Electrostatics Question 658
Question: In the given 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
Options:
A) $ {{( \frac{q^{2}L}{2\pi {\varepsilon _{0}}mg} )}^{\frac{1}{3}}} $
B) $ {{( \frac{{{q}^{{}}}L^{2}}{2\pi {\varepsilon _{0}}mg} )}^{\frac{1}{3}}} $
C) $ {{( \frac{q^{2}L^{2}}{4\pi {\varepsilon _{0}}mg} )}^{\frac{1}{3}}} $
D) $ {{( \frac{q^{2}{{L}^{{}}}}{4\pi {\varepsilon _{0}}mg} )}^{\frac{1}{3}}} $
Show Answer
Answer:
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{2q^{2}L}{4\pi {\varepsilon _{0}}mg}\Rightarrow x={{( \frac{q^{2}L}{2\pi {\varepsilon _{0}}mg} )}^{1/3}} $
