Charge Inside a Cavity in a Conductor

[Luke Tan]

Let us say we have a cavity inside a conductor. We then sprinkle some charge with density $\rho(x,y,z)$ inside this surface.

We have two equations for the electric field
$$\nabla\times\mathbf{E}=0$$
$$\nabla\cdot\mathbf{E}=\frac{\rho}{\epsilon_0}$$

We also have the boundary conditions
$$E=\frac{\sigma(x,y,z)}{\epsilon_0}\hat{n}$$
At the surface of the cavity.

However, the surface charge $\sigma(x,y,z)$ is an unknown function. Since we do not know a solution for $\mathbf{E}$, we are unable to determine this surface charge. However, this makes no sense.

If I were to sprinkle a charge density $\rho(x,y,z)$ that is known, surely there can only be one possible surface charge distribution. What equation am I missing?

I would think that it involves the geometry of the conductor outside the cavity, but I'm not sure how to add this in.

Thanks!

 

[vanhees71]

This is a very difficult to solve problem for a general shape of the cavity. A simple standard solution can be given for the sphere, because of its high symmetry. You can apply the method of an image charge to get the Green's function of this problem and then your question is answered by the corresponding folding of this Greens' function with the given charge distribution, $\rho$.

The influenced surface charge is then indeed given by the jump of the normal component of $\vec{E}$ along the surface (it's the "surface divergence" of $\vec{E}$), but you cannot get it before you have solved the full boundary problem of the Poisson equation.

 

[Luke Tan]

By "solving the full BVP of the poisson equation", do you mean inside the cavity only?

I'm not sure how we can do that, since we need the boundary conditions to get a unique solution to the Poisson equation, but I'm trying to do it in reverse here.

 

[sysprog]

There's a good discussion of the relevant electrostatics, using a Van De Graaff generator as a practical example, here: http://demoweb.physics.ucla.edu/sites/default/files/Physics6B_Exp4.pdf

 

[vanhees71]

By "solving the full BVP of the poisson equation", do you mean inside the cavity only?

I'm not sure how we can do that, since we need the boundary conditions to get a unique solution to the Poisson equation, but I'm trying to do it in reverse here.

As I said, in the general case it's a very hard problem. I'd say most of them can be handled only numerically. The one example you can do with standard textbook analytical methods is a spherical shell (or of course the infinite plane, but that's not really a cavity and has its own problems).

 

[Luke Tan]

As I said, in the general case it's a very hard problem. I'd say most of them can be handled only numerically. The one example you can do with standard textbook analytical methods is a spherical shell (or of course the infinite plane, but that's not really a cavity and has its own problems).

yea I know that it's really hard, but now I feel like I'm lacking equations (the solution to the poisson equation isn't unique, unless boundary conditions are specified, and in this case they are not - I am trying to find the boundary conditions from the solution), and so I can't even solve it computationally.

 

[vanhees71]

The boundary conditions are that the surfaces of conductors are equipotential surfaces (such that the tangent components of $\vec{E}$ vanish along the surface, such that you don't have surface currents, as is assumed for electrostatics). If your conductor is isolated then in addition you can also assume the total amount of surface charge on the conductors surface. These are the complete boundary conditions needed.