question_answer

Two small equal point charges of magnitude q are suspended from a common point on 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 centre of line joining them will be \[\left( \frac{1}{4\pi {{\in }_{0}}}=\operatorname{k} \right).\]     [JEE ONLINE 22-04-2013]

A) \[2\sqrt{\operatorname{k}\operatorname{mg}\tan \theta }\]

B) \[\sqrt{\operatorname{k}\operatorname{mg}\tan \theta }\]

C)             \[4\sqrt{\operatorname{k}\operatorname{mg}/\tan \theta }\]

D) \[4\sqrt{\operatorname{k}\operatorname{mg}\,\,\tan \theta }\]

Correct Answer: C

Solution :

[c]

In equilibrium, \[{{F}_{e}}=T\sin \theta \]

\[mg=\cos \theta \]

\[\tan \theta =\frac{{{F}_{e}}}{mg}=\frac{{{q}^{2}}}{4\pi {{\in }_{0}}{{x}^{2}}\times mg}\]

\[\therefore \]      \[x=\sqrt{\frac{{{q}^{2}}}{4\pi {{\in }_{0}}\tan \theta mg}}\]

Electric potential at the centre of the line

\[V=\frac{kq}{x/2}+\frac{kq}{x/2}=4\sqrt{kmg/\tan \theta }\]