1. A conducting sphere of radius R is kept in an electrostatic field given by
![](Selv Assesment_clip_image002.png)
. Obtain an expression for the potential in the region outside the sphere and obtain the charge density on the sphere
The potential corresponding to the electric field is
![](Selv Assesment_clip_image002_0002.png)
. Using the table for Spherical harmonics, the potential can be expressed as
![](Selv Assesment_clip_image004_0001.png)
The general expression for solution of Laplace’s equation is
The boundary conditions to be satisfied are :
(i) For all
, for
, the potential is constant, which we take to be zero. This implies
for all m. This makes the potential expression to be
![](Selv Assesment_clip_image014.png)
(ii )For large distances,
,
. in this limit,, the second term for the expression for potential goes to zero and we are left with,
![](Selv Assesment_clip_image020.png)
Comparing, we get only non-zero value of
to be 2 and the corresponding m values as
. Thus, the potential has the form
![](Selv Assesment_clip_image026.png)
The surface charge density is given by
![](Selv Assesment_clip_image028.png)
Because of symmetry along the z axis, the potential can only depend on the polar coordinates
![](Selv Assesment_clip_image002_0004.png)
. By using a separation of variables, we can show, as shown in the lecture, the angular part
![](Selv Assesment_clip_image004_0003.png)
The radial equation has the solution
![](Selv Assesment_clip_image006_0003.png)
, we have ignored n=0 solution because it gives rise to logarithm in radial coordinate which diverges at origin. Thus the general solution for the potential is
where K is a constant. Further, the constant D must be zero otherwise the potential would diverge at
. Let us now apply the boundary conditions, For ![](Selv Assesment_clip_image012_0001.png)
![](Selv Assesment_clip_image014_0001.png)
![](Selv Assesment_clip_image016_0000.png)
Integrate the first expression from
and the second expression from
and add. The integration over the second term on the right hand side of the two expressions is equal to an integration from 0 to 2p and the integral vanishes. We are left with
![](Selv Assesment_clip_image022_0000.png)
Thus the potential expression at
becomes
(I)
To evaluate the coefficients
, multiply both sides by
and integrate from 0 to 2p,
L.H.S. gives,
Similarly, the first term on the right also gives zero.
We are left with
![](Selv Assesment_clip_image034.png)
These integrals can be done by elementary methods,
If ![](Selv Assesment_clip_image036.png)
![](Selv Assesment_clip_image038.png)
If
,
![](Selv Assesment_clip_image042.png)
Thus,
To determine the coefficients
, multiply both sides of (I) by
and integrate from 0 to 2p,
L.H.S. gives,
![](Selv Assesment_clip_image050.png)
The first term on the right gives zero.
We are left with
![](Selv Assesment_clip_image054.png)
The integral can be easily done If ![](Selv Assesment_clip_image036_0000.png)
![](Selv Assesment_clip_image056.png)
If
,
![](Selv Assesment_clip_image058.png)
Thus,
![](Selv Assesment_clip_image060.png)
Let us write, ![](Selv Assesment_clip_image062.png)
We have the final expression for the potential,
![](Selv Assesment_clip_image064.png)
To verify that this is consistent with the given boundary conditions, let us evaluate the potential at
,
![](Selv Assesment_clip_image068.png)
![](Selv Assesment_clip_image070.png)
The infinite series has a value
. Adding , we get
. Similarly, one can check that
.