Two identical point charges placed at a distance 'r' apart exert a force F on each other. If each charge is halved and the distance between them is doubled, then the new force acting on each charge would be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → F/16
By Coulomb's law the force is proportional to the product of the charges and inversely proportional to the square of the distance.
\( \displaystyle F = \frac{1}{4\pi\varepsilon_0}\,\frac{q \cdot q}{r^{2}} \)
With each charge \( q/2 \) and the distance \( 2r \):
\( \displaystyle F' = \frac{1}{4\pi\varepsilon_0}\,\frac{(q/2)(q/2)}{(2r)^{2}} = \frac{1}{4\pi\varepsilon_0}\,\frac{q^{2}/4}{4r^{2}} = \frac{F}{16} \)
Halving both charges reduces the force 4 times, and doubling the distance reduces it another 4 times.