Solution to 1992 Problem 10

Use the method of images. There is an image charge -q along the x axis at x = -0.5 a and an image charge -2q at x = -1.5a along the x axis. The force on the charge q is therefore,
\begin{align*}\mathbf{F} = \frac{q^2}{4 \pi \epsilon_0} \left(\frac{-1}{a^2} + \frac{-2}{a^2} + \frac{-2}{\left(2a \right)^2}...
Therefore, answer (E) is correct.

return to the 1992 problem list

return to homepage

Please send questions or comments to where X = physgre.