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(C) The potential energy $U$ of a system of point charges is given by the sum of the potential energies of all pairs of charges: $U = \sum \frac{k q_i

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$1$ : $4$

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(C) The potential energy $U$ of a system of point charges is given by the sum of the potential energies of all pairs of charges: $U = \sum \frac{k q_i q_j}{r_{ij}}$.Given the charges $-q, Q, -q$ are placed at distances $x$ apart,the pairs are: $(-q, Q)$ at distance $x$,$(Q, -q)$ at distance $x$,and $(-q, -q)$ at distance $2x$.The total potential energy is:$U = \frac{1}{4 \pi \varepsilon_0} \left( \frac{(-q)(Q)}{x} + \frac{(Q)(-q)}{x} + \frac{(-q)(-q)}{2x} ight) = 0$Dividing by $\frac{1}{4 \pi \varepsilon_0 x}$ (assuming $x 
eq 0$):$-qQ - qQ + \frac{q^2}{2} = 0$$-2qQ + \frac{q^2}{2} = 0$$2qQ = \frac{q^2}{2}$Dividing both sides by $2q$ (assuming $q 
eq 0$):$Q = \frac{q}{4}$Therefore,the ratio $\frac{Q}{q} = \frac{1}{4}$.

Three charges $-q, Q$ and $-q$ are placed at equal distances on a straight line. If the total potential energy of the system of three charges is zero,then the ratio $\frac{Q}{q}$ is

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