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 \(\theta\) with the vertical?

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Answer & Solution

Correct Answer

(C) \({x}=\left(\frac{{q}^{2} l}{2 \pi \varepsilon_{0} {mg}}\right)^{1 / 3}\)

Step-by-step Solution

Detailed explanation

\(T \cos \theta= mg\)

\(T \sin \theta=\frac{ kq ^{2}}{ x ^{2}}\)

\(\tan \theta=\frac{k q^{2}}{x^{2} m g}\)

as \(\tan \theta \approx \sin \theta \approx \frac{x}{2 L}\)

\(\frac{ x }{2 L }=\frac{ Kq ^{2}}{ x ^{2} mg }\)

\(x=\left(\frac{q^{2} L}{2 \pi \varepsilon_{0} m g}\right)^{1 / 3}\)

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