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If the position of some charge $Q$ is known, the boundary condition is $u=0$ on some parabolic surface, and we know the image charge has its electric volume of $Q'$, then how can I determine the position of the image charge?

Same questions goes for the hyperbolic curve boundary: how can I determine the position of $Q'$?

I think may be there is a way to transform the coordinates to make everything into an easily-handled form, but I am not sure about it. Another solution I thought about is to put this question into a general question based on basic Poisson's equation and Laplace equation, but I do not have a specific idea on how to conduct this.

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What kind of parabolic boundary? A parabolic cylinder? An elliptical paraboloid? A hyperbolic paraboloid? – Dan Apr 6 '13 at 19:14
Also, these articles on parabolic coordinates might help: 1 2 – Dan Apr 6 '13 at 19:22
The basic concept is that you find an image charge placement that gives you parabolic equipotential surfaces. (so, it may help to transform all equations to parabolic coordinates) – Manishearth Apr 6 '13 at 19:26
Typical two dimension parabolic boundary like y=ax^2 is enough for me. Another boundary I deal with is x^2/a^2-y^2/b^2=1 – Emily Apr 7 '13 at 14:49
@MojtabaGolshani Right, the equipotentials of two (opposite sign) point charges are spheres, so you cannot get any other shape with a single image point charge. – Keith McClary Sep 6 '15 at 3:28

not an answer, but some thoughts:

first, we can't use a single charge $Q'$ to solve this. you will need an image "line charge" (you can think of a line charge as some very close small point charges).

lets say that the charge Q is located at $(0,a)$ on the y axis.

we are looking for some line charge shape, that we will call $f(x)$ (all image charges will be located on this line). this line charge will have a charge density dependent on x, which we will call $\lambda(x)$. we demand that the potential on the surface will be zero for every arbitrary point $x_0$ on the surface, so if our surface is of the form: $g(x)$, we need to find $\lambda(x)$, $f(x)$, $x_1, x_2$ so that: $$\int_{x_1}^{x_2}\frac {\lambda(x)}{\sqrt{(f(x)-g(x_0))^2+(x_0-x)^2}}dx+\frac {Q}{\sqrt{(g(x_0)-a)^2+x_0^2}}=0$$

in our case $$\int_{x_1}^{x_2}\frac {\lambda(x)}{\sqrt{(f(x)-cx_0^2)^2+(x_0-x)^2}}dx+\frac {Q}{\sqrt{(cx_0^2-a)^2+x_0^2}}=0$$

this is defiantly solvable, no sure if analytically....

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