Today in Physics 217: separation of variables III
A more complicated example:
‰ Concentric spheres, only one
a conductor
‰ Mixed boundary conditions
• specified V
• specified charge density:
apply though continuity
of V, discontinuity of
normal component of E
16 October 2002
Physics 217, Fall 2002
z
a
V0
b
σ (θ )
1
Solution for mixed boundary conditions
Example: Griffiths problem 3.37
A conducting sphere of radius a, at potential V0, is
surrounded by a thin concentric spherical shell of
radius b, over which someone has glued a surface
charge
σ (θ ) = σ 0 cosθ ,
where σ 0 is a constant.
a. Find the electrostatic potential in each region:
i)
r>b
ii)
a<r<b
b. Find the induced surface charge on the conductor.
c. What is the total charge of the system? Check that
your answer is consistent with the behavior of V at
large r.
16 October 2002
Physics 217, Fall 2002
z
a
V0
b
σ (θ )
2
Solution for mixed boundary conditions
(continued)
Before we start, a few remarks about what this looks like.
‰ Total charge on the shell is
z
Qshell = σ 0 b 2
2π
∫
0
π
dφ ∫ cosθ sin θ dθ
0
2
2 cos θ
= 2πσ 0 b
2
+
+ +++ +
π
+
a
=0 .
0
V0
b
‰ So the total charge on the system
- - - - - σ (θ )
is that on the sphere, which should
be Q = Va, though it won’t be distributed
uniformly on any of the surfaces.
‰ Potential should be proportional to Q/r at large r.
16 October 2002
Physics 217, Fall 2002
3
Solution for mixed boundary conditions
(continued)
This system is spherical and axisymmetric, and on Monday
we found that the most general solution for the potential in
this situation is
∞
BA 

A
V ( r ,θ ) = ∑  AA r + A + 1  PA ( cosθ )
r

A =0 
Boundary conditions:
1. σ (θ ) = σ 0 cosθ at r = b
2. V = V0 at r = a
This means:
3. V → 0 at r → ∞
V r =b + − V r =b − = 0
 ∂V
−
 ∂r
r =b +
16 October 2002
∂V
−
∂r

 = 4πσ 0 cosθ
r = b − 
Physics 217, Fall 2002
4
Solution for mixed boundary conditions
(continued)
Tactics for these mixed boundary conditions:
‰ Learn as much as you can first from the simpler
conditions, such as fixed potentials (e.g. conductors,
V → 0 at r → ∞ , V finite at r = 0 ). This may result in
elimination of large numbers of terms from the general
solution.
‰ Then apply continuity of V, discontinuity of E.
• Fourier’s Trick may have to be used here. (Not in this
particular problem, though.)
• This may generate a complicated web of relations
among the constants. Don’t despair, and try to match
up relations among those for the same index (A) for
quickest results.
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Physics 217, Fall 2002
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Solution for mixed boundary conditions
(continued)
First apply boundary condition 3, to r > b:
∞

A Bout, A 
r → ∞ : V ( r ,θ ) = ∑  Aout, A r + A + 1  PA ( cosθ ) → 0
r

A =0 
∴ Aout, A = 0
Then 2, to r = a:
∞

A Bin,A
V ( a ,θ ) =
 Ain,A a + A + 1
a
A =0 
∑
whence
Ain,0 +
Bin,0
Ain,A aA +
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
 PA ( cos θ ) = V0 = V0 P0 ( cos θ )

= V0
a
Bin,A
a
A+1
if A = 0
= 0 if A > 0
Physics 217, Fall 2002
6
Solution for mixed boundary conditions
(continued)
Now for boundary condition 1. Continuity of V:
∞
∞ B

out,A
=
θ
P
P ( cos θ )
cos
(
)
 A
A+1 A

A =0 b
Bout,A
A Bin,A
Ain,A b + A + 1 = A + 1 , or
b
b

A Bin,A
 Ain,A b + A + 1
b
A =0 
∑
∑
Bout,A − Bin,A = Ain,A b 2 A + 1
Discontinuity of normal derivative of V (electric field):
 ∂V
−
 ∂r
16 October 2002
r =b +
∂V
−
∂r

