- #1

andre220

- 75

- 1

## 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