What is the electric potential inside a finite box with a charge density?

Question

In order to find the potential inside a box, we use the Laplace equation if there is no charge. For example, for an infinitely long rectangular box along $z$-axis with sides at $x=0, \;x=a, \;y=0, \;y=b$, we can write that:

$$\frac{1}{X} \frac{\partial^2 X}{\partial x^2} + \frac{1}{Y} \frac{\partial^2 Y}{\partial y^2} = 0$$ when $V(x,y)=X(x)Y(y)$

We can use the boundary values(Let's say all the sides are at V=0) and solve this using separation of variable.

But my question is what happened when there is a charge density inside the box? Let's say I place a plane charge density $σ(y) = C\sin{(2πy/b)}$ at $x=a/2$ parallel to the y-z plane where $C$ is a constant.

Now I cannot use separation of variable anymore because there is a nonzero term on the right hand side of the Poisson equation. How can we solve this?

I found that we can solve this either numerically or using Green's Function method. I am asking if we can solve this problem other than these two methods.

Answer 1

Actually, you can still solve your problem using the separation of variables. A first method would be to solve Laplace’s equation in the two halves of the box using a Neumann boundary condition on the charged plane. Indeed, from symmetry considerations and the formula for discontinuity of electric field with surface charges, you know the electric field on the left/right limits of the plane charge.

Another method would be to use the separation of variables on the whole box which has the advantage of being more general. Indeed, you can always decompose the charge density in the eigenmodes of the Laplacian. Your equation becomes 1d linear so you find the correct coefficient in front of each mode of your potential.

Hope this helps.