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 $ heta$ from the vertical. If the mass of each charge is $m$, then the electrostatic potential…

Question

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 $	heta$ from the vertical. If the mass of each charge is $m$, then the electrostatic potential at the centre of line joining them will be $$ \left(\frac{1}{4 \pi \in_0}=kight) 	ext {. } $$

Aicharya · Accepted Answer

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} $

Two small equal point charges of magnitude $q$ are suspended from a common point on the ceiling by…

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