The force between two small charged spheres having charges of \( 0.2\,\mu\mathrm{C} \) and \( 0.6\,\mu\mathrm{C} \) placed 0.6 m apart in the air will be
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0.003 N
Small charged spheres act as point charges, so Coulomb's law gives the force; in air the permittivity is taken as that of vacuum.
\( q_1 = 0.2\,\mu\mathrm{C} = 0.2\times10^{-6}\,\mathrm{C} \)
\( q_2 = 0.6\,\mu\mathrm{C} = 0.6\times10^{-6}\,\mathrm{C} \)
\( r = 0.6\,\mathrm{m} \)
\( \displaystyle F = \frac{1}{4\pi\varepsilon_0}\,\frac{q_1 q_2}{r^{2}} = \frac{9\times10^{9} \times 0.2\times10^{-6} \times 0.6\times10^{-6}}{(0.6)^{2}} \)
\( \displaystyle F = \frac{1.08\times10^{-3}}{0.36} = 3\times10^{-3}\,\mathrm{N} = 0.003\,\mathrm{N} \)
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 – …
