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Homework Help: Potential and total charge on plane

  1. Mar 20, 2016 #1
    1. The problem statement, all variables and given/known data
    An infinite plane in z=0 is held in potential 0, except a square sheet -2a<x,y<2a which is held in potential V.
    Above it in z=d there is a grounded plane. Find:
    a) the potential in 0<z<d?
    b) the total induced charge on the z=0 plane.

    2. Relevant equations
    Green's function for a plane + Green's Theorem for the potential; method of images etc.

    3. The attempt at a solution
    a) First, I put an image square sheet on z=2d, then solved by superposition, calculating the potential for the z=0 plane and then adding a shifted result.
    For the z=0 plane, ##G = \frac{1}{|r-r'|}+\frac{1}{|r-r''|}## where ##r''=(x',y',-z')##, and due to Dirichle b.c. we get $$\phi_1(r)=-\frac{1}{4\pi}\iint\phi(r')\frac{\partial G}{\partial n'}dS=\frac{Vz}{2\pi}\iint_{-2a}^{2a}\frac{dx'dy'}{((x-x')^2+(y-y')^2+z^2)^{3/2}}$$ where the integrals are from -2a to 2a. This integral is a real pain, so I left it aside for now. I'm not sure, but we might be allowed to leave it in integral form. But still, is there a decent way to calculate it? Finally, $$\phi_2=\frac{V(2d-z)}{2\pi}\iint_{-2a}^{2a}\frac{dx'dy'}{((x-x')^2+(y-y')^2+(2d-z)^2)^{3/2}}$$ and ##\phi=\phi_1+\phi_2##. Thinking more about this integral, I thought maybe I could solve it using separation of variables, which I haven't tried yet, but I don't think it would make my life any easier in getting a closed form which doesn't contain series expansions or integrals.

    b) Here I'm kinda stuck... The induced charge can be attributed to the image square sheet. But now I find myself not knowing how to calculate total charge on a sheet given only its size and potential V. And also, I'm not sure if this is correct, since this is not a closed surface. If it were, I'd say from Gauss' Law we can replace the surface charge distribution with point charges inside (or outside depending on the original arrangement) the surface. Here, my equivalent argument is placing yet another image sheet in z=-2d with potential V to cancel out the potential on z=0 due to the sheet on z=2d.

    Summing up, any advice on the following would be greatly appreciated:
    1. Is there a more elegant way to solve (a), or a way to solve its integral?
    2. Is my argument in (b) about the induced charge of the plane correct?
    3. How can I calculate the charge given this potential?

  2. jcsd
  3. Mar 25, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
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