Q. Two positive point charges of 12 and 5 microcoulombs, are placed 10cm apart in air. The work needed to bring them 4cm closer is
A) $2.4\ J$
B) $3.6\ J$
C) $4.8\ J$
D) $6.0\ J$
Explanation
(b) Potential energy of charges $Q_{1}$ and $Q_{2}$ at 10cm apart
$= \frac{1}{4\pi\varepsilon_{0}}\frac{12 \times 10^{- 6} \times 5 \times 10^{- 6}}{0.1}$
$= \frac{9 \times 10^{9} \times 60 \times 10^{- 12}}{0.1} = 54 \times 10^{- 1} = 5.4\ J$
At 6 cm apart,
$P.E. = \frac{9 \times 10^{9} \times 60 \times 10^{- 12}}{0.06} = 9\ J$
$\therefore Work\ done = (9 - 5.4)J = 3.6\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.