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\begin{document}

method of images:

For a conducting surface in yz plane and a q somewhere along the x
axis, solution is the same as if there was no plane and a charge of
-q at -x along the axis. Solution is then: \( \phi =\frac{q}{|r-x|}-\frac{q}{|r+x|} \)

This can be generalized to 2 conducting surfaces at right angles.
... etc.

Method of images also works for spherical symmetry. Let sphere at
origin, grounded and charge q distance p outside of sphere. Then image
charge at q' at p' has a relationship: \( p'=\frac{q'^{2}}{q^{2}}p \) and \( p'=a^{2}/p \). So \( q'=\frac{-a}{p}q \). Can then calculate
a force: \( F^{i}=-qE^{i}=q\frac{q'}{|p-p'|^{2}}=\frac{q^{2}a}{p^{3}}\frac{1}{(1-\frac{a^{2}}{r^{2}})^{2}} \)



\end{document}
