Question

Two small metal spheres having equal charge and mass are suspended from some point on the ceiling of a damp room with silk threads of equal length. Let centre to centre distance between sphere be $x,x<<l,l$ is length of silk thread. Due to ionization of medium, charge leaks off from each sphere and they keep on coming closer to each other at a constant rate.
Let their approach velocity $v$ varies as $v\propto x^{-1/2}$.
If mass of each sphere is $m$ then the rate at which charge varies with respect to time is $\frac{dq}{dt}\propto \frac{N}{2}\sqrt{\frac{2\pi {\epsilon }_{0}mg}{l}}$. The value of $N$ is________

Answer 1

Answer: 3

At equilibrium, $\frac{q^2}{4\pi {\epsilon }_0{x}^2}\cos \theta =mg\sin \theta$
For small $\theta ,\cos \theta \to 1$ and $\sin \theta \approx \theta =\frac{x/2}{l}$
$\Rightarrow \frac{q^2}{4\pi {\epsilon }_0{x}^2}=\frac{mgx}{2l}$
$\Rightarrow q=x^{3/2}\cdot \sqrt{\frac{2\pi {\epsilon }_{0}mg}{l}}$
$\Rightarrow \frac{dq}{q}=\frac{3}{2}\left(x^{3/2}\cdot \sqrt{\frac{2\pi {\epsilon }_{0}mg}{l}}\right)\cdot \left(\frac{dx}{dt}\right)$
Put $v=\frac{dx}{dt}\propto x^{1/2}$
$\Rightarrow \frac{dq}{dt}\propto \frac{3}{2}\sqrt{\frac{2\pi {\epsilon }_{0}mg}{l}}$
Hence, $N=3$