question_answer

Two identical charged spheres suspended from a common point by two massless strings of length \[l\] are initially a distance d(d < < \[l\]) apart because of their mutual repulsion. The spheres begins to leak from both the spheres at a constant rate. As a result the charges approach each other with a velocity v. Then as a function of distance x between them

A) \[v\propto {{x}^{-1/2}}\]         

B) \[v\propto {{x}^{-1}}\]

C) \[v\propto {{x}^{1/2}}\]

D) \[v\propto x\]

Correct Answer: A

Solution :

[a] : Figure shows equilibrium positions of the two spheres.             \[\therefore \]\[T\cos \theta =mg\]and\[T\sin \theta =F\] \[=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}^{2}}}{{{d}^{2}}}\] \[\therefore \]\[\tan \theta =\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}^{2}}}{{{d}^{2}}mg}\] When charge begins to leak from both the spheres at a constant rate. Let x be the distance between them at any instant. \[\tan \theta =\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}^{2}}}{{{x}^{2}}mg}\] \[\frac{x}{2l}=\frac{{{q}^{2}}}{4\pi {{\varepsilon }_{0}}{{x}^{2}}mg}\]                    \[\left( \because \tan \theta =\frac{x}{2l} \right)\] or\[\frac{x}{2l}\propto \frac{{{q}^{2}}}{{{x}^{2}}}\]or\[{{q}^{2}}\propto {{x}^{3}}\Rightarrow q\propto {{x}^{3/2}}\] \[\frac{dq}{dt}\propto \frac{3}{2}{{x}^{1/2}}\frac{dx}{dt}\]or\[v\propto {{x}^{-1/2}}\]\[\left( \because \frac{dq}{dt}=\text{constant} \right)\]