JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Laws of Motion

The thread, of length 20 cm, makes an angle $\theta$ with the wall such that $\sin \theta =\frac{10}{20}=\frac{1}{2}$, so $\theta =30^{\circ }$. At equilibrium, the horizontal electric force on the particle balances the horizontal component of the tension, while the vertical component of tension balances gravity, giving $\tan \theta =\frac{qE}{mg}$. Substituting $\tan 30^{\circ }=\frac{1}{\sqrt{3}}$, $E=2\times 10^4$ N/C, $m=2\times 10^{-3}$ kg and $g=10$ m/s²: $\frac{1}{\sqrt{3}}=\frac{q\times 2\times 10^4}{2\times 10^{-2}}$, giving $q=\frac{1}{\sqrt{3}}\times 10^{-6}$ C $=\frac{1}{\sqrt{3}}\mu C$. Hence $x=3$.