Q. A charge q is fixed. Another charge $\mathcal{Q}$ is brought near it and rotated in a circle of radius r around it. Work done during rotation is
A) Zero
B) $\frac{\mathcal{Q}q}{4\pi\varepsilon_{0}r}$
C) $\frac{\mathcal{Q}q}{2\pi\varepsilon_{0}r}$
D) None of these
Explanation
(a) Potential due to charge (q) at point (r) is given by
$V = \frac{1}{4\pi e_{0}}.\frac{q}{r}$
Since, charge $\mathcal{Q}$ is rotated in a circle of radius $r$, hence its potential remains same at all points on the path, hence $\mathrm{\Delta}V = 0.$
Also, work done $= q\mathrm{\Delta}V$
Where q is charge and $\mathrm{\Delta}V = 0.$
$\therefore$ Work done =0.
[image:2_image139.emf]
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.