Electric potential inside a 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

Seperation of variables in 3D:
[itex]\frac{1}{X}[/itex][itex]\frac{\partial^{2} V}{\partial X}[/itex] + [itex]\frac{1}{Y}[/itex][itex]\frac{\partial^{2} V}{\partial Y}[/itex] + [itex]\frac{1}{Z}[/itex][itex]\frac{\partial^{2} V}{\partial Z}[/itex] =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

[itex]\sum[/itex] 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:

[itex]\frac{1}{X}[/itex][itex]\frac{\partial^{2} V}{\partial X}[/itex] + [itex]\frac{1}{Y}[/itex][itex]\frac{\partial^{2} V}{\partial Y}[/itex] + [itex]\frac{1}{Z}[/itex][itex]\frac{\partial^{2} V}{\partial Z}[/itex] =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?

 

[paranormal]

Your attempt at a solution is on the right track. To simplify the expression, you can use the boundary conditions to determine the values of A, B, C, and D. For example, from the boundary condition V(y,-d) = V0, you can set y=-d and solve for A and B. Then, using the other boundary conditions, you can determine the values of C and D. This will give you a specific expression for V(x,y) that satisfies all the boundary conditions.

As for the sum, it comes from the fact that there are an infinite number of possible solutions to the Laplace's equation (the equation you have for V(x,y)). Each solution will have its own set of coefficients (Cn) and functions (cosh and sin) that satisfy the boundary conditions. The sum represents the combination of all these solutions that, when summed together, give the final solution to the problem.

I hope this helps. Keep working on it and don't hesitate to ask for clarification if needed. Good luck!