Two identical tennis balls each having mass ‘m’ and charge ‘q’ are suspended from a fixed point by threads of length $l$ What is the equilibrium separation when each thread makes a small angle $' θ'$ with the vertical?

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$x = {[\frac{q^{2} l}{2 π ε_{0} m g}]}^{1 / 2}$

$x = {[\frac{q^{2} l^{2}}{2 π ε_{0} m^{2} g^{2}}]}^{1 / 3}$

$x = {[\frac{q^{2} l}{2 π ε_{0} m g}]}^{1 / 3}$

$x = {[\frac{q^{2} l^{2}}{2 π ε_{0} m^{2} g}]}^{1 / 3}$

answer is B.

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Detailed Solution

$T \cos θ = m g a n d T \sin θ = \frac{k q^{2}}{x^{2}}$

$\tan θ = \frac{k q^{2}}{x^{2} m g}$

As $\tan θ = \sin θ \approx \frac{x}{2 l}$

$\frac{x}{2 l} = \frac{k q^{2}}{x^{2} m g} x = {[\frac{q^{2} L}{2 π ε_{0}}]}^{1 / 3}$

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