The distance between two point charges \( q_1=5\,\mathrm{mC} \) and \( q_2=-10\,\mathrm{mC} \) in vacuum so that electrostatic force of attraction between them is 50 kN is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 3 m
Coulomb’s law gives the magnitude of the force between two point charges in vacuum, \( F = \displaystyle\frac{1}{4\pi\varepsilon_0}\frac{|q_1q_2|}{r^{2}} \), with \( \displaystyle\frac{1}{4\pi\varepsilon_0} = 9\times10^{9}\,\mathrm{N\,m^{2}\,C^{-2}} \).
Convert to SI units: \( q_1 = 5\times10^{-3}\,\mathrm{C} \), \( |q_2| = 10\times10^{-3}\,\mathrm{C} \), \( F = 50\,\mathrm{kN} = 5\times10^{4}\,\mathrm{N} \).
\( \displaystyle r^{2} = \frac{9\times10^{9}\times5\times10^{-3}\times10\times10^{-3}}{5\times10^{4}} = \frac{4.5\times10^{5}}{5\times10^{4}} = 9\,\mathrm{m^{2}} \)
\( r = 3\,\mathrm{m} \)