Two charged particles A and B repel each other by a force $\frac{k}{r^{2}}$ , where k is a constant and r is the separation between them. The particle A is clamped to a fixed point in the lab and the particle B which has a mass m, is released from rest with an initial separation $r_{0}$ from A. Find the change in the potential energy of the two particle system as the separation increases to a large value. What will be the speed of the particle B in this situation?

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Solution

The correct option is A
The situation is shown in figure. Take A + B as the system. The only external force acting on the system is that needed to hold A fixed. (You can imagine the experiment being conducted in a gravity free region or the particles may be kept and allowed to move on a smooth horizontal surface, so that the normal force balances the force of gravity). This force does no work on the system because it acts on the charge A which does not move. Thus, the external forces do no work and internal forces are conservative. The total mechanical energy must, therefore, remain constant. There are two internal forces; $F_{A B}$ acting on A and $F_{B A}$ acting on B. The force $F_{A B}$ does no work because it acts on A which does not move. The work done by $F_{B A}$ as the particle B istaken away is,
$W = \int \to F . d \to r = \int_{r_{0}}^{\infty} \frac{k}{r^{2}} d r = \frac{k}{r_{0}}$ .....(i)

The change in the potential energy of the system is
$U_{f} - U_{i} = - W = - \frac{k}{r_{0}}$
As the total mechanical energy is conserved,
$K_{f} + U_{f} = K_{i} + U_{i} o r, K_{f} = K_{i} - (U_{f} - U_{i}) o r, \frac{1}{2} m v^{2} = \frac{k}{r_{0}} o r, v = \sqrt{\frac{2 k}{m r_{0}}}$

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