Question

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?

• A. $\frac{k}{r_0},v=\sqrt{\frac{2k}{m{r}_0}}$
• B. $2\frac{k}{r_0},v=\sqrt{\frac{2k}{m{r}_0}}$
• C. $2\frac{k}{r_0},v=\sqrt{\frac{k}{2m{r}_0}}$
• D. $zero,v=\frac{k}{r_0},v=\sqrt{\frac{2k}{m{r}_0}}$

Answer 1

The correct option is A $\frac{k}{r_0},v=\sqrt{\frac{2k}{m{r}_0}}$

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; $F_{AB}$ acting on A and $F_{BA}$ acting on B. The force $F_{AB}$ does no work because it acts on A which does not move. The work done by $F_{BA}$ as the particle B istaken away is,
$W=\int \overrightarrow{F}.d\overrightarrow{r}={\int }_{r_0}^{\infty }\frac{k}{r^2}dr=\frac{k}{r_0}$ .....(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,
$K_f+U_f=K_i+U_{i}or,K_f=K_i-(U_f-U_i)or,\frac{1}{2}m{v}^2=\frac{k}{r_0}or,v=\sqrt{\frac{2k}{m{r}_0}}$