Q. The work done in carrying a charge of $5\mu C$ from a point $A$ to a point $B$ in an electric field is $10mJ.$ The potential difference $(V_{B} - V_{A})$ is then
A) $+ \ 2kV$
B) $- 2\ kV$
C) $+ 200\ V$
D) $- 200\ V$
Explanation
(a) Work done $W = Q\left( V_{B} - V_{A} \right) \Rightarrow \left( V_{B} - V_{A} \right) = \frac{W}{Q}$
$= \frac{10 \times 10^{- 3}}{5 \times 10^{- 6}}J/C = 2\ kV$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.