Two identical charged spheres suspended from a common point by two massless strings of length $l\ $ are initially a distance $d(d

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 charges approach each other with a velocity $v\ .$Then as a function of distance $x$ between them

Dr. Abubacker · Accepted Answer

(d) At an instants
$T\ \cos{	heta = mg}\ \ \ \ ..(i)\ $
$T\ sin	heta = F_{e}\ \ \ \ \ \ \ldots(ii)$
$= \frac{ka^{2}}{x^{2}}$
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From Eqs. (i) and (ii), we have
$\ \ \ \ \ \ \ \ \frac{ka^{2}}{x^{2}}$=$mg$ tan $	heta$
$\Longrightarrow \ \ q^{2} = \frac{mg}{k}\ \frac{x}{2l}\ x^{2}\ \ \ \ \left( 	an	heta \approx \frac{a}{2l} ight)$
$\Longrightarrow q^{2} = \frac{mg}{2kl}x^{3}\ \ \ \ \ \ \ \ldots(iii)$
$\Longrightarrow 2q\frac{dq}{dt} = \frac{3mg}{2kl}\ x^{2}\frac{dx}{dt}$
$\Longrightarrow 2\left( \frac{mg}{2kl}\ \ x^{3} ight)^{\frac{1}{2}}\frac{dq}{dt} = \frac{3mg}{2kl}\ \ \ x^{2}v$
$\left\lbrack \because q = \left( \frac{mg}{2kl}\ x^{3} ight)^{\frac{1}{2}} ightbrack$
$\Longrightarrow \ \ \ \ vx^{1/2}$=constant
$\Longrightarrow \ \ v\  \propto \ x^{- 1/2}$

Explanation

Reviewed by Dr. Abubacker

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