Question

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 $v$. Which of the following correctly represent the variation of $v$ as a function of the distance $x$ between the spheres.

• A. $v\propto x^{-\frac{1}{2}}$
• B. $v\propto x^{-1}$
• C. $v\propto x^{\frac{1}{2}}$
• D. $v\propto x^{-\frac{1}{2}}$

Answer 1

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


From figure,

$T\cos \theta =mg\dots \dots \left(1\right)$
$T\sin \theta =\frac{k{q}^2}{x^2}\dots \dots \left(2\right)$

From eqns. ($1$) and ($2$),

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

Since $d<<l$ so will be $x<<l$. so $\theta$ is small,
$\therefore \tan \theta \approx \sin \theta =\frac{x}{2l}$
$\therefore \frac{x}{2l}=\frac{k{q}^2}{x^{2}mg}\Rightarrow q^2=x^3\frac{mg}{2lk}$

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

Differentiating with respect to time on both sides we get,

$\frac{dq}{dt}\propto \frac{3}{2}\sqrt{x}\frac{dx}{dt}$

$\Rightarrow \frac{dq}{dt}\propto \frac{3}{2}\sqrt{x}v$

From the data given in the question,

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

$\therefore v\propto \frac{1}{\sqrt{x}}$

Hence, option (a) is the correct answer.