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 y 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 $\frac{d q}{d t} \propto \frac{N}{2} \sqrt{\frac{2 π ε_{0} m g}{l}}$ .
The value of N is _________.

Moderate

Solution

$\begin{array}{l} At equilibrium, \frac{q^{2}}{4 π ε_{0} x^{2}} \cos θ = m g \sin θ \\ For small θ, \cos θ \to 1 and \sin θ \approx θ = \frac{x / 2}{l} \\ \Rightarrow \frac{q^{2}}{4 π ε_{0} x^{2}} = \frac{m g x}{2 l} \Rightarrow q = x^{3 / 2} \cdot \sqrt{\frac{2 π ε_{0} m g}{l}} \\ \Rightarrow \frac{d q}{q} = \frac{3}{2} (x^{3 / 2} \cdot \sqrt{\frac{2 π ε_{0} m g}{l}}) \cdot (\frac{d x}{d t}) \\ Put v = \frac{d x}{d t} \propto x^{1 / 2} \Rightarrow \frac{d q}{d t} \propto \frac{3}{2} \sqrt{\frac{2 π ε_{0} m g}{l}} \end{array}$
$Hence, N = 3$

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