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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 :
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The correct answer is:
$l^{1 / 3}$
$l^{1 / 3}$
In equilibrium, $\mathrm{F}_{\mathrm{e}}=\mathrm{T} \sin \theta$
$m g=T \cos \theta$
$\tan \theta=\frac{F_e}{m g}=\frac{q^2}{4 \pi \epsilon_0 x^2 \times m g}$
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 $\mathrm{x} \propto \ell^{1 / 3}$
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