Empty conducting cube with 4 sides at ground and 2 sides at V

[randomchance]

Homework Statement

Empty cubic box of dimension "a" with conducting walls. Two opposite walls are held at a constant potential V, while the other walls are grounded. Find an expression for the electric potential inside the box. (Assume box centered on the origin, walls are all normal to one of the Cartesian axes, and that the non-grounded walls are normal to the x-axis.)
Now suppose that the two walls normal to the x-axis hare held at potentials plus/minusV. What now is the potential inside the box? Use separation of variables.

Homework Equations

So, we know that Poisson's equation inside the cube is =0

The Attempt at a Solution

I'm not sure how to type formula's in here, so take all of my "d"s to be partial "d"s
this means that d^2V/dx^2+d^2V/dy^2+d^2V/dz^2=0.
If we use separation of variables we can say that V(x,y,z)=X(x)Y(y)Z(z). I know that we can then plug this into Poisson's equation and then divide by XYZ to yield:
X''/X+Y''/Y+Z''/Z=0
I know that this means that each term needs to equal a constant and that the sum of those constants needs to equal 0. Now, this is where I start getting sketchy. If I say something like this:
X''=aX Y''=bY and Z''=cZ can I say that due to the symmetry of the cube and the fact that two endpoints (-a/2 and a/2) in both the y and z directions are equal to 0, that b=c and therefore a=(-2b)?
Also: I was trying to use Griffiths to help me, but I'm not sure why in some cases they had the solution be of the form Asin(kx)+Bcos(kx) and sometimes Aexp(kx)+Bexp(-kx). My instructor was vaguely said something to the effect that since our cube is centered at the origin that k can have both positive and negative values in the x direction and that means we need to consider both of those forms for the potential in the x direction (where I assume that we'd only need to consider one or the other in the other two directions??) Help!

 

[Mindscrape]

Hmm, is this a box or a cube? If it is a box, which is the impression I get (though you did say cube?), then you don't have the Z dimension in your separation of variables.

Basically, the two different solutions (complex exponentials vs. real exponentials) depend on the boundary conditions. You can assign the constant value from the separation of variables to be either positive or negative. If you make it negative you will get complex exponentials (sin and cos), and real exponentials if positive (sinh and cosh). The boundary conditions will give a trivial solution if you set that constant value to the wrong sign. If I remember my PDE class correctly, a dirichlet boundary condition always gives sin and cos (but I may be wrong, so look it up if you need). Griffiths does a pretty awful job explaining PDEs and this whole section is a blemish on a great book, in my opinion. Whether you use cosh and sinh or exponentials is also another story.

What are you trying to say by symmetry? Just go by the boundary conditions, and those will form your unique solution. Of course you can always guess at something (is that what you're trying to do?), and if it satisfies both the PDE and the BCs then it's the unique solution that you happened to get right.

 

[randomchance]

Sorry, this is a cube. I've never taken PDE's and Diff Eq was a few years ago, consequently my math kind of sucks (I'm trying to catch up as fast as possible). What I was trying to do with the symmetry thing was this:
If X''/X+Y''/Y+Z''/Z=0 then each term individually should equal a constant such that

X''/X= a Y''/Y=b and Z''/Z=c I know that a+b+c needs to equal 0. Since the boundary conditions are the same in the Y and Z directions, and the distance between the boundaries is the same in both directions I was trying to say that b=c so that now a+2b (or 2c) = 0 and a would need to equal -2b (or -2c). So instead of looking at 3 different constants, my equations would be in terms of 1:

X''=-2bX -> X=Acos(2bx)+Bsin(2bx)
Y''= bY -> Y=Cexp(by)+Dexp(-by)
Z''= bZ -> Z=Eexp(bz)+Fexp(-bz)

(*question here...on the exp, why is one positive and one negative?)
I dunno, I think maybe I'm just going about this in a screwy way.