Two point charges of 1 C each are separated from each other by a distance of 1 m in vacuum. The force of interaction between them is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 9\times10^{9}\,\mathrm{N} \)
The force between two point charges is given by Coulomb's law.
\( \displaystyle F = \frac{1}{4\pi\varepsilon_0}\,\frac{q_1 q_2}{r^{2}} \)
\( \displaystyle F = 9\times10^{9} \times \frac{1 \times 1}{1^{2}} = 9\times10^{9}\,\mathrm{N} \)
This huge force shows that the coulomb is a very large unit of charge; in practice charges are of the order of microcoulombs.
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):
Coulomb’s Law – “… Putting this value of k in Eq. (1.1), we see that for q1 = q2 = 1 C, r = 1 m
\( \displaystyle F = 9 \times 10^{9}\ \mathrm{N} \)
That is, 1 C is the charge that when placed at a distance of 1 m from another charge of the same magnitude in vacuum experiences an electrical force of repulsion of magnitude 9 × 109 N.”