Q 12-01-012NEETJEE MainMedium
Two small identical balls, each of mass $m$ and charge $q$, hang from a common point by threads of length $l$. In equilibrium each thread makes a small angle $\theta$ with the vertical. The separation $x$ between the balls varies with $q$ as
Answer: (D) $x \propto q^{2/3}$
For each ball, $\tan\theta = \dfrac{F_e}{mg} = \dfrac{kq^2}{x^2mg}$. For small angles, $\tan\theta \approx \dfrac{x/2}{l}$:
$$\frac{x}{2l} = \frac{kq^2}{mgx^2} \;\Rightarrow\; x^3 \propto q^2 \;\Rightarrow\; x \propto q^{2/3}$$
Solution by Sreeraj P, M.Sc Physics