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Homework Help: Method of Image Charges - Range of validity of solutions

  1. Sep 9, 2010 #1
    When using the method of image charges to solve a problem in electrostatics, how do we determine the volume of space in which our solution is valid? And how do we find the solution outside this volume?
    To be more specific, let's consider two examples.

    The first is the classic point charge above an infinite plane sheet of charge. The text-book solution is the field of an electric dipole symmetric about the plane of the sheet of charge, with the solution valid on the side of the plate that the charge is on. What is the electric field on the side opposite that of the point charge, and is there an easier way to find it other than integrating over the surface charge density found using the image charge?

    The second is a variation on the question posted here:

    The variation consists of two parallel grounded plates, (x=0 is the position of the left plate, x=L is the position of the right plate), with a point charge, q, at a distance d from the left plate, along the x axis.

    The potential along the x axis is given by the infinite sum:

    [tex]V(x) = kq \sum ^{\infty} _{n=-\infty}(\frac{1}{|x-(2nL+d)|}-\frac{1}{|x-(2nL-d)|})[/tex]

    Graphing this expression for x, we find that it diverges for x near d, as expected, 0 at x=0 and x=L, as expected, but it is obviously not valid outside the range 0<x<L since it diverges for x near -d.
    And the same question can be asked here as was asked for the single sheet of charge.

    With thanks in advance, Anatoli.
  2. jcsd
  3. Sep 10, 2010 #2


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    Homework Helper

    The field obtained with the method of image charges is valid in that volume of the space where the real charge is positioned. As for the field in the space at the other side: I think - I might be wrong- you can get it by assuming that the volume enclosed by the equipotential surface and containing the real charge is filled with an ideal metal: The outer surface charge is evenly distributed.

    See the following link which illustrates the method for a problem similar to the thread with two parallel plates.


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