Single Choice

Two identical charged spheres suspended from a common distance $$d(d < < 1)$$ apart because of their mutual repulsion. The charged begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velocity $$v$$. Then $$v$$ varies as a function of the distance $$x$$ between the spheres, as

A$$v\propto x^{\dfrac {1}{2}}$$
B$$v\propto x$$
C$$v\propto x^{-\dfrac {1}{2}}$$
Correct Answer
D$$v\propto x^{-1}$$

Solution

From figure $$\tan \theta = \dfrac {F_{e}}{mg} =\theta$$
$$\dfrac {kq^{2}}{x^{2}mg} = \dfrac {x}{2l}$$
or $$x^{3}\propto q^{2} .... (1)$$
or $$x^{3/2} \propto q ..... (2)$$
Differentiating eq. $$(1)$$ w.r.t time
$$3x^{2} \dfrac {dx}{dt}\propto 2q \dfrac {dq}{dt}$$ but $$\dfrac {dq}{dt}$$ is constant
so $$x^{2}(v) \propto q$$ Replace $$q$$ from eq. $$(2)$$
$$x^{2}(v) \propto x^{3/2}$$ or $$v\propto x^{-1/2}$$.

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