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If we put an external electron outside a elliptical metal described by: $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$, how do we determine the image charge or charges of that electron inside this ellipse?

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    Just a comment for those looking at this question who may not be familiar with the method of images. I assume the problem Rob is trying to solve is Poisson's equation for the potential in the region outside the (grounded) metal ellipse. By the uniqueness theorem, any solution for the potential that satisfies the boundary conditions for the surface of the ellipse being at 0 potential is a solution to the problem. Potential falls off as $\frac{1}{r}$ so by putting an "image charge" or charges inside the ellipse that together with the original charge to satisfy the conditions solves it.2012-05-29

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