1
PHYS809Class12Notes
Potential of grounded spherical conductor in a uniform external electric field
The potential for a uniform field
E
0
is
0
0
.
φ
= 
⋅
E
r
Since there is azimuthal symmetry about the axis through the center of sphere that is parallel to the
direction of the external field, we look for an external solution of form
(
)
(
)
(
)
0
0
1
1
1
0
0
cos
cos
cos
.
l
l
l
l
l
l
l
l
B
B
P
P
E rP
r
r
φ
θ
φ
θ
θ
∞
∞
+
+
=
=
=
+
=

∑
∑
Since the potential is zero on the sphere, which we take to have radius
a
, we see that
B
l
= 0
except for
B
1
. Hence
(
)
3
3
0
0
1
0
2
2
cos
cos
.
E a
a
E r
P
r
E
r
r
φ
θ
θ
=

=

This is the same as the solution found using the method of images.
Solution using expansion on the symmetry axis
An important result for an azimuthally symmetric solution of Laplace’s equation is that it completely
determined from the potential on the symmetry axis, which we will take to be the
z
axis. On the
symmetry axis
( )
(
)
(
)
(
)
1
0
1
0
,
0,
1
,
0.
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 Fall '11
 MacDonald
 Complex number, Legendre polynomials, AdrienMarie Legendre

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