Two identical charged spheres suspended from a common point by two massless strings of lengths , are initially at a distance apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the sph

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Solution

Q.
Two identical charged spheres suspended from a common point by two massless strings of lengths $l$ , are initially at a distance $d (d < < l)$ apart because of their mutual repulsion. The charges 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 [2016]

1 $v \propto x^{- 1 / 2}$

2 $v \propto x^{- 1}$

3 $v \propto x^{1 / 2}$

4 $v \propto x$

Ans.
(1)

From figure, $T \cos θ = m g$ ...(i)

   $T \sin θ = \frac{k q^{2}}{x^{2}}$ ...(ii)

From eqns. (i) and (ii), $\tan θ = \frac{k q^{2}}{x^{2} m g}$

Since $θ$ is small,    $∴$ $\tan θ \approx \sin θ = \frac{x}{2 l}$

$∴$    $\frac{x}{2 l} = \frac{k q^{2}}{x^{2} m g} \Rightarrow q^{2} = x^{3} \frac{m g}{2 l k} or q \propto x^{3 / 2}$

$\Rightarrow \frac{d q}{d t} \propto \frac{3}{2} \sqrt{x} \frac{d x}{d t} = \frac{3}{2} \sqrt{x} v$

Since, $\frac{d q}{d t} = constant$ ; $∴$ $v \propto \frac{1}{\sqrt{x}}$

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