Two similar balls of mass 'm' are hung by a silk thread of length L and carry similar charges Q as in figure
Assuming the separation to be small, the separation between the balls (denoted by x) is equal to

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${[\frac{Q^{2} \cdot 2 L}{4 {πε}_{0} \cdot mg}]}^{\frac{1}{3}}$

$\frac{QL}{mg}$

$\frac{Q^{2} L}{2 πmg}$

QLmg

answer is A.

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Detailed Solution

$\begin{array}{l} T \cos θ = m g \\ T \sin θ = F = \frac{1}{4 {πε}_{0}} \frac{Q^{2}}{x^{2}} \\ \tan θ = \frac{F}{m g} \\ \frac{x}{2 L} = \frac{1}{mg} [\frac{1}{4 {πE}_{0}} \frac{Q^{2}}{x^{2}}] \end{array}$

$x^{3} = [\frac{Q^{2} \times 2 L}{4 π \in_{0} mg}]; x = {[\frac{Q^{2} 2 L}{4 π ϵ_{0} \cdot mg}]}^{\frac{1}{3}}$

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