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 q is so small that \[\tan \theta \approx \sin \theta \], then for equilibrium x is equal to                                              [AMU 2000]

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 :

                   In equilibrium \[{{F}_{e}}=T\text{sin}\theta \]   ....... (i)                        \[mg=T\text{cos}\theta \]    ....... (ii) \[\tan \theta =\frac{{{F}_{e}}}{mg}=\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{o}}{{x}^{2}}\times mg}\] also \[\tan \theta \approx \sin \theta =\frac{x/2}{L}\] Hence \[\frac{x}{2L}=\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{o}}{{x}^{2}}\times mg}\] Þ \[{{x}^{3}}=\frac{2{{q}^{2}}L}{4\pi {{\varepsilon }_{o}}mg}\] Þ \[x={{\left( \frac{{{q}^{2}}L}{2\pi {{\varepsilon }_{o}}mg} \right)}^{1/3}}\]