 = 4πσ 0 cosθ = 4πσ 0 P1 ( cosθ )
r = b − 
Physics 217, Fall 2002
7
Solution for mixed boundary conditions
(continued)
Take the derivatives:
∂V
∂V
−
+
∂r r =b + ∂r
∞
r =b −
A+1
=
 A + 2 ( Bout,A − Bin,A )
A =0  b
∑
+AAin,A bA −1  PA ( cosθ )

= 4πσ 0 P1 ( cosθ )
Thus
2
3
b
A+1
b
16 October 2002
( Bout,1 − Bin,1 ) + Ain,1 = 4πσ 0
A −1
A
B
B
A
b
−
+
=0
(
)
out,A
in,A
in,A
A+2
Physics 217, Fall 2002
if A = 1
if A ≠ 1
8
Solution for mixed boundary conditions
(continued)
Substitute results from continuity of V into this latest result to
eliminate Bs:
A+1
2A+1
A −1
A −1
+
=
+
= 0 if A ≠ 1
A
A
A
b
A
b
A
b
2
1
(
)
in,A
in,A
A + 2 in,A
b
2
3
+ Ain,1 = 3 Ain,1 = 4πσ 0 if A = 1
A
b
3 in,1
b
4πσ 0
or
if A = 1
Ain,1 =
Ain,A = 0 if A ≠ 1
3
Put these back into the V continuity result, and get…
4πσ 0 3
Bout,1 − Bin,1 =
b if A = 1
3
Bout,A − Bin,A = 0 if A ≠ 1
16 October 2002
Physics 217, Fall 2002
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Solution for mixed boundary conditions
(continued)
Return to the result from r = a:
Bin,A = − Ain,A a2 A + 1 = 0 if A ≥ 2
4πσ 0 3
3
Bin,1 = − Ain,1 a = −
a if A = 1
3
Bin,0 = a (V0 − Ain,0 ) = aV0 if A = 0
These values for Bin,A immediately fix the values for Bout,A:
Bout,A = Bin,A = 0 if A ≥ 2
4πσ 0 3
4πσ 0 3 3
Bout,1 =
b + Bin,1 =
b −a
3
3
(
Bout,0 = Bin,0 = aV0
16 October 2002
Physics 217, Fall 2002
)
if A = 1
if A = 0
10
Solution for mixed boundary conditions
(continued)
The potential outside the shell is therefore
(
)
4πσ 0 1 3 3
aV0
Vr >b ( r ,θ ) =
P0 ( cosθ ) +
b − a P1 ( cosθ ) ,
2
3 r
r
and that between the shell and the sphere is
4πσ 0
aV0
Va<r <b ( r ,θ ) =
P0 ( cosθ ) +
3
r

a3
 r − 2
r


 P1 ( cosθ ) .

Of course it’s constant, and = V0, within the sphere.
16 October 2002
Physics 217, Fall 2002
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Solution for mixed boundary conditions
(continued)
The other two parts of the problem are easier: first the charge
density induced on the sphere:

∂V
1  ∂V
−
σ (θ ) = −


4π  ∂r r = a + ∂r r = a − 
1  V0
 V0
=−
−
+ 4πσ 0 cosθ  =
− σ 0 cosθ

4π  a
 4π a
And the total charge on the sphere:
Qa = ∫
π
0
2π
∫0
σ (θ ) a2 sin θ dθ dφ
π
V0

= 2π a  2
− σ 0 ∫ cosθ sin θ dθ  = aV0
0
 4π a

Oh yeah, I guess that’s what it should be.
2
16 October 2002
Physics 217, Fall 2002
12
Solution for mixed boundary conditions
(continued)
The total charge on the shell is zero (see above). Therefore
the total charge of the system is
Qtotal = aV0
Thus the electrostatic potential at large distances is
approximately equal to
Qtotal aV0
=
V=
r
r
as it should.
To convert to MKS answers: multiply RHS by 4πε 0
for charges and charge densities; divide RHS by 4πε 0
for potentials.
16 October 2002
Physics 217, Fall 2002
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