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Question
Three charges –q, Q and –q are placed at equal distances on a straight line. If the potential energy of the system of these charges is zero, then what is the ratio Q:q?
Solution
`(k(-q)Q)/x + (kQ(-q))/x + (k(-q)(-q))/(2x)` = 0
`(-2kqQ)/x + (kQ)^2/(2x)` = 0 or `(kq^2)/(2x) = (2kqQ)/x`
q = 4Q or `Q/q = 1/4`
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