Homework 9 Answers, 95.657 Electromagnetic Theory I
Dr. Christopher S. Baird, UMass Lowell
Jackson 4.10
Two concentric conducting spheres of inner and outer radii
a
and
b
, respectively, carry charges ±
Q
. The
empty space between the spheres is halffilled by a hemispherical shell of dielectric (of dielectric
constant
ε
/
ε
0
), as shown in the figure.
(a) Find the electric field everywhere between the spheres.
SOLUTION
We can split the region where we want to know the electric field into two regions, the left hemisphere
and the right hemisphere, solve for the field in each region separately and then apply the boundary
conditions to get the final solution. In the left region, there are no dielectrics and no charges. So the
electric potential obeys the Laplace equation:
∇
2
L
=
0
If we align the
z
axis pointing to the right, the problem has azimuthal symmetry. The general solution
to the Laplace equation in spherical coordinates for azimuthal symmetry is:
L
r ,
,
=
∑
l
A
l
r
l
B
l
r
−
l
−
1
P
l
cos
The potential on the surface of a conductor is always constant:
L
r
=
a
=
C
C
=
∑
l
A
l
a
l
B
l
a
−
l
−
1
P
l
cos
Because this equation must hold for all values of the independent polar variable and because the
Legendre polynomials are orthogonal, all coefficients must equate independently:
C
=
A
0
B
0
a
and
B
l
=−
A
l
a
2
l
1
for
l
> 0
a
b
+
Q
Q
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The solution now becomes:
L
r ,
,
=
C
−
B
0
a
B
0
r
∑
l
=
1
∞
A
l
r
a
l
−
a
r
l
1
P
l
cos
The other surface is also a conductor and must have a constant potential as well:
L
r
=
b
=
D
D
=
C
−
B
0
a
B
0
b
∑
l
=
1
∞
A
l
b
a
l
−
a
b
l
1
P
l
cos
Because this equation must hold for all values of the independent polar variable and because the
Legendre polynomials are orthogonal, all coefficients must equate independently:
B
0
=
ab
D
−
C
a
−
b
and
A
l
=
0
for
l
> 1
The solution now becomes:
L
r ,
,
=
C
ab
D
−
C
a
−
b
−
1
a
1
r
At this point
C
and
D
are undetermined, so we can redefine them to simplify the equation:
L
r ,
,
=
D
C
r
This leads to a total electric field of:
E
L
=
C
r
2
r
We could have guessed this form of the solution based on the symmetry of the problem, but it is often
safer and more instructive to go through all the steps.
The right hemisphere is a separate region and can be now solved separately. In the region between the
shells, we can tell that there are no free charges and no bound charges (the bound charges, or
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