1
$\begingroup$

I have a question where there is a grounded conducting sphere and inside that sphere is another sphere which is non-conducting and has a static surface charge density of $\sigma(\theta) = \sigma_0cos(\theta)$. If I want to find the electric potential inside the larger sphere, from looking at other questions I think I need to use Laplace's equation, but unsure why. I understand Laplace's equation is just a special case of Poisson's equation where $\rho$ = 0, and $\rho$ is the charge density. Here since we have a surface with surface charge density, isn't $\rho$ non zero? In general, in what scenarios is Laplace's equation used and when is Poisson's?

$\endgroup$
1
$\begingroup$

While $\rho$ is nonzero at the boundary of the region you're studying here, $\rho=0$ everywhere inside the region where you're trying to find the electric potential, namely in the region between the two spheres. Therefore the potential satisfies Laplace's equation everywhere in that region.

Poisson's equation would be useful in the instance where rather than having a surface charge density and looking for the electric potential outside of it, you were instead given a region with a specified volume charge density and asked for the electric displacement field ${\bf D}$ inside. Then there would be a charge density filling the space where you're looking for the solution, rather than just being present on the boundary.

$\endgroup$
0
$\begingroup$

You are in fact solving the Poisson's equation $$\nabla^2\Phi=-\rho/\epsilon_0$$

But since it is a surface charge density, in differential form the $\rho$ becomes $\infty$.

The solution is to go back to the integral form and integrate across the surface, which reduces to matching boundary conditions inside and outside the sphere, both regions inside which you solve the Laplace's equation since there is no charge.

$\endgroup$
0
$\begingroup$

This configuration is very special. I just wanted to tell you a beautiful way to solve this problem. Please read this only after trying to solve the problem without integration. You can treat this configuration as two uniformly charged solid spheres with equal and opposite charge density whose centers are a very small distance apart. Then it becomes a classic example of -ve charge superimposed on +ve charge, and you will find that the field inside is constant. The proof is sufficiently easy. You can try it if you don't know it already. I'm sorry if this is not what you asked, I just wanted to tell you. Hope it helps.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.