Electric Charge & Coulomb's Law question from JEE Main 2004 (AIEEE)

Answer: (b)

1. Consider one corner charge $-Q$, side $a$. Repulsion from the two adjacent $-Q$: resultant $\dfrac{\sqrt{2}kQ^2}{a^2}$ along the diagonal, outward.
2. Repulsion from the opposite corner (distance $\sqrt{2}a$): $\dfrac{kQ^2}{2a^2}$, outward along the same diagonal.
3. $q$ at the centre (distance $a/\sqrt{2}$) must attract: $\dfrac{2kQq}{a^2} = \dfrac{kQ^2}{a^2}\left(\sqrt{2} + \dfrac{1}{2}\right)$
4. $q = \dfrac{Q}{4}\left(1 + 2\sqrt{2}\right)$ (positive). So the answer is (b).