Q. Two point charges $A$ and $B$, having charges $+ Q$ and $- Q$ respectively, are placed at certain distance apart and force acting between them is $F$. If $25\%$ charge of $A$ is transferred to $B$. then force between the charges become
Solution:
$F =\frac{kQ^{2}}{r^{2}} $
If 25% of charge of A transferred to B then
$ q_{A} =Q - \frac{Q}{4} = \frac{3Q}{4} $ and $ q_{B} = -Q + \frac{Q}{4} = \frac{-3Q}{4} $
$ F_{1} = \frac{kq_{A}q_{B}}{r^{2}} $
$ F_{1} = \frac{k\left(\frac{3Q}{4}\right)^{2}}{r^{2}} $
$ F_{1} = \frac{9}{16} \frac{kQ}{r^{2}} $
$ F_{1} = \frac{9F}{16} $
