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

The correct option is C $v\propto x^{-\frac{1}{2}}$

The electrostatic force of repulsion between spheres is given as,
$F=\frac{k{q}^2}{d^2}$

Applying the equilibrium condition on the sphere,

$T\sin \theta =F....\left(1\right)$

$T\cos \theta =mg....\left(2\right)$

$\text{Or},\tan \theta =\frac{F}{mg}$

$\Rightarrow \tan \theta =\frac{k{q}^2}{d^{2}mg}$

When the charge starts to leak, the separation between spheres changes.

let it be ${}^{\prime }{x}^{\prime }$ at some point of time

Therefore,

$\tan \theta =\frac{k{q}^2}{x^{2}mg}....\left(3\right)$


From $△ABC$
$\tan \theta =\frac{\frac{x}{2}}{\sqrt{l^2-\frac{x^2}{4}}}$

$\because l>>d\text{hence}l>>x$

So $\tan \theta =\sin \theta =\frac{\left(\frac{x}{2}\right)}{l}=\frac{x}{2l}....\left(4\right)$

From Eq.$\left(3\right)$ and $\left(4\right)$,

$\frac{x}{2l}=\frac{k{q}^2}{x^{2}mg}$

$x^3=\frac{2kl{q}^2}{mg}....\left(5\right)$

$\Rightarrow q\propto x^{3/2}$

Differentiating equation $\left(5\right)$,

$\Rightarrow 3x^2\left(\frac{dx}{dt}\right)=\frac{2kl}{mg}\cdot 2q\frac{dq}{dt}$

Since the charged spheres are approaching each other with speed $v$, the velocity of approach will be

$v_r=v+v=2v$

That means the separation between them is reducing at a rate of $2v$

$\frac{dx}{dt}=2v$

Substituting it we get,

$\Rightarrow 6x^2\cdot v=\frac{2kl}{mg}\cdot 2q\cdot C$

Given that charge leaks at a constant rate,

$\frac{dq}{dt}$ $\text{= constant = C}$

$\Rightarrow v\propto (x^{-2}\cdot q)$

Substituting $q\propto x^{3/2}$ we get,

$v\propto x^{-2}\cdot x^{3/2}$

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