A system of three charges -q, + Q and -q are placed at three points at equal distances on a straight line in the same order. If the potential energy of the system is found to be zero, find the ratio Q/q.

$\frac{1}{4}$

The correct answer is Option (3) → $\frac{1}{4}$

Let the charges be $-q$, $+Q$, $-q$ placed on a straight line with equal separation $r$.

Potential energy of the system:

$U = \frac{k(-q)(+Q)}{r} + \frac{k(+Q)(-q)}{r} + \frac{k(-q)(-q)}{2r}$

$U = -\frac{k q Q}{r} - \frac{k q Q}{r} + \frac{k q^2}{2 r}$

$U = -\frac{2 k q Q}{r} + \frac{k q^2}{2 r}$

Given $U = 0$:

$-\frac{2 k q Q}{r} + \frac{k q^2}{2 r} = 0 \Rightarrow -2 q Q + \frac{q^2}{2} = 0$

$4 Q = q \Rightarrow Q/q = 1/4$

Answer: $Q/q = 1/4$