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  • In the given figure two tiny conducting balls of identical mass m and identical charge hang from non-conducting threads of eq

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 \tan\theta \approx \sin\theta is so small that, then for equilibrium x is equal to 

Option: 1

\left(\frac{q^2L}{2\pi\epsilon_0 mg} \right )^{\frac{1}{3}}


Option: 2

\left(\frac{qL^2}{2\pi\epsilon_0 mg} \right )^{\frac{1}{3}}


Option: 3

\left(\frac{q^2L^2}{4\pi\epsilon_0 mg} \right )^{\frac{1}{3}}


Option: 4

\left(\frac{q^2L}{4\pi\epsilon_0 mg} \right )^{\frac{1}{3}}


Answers (1)

best_answer

In equilibrium

   Fe = T sin\theta    ....... (i)

  mg = T cos\theta    ....... (ii)

\tan\theta = \frac{F_e}{mg} = \frac{q^2}{4\pi\epsilon_0 x^2\times mg}\; \&\; \tan\theta \approx \sin\theta = \frac{\frac{x}{2}}{L}

Hence

 \frac{x}{2L} = \frac{q^2}{4\pi\epsilon_0 x^2 \times mg} \\*\Rightarrow x^3 = \frac{2q^2L}{4\pi\epsilon_0 mg} \Rightarrow x = \left (\frac{2q^2L}{4\pi\epsilon_0 mg} \right )^\frac{1}{3}

Posted by

Divya Prakash Singh

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