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JEE Mains · Physics · STD 12 - 1. Electric charges and fields
Two balls of same mass and carrying equal charge are hung from a fixed support of length \(l\). At electrostatic equilibrium, assuming that angles made by each thread is small, the separation, \(x\) between the balls is proportional to
- A \(l\)
- B \(l^2\)
- C \({l^{2/3}}\)
- D \({l^{1/3}}\)
Answer & Solution
Correct Answer
(D) \({l^{1/3}}\)
Step-by-step Solution
Detailed explanation
\(\text { In equilibrium, } \mathrm{F}_{\mathrm{e}}=\mathrm{T} \sin \theta\) \(\mathrm{mg}=\mathrm{T} \cos \theta\) \(\tan \theta = \frac{{{F_e}}}{{mg}} = \frac{{{q^2}}}{{4\pi {_0}\,{x^2}}} \times mg\) \(\text { also } \tan \theta \approx \sin =\frac{x / 2}{\ell}\) Hence,…
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