Question

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. $x={\left(\frac{q^{2}l}{2\pi {\epsilon }_{0}mg}\right)}^{\frac{1}{2}}$
• B. $x={\left(\frac{q^{2}l}{2\pi {\epsilon }_{0}mg}\right)}^{\frac{1}{3}}$
• C. $x={\left(\frac{q^2{l}^2}{2\pi {\epsilon }_0{m}^{2}g}\right)}^{\frac{1}{2}}$
• D. $x={\left(\frac{q^2{l}^2}{2\pi {\epsilon }_0{m}^2{g}^2}\right)}^{\frac{1}{2}}$

Answer 1

Answer: (B)

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