Question

Two identical tennis balls each having mass ${}^{\prime }{m}^{\prime }$ and charge ${}^{\prime }{q}^{\prime }$ are suspended from a fixed point by threads of length ${}^{\prime }{l}^{\prime }$. 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. $x={\left(\frac{q^2{l}^2}{2\pi {\epsilon }_0{m}^{2}g}\right)}^{\frac{1}{3}}$
• C. $x={\left(\frac{q^{2}l}{2\pi {\epsilon }_{0}mg}\right)}^{\frac{1}{3}}$
• D. $x={\left(\frac{q^2{l}^2}{2\pi {\epsilon }_0{m}^2{g}^2}\right)}^{\frac{1}{3}}$

Answer 1

The correct option is C $x={\left(\frac{q^{2}l}{2\pi {\epsilon }_{0}mg}\right)}^{\frac{1}{3}}$

There are three forces, electostatic force $\left(F_e\right)$, tension $\left(T\right)$ and weight $(W=mg)$ acting on the charges.


The horizontal component of tension $T$ gets balanced the electrostatic force while the vertical component by weight.

$T\cos \theta =mg$

$T\sin \theta =\frac{k{q}^2}{x^2}$

Taking the ratio of these two equations,

$\tan \theta =\frac{k{q}^2}{x^{2}mg}$

Since, $\theta$ is small, so $\tan \theta \approx \sin \theta \approx \frac{x}{2l}$

$\Rightarrow \frac{x}{2l}=\frac{k{q}^2}{x^{2}mg}$

Plugging the value of $k=\frac{1}{4\pi {\epsilon }_0}$, we get,

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

Hence, option $\left(\text{B}\right)$ is correct.