The force acting between two point charges q1 and q2 kept at some distance in air, would be-
(A) repulsive if q1 q2 > 0
(B) attractive if q1 q2 > 0
(C) repulsive if q1 q2 < 0
(D) attractive if q1 q2 < 0
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) and (D) only
By Coulomb’s law, the force is along the line joining the charges and its sign is set by the product \( q_1 q_2 \). A positive product means like charges, which repel; a negative product means unlike charges, which attract.
\( \displaystyle F = \frac{1}{4\pi\varepsilon_0}\,\frac{q_1 q_2}{r^{2}} \)
(A) repulsive if q1 q2 > 0 — Correct. Both charges have the same sign, so they repel.
(B) attractive if q1 q2 > 0 — Incorrect. Like charges never attract.
(C) repulsive if q1 q2 < 0 — Incorrect. Unlike charges never repel.
(D) attractive if q1 q2 < 0 — Correct. The charges have opposite signs, so they attract.
NCERT (2026-27):
Coulomb’s Law – “… If \( q_1 \) and \( q_2 \) are of the same sign (either both positive or both negative), \( \mathbf{F}_{21} \) is along \( \hat{\mathbf{r}}_{21} \), which denotes repulsion, as it should be for like charges. If \( q_1 \) and \( q_2 \) are of opposite signs, \( \mathbf{F}_{21} \) is along \( -\hat{\mathbf{r}}_{21} (= \hat{\mathbf{r}}_{12}) \), which denotes attraction, as expected for unlike charges.”