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JEE Mains · Physics · STD 12 - Electric charges and fields
- JEE Mains 2013
- Hard
- MCQ
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
Pick an option to check your answer.
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, \(\frac{x}{2 \ell}=\frac{q^{2}}{4 \pi \epsilon_{0} x^{2} \times m g}\)
\(\Rightarrow x^{3}=\frac{2 q^{2} \ell}{4 \pi \epsilon_{0} m g}\)
\(\therefore x=\left(\frac{q^{2} \ell}{2 \pi \epsilon_{0} m g}\right)^{1 / 3}\)
Therefore \(x \propto \ell^{1 / 3}\)
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