Read the passage and answer the question.
Two spherical conductors A and B having equal radii and carrying equal charges repel each other with a force F when kept apart at some distance. A third spherical conductor P having the same radius as that of A, but uncharged, is brought in contact with A and then removed.
The new force of repulsion between A and B is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( F/2 \)
When two identical conducting spheres touch, their total charge is shared equally. By Coulomb’s law, at a fixed distance the force is proportional to the product of the two charges.
Let each of A and B carry charge \( q \) at first, so \( F=kq^{2}/r^{2} \). When the uncharged sphere P touches A, each gets
\( \displaystyle \frac{q+0}{2}=\frac{q}{2} \)
So A is left with \( q/2 \), while B still has \( q \). The new force is
\( \begin{aligned} F' &= \frac{k\,(q/2)\,q}{r^{2}} \\ &= \frac{1}{2}\cdot\frac{kq^{2}}{r^{2}} \\ &= \frac{F}{2} \end{aligned} \)