 1

but 2G = 0 at points other than (, ). These conditions are met if

2G(, ; , ) = 1

( )( ). (20)

You need the on the bottom there to cancel out the in the area element inthe integral. The Laplacian in two-dimensional cylindrical coordinates is

2 = 1

(

) 1

2

2.

Applying this to the suggested expansion for G gives

2G(, ; , ) = 12pi

{1

(

gm

) m

2

2gm

}eim(

).

If gm satisfies its differential equation as specified in the problem, the term inbrackets equals ()/ for all m and may be removed from the sum, leaving

2G(, ; , ) =(

( )

) 12pi

eim()

=

(( )

)( ).


(c) As in Problem 2.15, well construct the functions gm by finding solutions ofthe homogenous radial differential equation in the two regions and piecing themtogether at = such that the function is continuous but its derivative has afinite jump of magnitude 1/.

For m 1, the solution to the homogenous equation{1

(

) m

2

2

}f() = 0

isf() = Am

m + Bmm.

Thus we take

gm =

{A1m

m + B1mm , <

A2mm + B2m

m , > .

In order that the first solution be finite at the origin, and the second solutionbe finite at infinity, we have to take B1m = A2m = 0. Then the condition thatthe two solutions match at = is

A1mm = B2m

m

which requiresA1m = m

m B2m = mm

for some constant m. Now we have

gm =

m

(

)m, <

m

(

)m, >

The finite-derivative step condition is

dgmd

=+

dgmd

=

=1

or

mm(

1

+

1

)=

1

so

m = 12m

.

Then

gm =

12m(

)m, <

12m(

)m, >

= 12m

(

)m.


Plugging this back into the expansion gives

G = 14pi

1

m

(

)meim(

)

= 12pi

1

1

m

(

)mcos[m( )].

Jackson seems to be adding a ln term to this, which comes from the m = 0solution of the radial equation, but I have left it out because it doesnt vanishas .

Problem 2.18

(a) By finding appropriate solutions of the radial equation in part b ofProblem 2.17, find the Green function for the interior Dirichlet prob-lem of a cylinder of radius b [gm(,

= b) = 0. See (1.40)]. Firstfind the series expansion akin to the free-space Green function ofProblem 2.17. Then show that it can be written in closed form as

G = ln

[22 + b4 2b2 cos( )b2(2 + 2 2 cos( ))

]

or

G = ln

[(b2 2)(b2 2) + b2| |2

b2| |2]

.

(b) Show that the solution of the Laplace equation with the potentialgiven as (b, ) on the cylinder can be expressed as Poissons inte-gral of Problem 2.12.

(c) What changes are necessary for the Green function for the exteriorproblem (b < < ), for both the Fourier expansion and theclosed form? [Note that the exterior Green function is not rigorouslycorrect because it does not vanish for or . For situationsin which the potential falls of fast enough as , no mistake ismade in its use.]

(a) As before, we write the general solution of the radial equation for gm in thetwo distinct regions:

gm(, ) =

{A1m

m + B1mm , <

A2mm + B2m

m , > .(21)

The first boundary conditions are that gm remain finite at the origin andvanish on the cylinder boundary. This requires that

B1m = 0


andA2mb

m + B2mbm = 0

soA2m = mb

m B2m = mbm

for some constant m.Next, gm must be continuous at =

:

A1mm = m

[(b

)m

(b

)m]

A1m =mm

[(b

)m

(b

)m].

With this we have

gm(, ) = m

[(b

)m

(b

)m] (

)m, <

= m

[(

b

)m

(b

)m], > .

Finally, dgm/d must have a finite jump of magnitude 1/ at = .

1

=

dgmd

=+

dgmd

=

= mm

[m1

bm+

bm

m+1

]mm

[(b

)m

(b

)m]1

= 2mm

(b

)m1

so

m =1

2m

(b

)mand

gm(, ) =

1

2m

[(

b2

)m

(

)m], <

=1

2m

[(

b2

)m

(

)m], > .

or

gm(, ) =

1

2m

[(

b2

)m

(

)m].

Plugging into the expansion for G gives

G(, , , ) =1

2pi

n=1

1

m

[(

b2

)m

(

)m]cosm( ). (22)


Here we need to work out an auxiliary result:

n=1

1

nxn cosn( ) =

n=1

[ x0

un1du

]cosm( )

=

x0

{1

u

n=1

un cosn( )}

du

=

x0

{cos( ) u

1 + u2 2u cos( )}

du

= 12

ln(1 2u cos( ) + u2)x0

= 12

ln[1 2x cos( ) + x2].

(I summed the infinite series here back in Problem 2.12. The integral in thesecond-to-last step can be done by partial fraction decomposition, although Icheated and looked it up on www.integrals.com). We can apply this resultindividually to the two terms in (22):

G(, ; , ) = 14pi

ln

[1 + (/b2)2 2(/b2) cos( )1 + ()2 2() cos( )

]

= 14pi

ln

[(2>b4

)b4 + 22 2b2 cos( )2> +

2< 2 cos( )

]

= 14pi

ln

[(2>b4

)b4 + 22 2b2 cos( )2> +

2< 2 cos( )

]

= 14pi

ln

(2>b2

)

14pi

ln

[b4 + 22 2b2 cos( )b2(2 + 2 2 cos( ))

](23)

This is Jacksons result, with an additional ln term thrown in for good measure.Im not sure why Jackson didnt quote this term as part of his answer; he didinclude it in his answer to problem 2.17 (c). Did I do something wrong?

(b) Now we want to plug the expression for G above into (16) to computethe potential within the cylinder. If there is no charge inside the cylinder, thevolume integral vanishes, and we are left with the surface integral:

(, ) =

(b, )

G

=b

dA. (24)

where the integral is over the surface of the cylinder.For this we need the normal derivative of (23) on the cylinder:

G

= 1

4pi

{22 2b2 cos( )

b4 + 22 2b2 cos( ) 2 2 cos( )

2 + 2 2 cos( )}

.


Evaluated at = b this is

G

=b

= 12pi

{2 b2

b(2 + b2 2b cos( ))}

.

In the surface integral, the extra factor of b on the bottom is cancelled by thefactor of b in the area element dA, and (24) becomes just the result of Problem2.12.

(c) For the exterior problem we again start with the solution (21). Now theboundary conditions are different; the condition at gives A2m = 0, while thecondition at b gives

A1m = mbm B1m = mbm.

