The force between two small charged spheres having charges of \( 2 \times 10^{-7}\,\mathrm{C} \) and \( 3 \times 10^{-7}\,\mathrm{C} \), respectively placed 30 cm apart in air would be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 6 \times 10^{-3}\,\mathrm{N} \)
Small charged spheres act as point charges, so Coulomb's law gives the force.
\( \displaystyle F = \frac{1}{4\pi\varepsilon_0}\,\frac{q_1 q_2}{r^{2}} \)
\( k = 9\times10^{9}\,\mathrm{N\,m^2\,C^{-2}} \)
\( r = 30\,\mathrm{cm} = 0.3\,\mathrm{m} \)
\( \displaystyle F = \frac{9\times10^{9} \times 2\times10^{-7} \times 3\times10^{-7}}{(0.3)^{2}} = \frac{5.4\times10^{-4}}{0.09} = 6\times10^{-3}\,\mathrm{N} \)
Both charges are positive, so the force is repulsive.
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
