Two equal and similar charges 3 cm apart in air repel each other with a force of 1.6 N. What will be the magnitude of each charge?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 4\times10^{-7}\,\mathrm{C} \)
The force between two point charges in air is given by Coulomb's law. With equal charges \( q \) at distance \( r \):
\( \displaystyle F = \frac{1}{4\pi\varepsilon_0}\,\frac{q^{2}}{r^{2}} \;\Rightarrow\; q^{2} = \frac{F\,r^{2}}{9\times10^{9}} \)
\( r = 3\,\mathrm{cm} = 3\times10^{-2}\,\mathrm{m} \)
\( F = 1.6\,\mathrm{N} \)
\( \displaystyle q^{2} = \frac{1.6 \times 9\times10^{-4}}{9\times10^{9}} = 1.6\times10^{-13} = 16\times10^{-14} \)
\( \displaystyle q = 4\times10^{-7}\,\mathrm{C} \)