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Potential anywhere inside a cube

  1. Mar 5, 2015 #1
    1. The problem statement, all variables and given/known data
    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?

    2. Relevant 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}##


    3. 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
     
  2. jcsd
  3. Mar 5, 2015 #2

    Orodruin

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    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.
     
  4. Mar 5, 2015 #3
    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.
     
  5. Mar 6, 2015 #4

    Orodruin

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    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.
     
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