From the continuity condition at = we find

A2m = mm

[(b

)m

(b

)m].

The finite derivative jump condition gives

mm[(

b

)m

(b

)m]1

mm

[(b

)m+

(b

)m]1

=

1

or

m = 12m

(b

)m.

Putting it all together we have for the exterior problem

gm =1

2m

[(b2

)m

(

)m].

This is the same gm we came up with before, but with b2 and terms flipped

in first term. But the closed-form expression was symmetrical in those twoexpressions (except for the mysterious ln term) so the closed-form expressionfor the exterior Greens function should be the same as the interior Greensfunction.



Homer Reid

June 15, 2000


Problem 3.1

Two concentric spheres have radii a, b(b > a) and each is divided into two hemi-spheres by the same horizontal plane. The upper hemisphere of the inner sphereand the lower hemisphere of the outer sphere are maintained at potential V . Theother hemispheres are at zero potential.Detemine the potential in the region a r b as a series in Legendre polynomials.Include terms at least up to l = 4. Check your solution against known results inthe limiting cases b and a 0.

The expansion of the electrostatic potential in spherical coordinates for prob-lems with azimuthal symmetry is

(r, ) =

l=0

[Alr

l + Blr(l+1)

]Pl(cos ). (1)

We find the coefficients Al and Bl by applying the boundary conditions. Mul-tiplying both sides by Pl(cos ) and integrating from -1 to 1 gives 1

1

(r, )Pl(cos )d(cos ) =2

2l + 1

[Alr

l + Blr(l+1)

].

At r = a this yields

V

10

Pl(x)dx =2

2l + 1

[Ala

l + Bla(l+1)

],

1


and at r = b,

V

01

Pl(x)dx =2

2l + 1

[Alb

l + Blb(l+1)

].

The integral from 0 to 1 vanishes for l even, and is given in the text for l odd: 10

Pl(x)dx = (12)(l1)/2

(l 2)!!2(

l+12

)!.

The integral from -1 to 0 also vanishes for l even, and is just the above resultinverted for l odd. This gives

V (12)(l1)/2

(l 2)!!2(

l+12

)!

=2

2l + 1

[Ala

l + Bla(l+1)

]

V (12)(l1)/2

(l 2)!!2(

l+12

)!

=2

2l + 1

[Alb

l + Blb(l+1)

].

or

l = Alal + Bla

(l+1)

l = Albl + Blb(l+1)

with

l = V (12)a(l1)/2

(2l + 1)(l 2)!!4(

l+12

)!

.

The solution is

Al = l

[bl+1 + al+1

a2l+1 b2l+1]

Bl = l[al+1bl+1(bl + al)

a2l+1 b2l+1]

The first few terms of (1) are

(r, ) =3

4V

[(a2 + b2)r

a3 b3 a2b2(a + b)

r2(a3 b3)]

P1(cos ) 716

[(a4 + b4)r3

a7 b7 a4b4(a3 + b3)

r4(a7 b7)]

P3(cos )+

In the limit as b , the problem reduces to the exterior problem treatedin Section 2.7 of the text. In that limit, the above expression becomes

(r, ) 34V(a

r

)2P1(cos ) 7

16V(a

r

)4P3(cos ) +

in agreement with (2.27) with half the potential spacing. When a 0, theproblem goes over to the interior version of the same problem, as treated insection 3.3 of the text. In that limit the above expression goes to

(r, ) 34V(r

b

)P1(cos ) +

7

16V(r

b

)3P3(cos ) +

This agrees with equation (3.36) in the text, with the sign of V flipped, becausehere the more positive potential is on the lower hemisphere.


Problem 3.2

A spherical surface of radius R has charge uniformly distributed over its surfacewith a density Q/4piR2, except for a spherical cap at the north pole, defined by thecone = .

(a) Show that the potential inside the spherical surface can be expressed as

=Q

8pi0

l=0

1

2l + 1[Pl+1(cos) Pl1(cos)] r

l

Rl+1Pl(cos )

where, for l = 0, Pl1(cos ) = 1. What is the potential outside?(b) Find the magnitude and direction of the electric field at the origin.

(c) Discuss the limiting forms of the potential (part a) and electric field (part b) asthe spherical cap becomes (1)very small, and (2) so large that the area withcharge on it becomes a very small cap at the south pole.

(a) Lets denote the charge density on the sphere by (). At a point infinites-imally close to the surface of the sphere, the electric field is

F = = 0

r

so

r

r=R

=

0. (2)

The expression for the potential within the sphere must be finite at the origin,so the Bl in (1) are zero. Differentiating that expansion, (2) becomes

r(r, ) =

l=1

lAlrl1Pl(cos )

Multiplying by Pl and integrating at r = R gives

1

0

11

()Pl(cos )d(cos ) =2l

2l + 1AlR

l1

so

Al =2l + 1

2lRl1(

Q

4piR20

) cos 1

Pl(x)dx.

To evaluate the integral we use the identity (eq. 3.28 in the text)

Pl(x) =1

(2l + 1)

d

dx[Pl+1(x) Pl1(x)]


so cos 1

Pl(x)dx =1

2l + 1[Pl+1(cos) Pl1(cos)] .

(We used the fact that Pl+1(1) = Pl1(1) for all l.) With this we have

Al =Q

8pi0lRl+1[Pl+1(cos) Pl1(cos)]

so the potential expansion is

(r, ) =Q

8pi0

l=1

1

l[Pl+1(cos) Pl1(cos)] r

l

Rl+1Pl(cos ).

Within the body of the sum, I have an l where Jackson has a 2l + 1. Also,he includes the l = 0 term in the sum, corresponding to a constant term inthe potential. I dont understand how he can determine that constant from theinformation contained in the problem; the information about the charge densityonly tells you the derivative of the potential. Theres nothing in this problemthat fixes the value of the potential on the surface beyond an arbitrary constant.

(b) The field at the origin comes from the l = 1 term in the potential:

E(r = 0) = |r=0 = r r + 1r

r=0

= Q8pi0R2

[P2(cos) 1][P1(cos )r +

d

dP1(cos )

]

= Q8pi0R2

[3

2cos2 3

2

] [cos r sin

]

=3Q sin2

16pi0R2k.

The field points in the positive z direction. That makes sense, since a positivetest charge at the origin would sooner fly up out through the uncharged capthan through any of the charged surface.


