Two identical charged spheres suspended from a common point by two massless strings of length I are…

Two identical charged spheres suspended from a common point by two massless strings of length I are initially a distance $d(d< < 1)$ 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 $v$. Then as a function of distance $x$ between them,
  1. $v \propto x^{-1}$
  2. $v \propto x^{1 / 2}$
  3. $v \propto x$
  4. $v \propto x^{-1 / 2}$

Solution

At any instant of separation between charges is $x$. $ \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} $

Asked in: JEE Main 2011

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