# Electrostatic boundary value problem with dielectrics

1. Sep 3, 2011

### Fernbauer

I am trying to understand how to deal with a boundary value problem when there are multiple dielectrics inside the volume.

We want to solve for the potential inside a charge-free axis-aligned 3D rectilinear volume.
5 of the 6 sides of the volume are at 0 potential. The sixth side (call it the bottom, with Z=0) has a known fixed potential V(x, y, 0). Solve for V(x, y, z) inside the volume.

This is a classic boundary value problem, and nearly every electrostatics text shows the solution by separation of variables, building up the base potential by a Fourier series, with each term decaying in Z based on its frequency.

Now to involve dielectrics. Take the same problem, but now the volume X<C is filled with an ideal linear dielectric with permittivity $\epsilon_A$ and the volume X>C is filled filled with an ideal linear dielectric with permittivity $\epsilon_B$.

What is the potential everywhere in the volume?

I see two ways of attacking this. My first thought is to take the original, known, solution for constant $\epsilon$ and look at the plane V(x=c, y, z), specifically at the Ex component of it (just $\frac{\partial V}{\partial x}$). From dielectrics, we know the real solution is continuous in V here but discontinuous in Ex, so we can determine the amount of induced charge needed so it creates the required Ex ratio. Then use that induced charge to solve two more boundary value problem for each of the two halves independently, this time dealing with constant $\epsilon$. Sum the three solutions for the net solution. This approach feels ugly, and I'm sure the algebra will explode into many lines of fiddly nested integrals. And I'm not even sure if it's a way to actually make the proper answer.

The other way to attack the problem is by images. All the electrodynamics texts talk about solving planar dielectric boundary value problems like this (for example, in Jackson, page 155, and Jackson, page 189) so it's natural to think that maybe V(x, y, z) could be represented by some weighted sum of the known constant $\epsilon$ solution, like A*V(x, y, z) + B*V(2C-x, y, z) for computable weights A and B (which would be different for the two $\epsilon$ regions.)

The image approach is a lot more elegant in that it is easy to evaluate, no pages of algebra, but it also can't be correct, because what if the image point is reflected across X=C to a point that's outside the bounding region? That makes no sense, so this simple guess can't be right.

But then maybe the images aren't using the same bounding region.. maybe for the left side, we reflect the left part of the bounding region itself over X=C to make it symmetric, solve that.. similarly for the right side, then compute some weighted sum of the two solutions... but now I'm just making up strategies with no justification. But as Griffith himself says on page 190, "like all image solutions, this owes its justification to the fact that it works... image solutions are always a matter of guesswork."

So, I open the question to the experts: what's the best way to attack this dielectric boundary value problem?

Thanks! It's a fun puzzle I think!