JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Electrostatics

Place the three charges on the equilateral triangle and treat the two $+q$ and $-q$ pairs as two dipoles, each of magnitude $q\ell$. From the geometry each dipole vector points $30^{\circ }$ on either side of the downward vertical, that is of $-\hat{j}$. When you add them, the horizontal ($\hat{i}$) components are equal and opposite and cancel, while the vertical components reinforce. The resultant is $2(q\ell )\cos 30^{\circ }$ along $-\hat{j}$, and since $\cos 30^{\circ }=\frac{\sqrt{3}}{2}$ this is $2q\ell \cdot \frac{\sqrt{3}}{2}(-\hat{j})=-\sqrt{3}q\ell \hat{j}$.