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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
$l$
$l^2$
${l^{2/3}}$
${l^{1/3}}$
Solution
$\text { In equilibrium, } \mathrm{F}_{\mathrm{e}}=\mathrm{T} \sin \theta$
$\mathrm{mg}=\mathrm{T} \cos \theta$
$\tan \theta = \frac{{{F_e}}}{{mg}} = \frac{{{q^2}}}{{4\pi {_0}\,{x^2}}} \times mg$
$\text { also } \tan \theta \approx \sin =\frac{x / 2}{\ell}$
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 $x \propto \ell^{1 / 3}$
