What Is the Electric Potential Inside an Insulated Metal Cube?

[jerro]

Homework Statement

A metal box has 6 walls, all insulated from one another. The left and right wall are held at V= V0, which are at y=-d and y=d. All the other walls are grounded.

The cube has dimensions where walls run from x=0 to x=2d, z=0 to z=2d, and y=-d to y=d.

Homework Equations

separation of variables in 3D:
\frac{1}{X}\frac{\partial^{2} V}{\partial X} + \frac{1}{Y}\frac{\partial^{2} V}{\partial Y} + \frac{1}{Z}\frac{\partial^{2} V}{\partial Z} =0

Boundary conditions:
V(x,0) = 0
V(x,2d) = 0
V(y, -d) = V0
V(y,d) = V0
V(z,0) = 0
V(z,2d) = 0

The Attempt at a Solution

I am thinking that there is symmetry around z, so we can only worry about x and y.

V(x,y) = (Asinh(kx) + Bcosh(kx))(Ce^{ky} + De^{-ky})
which simplifies to (2cosh(ky)(Asinh(kx) + Bcosh(kx)).

Assuming all the above is correct, I am having issues at this point. I need to put this into a sum, which according to Griffiths should look something like

\sum Cn cosh( npiy/a)sin(npix/a) = V0.

I don't understand how to simplify this. I have limits of integration, but I am not sure where an integral can arise from this.

Thank you.

 

[haruspex]

Seems more natural to me to have all three coordinates run from -d to +d.

I am thinking that there is symmetry around z, so we can only worry about x and y.

Yes, there are symmetries, but I don't see this allows you to write the potential as a function of only two coordinates. It should be an even function wrt x, y and z, and there should be a symmetry between x and z.

 

[jerro]

Ok, that makes sense. So, with all three coordinates, the expression should be:

∑ Cn cosh( npiy/a)sin(npix/a)sin(npiz) = V0.

But what's next?

 

[haruspex]

Just noticed this was wrong in the OP:

\frac{1}{X}\frac{\partial^{2} V}{\partial X} + \frac{1}{Y}\frac{\partial^{2} V}{\partial Y} + \frac{1}{Z}\frac{\partial^{2} V}{\partial Z} =0

You mean (1)$\frac{\partial^2 V}{\partial X^2}+\frac{\partial^2 V}{\partial Y^2}+\frac{\partial^2 V}{\partial Z^2} =0$and by separation of variables to V=F(x)G(y)H(z):
(2) $\frac1F\frac{\partial^2 F}{\partial X^2}=a_n, \frac1G\frac{\partial^2 G}{\partial Y^2}=b_n, \frac1H\frac{\partial^2 H}{\partial Z^2} =c_n$ where $a_n+b_n+c_n=0$.

∑ Cn cosh( npiy/a)sin(npix/a)sin(npiz) = V0.

Surely that should be something like V(x,y,z) = ∑ Cn cosh( nπy/d)sin(nπx/d)sin(nπz/d). V0 applies whenever |y|=d:
∑ Cn cosh( npiy/a)sin(npix/a)sin(npiz) = V0.
Even then, doesn't seem to me that it satisfies (1). Maybe need a √2 factor inside the cosh?