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 $\frac{d q}{d t} \propto \frac{N}{2} \sqrt{\frac{2 \pi \varepsilon_{0} m g}{l}}$. The value of $N$ is________

Answer 1

At equilibrium, $\frac{q^{2}}{4 \pi \varepsilon_{0} x^{2}} \cos \theta=m g \sin \theta$
For small $\theta, \cos \theta \rightarrow 1$ and $\sin \theta \approx \theta=\frac{x / 2}{l}$
$\Rightarrow \frac{q^{2}}{4 \pi \varepsilon_{0} x^{2}}=\frac{m g x}{2 l}$
$ \Rightarrow q=x^{3 / 2} \cdot \sqrt{\frac{2 \pi \varepsilon_{0} m g}{l}}$
$\Rightarrow \frac{d q}{q} =\frac{3}{2}\left(x^{3 / 2} \cdot \sqrt{\frac{2 \pi \varepsilon_{0} m g}{l}}\right) \cdot\left(\frac{d x}{d t}\right)$
Put $v=\frac{d x}{d t} \propto x^{1 / 2} $
$\Rightarrow \frac{d q}{d t} \propto \frac{3}{2} \sqrt{\frac{2 \pi \varepsilon_{0} m g}{l}}$
Hence, $N=3$