Q. Two positive point charges of 12 $\mu$ C and 8 $\mu$C are placed 10 cm, apart in air. The work done to bring them 4 cm closer is
A) Zero
B) 3.5 J
C) 4.8 J
D) 5.8 J
Explanation
(d) Work done $= U_{2} - U_{1} = \frac{q_{1}q_{2}}{4\pi\varepsilon_{0}}\left\lbrack \frac{1}{r_{2}} - \frac{1}{r_{1}} \right\rbrack$
$= 12 \times 10^{- 6} \times 8 \times 10^{- 6} \times 9 \times 10^{9}\left\lbrack \begin{array}{r}
\frac{10^{2}}{6} - \frac{10^{2}}{10} \\
\
\end{array} \right\rbrack$
$W = 96 \times 9 \times 10^{- 3} \times 10^{2} \times \frac{4}{60} = 5.8J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.