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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. –  Bitrex May 29 '12 at 15:49
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1 Answer 1

I don't know that there's an easy way to write an answer to this problem in closed form - at least's it's not obvious to me! I think you may be able to get an approximate solution, based on a problem that we already know how to solve.

Instead of an elliptical hoop, consider the problem of finding the potential outside of a grounded circular hoop in the $\rho$-$\theta$ plane. The potential due to two charged particles $q_1$ and $q_2$, at positions $(R_1, \theta_1)$ and $(R_2, \theta_2)$ at any point $(\rho, \theta)$ in the plane is given by:

$4\pi\epsilon_0V = \frac{q_1}{\sqrt{R_1^2 +\rho^2 - 2R_1\rho\cos(\theta_1 - \theta)}} + \frac{q_2}{\sqrt{R_2^2 +\rho^2 - 2R_2\rho\cos(\theta_2 - \theta)}}$

If we let $q_2 = -\frac{q_1R}{R_1}$, $R_2 = \frac{R^2}{R_1}$, and $\theta_1 = \theta_2$, we find:

$4\pi\epsilon_0V = \frac{q_1}{\sqrt{R_1^2 +\rho^2 - 2R_1\rho\cos(\theta_1 - \theta)}} - \frac{\frac{q_1R}{R_1}}{\sqrt{\frac{R^4}{R_1^2} +\rho^2 - 2\frac{R^2}{R_1}\rho\cos(\theta_1 - \theta)}}$

From which it can be easily seen that the potential vanishes on the circle (i.e. when $\rho = R$).

Now we want to find a way to apply our knowledge of how to solve this problem to the more complicated problem of the potential in the space outside a grounded elliptical hoop. Consider the Joukowski mapping: $z = \zeta + \frac{c^2}{\zeta}$. Plugging in $R_{\zeta}e^{i\theta_\zeta}$ for $\zeta$ in this equation gives us:

$z = (R_\zeta + \frac{c^2}{R_\zeta})\cos(\theta_\zeta) + (R_\zeta - \frac{c^2}{R_\zeta})i\sin(\theta_\zeta)$

Which is the equation of an ellipse with semiaxes $a = (R_\zeta + \frac{c^2}{R_\zeta})$, $b = (R_\zeta - \frac{c^2}{R_\zeta})$. Given the a and b of the ellipse in the problem, $R_{\zeta}$ and the parameter $c$ can be solved for, giving $c = \frac{\sqrt{a^2 - b^2}}{2}$, $R_\zeta = \frac{a + b}{2}$, $ a > b$.

This mapping takes points in the $z$ plane where our ellipse lives, and maps them to points in the $\zeta$ plane. If we have a test charge $q$ at a point $R_{z1}e^{i\theta_{z1}}$ outside the ellipse in this plane, we can use the inverse mapping $\zeta(z) = \frac{z + \sqrt{z^2 - 4c^2}}{2}$ to find its mirror position in the $\zeta$ plane. Once in the $\zeta$ plane, we can find the position $(R_{\zeta2}, \theta_{\zeta2})$ of the image charge $q_2 = -q_1\frac{R_\zeta}{R_\zeta1}$ by noting that $R_{\zeta2} = \frac{R_\zeta^2}{R_{\zeta1}}$, $\theta_{\zeta2} = \theta_{\zeta1}$. Then, since the transformation preserves angle, we can go back to the $z$ plane (where the ellipse is) and find $(R_{z2}, \theta_{z2})$ by scaling $R_{\zeta2}$ by the amount $\sqrt{(R_{\zeta2} + \frac{c^2}{R_{\zeta2}})^2 + (R_{\zeta2} - \frac{c^2}{R_{\zeta2}})^2}$, with $\theta_{z2} = \theta_{\zeta2}$.

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But does the image charge found by using this Joukowski transformation necessarily guarantee that the potential also vanishes at the boundary of the elliptical metal? Or is it just a approximation? And do you know how good is this approximation? –  Rob Jun 6 '12 at 12:59
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