What is the force between two small charged spheres having charges of \( 4\times10^{-7}\,\mathrm{C} \) and \( 6\times10^{-7}\,\mathrm{C} \) placed 60 cm apart in air?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 6\times10^{-3}\,\mathrm{N} \)
The force between two small charged spheres (point charges) is given by Coulomb’s law, \( F = \frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r^{2}} \), with \( \frac{1}{4\pi\varepsilon_0} = 9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}} \) for air. The distance is \( r = 60\,\mathrm{cm} = 0.6\,\mathrm{m} \).
\( \displaystyle F = \frac{9\times10^{9}\times4\times10^{-7}\times6\times10^{-7}}{(0.6)^{2}} = \frac{9\times10^{9}\times24\times10^{-14}}{0.36}\,\mathrm{N} \)
\( F = 6\times10^{-3}\,\mathrm{N} \)
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 – …
