Coulomb used a torsion balance to measure the force \(F\) between two small charged spheres. He kept the charges fixed and changed the distance \(r\) between them. If the product \(Fr^{2}\) is plotted against \(r\) for these readings, the graph is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → a straight line parallel to the \(r\)-axis
By Coulomb’s law, \( F=kq_1q_2/r^{2} \), so \( Fr^{2}=kq_1q_2 \). With the charges fixed, this product is a constant. It does not change with \( r \).
Option (1) a straight line passing through the origin — Incorrect. This would need \( Fr^{2}\propto r \), that is, \( F\propto1/r \). The force falls faster than this, as the square of the distance.
Option (2) a rectangular hyperbola — Incorrect. This would need \( Fr^{2}\propto1/r \), that is, \( F\propto1/r^{3} \). Such a force does not follow the inverse square law.
Option (3) a straight line parallel to the \(r\)-axis — Correct. \( Fr^{2} \) has the same value for every reading. So the points lie on a line parallel to the \( r \)-axis, at a height \( kq_1q_2 \).
Option (4) a parabola passing through the origin — Incorrect. This would need \( Fr^{2}\propto r^{2} \), that is, a force that does not depend on distance. Coulomb found that the force changes with distance.
NCERT (2026-27):
Coulomb’s Law – “… Coulomb measured the force between two point charges and found that it varied inversely as the square of the distance between the charges and was directly proportional to the product of the magnitude of the two charges and acted along the line joining the two charges.”