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

Answer 1

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


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