Problem 3.3

A thin, flat, conducting, circular disk of radius R is located in the x y planewith its center at the origin, and is maintained at a fixed potential V . With theinformation that the charge density on a disc at fixed potential is proportional to(R2 2)1/2, where is the distance out from the center of the disc,

(a) show that for r > R the potential is

(r, , ) =2V

pi

R

r

l=0

(1)l2l + 1

(R

r

2l)

P2l(cos )

(b) find the potential for r < R.

(c) What is the capacitance of the disk?

We are told that the surface charge density on the disk goes like

(r) = K(R2 r2)1/2

=K

R

[1 +

1

2

( rR

)2+

3 1(2!)(2 2)

( rR

)4+

5 3 1(3!)(2 2 2)

( rR

)6+

]

=K

R

n=0

(2n 1)!!n! 2n

( rR

)2n(3)

for some constant K. From the way the problem is worded, I take it werenot supposed to try to figure out what K is explicitly, but rather to work theproblem knowing only the form of (3).

At a point infinitesimally close to the surface of the disk (i.e., as pi/2),the component of in the direction normal to the surface of the disk mustbe proportional to the surface charge. At the surface of the disk, the normaldirection is the negative direction. Hence

1

r

(r, )

=(pi/2)

= 0

. (4)

with the plus (minus) sign valid for above (below) the disc.For r < R the potential expansion is

(r, ) =

l=0

AlrlPl(cos ). (5)

Combining (3), (4), and (5) we have

l=0

Alrl1 d

dPl(cos )

cos =0

= KR0

n=0

(2n 1)!!n! 2n

( rR

)2n. (6)


For l even, dPl/dx vanishes at x = 0. For l odd, I used some of the Legendrepolynomial identities to derive the formula

d

dxP2l+1(x)

x=0

= (1)l(2l + 1)(2l 1)!!l! 2l .

This formula reminds one strongly of expansion (3). Plugging into (6) andequating coefficents of powers of r, we find

A2l+1 = (1)lK

(2l + 1)R2l+10

so

(r, ) = A0 K0

l=1

(1)l2l + 1

( rR

)2l+1P2l+1(cos ).

I wrote A0 explicitly because we havent evaluated it yetthe derivative conditionwe used earlier gave no information about it. To find A0, observe that, on thesurface of the disk (cos = 0), all the terms in the above sum vanish ( becausePl(0) is 0 for odd l) so = A0 on the disk. But = V on the disk. Therefore,A0 = V . We have

(r, ) = V K0

l=1

(1)l2l + 1

( rR

)2l+1P2l+1(cos ) (7)

where the plus (minus) sign is good for less than (greater than)pi/2. Note thatthe presence of that sign preserves symmetry under reflection through the zaxis, a symmetry that is clearly present in the physical problem.

(a) For r > R, there is no charge. Thus the potential and its derivative must becontinuous everywherewe cant have anything like the derivative discontinuitythat exists at = pi/2 for r < R. Since the physical problem is symmetric undera sign flip in cos , the potential expansion can only contain Pl terms for l even.The expansion is

(r, ) =

l=0

B2lr(2l+1)P2l(cos ).

At r = R, this must match up with (7):

V K0

l=1

(1)l2l + 1

P2l+1(cos ) =

l=0

B2lR(2l+1)P2l(cos ).

Multiplying both sides by P2l(cos ) sin() and integrating gives

B2l2R(2l+1)

4l + 1= V

11

Pl(x)dx +K

0

l=1

(1)l2l + 1

{ 01

P2l+1(x)P2l(x)dx +

10

P2l+1(x)Pl(x)dx

}

= 2V l,0 +2K

0

l=1

(1)l2l + 1

10

P2l+1(x)P2l(x)dx.


but I cant do this last integral.

Problem 3.4

The surface of a hollow conducting sphere of inner radius a is divided into an evennumber of equal segments by a set of planes; their common line of intersection isthe z axis and they are distributed uniformly in the angle . (The segments are likethe skin on wedges of an apple, or the earths surface between successive meridiansof longitude.) The segments are kept at fixed potentials V , alternately.(a) Set up a series representation for the potential inside the sphere for the general

case of 2n segments, and carry the calculation of the coefficients in the seriesfar enough to determine exactly which coefficients are different from zero. Forthe nonvanishing terms exhibit the coefficients as an integral over cos .

(b) For the special case of n = 1 (two hemispheres) determine explicitly the poten-tial up to and including all terms with l = 3. By a coordinate transformationverify that this reduces to result (3.36) of Section 3.3.

(a) The general potential expansion is

(r, , ) =

l=0

lm=l

[Almr

l + Blmr(l+1)

]Ylm(, ). (8)

For the solution within the sphere, finiteness at the origin requires Blm = 0.Multiplying by Y lm and integrating over the surface of the sphere we find

Alm =1

al

(a, , ) Y lm(, ) d

=V

al

nk=1

(1)k pi

0

2kpi/n2(k1)pi/n

Y lm(, ) sin d d

=V

al

[2l + 1

4pi

(l m)!(l + m)!

]1/2{ 11

P ml (x) dx

} nk=1

(1)k{ 2kpi/n

2(k1)pi/n

eim d

}.

(9)

The integral is easy: 2kpi/n2(k1)pi/n

eim d = 1im

[e2imkpi/n e2im(k1)pi/n

].

This is to be summed from k = 1 to n with a factor of (1)k thrown in:= 1

im

[(e2mpii(1/n) 1) (e2mpii(2/n) e2mpii(1/n)) + (1 e2mpii((n1)/n))

]=

2

im

{1 e2mpii/n + e2(2mpii/n) e3(2mpii/n) + + e(n1)(2mpii/n)

}. (10)


Putting x = exp(2mpii/n), the thing in braces is

1 + x + x2 + x3 + + xn1 = 1 xn

1 x =1 e2mpii

1 + e2impi/n,

Note that the numerator vanishes. Thus the only way this thing can be nonzerois if the denominator also vanishes, which only happens if the exponent in thedenominator equates to -1. This only happens if m/n = 1/2, 3/2, 5/2, . Inthat case, the 2mpii/n term in the exponent of the terms in (10) equates to pii,so all the terms with a plus sign in (10) come out to +1, while all the termswith a minus sign come out to -1, so all n terms add constructively, and (10)equates to

=

{2nim , m = n/2, 3n/2, 5n/2, 0, otherwise.

