Electrostatics

[Arjuna]

Hi!

Need help with this problem. I tried it out but not sure about what I've done.

The potential at the surface of a sphere (radius R) is given by V_0 = k cos 3 , where k is a constant. Find the potential inside and outside the sphere as well as the charge density on the sphere. (Assume there's no charge inside or outside the sphere.)


Can someone help me with this one?

[avinash]

In the following replace by +. There is some problem with using + with math symbols. That is why I write

The general solution of Laplace's equation in spherical coordinates is:
V(r, ) =

For a region inside the sphere(r < R), the value of Bk must be 0 as otherwise will become infinitely large at r=0
Therefore
V(r, ) =
The potential at the surface(r=R) is therefore
V(R, ) = (given)


Substituting the above expression in the equation for V(R, ), we get
V(R, ) =
The above equation shows that
A1 * R = -3k/5 OR A1 =
A3 * R^3 = 8k/5 OR A3 =
Ak = 0 for any value of k except 1 and 3
The electrostatic potential inside the sphere is therefore


Consider a region outside the sphere (r > R). In this region Ak = 0 as otherwise Ak r^k will become infinitely large as r->infinity
Therefore
V(r, ) =

From the above equation
B1 = {-3/5} k R^2
B3 = {8/5} k R^4
Bk = 0 for any value of k except 1 and 3

The electrostatic potential inside the sphere is therefore
V(r, ) =

Will add derivation for charge density later.

[Arjuna]

Thanks a lot.

I was trying to find the co-efficients, the A's and B's by Fourier's formula.

A_l = 1 / 2*R^l-1 * }{V(r,) P_l(cos) sin d}[/mathSymbols]

Integral 0 to pi V(r, theta)*P_l*cos (theta)*sin(theta)

Similarly for B_l's, (2l+1) / 2 R^(l+1) followed by the same integral.


But there is one thing in the solution that I don't understand. The potential at the surface is given and the expansion is as you have given. But how can we substitute that for the potential inside and outside the sphere as you have explained?



When I tried out this problem, I got a value for A_0 to be -3*k*R /

[avinash]

I used general Laplace solution for inside and outside the sphere.
The potential just inside the sphere = the potential just outside the sphere = the potential at the surface = the potential given in the question
So in the formula for the potential inside the sphere if we substitute r = R, then we get the expression given in the question.
Likewise, in the formula for the potential outside the sphere if we substitute r = R, then also we get the expression given in the question.

[Arjuna]

I get it now. Thanks - after finding the two coefficients, it is simple to solve the problem.