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 $\frac{q}{Q}$.

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

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

$E=E_C-E_D-E_A\cos θ-E_B\cos θ$

$=\frac{K(2Q)}{b^2}-\frac{KQ}{b^2}⇒2×\frac{Kqb}{(b^2+4b^2)^{3/2}}$

$=\frac{2}{5\sqrt{5}}E$

$⇒\frac{E}{E'}=\frac{5\sqrt{5}}{2}$