Point charges $-3 \mathrm{Q},-\mathrm{q}, 2 \mathrm{q}$ and 2 Q are placed, one at each corner of a square. The relation between Q and q for which the potential at the centre of the square is zero is

Let the side length of the square be $a$. The distance from each corner to the center of the square is $r = \dfrac{a}{\sqrt{2}}$.

The electric potential $V$ at the center due to a point charge $q_i$ at distance $r$ is given by $V_i = \dfrac{k q_i}{r}$, where $k = \dfrac{1}{4 \pi \epsilon_0}$.

The total potential at the center is the sum of the potentials due to the four charges $-3Q, -q, 2q, 2Q$ placed at the corners:

$V_{total} = \dfrac{k}{r} [(-3Q) + (-q) + (2q) + (2Q)]$

Setting the total potential to zero:

$\dfrac{k}{r} [-3Q + 2Q - q + 2q] = 0$

Since $\dfrac{k}{r} \neq 0$, we have:

$-Q + q = 0$

$Q = q$

Answer: $Q=q$