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 $\theta$ from the vertical. If the mass of each charge is $m$, then the electrostatic potential at the centre of line joining them will be $\left(\frac{1}{4\pi\,\in_{0}}=k\right).$

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

Answer 1

Answer: (C)

$4\sqrt{k\,mg/tan\,\theta}$

Answer 2

In equilibrium, $Fe = T\, sin\, \theta$ $mg=T\,cos\,\theta$ $tan\,\theta=\frac{F_{e}}{mg}=\frac{q^{2}}{4\pi\,\in_{0}\,x^{2}\times mg}$ $\therefore x=\sqrt{\frac{q^{2}}{4\pi \,\in_{0}\,tan\,\theta\, mg}}$ Electric potential at the centre of the line $V=\frac{kq}{x/2}+\frac{kq}{x/2}=4 \sqrt{kmg/tan\,\theta}$