Two small metal spheres having equal charge and mass are suspended from same point on the ceiling of a damp room with silk threads of equal length. Let center to center 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 $\frac{d q}{d t} \propto \frac{N}{2} \sqrt{\frac{2 π ε_{0} m g}{l}}$ . Find the value of N.

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answer is 3.

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Detailed Solution

At equilibrium

Question Image

$T \sin θ = \frac{R q^{2}}{x^{2}} T \cos θ = m g \tan θ = \frac{R q^{2}}{m g x^{2}}$

For small angle

$θ = \frac{R q^{2}}{m g x^{2}} \frac{x}{2 l} = \frac{R q^{2}}{m g x^{2}} q = x^{3 / 2} \frac{\sqrt{2 π ε_{0} m g}}{l} \frac{d q}{d t} = \frac{3}{2} x^{1 / 2} \frac{\sqrt{2 π ε_{0} m g}}{l} (\frac{d x}{d t}) \frac{d q}{d t} = \frac{3}{2} x^{1 / 2} V . \frac{\sqrt{2 π ε_{0} m g}}{l} N = 3$