Q. Two point charges $100\ \mu C\ and\ 5\ \mu C$ are placed at points $A\ and\ B$ respectively with $AB = 40\ cm.$ The work done by external force in displacing the charge $5\ \mu C$ from $B\ to\ C,$ where $BC = 30cm$, angle $ABC = \frac{\pi}{2}$ and $\frac{1}{4\pi\varepsilon_{0}} = \ 9 \times 10^{9}Nm^{2}/C^{2}$
A) $9\ J$
B) $\frac{81}{20}J$
C) $\frac{9}{25}J$
D) $- \frac{9}{4}J$
Explanation
(d) Work done in displacing charge of $5\mu C$ from $B$ to $C$ is $W = 5 \times 10^{- 6}(V_{C} - V_{B})$ where
$V_{B} = 9 \times 10^{9} \times \frac{100 \times 10^{- 6}}{0.4} = \frac{9}{4} \times 10^{6}V$
and $V_{C} = 9 \times 10^{9} \times \frac{100 \times 10^{- 6}}{0.5} = \frac{9}{5} \times 10^{6}V$
so $W = 5 \times 10^{- 6} \times \left( \frac{9}{5} \times 10^{6} - \frac{9}{4} \times 10^{6} \right) = \ - \frac{9}{4}\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.