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 \[V=\frac{kq}{x/2}+\frac{kq}{x/2}=4\sqrt{kmg/\tan \theta }\]You need to login to perform this action.
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