JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Electrostatics

Because $A$, $C$ and the centre $O$ form an equilateral triangle, each field at $O$ makes $30^{\circ }$ with the $x$-axis. The charge at $A$ ($-4q$) sets up a field $E_A=\frac{4kq}{d^2}$ and the charge at $C$ ($-2q$) a field $E_C=\frac{2kq}{d^2}$, each pointing toward its negative charge so that their $x$-components reinforce. Adding those projections with $k=\frac{1}{4\pi {\epsilon }_0}$, the net field at $O$ along the $x$-direction is $\frac{\sqrt{3}q}{\pi {\epsilon }_0{d}^2}$.