91Ó°ÊÓ

The charge at point$\mathrm{P}$is$\mathrm{Q}$.

Mass of the second particle is$\mathrm{m}$.

Charge of the second particle is $−\mathrm{q}$.

After reading the question, the second particle has both potential and kinetic energy and both of them change on changing the radius of the orbit. The difference in the total energy of the second particle is the work done by the external agent to change the radius.

Formulae:

The centripetal force is define by,

${\mathrm{F}}_{\mathrm{c}}=\frac{{\mathrm{mv}}^2}{\mathrm{r}}$

The kinetic energy is define by,

$\mathrm{K}=\frac{1}{2}{\mathrm{mv}}^2$

Where, ${\mathrm{F}}_{\mathrm{c}}$is centripetal force, $\mathrm{K}$ is kinetic energy, $\mathrm{m}$ is the mass, $\mathrm{v}$ is the velocity, and $\mathrm{r}$ is the distance.

Radius of the circle centered at$\mathrm{P}$is${\mathrm{r}}_1$.

The charge $\mathrm{Q}$at point $\mathrm{P}$provides the necessary centripetal force required for $−\mathrm{q}$to move in a circular orbit. This centripetal force is equal to the electrostatic force between the two charges.

The magnitude of the force between the two charges is,

$\mathrm{F}=\mathrm{k}\frac{\mathrm{Qq}}{{\mathrm{r}}_1{}^2}$

The centripetal force required for $-\mathrm{q}$is,

${\mathrm{F}}_{\mathrm{c}}=\frac{{\mathrm{mv}}^2}{{\mathrm{r}}_1}$

According to the problem,

$\mathrm{F}={\mathrm{F}}_{\mathrm{c}}$

$\left(\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\right)\frac{\mathrm{Qq}}{{\mathrm{r}}_1^2}=\frac{{\mathrm{mv}}^2}{{\mathrm{r}}_1}$

${\mathrm{mv}}^2=\frac{\mathrm{Qq}}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}_1}$ …. (1)

The initial kinetic energy of charge $−\mathrm{q}$ is,

$\begin{array}{c}{\mathrm{K}}_1=\frac{1}{2}{\mathrm{mv}}^2\\ =\frac{1}{2}\left(\frac{\mathrm{Qq}}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}_1}\right)\\ =\frac{\mathrm{Qq}}{8{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}_1}\end{array}$

The initial potential energy of the charge$-q$is,

${\mathrm{U}}_1=−\frac{\mathrm{Qq}}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}_1}$

The total energy of the system is,

$\begin{array}{c}{\mathrm{E}}_1={\mathrm{K}}_1+{\mathrm{U}}_1\\ =\frac{\mathrm{Qq}}{8{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}_1}−\frac{\mathrm{Qq}}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}_1}\\ =−\frac{\mathrm{Qq}}{8{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}_1}\end{array}$

If the radius of the orbit is changed to${\mathrm{r}}_2$, the total energy of the system according to equation (4), write as,

${\mathrm{E}}_2=−\frac{\mathrm{Qq}}{8{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}_2}$

Work done $\left(\mathrm{W}\right)$ by the external agent is the difference in the total energy $({\mathrm{E}}_2−{\mathrm{E}}_1)$, that is,

$\begin{array}{c}\mathrm{W}={\mathrm{E}}_2−{\mathrm{E}}_1\\ =−\frac{\mathrm{Qq}}{8{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}_2}−\left(−\frac{\mathrm{Qq}}{8{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}_1}\right)\\ =\frac{\mathrm{Qq}}{8{\mathrm{Ï€Î\mu }}_0}\left(\frac{1}{{\mathrm{r}}_1}−\frac{1}{{\mathrm{r}}_2}\right)\end{array}$