Coulomb's law states $F = \frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r^2} = \frac{kq_1q_2}{r^2}$, where $k = 9 \times 10^9 \, \text{N m}^2 \text{C}^{-2}$. The force is attractive for unlike charges and repulsive for like charges.
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JEE Main 2024 — Physics Electromagnetism
easy
mcq
2024
Official previous-year question
Verified 30 May 2026.
Question
The electrostatic force between two point charges $q_1$ and $q_2$ separated by distance $r$ is given by Coulomb's law as:
Options
- A
$F = \frac{kq_1q_2}{r^2}$
- B
$F = \frac{kq_1q_2}{r}$
- C
$F = kq_1q_2r^2$
- D
$F = kq_1q_2r$
Solution
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