Two charged particles are placed at a distance \( 0.01 \) m apart. The minimum possible magnitude of the electric force acting on each charge will be
(Take \( k = 9\times10^{9} \) SI unit)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 2.3\times10^{-24} \) N
Charge is quantised: the smallest charge a free particle can carry is the electronic charge \( e = 1.6\times10^{-19}\,\mathrm{C} \). So the force is least when each particle carries \( q = e \), and Coulomb’s law gives it.
\( q_{1} = q_{2} = e = 1.6\times10^{-19}\,\mathrm{C} \)
\( r = 0.01\,\mathrm{m} \)
\( \displaystyle F_{\min} = \frac{ke^{2}}{r^{2}} = \frac{(9\times10^{9})(1.6\times10^{-19})^{2}}{(0.01)^{2}} \)
\( \displaystyle F_{\min} = \frac{9\times10^{9}\times2.56\times10^{-38}}{10^{-4}}\ \mathrm{N} \)
\( F_{\min} = 2.304\times10^{-24}\,\mathrm{N} \approx 2.3\times10^{-24}\,\mathrm{N} \)