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Thin disc above grounded plane

  1. Oct 4, 2011 #1
    1. The problem statement, all variables and given/known data

    A thin disk of radius R consists of a uniformly distributed total charge Q. The disk lies a distance D above a grounded perfectly conducting plane. The disk and the plane are parallel. Set the conducting plane in the x-y axis, and the z axis through the center of the disk.
    R=0.1 m, the distance D = 2R = 0.2 m, and the charge Q=1/(9E9) Coulomb

    Compute the potential along the axis through the center of the charged disk

    2. Relevant equations

    If it was a single point charge:
    V(x,y,z) = 1/4*∏*[itex]\epsilon * [ q/ \sqrt{x^2+y^2+(z-d)^2} - q/ \sqrt{x^2+y^2+(z+d)^2}[/itex] ]

    3. The attempt at a solution

    I have a feeling I would solve this using the method of images with an imaginary disc of -q charge below the conducting plane. Would this be the same for the thin disc? With the given value turning it into this?

    [itex][ 1/ \sqrt{x^2+y^2+(z-d)^2} - 1/ \sqrt{x^2+y^2+(z+d)^2}[/itex] ]

    My second problem is to to calculate numerically the potential everywhere in space above the conducting plane. Could I get any hints as to how to do this?
    Last edited: Oct 4, 2011
  2. jcsd
  3. Oct 4, 2011 #2


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    That does seem like the way to go. Just keep in mind that the charge is a disc, not a point, so you can't just use the point charge formula unaltered. You'll have to do an integral.
  4. Oct 5, 2011 #3

    rude man

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    I would run a Gaussian surface, a right circular cylinder, with one flat surface located within the disk (and parallel to it) to a plane a distance r between the disk and the ground plane, compute E(r), integrate E(r) from the disk to the ground plane, with the constant of integration such that V(D-r) = 0.

    You can also do it by the image method, you'd get the same result. Since you would not extend the Gaussian variable surface beyond r = D there would be no difference in the equations.
  5. Oct 6, 2011 #4
    I figured out the z axis part since it forms a right triangle. I can't wrap my head around how I would measure the potential at a point not on the z axis.
  6. Oct 6, 2011 #5

    rude man

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    I didn't see that part of the problem. That's a lot more difficult all right.

    I would go with V(r) = k∫σ*dA/r integrated over disk surface A
    r = distance from element of charge σ*dA
    k = 9e9 SI
    σ = surface charge density = constant thruout disk = Q/πR^2
    A = disk area

    But you've probably figured that far already.
    r = distance from point (x0, y0) on disk defined by surface (x^2 + y^2 = R^2, D)
    to arbitrary poit (x,y,z).

    Just a math problem ... :-)
    Last edited: Oct 6, 2011
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