Then the expression (9) for the coefficients becomes

Alm =2nV

imal

[2l + 1

4pi

(l m)!(l + m)!

]1/2 11

P ml (x)dx, m =n

2,3n

2, = 0, otherwise.

(b) As shown above, the only terms that contribute are those with m = n/2,m = 3n/2, et cetera. Of course there is also the constraint that m < l. Then,with n = 2, up to l = 3 the only nonzero terms in the series (9) are those withl = 1, m = 1, and l = 3, m = 1 or 3. We need to evaluate the integralfor these terms. We have 1

1

P 11 (x) dx = 11

(1 x2)1/2 dx = pi 11

P 13 (x) dx = 11

(1 x2)1/2[15

2x2 3

2

]dx = 3pi

8 11

P 33 (x) dx = 15 11

(1 x2)3/2 dx = 15pi4

.

Using these results in (??), we have

A11 = 4piV ia

[3

4pi 2]1/2

A31 = 3piV i2a3

[7 2

4pi 4!]1/2

A33 = 5piV ia3

[7

4pi 6!]1/2

Now we can plug these coefficients into (8) to piece together the solution.This involves some arithmetic in combining all the numerical factors in each


coefficient, which I have skipped here.

(r, , ) = V[3(r

a

)sin sin +

7

16

( ra

)3sin (5 cos2 1) sin

+7

144

(ra

)3sin3 sin 3 +

]

Problem 3.6

Two point charges q and q are located on the z azis at z = +a and z = a,respectively.

(a) Find the electrostatic potential as an expansion in spherical harmonics andpowers of r for both r > a and r < a.

(b) Keeping the product qa = p/2 constant, take the limit of a 0 and find thepotential for r 6= 0. This is by definition a dipole along the z azis and itspotential.

(c) Suppose now that the dipole of part b is surrounded by a grounded sphericalshell of radius b concentric with the origin. By linear superposition find thepotential everwhere inside the shell.

(a) First of all, for a point on the z axis the potential is

(z) =q

4pi0

[1

|z a| 1

z + a

]

=q

4pi0z

[1 +

(az

)+(a

z

)2+

(1

(az

)+(a

z

)2 )]

=q

2pi0z

[(az

)+(a

z

)3+

]

for z > a. Comparing this with the general expansion =

Blr(l+1)Pl(cos )

at = 0 we can identify the Bls and write

(r, ) =q

2pi0r

[(ar

)P1(cos ) +

(ar

)3P3(cos ) +

]for r > a. For r < a we can just swap a and r in this equation.

(b)

(r, ) =qa

2pi0r2

[P1(cos ) +

(ar

)2P3(cos ) +

]

=p

4pi0r2

[P1(cos ) +

(ar

)2P3(cos ) +

]

p4pi0r2

cos as a 0.


(c) When we put the grounded sphere around the two charges, a surface chargedistribution forms on the sphere. Lets denote by s the potential due to thischarge distribution alone (not including the potential of the dipole) and by dthe potential due to the dipole. To calculate s, we pretend there are no chargeswithin the sphere, in which case we have the general expansion (1), with Bl = 0to keep us finite at the origin. The total potential is just the sum s + d:

(r, ) =p

4pi0r2cos +

l=0

AlrlPl(cos ).

The condition that this vanish at r = b ensures, by the orthogonality of the Pl,that only the l = 1 term in the sum contribute, and that

A1 = p4pi0b3

.

The total potential inside the sphere is then

(r, ) =p

4pi0b2

(1 r

b

)P1(cos ).

Problem 3.7

Three point charges (q,2q, q) are located in a straight line with separation a andwith the middle charge (2q) at the origin of a grounded conducting spherical shellof radius b, as indicated in the figure.

(a) Write down the potential of the three charges in the absence of the groundedsphere. Find the limiting form of the potential as a 0, but the productqa2 = Q remains finite. Write this latter answer in spherical coordinates.

(b) The presence of the grounded sphere of radius b alters the potential for r < b.The added potential can be viewed as caused by the surface-charge densityinduced on the inner surface at r = b or by image charges located at r > b. Uselinear superposition to satisfy the boundary conditions and find the potentialeverywhere inside the sphere for r < a and r > a. Show that in the limita 0,

(r, , ) Q2pi0r3

(1 r

5

b5

)P2(cos ).

(a) On the z axis, the potential is

(z) =q

4pi0

[2

z+

1

|z a| +1

z + a

]

=q

4pi0r

[2 +

(1 +

(az

)+(a

z

)2 )

+

(1

(az

)+(a

z

)2+

)]

=q

2pi0z

[(az

)2+(a

z

)4+

].


As before, from this result we can immediately infer the expression for thepotential at all points:

(r, ) =q

2pi0r

[(ar

)2P2(cos ) +

(ar

)4P4(cos ) +

]

=qa2

2pi0r3

[P2(cos ) +

(ar

)2P4(cos ) +

]

Q2pi0r3

P2(cos ) as a 0 (11)

(b) As in the previous problem, the surface charges on the sphere produce anextra contribution s to the potential within the sphere. Again we can expresss with the expansion (1) (with Bl = 0), and we add s to (11) to get the fullpotential within the sphere:

(r, ) =Q

2pi0r3P2(cos ) +

l=0

AlrlPl(cos )

From the condition that vanish at r = b, we determine that only the l = 2term in the sum contributes, and that

A2 = Q2pi0b5

.

Then the potential within the sphere is

(r, ) =Q

2pi0r3

[1

(rb

)5]P2(cos ).

Problem 3.9

A hollow right circular cylinder of radius b has its axis coincident with the z axisand its ends at z = 0 and z = L. The potential on the end faces is zero, while thepotential on the cylindrical surface is given as V (, z). Using the appropriate sepa-ration of variables in cylindrical coordinates, find a series solution for the potentialanywhere inside the cylinder.

The general solution of the Laplace equation for problems in cylindrical coordi-nates consists of a sum of terms of the form

R()Q()Z(z).

The function is of the form

Q() = A sin + B cos


with an integer. The z function is of the form

Z(z) = Cekz + Dekz .

In this case, Z must vanish at z = 0 and z = L, which means we have to takek imaginary, i.e.

