Electrostatics

JEE Main level, single correct, medium

47 of 311

Question 47 of 311

Two identical charged spheres suspended from a common point by two massless strings of length I are initially a distance d(d<<1)d(d\lt \lt 1) 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 vv. Then as a function of distance xx between them,

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Answer: (D) v∝x−1/2v \propto x^{-1 / 2}

At any instant of separation between charges is xx.
 equilibrium condition =KQ2x2=ωx2ℓ⇒Q2=Cx3⇒2QdQdt=C3x2dxdt⇒dxdt∝x3/2x2∝x−1/2\begin{aligned} & \text { equilibrium condition }=\mathrm{K} \frac{\mathrm{Q}^2}{\mathrm{x}^2}=\omega \frac{\mathrm{x}}{2 \ell} \\ & \Rightarrow \mathrm{Q}^2=\mathrm{Cx}^3 \\ & \Rightarrow 2 \mathrm{Q} \frac{\mathrm{dQ}}{\mathrm{dt}}=\mathrm{C} 3 \mathrm{x}^2 \frac{\mathrm{dx}}{\mathrm{dt}} \\ & \Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}} \propto \frac{\mathrm{x}^{3 / 2}}{\mathrm{x}^2} \propto \mathrm{x}^{-1 / 2} \end{aligned}

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