Two identical charged spheres suspended from a common point by two… | NEET 2016 (Phase 1)

Q68NEET 2016 · PHASE 1Difficult

Electric Charge & Coulomb's Law

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. Then v varies as a function of the distance x between the spheres, as

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Answer: (a)

Equilibrium of each sphere: $\tan\theta = \dfrac{F}{mg}$ , and for small angles $\tan\theta \approx \dfrac{x/2}{l}$ .
$\dfrac{x}{2l} = \dfrac{kq^2}{x^2 mg} \Rightarrow q^2 \propto x^3 \Rightarrow q \propto x^{3/2}$
Charge leaks at a constant rate: $\dfrac{dq}{dt} \propto \dfrac{3}{2}x^{1/2}\dfrac{dx}{dt} = \text{constant}$
So $v = \dfrac{dx}{dt} \propto x^{-1/2}$ .
So the answer is (a) $v \propto x^{-1/2}$ .

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