Question

In the given figure two tiny conducting balls of identical mass m and identical charge q hang from non-conducting threads of equal length L. Assume that $\theta$ is so small that $\tan \theta \approx sin\theta$, then for equilibrium x is equal to

• A. ${\left(\frac{q^{2}L}{2\pi {ϵ}_{0}mg}\right)}^{\frac{1}{3}}$
• B. ${\left(\frac{q{L}^2}{2\pi {ϵ}_{0}mg}\right)}^{\frac{1}{3}}$
• C. ${\left(\frac{q^2{L}^2}{4\pi {ϵ}_{0}mg}\right)}^{\frac{1}{3}}$
• D. ${\left(\frac{q^{2}L}{4\pi {ϵ}_{0}mg}\right)}^{\frac{1}{3}}$

Answer 1

The correct option is A ${\left(\frac{q^{2}L}{2\pi {ϵ}_{0}mg}\right)}^{\frac{1}{3}}$


In equilibrium $F_e=Tsin\theta$ .......(i)
$mq=Tcos\theta$ ........(ii)
$tan\theta =\frac{F_e}{mg}=\frac{q^2}{4\pi {ϵ}_0{x}^2\times mg}$ also $tan\theta \approx sin\theta =\frac{x/2}{L}$
Hence $\frac{x}{2L}=\frac{q^2}{4\pi {ϵ}_0{x}^2\times mg}$
$\Rightarrow x^3=\frac{2q^{2}L}{4\pi {ϵ}_{0}mg}\Rightarrow x={\left(\frac{q^{2}L}{2\pi {ϵ}_{0}mg}\right)}^{1/3}$