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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Aicharya · Accepted Answer

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 $\mathrm{x} \propto \ell^{1 / 3}$

Two balls of same mass and carrying equal charge are hung from a fixed support of length $l$. At…

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