Method of image charge for ellipse

Question

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?

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 $\theta_{z2} = \theta_{\zeta2}$.

Answer 2

This problem is treated in a paper by J. Sten in 1996: "Focal image charge singularities for the dielectric elliptic cylinder".

The image solution basically involves a strip with varying charge and dipole density, plus (depending on the location of the source and the cylinder's shape) an image line charge similar to the one in the case of the circular cylinder.