Two point charges \( +4\,\mu\mathrm{C} \) and \( -2\,\mu\mathrm{C} \) are placed at a certain distance in air. They experience an attractive force F. If they are brought in contact with each other and then separated to the same distance, the new repulsive force between them will be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → F/8
By Coulomb's law, at a fixed distance the force is proportional to the product of the magnitudes of the charges, \( F \propto q_1 q_2 \).
When two identical conductors touch, the total charge is conserved and shared equally.
\( \displaystyle q' = \frac{(+4) + (-2)}{2} = +1\,\mu\mathrm{C}\ \text{each} \)
Both charges are now positive, so the force is repulsive.
\( \displaystyle \frac{F'}{F} = \frac{1 \times 1}{4 \times 2} = \frac{1}{8} \;\Rightarrow\; F' = \frac{F}{8} \)