Two identical charged spheres suspended from a common point by two… | JEE Main 2011 (AIEEE)

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Electric Charge & Coulomb's Law

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,

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Answer: (d)

For small angles: $\dfrac{x}{2l} = \dfrac{F}{mg} = \dfrac{kq^2}{x^2 mg} \Rightarrow q \propto x^{3/2}$ .
Constant leakage rate: $\dfrac{dq}{dt} \propto x^{1/2}\dfrac{dx}{dt} = \text{constant}$
So $v = \dfrac{dx}{dt} \propto x^{-1/2}$ .
So the answer is (d) $v \propto x^{-1/2}$ .

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