1. A conducting sphere of radius R is kept in an electrostatic field given by

. 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

. Using the table for Spherical harmonics, the potential can be expressed as

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

(ii )For large distances, , . in this limit,, the second term for the expression for potential goes to zero and we are left with,

Comparing, we get only non-zero value of to be 2 and the corresponding *m* values as . Thus, the potential has the form

The surface charge density is given by

Because of symmetry along the z axis, the potential can only depend on the polar coordinates

. By using a separation of variables, we can show, as shown in the lecture, the angular part

The radial equation has the solution

, 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

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

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

These integrals can be done by elementary methods,

If

If ,

Thus,

To determine the coefficients , multiply both sides of (I) by and integrate from 0 to 2p,

L.H.S. gives,

The first term on the right gives zero.

We are left with

The integral can be easily done If

If ,

Thus,

Let us write,

We have the final expression for the potential,

To verify that this is consistent with the given boundary conditions, let us evaluate the potential at ,

The infinite series has a value . Adding , we get . Similarly, one can check that .