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 Main  Online Paper (Held On 22 April 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 :

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