Question

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 charge 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. $v\propto x$

Answer 1

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

From the figure it is clear that tension is acting on the string.

So,

$T\cos \theta =mg.........\left(1\right)$

$T\sin \theta =F$

$T\sin \theta =\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{d^2}.................\left(2\right)$

From the figure, $tan\theta =\frac{x}{2l}$

From equation (1) and (2)

$\tan \theta =\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{d^{2}mg}$

When the charges starts approaching each other,

$\frac{x}{2l}=\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{x^{2}mg}$

$\frac{x^{3}mg2\pi {\epsilon }_0}{l}=q^2$

Taking derivation on both sides:

$3x^2\frac{dx}{dt}\frac{mg2\pi {\epsilon }_0}{l}=2q\frac{dq}{dt}$

$\frac{dx}{dt}=vand\frac{dq}{dt}=constant$

$\frac{dq}{dt}\propto \frac{3}{2}{x}^{\frac{1}{2}}\frac{dx}{dt}$

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