Q 12-01-106JEE MainJEE Main 2021 (27 Jul, Shift 1)Medium
Two identical tennis balls each having mass $m$ and charge $q$ are suspended from a fixed point by threads of length $l$. What is the equilibrium separation when each thread makes a small angle $\theta$ with the vertical?
Answer: (B) $x = \left(\dfrac{q^2 l}{2\pi\varepsilon_0 mg}\right)^{1/3}$
For each ball, $T\cos\theta = mg$ and $T\sin\theta = F = \dfrac{q^2}{4\pi\varepsilon_0 x^2}$, so $\tan\theta = \dfrac{F}{mg}$.
For small $\theta$: $\tan\theta \approx \sin\theta = \dfrac{x/2}{l}$.
$$\frac{x}{2l} = \frac{q^2}{4\pi\varepsilon_0 x^2 mg} \Rightarrow x^3 = \frac{q^2 l}{2\pi\varepsilon_0 mg}$$
Solution by Sreeraj P, M.Sc Physics