Q. Two identical charged spheres suspended from a common point by two massless stings 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) $v \propto x$
Explanation
(a) $T\sin\theta = F = \frac{q^{2}}{4\pi\varepsilon_{0}d^{2}}$
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$T\cos\theta = mg$
$\tan\theta = \frac{1}{4\pi\varepsilon_{0}}\frac{q^{2}}{x^{2}mg}$
$\frac{x}{2l} = \frac{1.q^{2}}{4\pi\varepsilon_{0}x^{2}mg}$
or $\frac{x}{2l} \propto \frac{q^{2}}{x^{2}},$ or $q^{2} \propto x^{3} \Rightarrow q \propto x^{\frac{3}{2}}$
$\frac{dq}{dt} \propto \frac{3}{2}x^{\frac{1}{2}}\frac{dx}{dt}$
$v \propto x^{\frac{1}{2}}\ \ \ \ \ \ \ \ \ \ \left\lbrack \frac{dq}{dt} = cons. \right\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.