Read the passage and answer the question.
Two spherical conductors A and B having equal radii and carrying equal charges in them 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 (or B) but uncharged, is brought in contact with A and then removed away from both.
What is the new force of repulsion between A and P ?
Answer & explanation
Correct answer: option 1
The charge on each of the A and P is half the original charge. So the force gets reduced by a factor of 4.
$\text{Initially: charges on A and B are } q,\ q$
$F = \frac{kq^2}{r^2}$
$\text{P (uncharged) touches A}$
$\text{Since equal radii: charge divides equally}$
$\text{Charge on A} = \frac{q}{2},\ \text{on P} = \frac{q}{2}$
$\text{P removed, B remains with } q$
$\text{New force between A and B}$
$F' = \frac{k(\frac{q}{2})(q)}{r^2} = \frac{1}{2}\frac{kq^2}{r^2}$
$F' = \frac{F}{2}$
The new force is $\frac{F}{2}$.