JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Electrostatics

Place $O$ at the origin with the arms along the axes. The charges $+q$ at $B(-\ell ,0)$, $2q$ at $C(0,\ell )$, $q$ at $F(0,-\ell )$ and $2q$ at $G(\ell ,0)$ lie at distance $\ell$, while $-q$ at $A(-\ell ,\ell )$, $+q$ at $D(\ell ,\ell )$, $2q$ at $E(-\ell ,-\ell )$ and $-q$ at $H(\ell ,-\ell )$ lie at distance $\sqrt{2}\ell$. Writing $k=\frac{1}{4\pi {\epsilon }_0}$, the four nearer charges contribute $\frac{kq}{{\ell }^2}\hat{i}$, $-\frac{2kq}{{\ell }^2}\hat{j}$, $\frac{kq}{{\ell }^2}\hat{j}$ and $-\frac{2kq}{{\ell }^2}\hat{i}$, so

Each diagonal charge of magnitude $|Q|$ contributes $\frac{k|Q|}{2{\ell }^2}$ along its own diagonal, and adding the four gives

The two resultants are antiparallel, so

The magnitude of the field at $O$ is $\frac{1}{4\pi {\epsilon }_0}\frac{q}{2{\ell }^2}\left(2\sqrt{2}-1\right)$.