question_answer

In the given figure two tiny conducting balls of identical mass \[m\] and identical charge \[q\] hang from non-conducting threads of equal length\[L\]. Assume that \[\theta \] is so small that\[\tan \theta \]\[\approx \]\[\sin \theta \], then for equilibrium \[x\] is equal to

A) \[{{\left( \frac{{{q}^{2}}L}{2\pi {{\varepsilon }_{0}}mg} \right)}^{\frac{1}{3}}}\]       

B) \[{{\left( \frac{{{q}^{{}}}{{L}^{2}}}{2\pi {{\varepsilon }_{0}}mg} \right)}^{\frac{1}{3}}}\]

C) \[{{\left( \frac{{{q}^{2}}{{L}^{2}}}{4\pi {{\varepsilon }_{0}}mg} \right)}^{\frac{1}{3}}}\]    

D) \[{{\left( \frac{{{q}^{2}}{{L}^{{}}}}{4\pi {{\varepsilon }_{0}}mg} \right)}^{\frac{1}{3}}}\]

Correct Answer: A

Solution :

[a] In equilibrium \[{{F}_{e}}=T\sin \theta \] \[mg=T\cos \theta \] \[\tan \theta =\frac{{{F}_{e}}}{mg}=\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}{{x}^{2}}\times mg}\] Also \[\tan \theta \approx \sin \theta =\frac{x/2}{L}\] Hence \[\frac{x}{2L}=\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}{{x}^{2}}\times mg}\] \[\Rightarrow {{x}^{3}}=\frac{2{{q}^{2}}L}{4\pi {{\varepsilon }_{0}}mg}\Rightarrow x={{\left( \frac{{{q}^{2}}L}{2\pi {{\varepsilon }_{0}}mg} \right)}^{1/3}}\]