Question

Two charged balls are attached by silk threads of length $l$ to the same point. Their velocity is $\frac{K}{\sqrt{x}}$, where $K$ is a constant and $x$ is the distance between the balls, $x$ is very small in comparison to $l$. The rate of leakage of charge in ${10}^{-5}\text{C/s}$ is (take $\frac{l}{mg}=10,K=4\sqrt{2}$)

• A. ${10}^{-5}\text{C/sec}$
• B. $2\times {10}^{-5}\text{C/sec}$
• C. $3\times {10}^{-5}\text{C/sec}$
• D. ${10}^{-3}\text{C/sec}$

Answer 1

The correct option is B $2\times {10}^{-5}\text{C/sec}$


Let $T$ be the tension in each of the silk threads.
$T\sin \theta =F,T\cos \theta =mg$
$\tan \theta =\frac{F}{mg}=\frac{q}{4\pi {ϵ}_0{x}^2}.\frac{1}{mg}$
Since $\theta$ is small, $\tan \theta =\sin \theta =\frac{x}{2l}$
$x=\frac{2lF}{mg}=\frac{2l}{mg}.\frac{q^2}{4\pi {ϵ}_0{x}^2}$
$\Rightarrow x^3=\frac{2l}{mg}.\frac{q^2}{4\pi {ϵ}_0}$
$\Rightarrow x^3=\frac{l}{2\pi {ϵ}_{0}mg}.q^2$
$x={\left(\frac{l}{2\pi {ϵ}_{0}mg}\right)}^{\frac{1}{3}}{q}^{\frac{2}{3}}\dots \dots \left(i\right)$

$\frac{dx}{dq}=\frac{\frac{dx}{dt}}{\frac{dq}{dt}}={\left(\frac{l}{2\pi {ϵ}_{0}mg}\right)}^{\frac{1}{3}}\times \frac{2}{3}{q}^{-\frac{1}{3}}$

$\frac{dq}{dt}=q^{\frac{1}{3}}\frac{\frac{dx}{dt}}{\frac{2}{3}{\left(\frac{l}{2\pi {ϵ}_{0}mg}\right)}^{\frac{1}{3}}}\dots \dots \left(ii\right)$
It is given, $\frac{dx}{dt}=\frac{K}{\sqrt{x}}=\frac{K}{{\left(\frac{l}{2\pi {ϵ}_{0}mg}\right)}^{\frac{1}{6}}{q}^{\frac{1}{3}}}\dots \dots \left(iii\right)$
From equations (ii) and (iii), we get, $\frac{dq}{dt}=\frac{K}{\frac{2}{3}{\left(\frac{l}{2\pi {ϵ}_{0}mg}\right)}^{\frac{1}{2}}}$
$\Rightarrow \frac{dq}{dt}=\frac{3K}{2}{\left[\frac{2\pi {ϵ}_{0}mg}{l}\right]}^{\frac{1}{2}}$
$\Rightarrow \frac{dq}{dt}=2\times {10}^{-5}\text{C/s}$.

Hence, option (a) is the correct answer.