# Potential inside NON-Conducting hollow spherical shell

1. Sep 15, 2011

### dikmikkel

Hi Guys,
Suppose we have a spherical shell with charge density on the surface $\sigma$ and radius R. The potential inside the shell is given by:

V_(x,y,z) = $\frac{V0}{R^{2}}(6z^2+ax^2+by^2)$
It is assumed, that the potential is rotational symmetric around the z-axis inside and outside the shell, and goes to 0 far away from the shell. There's no charge inside and outside the shell and no outer field.

How do i determine the constants a and b?

Mabye change to spherical coordinates and solve the equation:
$\frac{\partial{V}}{\partial{\theta}}=0$
for a or b. But i can't figure out any other conditions if this is right.

2. Sep 15, 2011

### Born2bwire

The Laplacian of the voltage is related to the charge. So your boundary conditions would be:

$$\nabla^2 V = -\frac{\sigma}{\epsilon}$$

On the surface of the shell. So for ease of analysis you would want to do this in spherical coordinates.

3. Sep 15, 2011

### dikmikkel

I can't really see that this is a boundary condition? And would solving possions equation on the surface help me determine a and b, i can't really follow?

4. Sep 15, 2011

### dikmikkel

Aah sorry im really tired now. I take the laplacian, and this would relate it all, beautiful, Thanks. Just so used to try finding the potential :)
And sorry for the language. I'm from Denmark.

5. Sep 15, 2011

### clem

If the pot inside is axially symjmetric, and b must be equal.
Since DEl^2 V=0 inside, the angular distribution must be a Legendre polynomial.
Since the power of x,y,and z is 2, it must be
$$P_2(\cos\theta)=(3\cos^2\theta-1)/2$$.
This means a=b=-3.

Last edited: Sep 15, 2011
6. Sep 15, 2011

### dikmikkel

That's gold thank you.