Question

A particle of positive charge $Q$ is fixed at point $P$. A second particle of mass m and negative charge $-q$ moves at constant speed in a circle of radius $r_1$ , centered at $P$. The work $W$ that must be done by an external agent on the second particle to increase the radius of the motion to $r_2$ , is given by

• A. $W=\frac{Qq}{8\pi {\epsilon }_0}\left(\frac{1}{r_2}-\frac{1}{r_1}\right)$
• B. $W=\frac{Qq}{8\pi {\epsilon }_0}\left(\frac{1}{r_1}-\frac{1}{r_2}\right)$
• C. $W=\frac{Qq}{4\pi {\epsilon }_0}\left(\frac{1}{r_2}-\frac{1}{r_1}\right)$
• D. $W=\frac{Qq}{4\pi {\epsilon }_0}\left(\frac{1}{r_1}-\frac{1}{r_2}\right)$

Answer 1

Answer: (D)

Force at $P$ due to $q$ when the radius of circle is
$F_1=\frac{Qq}{4\pi {\epsilon }_0{r}_1^2}$.... (i)
Force at $P$ due to $q$ when the radius of circle is
$F_2=\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{r_1^2}\dots$.. (ii)
Increase in radius $\left(r_2-r_1\right)$ (distance)
Resultant force $=\sqrt{F_1\times F_2}=\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{r_1{r}_1^2}$
$\therefore$ Work done $W=F\times dr$
$=\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{r_1{r}_2}\times \left(r_2-r_1\right)$
$W=\frac{Qq}{4\pi {\epsilon }_0}\left(\frac{1}{r_1}-\frac{1}{r_2}\right)$