Two small equally charged spheres each of Mass $\text{'}m\text{'}$  are suspended from the same point by silk threads of length L. The distance between the spheres $x<<L$ . Find the rate $\left(\frac{dq}{dt}\right)$  with which the charges leak off each sphere if their approach velocity varies as  $V=\frac{a}{\sqrt{x}}$ when $\text{'}a\text{'}$  is constant

 

For momentary equilibrium of the spheres

                                        $$\begin{gathered} T\cos \theta =mg and T\sin \theta =F \\ \Rightarrow tan\theta =\frac{F}{mg} \\ for x<<L , tan\theta =\frac{x}{2L} \\ So \frac{F}{mg}=\frac{x}{2L} \\ \Rightarrow F=\frac{mgx}{2L} \\ But F=\frac{1}{4\pi {\in }_0}\frac{q^2}{x^2} \\ \therefore \frac{1}{4\pi {\in }_0}\frac{q^2}{x^2}=\frac{mgx}{2L} \\ \Rightarrow q^2=\frac{2\pi {\in }_{0}mg{x}^3}{L} \\ Differentiate q wrt t \\ \Rightarrow 2q\frac{dq}{dt}=\frac{2\pi {\in }_{0}mg3x^2}{L}\frac{dx}{dt} \\ given \frac{dx}{dt}=V=\frac{a}{\sqrt{x}} \\ \therefore \frac{dq}{dt}=\frac{3a}{2}\sqrt{\frac{2\pi {\in }_{0}mg}{L}} \end{gathered}$$