# Electromagnetics - Surface Charge Density

1. Mar 15, 2012

### BigD959919

I answered my own question but thank you for the help.

Last edited: Mar 15, 2012
2. Mar 15, 2012

### dipole

I can't read your equations because you're not using LaTeX, but the general way to approach this would be to solve for the electric field at the surface of the sphere (only the normal component is non-vanishing, because it is a conductor) by first writing down boundary conditions on the potential, and then solving for the potential and finding the field from this.

Since the sphere is a conductor, it's surface forms and equipotential, so let the radius of the sphere be R, then you have two boundary conditions:

$\Phi(R) = \Phi_{0}$ where $\Phi_{0}$ is some constant which must be specified in the problem, if the sphere is grounded then it is zero. The other B.C. is:

$\Phi(\infinity) = E_{0}r\cos(\theta)$ this comes from the fact that at infinity the field behaves as if the sphere isn't there.

This is enough to get the potential, and the field at the surface. Once you have these things, then use gauss's law:

$$\nabla*E = \frac{\rho}{\epsilon_{0}}$$

to find the charge distribution.

(hint: You'll find that the sphere produces a dipole)