Two point charges repel each other with a force of 100 N. One of the charges is increased by 10% and the other is decreased by 10%. The new force of repulsion at the same distance will be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 99 N
By Coulomb’s law, \( F=kq_1q_2/r^{2} \), so at a fixed distance the force is proportional to the product of the charges. The new charges are \( 1.1\,q_1 \) and \( 0.9\,q_2 \).
The new force of repulsion is:
\( \begin{aligned} F' &= \frac{k\,(1.1\,q_1)(0.9\,q_2)}{r^{2}} \\ &= 0.99\,F \\ &= 0.99\times100\,\mathrm{N} \\ &= 99\,\mathrm{N} \end{aligned} \)