A point charge of \(+2\ \mu\mathrm{C}\) is kept at the origin. Another point charge of \(+5\ \mu\mathrm{C}\) is kept at the point (3 m, 4 m) in the x-y plane. The magnitude of the electrostatic force between them is (Take \(\frac{1}{4\pi\varepsilon_0}=9\times10^{9}\ \mathrm{N\,m^{2}\,C^{-2}}\))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(3.6\times10^{-3}\ \mathrm{N}\)
Coulomb’s law needs the straight-line distance between the two charges. Here it is the distance of the point (3 m, 4 m) from the origin.
\( \begin{aligned} r &= \sqrt{3^{2}+4^{2}} \\ &= 5\ \mathrm{m} \end{aligned} \)
\( \begin{aligned} F &= k\frac{q_1q_2}{r^{2}} \\ &= \frac{9\times10^{9}\times2\times10^{-6}\times5\times10^{-6}}{5^{2}} \\ &= 3.6\times10^{-3}\ \mathrm{N} \end{aligned} \)
NCERT (2026-27):
Coulomb’s Law – “… Thus, if two point charges q1, q2 are separated by a distance r in vacuum, the magnitude of the force (F) between them is given by”
\( \displaystyle F=k\frac{q_1q_2}{r^{2}} \)\( \qquad (1.1) \)