Question 160 of 194

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?

A$x={\left(\dfrac{{q}^{2}l}{2\pi {\epsilon }_{0}mg}\right)}^{\dfrac{1}{2}}$B$x={\left(\dfrac{{q}^{2}l}{2\pi {\epsilon }_{0}mg}\right)}^{\dfrac{1}{3}}$C$x={\left(\dfrac{{q}^{2}{l}^{2}}{2\pi {\epsilon }_{0}{m}^{2}g}\right)}^{\dfrac{1}{3}}$D$x={\left(\dfrac{{q}^{2}{l}^{2}}{2{\pi \epsilon }_{0}{m}^{2}{g}^{2}}\right)}^{\dfrac{1}{3}}$

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Answer: (B) $x={\left(\dfrac{{q}^{2}l}{2\pi {\epsilon }_{0}mg}\right)}^{\dfrac{1}{3}}$

$T\cos \theta =mg$
$T\sin \theta =\dfrac{k{q}^{2}}{{x}^{2}}$
$\tan \theta =\dfrac{k{q}^{2}}{{x}^{2}mg}$
as $\tan \theta \approx \sin \theta \approx \dfrac{x}{2L}$
$\dfrac{x}{2L}=\dfrac{K{q}^{2}}{{x}^{2}mg}$
$x={\left(\dfrac{{q}^{2}L}{2\pi {\epsilon }_{0}mg}\right)}^{\dfrac{1}{3}}$