Question 133 of 311

Two small equal point charges of magnitude $q$ are suspended from a common point on the ceiling by insulating mass less strings of equal lengths. They come to equilibrium with each string making angle $\theta$ from the vertical. If the mass of each charge is $m$, then the electrostatic potential at the centre of line joining them will be
$$\left(\frac{1}{4 \pi \in_0}=k\right) \text {. }$$

A$2 \sqrt{k m g \tan \theta}$B$\sqrt{k m g \tan \theta}$C$4 \sqrt{k m g / \tan \theta}$D$4 \sqrt{k m g / \tan \theta}$

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Answer: (C) $4 \sqrt{k m g / \tan \theta}$

In equilibrium, $\mathrm{F}_{\mathrm{e}}=\mathrm{T} \sin \theta$
$$\begin{aligned} & m g=T \cos \theta \\ & \tan \theta=\frac{F_e}{m g}=\frac{q^2}{4 \pi \epsilon_0 x^2 \times m g} \\ & \therefore x=\sqrt{\frac{q^2}{4 \pi \epsilon_0 \tan \theta m g}} \end{aligned}$$
Electric potential at the centre of the line
$$\mathrm{V}=\frac{\mathrm{kq}}{\mathrm{x} / 2}+\frac{\mathrm{kq}}{\mathrm{x} / 2}=4 \sqrt{\mathrm{kmg} / \tan \theta}$$