Three charges -q, Q and -q are placed respect | vedclass

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(D) The potential energy $U$ of a system of point charges is given by $U = \sum \frac{k q_i q_j}{r_{ij}}$.For the three charges $-q, Q, -q$ placed at

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$1: 4$

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(D) The potential energy $U$ of a system of point charges is given by $U = \sum \frac{k q_i q_j}{r_{ij}}$.For the three charges $-q, Q, -q$ placed at distances $x$ apart as shown in the figure,the pairs are $(-q, Q)$ at distance $x$,$(Q, -q)$ at distance $x$,and $(-q, -q)$ at distance $2x$.The total potential energy is:$U = \frac{k(-q)(Q)}{x} + \frac{k(Q)(-q)}{x} + \frac{k(-q)(-q)}{2x} = 0$$-\frac{kqQ}{x} - \frac{kqQ}{x} + \frac{kq^2}{2x} = 0$$-\frac{2kqQ}{x} + \frac{kq^2}{2x} = 0$$\frac{kq^2}{2x} = \frac{2kqQ}{x}$$q^2 = 4qQ$$q = 4Q$Therefore,the ratio $Q: q = 1: 4$.

Three charges $-q, Q$ and $-q$ are placed respectively at equal distances on a straight line. If the potential energy of the system of three charges is zero,then what is the ratio of $Q: q$ ?

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