Q. A charge of 1 $\mu$C is divided into two parts such that their charges are in the ratio of 2:3. These two charges are kept at a distance 1 m apart in vacuum. Then, the electric force between them (in N) is
A) 0.216
B) 0.00216
C) 0.0216
D) 2.16
Explanation
(b) Ratio of charges =2:3
$\therefore$ $q_{1} = \frac{2}{5} \times 1\mu\complement$ and $q_{2} = \frac{3}{5} \times 1\mu\complement$
Electrostatic force between the two charges
$F = \frac{1}{4\pi\varepsilon_{0}}\ \frac{q_{1}q_{2}}{r^{2}}$
$= \frac{9 \times 10^{9} \times 2 \times 10^{- 6} \times 3 \times 10^{- 6}}{5 \times 5 \times (1)^{2}}$
= $2.16 \times 10^{- 3}$N
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.