question_answer

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:                     [JEE ONLINE 09-04-2013]

A) \[l\]

B) \[{{l}^{2}}\]

C)             \[{{l}^{2/3}}\]

D) \[{{l}^{1/3}}\]

Correct Answer: D

Solution :

[d]

\[T\,\cos \theta =mg\]

\[T\,\sin \theta =\frac{k{{q}^{2}}}{{{x}^{2}}}\]

\[\tan \theta =\frac{k{{q}^{2}}}{mg\,{{x}^{2}}}\]                                          (i)

From OBC

\[\Rightarrow \] \[\tan \theta =\frac{CB}{OB}=\frac{X/2}{{{\left( {{l}^{2}}-\frac{{{x}^{2}}}{4} \right)}^{1/2}}}\]       (ii)

Form Eqs. (i) and (ii)

\[\frac{x}{2{{\left( {{l}^{2}}-\frac{{{x}^{2}}}{4} \right)}^{1/2}}}=\frac{k{{q}^{2}}}{mg\,{{x}^{2}}}\]

\[{{x}^{3}}=\frac{2k{{q}^{2}}}{mg}\,{{\left( {{l}^{2}}-\frac{{{x}^{2}}}{4} \right)}^{1/2}}\]

If \[\theta \] is small, then

\[\frac{{{x}^{2}}}{4}<<{{l}^{2}}\,\,\,\Rightarrow \,\,{{x}^{3}}=\frac{2k{{q}^{2}}}{mg}{{({{l}^{2}})}^{1/2}}\]

\[x=\frac{2k{{q}^{2}}}{mg}{{l}^{1/3}}\]

\[x\propto {{l}^{1/3}}\]