Z(z) = C sin(knz) with kn =pin

L, n = 1, 2, 3,

With this form for Z, R must be taken to be of the form

R() = EI(kn) + FK(kn).

Since were looking for the potential on the inside of the cylinder and there isno charge at the origin, the solution must be finite as 0, which requiresF = 0. Then the potential expansion becomes

(, , z) =

n=1

=0

[An sin + Bn cos ] sin(knz)I(kn). (12)

Multiplying by sin sin knz and integrating at r = b, we find L0

2pi0

V (, z) sin sin(knz) d dz =piL

2I(knb)An

so

An =2

piLI(knb)

L0

2pi0

V (, z) sin() sin(knz) d dz. (13)

Similarly,

Bn =2

piLI(knb)

L0

2pi0

V (, z) cos() sin(knz) d dz. (14)

Problem 3.10

For the cylinder in Problem 3.9 the cylindrical surface is made of two equal half-cylinders, one at potential V and the other at potential V , so that

V (, z) =

{V for pi/2 < < pi/2

V for pi/2 < < 3pi/2

(a) Find the potential inside the cylinder.

(b) Assuming L >> b, consider the potential at z = L/2 as a function of and and compare it with two-dimensional Problem 2.13.

The potential expansion is (12) with coefficients given by (13) and (14). Therelevant integrals are L

0

2pi0

V (, z) sin() sin(knz) d dz


= V

{ L0

sin(knz) dz

}{ pi/2pi/2

sin() d 3pi/2

pi/2

sin() d

}

= 0 L0

2pi0

V (, z) cos() sin(knz) d dz

= V

{ L0

sin(knz) dz

}{ pi/2pi/2

cos() d 3pi/2

pi/2

cos() d

}

=2V

kn

{|sin |pi/2

pi/2 |sin |3pi/2pi/2

}(n odd)

=

0 , n or even8V/kn , n odd, = 1, 5, 9,

8V/kn , n odd, = 3, 7, 11, Hence, from (13) and (14),

An = 0Bn = 0, n or even

= (1)(1)/2 16V/(npi2I(knb)), n and oddThe potential expansion is

(, , z) =16V

pi2

n,

(1)(1)/2nIv(knb)

cos() sin(knz)I(kn) (15)

where the sum contains only terms with n and odd.

(b) At z = L/2 we have

(, , L/2) =16V

pi2

n,

(1)(n+2)/2n

cos()I (kn)

I (knb).

As L , the arguments to the I functions become small. Using the limitingform for I quoted in the text as equation (3.102), we have

(, ) =16V

pi2

n,

(1)(n+2)/2n

cos()(

b

).

The sums over n and are now decoupled:

(, ) =16V

pi2

{

n=0

(1)n2n + 1

}{

=0

(1)2 + 1

cos()(

b

)}

=16V

pi2

{pi4

}{ =0

(1)2 + 1

cos()(

b

)}

=4V

pitan1

(2b cos

b2 2)


This agrees with the result of Problem 2.13, with V1 = V2 = V . The firstseries is just the Taylor series for tan1(x) at x = 1, so it sums to pi/4. Thesecond series can also be put into the form of the Taylor series for tan1(x),using tricks exactly analogous to what I did in my solution for Problem 2.13.



Homer Reid

June 15, 2000


Problem 3.11

A modified Bessel-Fourier series on the interval 0 a for an arbitrary functionf() can be based on the homogenous boundary conditions:

At = 0, J(k)d

dJ(k

) = 0

At = a,d

dln[J(k)] =

a( real)

The first condition restricts . The second condition yields eigenvalues k = yn/a,where yn is the nth positive root of x dJ(x)/dx + J(x) = 0.

(a) Show that the Bessel functions of different eigenvalues are orthogonal in theusual way.

(b) Find the normalization integral and show that an arbitrary function f() canbe expanded on the interval in the modified Bessel-Fourier series

f() =

n=1

AnJ

(yna

)

with the coefficients An given by

An =2

a2

[(1

2

y2n

)J2 (yn) +

(dJ(yn)

dyn

)2]1 a0

f()J

(yna

)d.

1


(a) The function J(k) satisfies the equation

1

d

d

[

d

dJ(k)

]+

(k2

2

2

)J(k) = 0. (1)

Multiplying both sides by J(k) and integrating from 0 to a gives a

0

{J(k

)d

d

[

d

dJ(k)

]+

(k2

2

)J(k

)J(k)

}d = 0. (2)

The first term on the left can be integrated by parts: a0

J(k)

d

d

[

d

dJ(k)

]d

=

J(k) ddJ(k)a

0

a

0

[d

dJ(k

)

] [d

dJ(k)

]d. (3)

One of the conditions were given is that the thing in braces in the first termhere vanishes at = 0. At = a we can invoke the other condition:

d

dln[J(k)]

=a

=1

J(k)

d

dJ(k)

=a

= a

a dd

J(ka) = J(ka).

Plugging this into (3), we have a0

J(k)

d

d

[

d

dJ(k)

]d

= J(k)J(k) a

0

[d

dJ(k

)

] [d

dJ(k)

]. (4)

This is clearly symmetric in k and k, so when we write down (2) with k andk switched and subtract from (2), the first integral (along with the 2/ term)vanishes, and we are left with

(k2 k2) a

0

J(k)J(k) d = 0

proving orthogonality.

(b) If we multiply (1) by 2J (k) and integrate, we find a0

J (k)d

d[J (k)]d+k

2

a0

2J(k)J

(k)d2 a

0

J(k)J

(k)d = 0.

(5)


The first and third integrals are of the form

f(x)f (x)dx and can be doneimmediately. In the second integral we put f() = 2J(k), g

() = J (k)and integrate by parts: a

0

2J(k)J

(k)d =2J2 (k)a0 2

a0

J2 (k)d a

0

2J(k)J

(k)d

a

0

2J(k)J

(k)d =1

2a2J2 (ka)

a0

J2 (k)d.

Using this in (5),

a2

2J 2 (ka) +

(ak)2

2aJ2 (ka) k2

a0

J2 (k)d2

2J2 (ka) = 0

so a0

J2 (k)d =

(a2

2

2

2k2

)J2 (ka) +

a2

2k2J 2 (ka)

=a2

2

{(1

2

(ka)2

)J2 (ka) +

[d

d(ka)J(ka)

]2}

This agrees with what Jackson has if you note that k is chosen such that ka =ynm.

