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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?
${x}=\left(\frac{{q}^{2} l}{2 \pi \varepsilon_{0} {mg}}\right)^{1 / 2}$
${x}=\left(\frac{{q}^{2} l^{2}}{2 \pi \varepsilon_{0} {m}^{2} {g}^{2}}\right)^{1 / 3}$
${x}=\left(\frac{{q}^{2} l}{2 \pi \varepsilon_{0} {mg}}\right)^{1 / 3}$
${x}=\left(\frac{{q}^{2} l^{2}}{2 \pi \varepsilon_{0} {m}^{2} {g}}\right)^{1 / 3}$
Solution
$T \cos \theta= mg$
$T \sin \theta=\frac{ kq ^{2}}{ x ^{2}}$
$\tan \theta=\frac{k q^{2}}{x^{2} m g}$
as $\tan \theta \approx \sin \theta \approx \frac{x}{2 L}$
$\frac{ x }{2 L }=\frac{ Kq ^{2}}{ x ^{2} mg }$
$x=\left(\frac{q^{2} L}{2 \pi \varepsilon_{0} m g}\right)^{1 / 3}$
