Question

Two small equal point charges of magnitude $q$ are suspended from a common point $O$ the ceiling by insulating massless 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 center of line joining them will be $(\frac{1}{4\pi {ϵ}_0}=K)$

• A. $2\sqrt{k/mg\tan \theta }$
• B. $\sqrt{kmg\tan \theta }$
• C. $4\sqrt{k/mg\tan /\theta }$
• D. $4\sqrt{kmg\tan \theta }$

Answer 1

The correct option is D $4\sqrt{kmg\tan \theta }$

In equilibrium, the expressions are given as,

$F=T\sin \theta$ (1)

$mg=T\cos \theta$ (2)

From above equations, it can be written as,

$\tan \theta =\frac{F}{mg}$ (3)

The electric force is given as,

$F=\frac{q^2}{4\pi {\epsilon }_0{x}^2}$

Substitute the value of $F$ in equation (3), we get

$\tan \theta =\frac{q^2}{4\pi {\epsilon }_{0}mg{x}^2}$

$x=\sqrt{\frac{q^2}{4\pi {\epsilon }_{0}mg\tan \theta }}$

$x=\sqrt{\frac{k{q}^2}{mg\tan \theta }}$

The electric potential is given as,

$V=\frac{kq}{\frac{x}{2}}+\frac{kq}{\frac{x}{2}}$

$V=\frac{4kq}{x}$

$V=4\sqrt{kmg\tan \theta }$

Thus, the electric potential at the centre of the line is $4\sqrt{kmg\tan \theta }$.