Potential anywhere inside a cube

[andre220]

Homework Statement

All six faces of a cube, of side length $L$, are maintained at constant, but different potentials. The left and right faces are at $V_1$ each. The top and bottom are at $V_2$ each. The front and back are at $V_3$. Determine the electrostatic potential $\Phi(x,y,z)$ at any point inside the cube. What is the value of the electrostatic potential at the center of the cube?

Homework Equations

Solutions are of the form:

$X(x) = A \sin{k_1 x} + B\cos{k_1 x}$
$Y(y) = C\sin{k_2 y} + D\cos{k_2 y}$
$Z(z) = E\sinh{k_3 z} + F\cosh{k_3 z}$
where $k_3 = \sqrt{k_1 + k_2}$

The Attempt at a Solution

Normally, the case in which one or two sides are held at some potential, the symmetry is exploited to get $k_i$ and the prevailing constants. However, here, that does not seem to be possible here.

When: $x =0, x= L$, $\Phi = V_1$
$y =0, y= L$, $\Phi = V_2$
$z =0, z= L$, $\Phi = V_3$

However, I do not see how this get me any closer to finding, any of the constants $A,B,C,D,E,F$. $\Phi = X(x)Y(y)Z(z)$.
Thus for the first B.C.:

$\Phi(0,y,z) = (A\sin{0} + B\cos{0})(C\sin{k_2 y} + D \cos{k_2 y})(E\sinh{k_3 z} + F\cosh{k_3 z})$
$= B (C\sin{k_2 y} + D \cos{k_2 y})(E\sinh{k_3 z} + F\cosh{k_3 z}) = V_1$

and so do I just keep doing this and will eventually, get my coefficients? Regardless, I don't necessarily see how $k_i$ would be found.

Thank you

 

[Orodruin]

The problem is a linear PDE. You can solve for one inhomogeneity at a time and make a superposition of the solutions to find the full solution.

 

[andre220]

Okay, I take when $x=0\implies$ $V_1$:
$A\sin{k_x 0} + B\cos{k_x 0} = V_1 \implies B = V_1$ and I could keep going on like that for each of the six boundary conditions
But I am still not seeing how that would work. Plus I don't see how I could get the $k$'s from this.

 

[Orodruin]

No. You need to split the entire problem into several parts, then solve each part separately, including the now homogeneous boundary conditions on the sides.