Q. The work of electric field done during the displacement of a negatively charged particle towards a fixed positively charged particle is 9 J. 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?
Explanation
(a) Here, $U_{1} = \frac{Q( - q)}{4\pi\varepsilon_{0}r};U_{2} = \frac{Q( - q)}{4\pi\varepsilon_{0}(r/2)}$
$U_{1} - U_{2} = \frac{Q( - q)}{4\pi\varepsilon_{0}}\left\lbrack \frac{1}{r} - \frac{2}{r} \right\rbrack$
$= \frac{Qq}{4\pi\varepsilon_{0}}$=9 …(i)
When negative charge travels first half of distance, $ie,r/4,$ potential energy of the system
$U_{3} = \frac{Q( - q)}{4\pi\varepsilon_{0}\left( \frac{3r}{4} \right)} = - \frac{Qr}{4\pi\varepsilon_{0}r} \times \frac{4}{3}$
$\therefore$ work done $= U_{1} - U_{3}$
$= \frac{Q( - q)}{4\pi\varepsilon_{0}r} + \frac{Qr}{4\pi\varepsilon_{0}r} \times \frac{4}{3}$
$= \frac{Qq}{4\pi\varepsilon_{0}r} \times \frac{1}{3} = \frac{9}{3} = 3J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.