Problem 3.12

An infinite, thin, plane sheet of conducting material has a circular hole of radius acut in it. A thin, flat, disc of the same material and slightly smaller radius lies inthe plane, filling the hole, but separated from the sheet by a very narrow insulatingring. The disc is maintained at a fixed potential V , whilc the infinite sheet is keptat zero potential.

(a) Using appropriate cylindrical coordinates, find an integral expression involvingBessel functions for the potential at any point above the plane.

(b) Show that the potential a perpendicular distance z above the center of the discis

0(z) = V

(1 z

a2 + z2

)

(c) Show that the potential a perpendicular distance z above the edge of the discis

a(z) =V

2

[1 kz

piaK(k)

]

where k = 2a/(z2 + 4a2)1/2, and K(k) is the complete elliptic integral of thefirst kind.


(a) As before, we can write the potential as a sum of terms R()Q()Z(z). Inthis problem there is no dependence, so Q = 1. Also, the boundary conditionson Z are that it vanish at and be finite at 0, whence Z(z) exp(kz) forany k. Then the potential expansion becomes

(, z) =

0

A(k)ekzJ0(k) dk. (6)

To evaluate the coefficients A(k), we multiply both sides by J0(k) and inte-

grate over at z = 0:

0

(, 0)J0(k) d =

0

A(k)

{

0

J0(k)J0(k) d

}dk

=A(k)

k

so

A(k) = k

0

(, 0)J0(k) d

= kV

a0

J0(k)d.

Plugging this back into (6),

(, z) = V

0

a0

kekzJ0(k)J0(k) d dk. (7)

The integral can be done right away. To do it, I appealed to the differentialequation for J0:

J 0 (u) +1

uJ 0(u) + J0(u) = 0

so x0

uJ0(u) du = x

0

uJ 0 du x

0

J 0(u) du

= |uJ 0(u)|x0 + x

0

J 0(u) du x

0

J 0(u) du

= |uJ 0(u)|x0 = xJ 0(x) = xJ1(x).

(In going from the first to second line, I integrated by parts.) Then (7) becomes

(, z) = aV

0

J1(ka)J0(k)ekz dk. (8)


(b) At = 0, (7) becomes

(0, z) = V J0(0)

a0

{

0

kekzJ0(k)dk

}d

= V

a0

{

z

0

ekzJ0(k)dk

}d

= V

a0

{

z

(1

2 + z2

)}d

= V

a0

z

(2 + z2)3/2d

Here we substitute u = 2 + z2, du = 2d:

(0, z) =V zJ0(0)

2

a2+z2z2

u3/2 du

= V z 1u1/2

a2+z2

z2

= V z

[1

z 1

z2 + z2

]

= V

[1 z

a2 + z2

]

(b) At = a, (8) becomes

(a, z) = aV

0

J1(ka)J0(ka)ekz dk

Problem 3.13

Solve for the potential in Problem 3.1, using the appropriate Greenfunction obtained in the text, and verify that the answer obtainedin this way agrees with the direct solution from the differentialequation.

For Dirichlet boundary value problems, the basic equation is

(x) = 10

V

G(x;x)(x) dV +

S

(x)G(x;x)

n

x

dA. (9)

Here there is no charge in the region of interest, so only the surface integralcontributes. The Greens function for the two-sphere problem is

G(x;x) =

l=0

lm=l

Y lm(, ) Ylm(, )

2l + 1Rl(r; r

) (10)


with

Rl(r; r) =

1[1 (ab )]2l+1

(rl<

a2l+1

rl+1 r

l>

b2l+1

). (11)

Actually in this case the potential cannot have any dependence, so all termswith m 6= 0 in (10) vanish, and we have

G(x;x) = 14pi

l=0

Pl(cos )Pl(cos )Rl(r; r

).

In this case the boundary surfaces are spherical, which means the normal to asurface element is always in the radial direction:

nG(x;x) = 1

4pi

l=0

Pl(cos )Pl(cos )

nRl(r; r

).

The surface integral in (9) has two parts: one integral S1 over the surface ofthe inner sphere, and a second integral S2 over the surface of the outer sphere:

S1 = 14pi

l=0

Pl(cos )Rln

r=a

{ pi0

2pi0

(a, )Pl(cos )a2 sin d d

}

= V2

l=0

a2Pl(cos )Rln

r=a

{ 10

Pl(x) dx

}

= V2

l=0

a2lPl(cos ) Rln

r=a

where

l =

10

Pl(x) dx

= (12)(l1)/2

(l 2)!!2[(l + 1)/2]!

, l odd

= 0, l even.

A similar calculation gives

S2 = V2

l=0

b2Pl(cos )Rln

r=b

{ 01

Pl(x) dx

}

=V

2

l=0

b2lPl(cos )Rln

r=b

because Pl is odd for l odd, so its integral from -1 to 0 is just the negative ofthe integral from 0 to 1. The final potential is the sum of S1 and S2:

(r, ) =V

2

l=0

lPl(cos )

r2 Rlnr=b

r=a

(12)


Since the point of interest is always between the two spheres, to find thenormal derivative at r = a we differentiate with respect to r. Also, at r = a the normal is in the +r direction, while atr = b the normal is in the negative r direction.

a2

nRl(r; r

)

r=a

= (2l + 1)a2al1[

1 (ab )]2l+1(

1

rl+1 r

l

b2l+1

)

b2

nRl(r; r

)

r=b

= (2l + 1)b2b(l+2)[

1 (ab )]2l+1(

rl a2l+1

rl+1

)

Combining these with some algebra gives

(r, ) =V

2

l=0

(2l + 1)lPl(cos )

[(ab)l+1(bl + al)r(l+1) (al+1 + bl+1)rl

b2l+1 a2l+1]

in agreement with what we found in Problem 3.1.

Problem 3.14

A line charge of length 2d with a total charge Q has a linear chargedensity varying as (d2 z2), where z is the distance from the mid-point. A grounded, conducting spherical shell of inner radius b > dis centered at the midpoint of the line charge.

(a) Find the potential everywhere inside the spherical shell as anexpansion in Legendre polynomials.

(b) Calculate the surface-charge density induced on the shell.

(c) Discuss your answers to parts a and b in the limit that d


with

Rl(r; r) = rl r

l>

b2l+1

).

Since the potential vanishes on the boundary surface, the potential inside thesphere is given by

(r, ) = 10

V

G(r, ; r, )(r, )dV.

