Two negative charges,each of magnitude q,are | vedclass

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(C) The potential energy of a system of charges is given by $U = \sum \frac{k q_i q_j}{r_{ij}}$.For the first configuration ($-q, q, -q$ with spacing

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$3$

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(C) The potential energy of a system of charges is given by $U = \sum \frac{k q_i q_j}{r_{ij}}$.For the first configuration ($-q, q, -q$ with spacing $r$ between adjacent charges):$U_1 = \frac{k(-q)(q)}{r} + \frac{k(q)(-q)}{r} + \frac{k(-q)(-q)}{2r} = -\frac{kq^2}{r} - \frac{kq^2}{r} + \frac{kq^2}{2r} = -\frac{2kq^2}{r} + \frac{kq^2}{2r} = -\frac{3kq^2}{2r}$.For the second configuration ($-q, -q, q$ with spacing $r$ between adjacent charges):$U_2 = \frac{k(-q)(-q)}{r} + \frac{k(-q)(q)}{r} + \frac{k(-q)(q)}{2r} = \frac{kq^2}{r} - \frac{kq^2}{r} - \frac{kq^2}{2r} = -\frac{kq^2}{2r}$.Therefore,the ratio is:$\frac{U_1}{U_2} = \frac{-3kq^2 / 2r}{-kq^2 / 2r} = 3$.

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