JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Electrostatics

For small angle θ, the equilibrium condition balances the electrostatic repulsion with the horizontal component of tension. The electrostatic force is $F_e=\frac{k{q}^2}{x^2}$ where $k=\frac{1}{4\pi {\epsilon }_0}$. For small angles, $\tan \theta \approx \theta =\frac{x/2}{\ell }$, and the horizontal component of tension equals $T\sin \theta \approx mg\theta =\frac{mgx}{2\ell }$. Setting $F_e=\frac{mgx}{2\ell }$ and solving: $\frac{q^2}{4\pi {\epsilon }_0{x}^2}=\frac{mgx}{2\ell }$, which gives $x^3=\frac{q^2\ell }{2\pi {\epsilon }_{0}mg}$, so $x={\left(\frac{q^2\ell }{2\pi {\epsilon }_{0}mg}\right)}^{1/3}$.