Question

The work done by electric field during the displacement of a negatively charged particle towards a fixed positively charged particle is $9J$. As a result the distance between the charges has been decreased by half. What work is done by the electric field over the first half of this distance?

• A. 3 J
• B. 6 J
• C. 1.5 J
• D. 9 J

Answer 1

Answer: (A)

Here, $U_1=\frac{Q\left(-q\right)}{4\pi {\epsilon }_{0}r};U_2=\frac{Q\left(-q\right)}{4\pi {\epsilon }_0(r/2)}$
$U_1-U_2=\frac{Q\left(-q\right)}{4\pi {\epsilon }_0}\left[\frac{1}{r}-\frac{2}{r}\right]$
= $\frac{Qq}{4\pi {\epsilon }_{0}r}=9$ ..(i)
When negative charge travels first half of distance, $i.e.,r/4$, potential energy of the system
$U_3=\frac{Q\left(-q\right)}{4\pi {\epsilon }_0(3r/4)}=\frac{-Qq}{4\pi {\epsilon }_{0}r}\times \frac{4}{3}$
$\therefore$ Work done = $U_1-U_3$
$=\frac{Q\left(-q\right)}{4\pi {\epsilon }_{0}r}+\frac{Qr}{4\pi {\epsilon }_{0}r}\times \frac{4}{3}$
$=\frac{Qq}{4\pi {\epsilon }_{0}r}\times \frac{1}{3}=\frac{9}{3}=3J$ (Using (i))