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13.3) We would like to find the potential created by a thin, flat conducting circular disc of radius R located in thex y plane, mantained at fixed potential V . The charge density on a disc is proportional to 1/pR2 r2, where r isthe distance from the center of the disc.WARNING: Unfortunately the solution in the Jackson is wrong. Nevertheless, we will give credit

to everyone who solved the problem according to Jackson!To show that the Jacksons solution is wrong, we note that there is no charge on the sphere at r = R in 3D, hence

we can use analytic continuation of from r > R to r < R at finite ( = 0 might be problematic because the discis located in the x y plane). Using analytic continuation, we conclude that Jackson proposes solution

(r < R) =2V

1Xl=0

(1)l2l + 1

rR

2lP2l(cos ) (1)

which gives at = /2 non-constant potential, i.e.,

(r < R, = /2) =2V

+

2V

1Xl=1

(1)l2l + 1

rR

2lP2l(0) 6= V (2)

Correct solution of 3.3All points in 3D, except at ( = /2, r < R), satisfy the Laplace Eq. r2 = 0. The solution at all points except

at /2 has to satisfy

inside(r) =Xl

Bl rR

lPl(cos ) (3)

outside(r) =Xl

Bl

R

r

l+1Pl(cos ) (4)

We already took into account the analytic continuation across the boundary at r = R. We might have dierent Blsfor < /2 and > /2, because there is a charged disc present at = /2. At = /2 the potential is continuous,but it must have discontinuous derivative (because electric field jumps). Let us concentrate on < /2 for now.Approaching point /2 from above, we have the condition of a constant potential on the surface of the conducting

disc, i.e.,

V = inside(r, = /2) = B0 +1Xl=1

Bl rR

lPl(0) (5)

This is satisfied for all r < R only if B0 = V and B2n = 0 (n integer) because P2n(0) 6= 0. On the other handP2n+1(0) = 0, hence arbitrary coecients B2n+1 are possible. We thus have

inside(r) = V +1Xn=0

B2n+1 rR

2n+1P2n+1(cos ) (6)

outside(r) = VR

r+1Xn=0

B2n+1

R

r

2n+2P2n+1(cos ) (7)

To determine coecients B2n+1, we note that the electric field needs to be such that it gives rise to the given surfacecharge distribution = p

R2r2 . The electric field just above the conducting disc is

Er = 0 (8)

E = 1r

@(r, = /2)

@=

1Xn=0

B2n+1r2n

R2n+1dP2n+1dx

(0) (9)

(10)

The normal component of the electric field is proportional to the surface charge, hence

/2pR2 r2 = 0E

inside ( = /2), (11)

2where we took into account that half of the charge appears on the upper part of the conducting disc, and half on thelower part. This equation has to be satisified for every r < R, hence every power of r has to match. Note that onlyeven powers in r appear on both the right and the left hand side. We thus perform power expansion and compareterm by term:

/2

R

1Xn=0

(n+ 12 )

(n+ 1)(1/2)

rR

2n= 0

1Xn=0

B2n+1r2n

R2n+1dP2n+1dx

(0) (12)

(13)

We thus conclude

B2n+1 =

20P 02n+1(0)(n+ 12 )

(n+ 1)(1/2)(14)

which can be simplified to

B2n+1 =

20

(1)n2n+ 1

(15)

We finally have for < /2 the following solution

inside(r) = V +

20

1Xn=0

(1)n2n+ 1

rR

2n+1P2n+1(cos ) (16)

outside(r) = VR

r+

20

1Xn=0

(1)n2n+ 1

R

r

2n+2P2n+1(cos ) (17)

To continue the solution below the x y plane (to > /2) we notice that E must change sign across x y plane,but it needs to be of equal strength above and below the plane. Lets check the first few terms in inside close to thex y plane:

inside V + 20

z

R

120

(5z3 3zr2)R3

+ (18)

which gives

E = ez

20

1

R+

40

r2

R3+

(19)

The electric field should change sign, and therefore should be of the form

E = ezsign(z)

20

1

R+

40

r2

R3+

(20)

therefore potential must also contain sign(cos ), i.e.,

inside(r) = V +

20

1Xn=0

(1)n2n+ 1

rR

2n+1P2n+1(cos )sign(cos ) (21)

outside(r) = VR

r+

20

1Xn=0

(1)n2n+ 1

R

r

2n+2P2n+1(cos )sign(cos ) (22)

An interesting question is why does the integral of the charge not give correct answer.We would expect

(r) =1

40

Z(r0)

|r r0|dr0 (23)

Taking into account the expansion of 1|rr0| in terms of spherical harmonics, and noting that the problem is independentof azimuthal angle ', we conclude

(r, r > R) =1

40

Xl

Pl(cos )

rl+1

Z(r0)Pl(cos 0)r0

ld3r0 (24)

3Naively, one would expect (r, ) = (cos )rpR2r2 . This charge density leads to the solution in the Jacksons book, which

is wrong. The problem is that the conducting disc does not contain only one thin layer of charge, but actually two

thin layers, one above and one below, which give the charge density of the form (r, ) = 0(cos )sign(cos )rpR2r2 , where

0(x) is the derivative of the -function. Only this form leads to the correct solution derived above.The wrong solution follows

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