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) $x \propto q^{2/3}$
Concept. Each ball is in equilibrium under weight, tension and Coulomb repulsion: $\tan\theta = \dfrac{F_e}{mg}$.
Step 1. Equilibrium. $\tan\theta = \dfrac{kq^2}{x^2mg}$.
Step 2. Small angle. $\tan\theta \approx \dfrac{x/2}{l}$:
$$\frac{x}{2l} = \frac{kq^2}{mgx^2} \;\Rightarrow\; x^3 \propto q^2$$
Step 3. Result. $x \propto q^{2/3}$.
Answer. $x \propto q^{2/3}$ (Option D).
Solution by Sreeraj P, M.Sc Physics