Two point charges \( 2\,\mu\mathrm{C} \) and \( 3\,\mu\mathrm{C} \) are placed at a distance of 5 cm from each other. The magnitude of the electrostatic force between them is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 21.6 N
The force between two point charges is given by Coulomb's law (charges in air); it falls as the inverse square of the separation.
\( \displaystyle F = \frac{1}{4\pi\varepsilon_0}\,\frac{q_1 q_2}{r^{2}} \)
\( q_1 = 2\times10^{-6}\,\mathrm{C} \)
\( q_2 = 3\times10^{-6}\,\mathrm{C} \)
\( r = 5\,\mathrm{cm} = 0.05\,\mathrm{m} \)
\( 1/4\pi\varepsilon_0 = 9\times10^{9}\,\mathrm{N\,m^2\,C^{-2}} \)
\( \displaystyle F = \frac{9\times10^{9} \times 2\times10^{-6} \times 3\times10^{-6}}{(0.05)^{2}} \)
\( \displaystyle F = \frac{5.4\times10^{-2}}{2.5\times10^{-3}} = 21.6\,\mathrm{N} \)
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