In this case is only nonzero on the z axis, where r = z. Also, Pl(cos )=1 forz > 0, and (1)l for z < 0. This means that the contributions to the integralfrom the portions of the line charge for z > 0 and z < 0 cancel out for odd l,and add constructively for even l:

(r, ) =1

4pi0

l=0,2,4,...

Pl(cos )

[2

d0

Rl(r; z)(z) dz

]

We have d0

Rl(r; z)(z) dz =

d0

rl r

l>

b2l+1

)(d2 z2) dz

This is best split up into two separate integrals:

=

d0

rl

(d2 z2) dz b2l+1

d0

rl(d

2 z2) dz

The second integral is symmetric between r and r, so we may integrate itdirectly:

b2l+1

d0

rl(d

2 z2) dz = rl

b2l+1

d0

zl(d2 z2) dz

= rl

b2l+1

[dl+3

l + 1 d

l+3

l + 3

]

= rldl+3

(l + 1)(l + 3)b2l+1(14)

The first integral must be further split into two:

d0

rl

(d2 z2) dz


=

{1

rl+1

r0

zl(d2 z2) dz + rl d

r

d2 z2zl+1

dz

}

=

{1

rl+1

[d2rl+1

l + 1 r

l+3

l + 3

]+ rl

d2lzl + 1(l 2)zl2d

r

}

=

{d2

l + 1 r

2

l + 3+(r

d

)ld2

2

l(l + 2) d

2

l+

r2

l + 2

}

=

{r2

(l + 2)(l + 3) d

2

l(l + 1)+(r

d

)ld2

2

l(l + 2)

}

Combining this with (14), we have

d0

Rl(r; z)(z) dz =

{r2

(l + 2)(l + 3) d

2

l(l + 1)+(r

d

)ld2

2

l(l + 2) r

ldl+3

(l + 1)(l + 3)b2l+1

}(15)

But something is wrong here, because with this result the final potential willcontain terms like r0Pl(cos ) and r

2Pl(cos ), which do not satisfy the Laplaceequation.


Problem 3.15

Consider the following spherical cow model of a battery con-nected to an external circuit. A sphere of radius a and conductiv-ity is embedded in a uniform medium of conductivity . Insidethe sphere there is a uniform (chemical) force in the z directionacting on the charge carriers; its strength as an effective electricfield entering Ohms law is F . In the steady state, electric fieldsexist inside and outside the sphere and surface charge resides onits surface.

(a) Find the electric field (in addition to F ) and current densityeverywhere in space. Determine the surface-charge densityand show that the electric dipole moment of the sphere isp = 4pi0a

3F/( + 2).

(b) Show that the total current flowing out through the upperhemisphere of the sphere is

I =2

+ 2 pia2F

Calculate the total power dissipation outside the sphere. Us-ing the lumped circuit relations, P = I2Re = IVe, find theeffective external resistance Re and voltage Ve.

(c) Find the power dissipated within the sphere and deduce theeffective internal resistance Ri and voltage Vi.

(d) Define the total voltage through the relation Vt = (Re + Ri)Iand show that Vt = 4aF/3, as well as Ve + Vi = Vt. Showthat IVt is the power supplied by the chemical force.

(a) Whats going on in this problem is that the conductivity has a discontinu-ity going across the boundary of the sphere, but the current density must beconstant there, which means there must an electric field discontinuity in inverseproportion to the conductivity discontinuity. To create this electric field discon-tinuity, there has to be some surface charge on the sphere, and this charge givesrise to extra fields both inside and outside the sphere.

Since there is no charge inside or outside the sphere, the potential in thosetwo regions satisfied the Laplace equation, and may be expanded in Legendrepolynomials:


for r < a, (r, ) = in(r, ) =

l=0

AlrlPl(cos )

for r > a, (r, ) = out(r, ) =

l=0

Blr(l+1)Pl(cos )

Continuity at r = a requires that

Alal = Bla

l+1 Bl = a2l+1Also

(r, ) =

{in(r, ) =

l=0 AlrlPl(cos ), r < a

out(r, ) =

l=0 Ala2l+1r(l+1)Pl(cos ), r > a.

(16)

Now, in the steady state there can be no discontinuities in the current den-sity, because if there were than there would be more current flowing into someregion of space than out of it, which means charge would pile up in that region,which would be a growing source of electric field, which would mean we arentin steady state. So the current density is continuous everywhere. In particular,the radial component of the current density is continuous across the boundaryof the sphere, i.e.

Jr(r = a, ) = Jr(r = a+, ). (17)

Outside of the sphere, Ohms law says that

J = E = out.

Inside the sphere, there is an extra term coming from the chemical force:

J = (E + F k) = (in + F k).

Applying (17) to these expressions, we have

(

rin

r=a

+ F cos

)=

rout

r=a

Using (16), this is

FP1(cos )l=0

lAlal1Pl(cos ) =

(

) l=0

(l + 1)Alal1Pl(cos ).

Multiplying both sides by Pl (cos ) and integrating from pi to pi, we find

F A1 =(

)2A1 (18)


for l=1, and

lAl =(

)(l + 1)Al (19)

(20)

for l 6= 1. Since the conductivity ratio is positive, the second relation is impos-sible to satisfy unless Al = 0 for l 6= 1. The first relation becomes

A1 =

+ 2F.

Then the potential is

(r, ) =

{

+2 Fr cos , r < a

+2 Fa

3r2 cos , r > a(21)

The dipole moment p is defined by

(r, ) 14pi0

p rr3

as r . (22)

The external portion of (21) can be written as

(r, ) =

+ 2Fa3z

r3

and comparing this with (22) we can read off

p = 4pi0

+ 2Fa3k.

The electric field is found by taking the gradient of (21):

E(r, ) =

{ +2 F k, r < a

+2 F(

ar

)3(2 cos r + sin ), r > a

The surface charge s() on the sphere is proportional to the discontinuityin the electric field:

s() = 0[Er(r = a+)Er(r = a)]

=30

+ 2F cos .

