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?(adsbygoogle = window.adsbygoogle || []).push({});

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:

https://www.physicsforums.com/showthread.php?t=426237

The variation consists of two parallel grounded plates, (x=0 is the position of the left plate, x=Lis the position of the right plate), with a point charge,q, at a distancedfrom 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.

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

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