ExamBro
ExamBro
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.
Verified Solution

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 \(x \propto \ell^{1 / 3}\)

Same subject
Explore more questions on app