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Homework Help: Electric potential inside a cube

  1. Mar 5, 2013 #1
    1. The problem statement, all variables and given/known data

    A metal box has 6 walls, all insulated from one another. The left and right wall are held at V= V0, which are at y=-d and y=d. All the other walls are grounded.

    The cube has dimensions where walls run from x=0 to x=2d, z=0 to z=2d, and y=-d to y=d.

    2. Relevant equations

    Seperation of variables in 3D:
    [itex]\frac{1}{X}[/itex][itex]\frac{\partial^{2} V}{\partial X}[/itex] + [itex]\frac{1}{Y}[/itex][itex]\frac{\partial^{2} V}{\partial Y}[/itex] + [itex]\frac{1}{Z}[/itex][itex]\frac{\partial^{2} V}{\partial Z}[/itex] =0

    Boundary conditions:
    V(x,0) = 0
    V(x,2d) = 0
    V(y, -d) = V0
    V(y,d) = V0
    V(z,0) = 0
    V(z,2d) = 0

    3. The attempt at a solution

    I am thinking that there is symmetry around z, so we can only worry about x and y.

    V(x,y) = (Asinh(kx) + Bcosh(kx))(Ce^{ky} + De^{-ky})
    which simplifies to (2cosh(ky)(Asinh(kx) + Bcosh(kx)).

    Assuming all the above is correct, I am having issues at this point. I need to put this into a sum, which according to Griffiths should look something like

    [itex]\sum[/itex] Cn cosh( npiy/a)sin(npix/a) = V0.

    I don't understand how to simplify this. I have limits of integration, but I am not sure where an integral can arise from this.

    Thank you.
  2. jcsd
  3. Mar 5, 2013 #2


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    Seems more natural to me to have all three coordinates run from -d to +d.
    Yes, there are symmetries, but I don't see this allows you to write the potential as a function of only two coordinates. It should be an even function wrt x, y and z, and there should be a symmetry between x and z.
  4. Mar 5, 2013 #3
    Ok, that makes sense. So, with all three coordinates, the expression should be:

    ∑ Cn cosh( npiy/a)sin(npix/a)sin(npiz) = V0.

    But what's next?
  5. Mar 5, 2013 #4


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    Just noticed this was wrong in the OP:
    You mean (1)## \frac{\partial^2 V}{\partial X^2}+\frac{\partial^2 V}{\partial Y^2}+\frac{\partial^2 V}{\partial Z^2} =0##and by separation of variables to V=F(x)G(y)H(z):
    (2) ##\frac1F\frac{\partial^2 F}{\partial X^2}=a_n, \frac1G\frac{\partial^2 G}{\partial Y^2}=b_n, \frac1H\frac{\partial^2 H}{\partial Z^2} =c_n## where ##a_n+b_n+c_n=0##.
    Surely that should be something like V(x,y,z) = ∑ Cn cosh( nπy/d)sin(nπx/d)sin(nπz/d). V0 applies whenever |y|=d:
    ∑ Cn cosh( npiy/a)sin(npix/a)sin(npiz) = V0.
    Even then, doesn't seem to me that it satisfies (1). Maybe need a √2 factor inside the cosh?
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