Four charges q, -q, 2Q and Q are placed at the corners of a square ABCD of side '2a' as shown. The field at the midpoint CD is zero. Find the ratio of $\frac{q}{Q}$.

$\frac{15\sqrt{5}}{2}$

The correct answer is Option (3) → $\frac{15\sqrt{5}}{2}$

At the midpoint of side $CD$, the fields from the bottom charges point in the same direction:

$E_{CD} = \frac{kQ}{a^2} + \frac{k(2Q)}{a^2} = \frac{3kQ}{a^2}$

For the top charges, the distance to the midpoint is $a\sqrt{5}$. Only the horizontal components add up:

$E_{AB} = 2 \times \frac{kq}{(a\sqrt{5})^2} \times \frac{1}{\sqrt{5}} = \frac{2kq}{5\sqrt{5}a^2}$$

Setting the net field to zero:

$\frac{3kQ}{a^2} = \frac{2kq}{5\sqrt{5}a^2}$

$15\sqrt{5}Q = 2q$

$\frac{q}{Q} = \frac{15\sqrt{5}}{2}$