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

Answer 1

Answer: (B)

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