In the given figure two tiny conducting balls of identical mass m and identical charge q hang from non-conducting threads of equal length L. Assume that θ θ is so small that $\tan \theta \approx \sin \theta$ , then for equilibrium x is equal to
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In equilibrium F e = T sin θ θ ....... (i)
mg = T cos θ θ ....... (ii)
$\tan \theta = \frac{F_e}{mg} = \frac{q^2}{4 \pi \varepsilon_0 x^2 mg}$ also $\tan \theta \approx \sin \theta = \frac{x/2}{L}$
Hence $\frac{x}{2L} = \frac{q^{2}}{4 \pi \epsilon_{0} x^{2} \times mg}$
⇒ ⇒ $x^{3} = \frac{2q^{2}L}{4 \pi \epsilon_{o} mg}$ ⇒ ⇒ $x=\left(\frac{q^{2}L}{2\pi\varepsilon_{0}mg}\right)^{1/3}$
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