A point charge q is placed at a distance r from another point charge Q. If the distance between the charges is doubled then :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → the force between the charges becomes one-fourth and the potential energy becomes half.
By Coulomb's law the force falls as the square of the distance, while the electrostatic potential energy falls only as the first power:
\( \displaystyle F = \frac{1}{4\pi\varepsilon_0}\frac{qQ}{r^{2}} \propto \frac{1}{r^{2}}, \qquad U = \frac{1}{4\pi\varepsilon_0}\frac{qQ}{r} \propto \frac{1}{r} \)
Doubling the distance:
\( F' = F/2^{2} = F/4 \)
\( U' = U/2 \)
So the force becomes one-fourth and the potential energy becomes half.