(b) The current flowing out of the upper hemisphere is justJ dA =

(Ein + F k) dA

=

(1

+ 2

)F

pi/20

2pi0

cos sin a2 d d

= 2

+ 2 pia2F (23)


The Ohmic power dissipation in a volume dV is

dP = E2dV (24)

To see this, suppose we have a rectangular volume element with sides dx, dy,and dz. Consider first the current flowing in the x direction. The current densitythere is Ex and the cross-sectional area is dydz, so I = Exdydz. Also, thevoltage drop in the direction of current flow is V = Exdx. Hence the powerdissipation due to current in the x direction is IV = E2xdV . Adding in thecontributions from the other two directions gives (24).

For the power dissipated outside the sphere we use the expression for theelectric field we found earlier:

Pout =

a

pi0

2pi0

E2(r, , )r2 sin d d dr

= 2pi(

+ 2

)2F 2a6

a

pi0

1

r4(4 cos2 + sin2 ) sin d dr

=8pi

3(

+ 2

)2F 2a3

Dividing by (23), we find the effective external voltage Ve:

Ve = Pout/I =4

3aF

+ 2

and the effective external resistance:

Re = Pout/I2 =

2

3pia.

(c) The power dissipated inside the sphere is

Pin =

(E + F k)2dV =

42

( + 2)2F 2

dV

=162

3( + 2)2pia3F 2

Since were in steady state, the current flowing out through the upper hemi-sphere of the sphere must be replenished by an equal current flowing in throughthe lower half of the sphere, so to find the internal voltage and resistance wecan just divide by (23):

Vi = Pin/I =8

3

+ 2aF

Ri = Pin/I2 =

4

3pia.


(c)

(Re + Ri)I =2

3pia

(1

+

2

) 2

+ 2pia2F =

4

3aF

(Vi + Ve) =4aF

3( + 2) + 2 =

4

3aF

Problem 3.17

The Dirichlet Green function for the unbounded space between the planes at z = 0and z = L allows discussion of a point charge or a distribution of charge betweenparallel conducting planes held at zero potential.

(a) Using cylindrical coordinates show that one form of the Green function is

G(x,x) = 1piL

n=1

m=

eim() sin

(npizL

)sin

(npiz

L

)Im

(npiL

).

(b) Show that an alternative form of the Green function is

G(x,x) = 12pi

m=

0

dk eim()Jm(k)Jm(k

)sinh(kz)]

sinh(kL).

In cylindrical coordinates, the solutions of the Laplace equation look likelinear combinations of terms of the form

Tmk(, , z) = eimZ(kz)Rm(k). (25)

There are two possibilities for the combination Z(kz)Rm(k), both of whichsolve the Laplace equation:

Z(kz)Rm(k) = (Aekz + Bekz)[CJm(k) + DNm(k)] (26)

or

Z(kz)Rm(k) = (Aeikz + Beikz)[CIm(k) + DKm(k)]. (27)

The Greens function G(x;x) must be a solution of the Laplace equation,and must thus take one of the above forms, at all points x 6= x. At x = x,G must be continuous, but have a finite discontinuity in its first derivative.


Furthermore, G must vanish on the boundary surfaces. These conditions maybe met by dividing space into two regions, one on either side of the source pointx, and taking G to be different linear combinations of terms T (as in (25)) inthe two regions. The question is, in which dimension (i.e., , z, or ) do wedefine the two sides of the source point?

(a) The first option is to imagine a cylindrical boundary at = , i.e. at theradius of the source point, and take the inside and outside of the cylinder (i.e., < and > ) as the two distinct regions of space. Then, within eachregion, the entire range of z must be handled by one function, which means thisone function must vanish at z = 0 and z = L. This cannot happen with terms ofthe form (26), so we are forced to take Z and R as in (27), with B = A and krestricted to the discrete values kn = npi/L. Next considering the singularitiesof the functions in (27), we see that, to keep G finite everywhere, for theinner region ( < ) we can only keep the Im(k) term, while for the outerregion we can only keep the Km(k) term. Then G(x;x

) will consist of linearcombinations of terms T as in (25) subject to the restrictions discussed above:

G(x;x) =

{mn Amn(x)e

im sin(knz)Im(kn

), < mn Bmn(x)e

im sin(knz)Km(kn

), > .

Clearly, to establish continuity at = , we need to take Amk(x) = mk(z, )Km(k)and Bmk(x) = mk(z, )Im(k), where mk is any function of z and . Thenwe can write G as

G(x;x) =mk

mk(z, )eim sin(kz)Im(k).

The obvious choice of mk needed to make this a delta function in z and ismk = (4/L)e

im sin(kz). Then we have

G(x;x) =4

L

mk

eim() sin(kz) sin(kz)Im(k).

What I dont quite understand is that this expression already has the correctdelta function behavior in , even though I never explicitly required this. Toobtain this expression I first demanded that it satisfy the Laplace equation forall points x 6= x, that it satisfy the boundary conditions of the geometry, andthat it have the right delta function behavior in z and . But I never demandedthat it have the correct delta function behavior in , and yet it does. I guessthe combination of the requirements that I did impose on this thing is alreadyenough to ensure that it meets the final requirement.

(b) The second option is to imagine a plane boundary at z = z, and take thetwo distinct regions to be the regions above and below the plane. In other words,the first region is that for which 0 z z, and the second region that for whichz z L. In this case, within each region the entire range of (from 0 to )must be handled by one function. This requirement excludes terms of the form


(27), because Km is singular at the origin, while Im is singular at infinity, andthere is no linear combination of these functions that will be finite over the wholerange of . Hence we must use terms of the form (26). To ensure finiteness atthe origin we must exlude the Nm term, so D = 0. To ensure vanishing at z

= 0we must take A = B, so the z function in the region 0 z z is proportionalto sinh(kz). To ensure vanishing at z = L we must take A = Be2kL, sothe z function in the region z z L is proportional to sinh[k(z L)].With these restrictions, the differential equation and the boundary conditionsare satisfied for all terms of the form (25) with no limitation on k. Hence theGreens function will be an integral, not a sum, over these terms:

G(x;x) =

{

m=0

0Am(k, , , z)e

im sinh(kz)Jm(k) dk, 0 z z

m=0

0 Bm(k, , , z)eim sinh[k(z L)]Jm(k) dk, z z L

Problem 3.18

The configuration of Problem 3.12 is modified by placing a conducting plane heldat zero potential parallel to and a distance L away from the plane with the discinsert in it. For definiteness put the grounded plane at z = 0 and the other planewith the center of the disc on the z axis at z = L.

(a) Show that the potential between

Classical Electrodynamics - Jackson (Solucionario 1)

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