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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!