Electric Charge & Coulomb's Law question from JEE Main 2019 (9 Jan, Shift 1)

Answer: (a)

1. On $+Q$ at $x=0$: force from $+Q$ at $d$ is $\dfrac{kQ^2}{d^2}$; from $q$ at $d/2$ is $\dfrac{kQq}{(d/2)^2} = \dfrac{4kQq}{d^2}$.
2. Net zero: $\dfrac{kQ^2}{d^2} + \dfrac{4kQq}{d^2} = 0 \Rightarrow q = -\dfrac{Q}{4}$
3. So the answer is (a) $-\dfrac{Q}{4}$.