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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 becomes:
A
$F$
B
$\frac{9F}{16}$
C
$\frac{16F}{9}$
D
$\frac{4F}{3}$
Detailed Solution
Initially: $F=\frac{kQ^2}{r^2}$
If 25% of the charge of A is transferred to B:
$q_A=Q-\frac{Q}{4}=\frac{3Q}{4}$, $q_B=-Q+\frac{Q}{4}=-\frac{3Q}{4}$
$F_1=\frac{k\left(\frac{3Q}{4}\right)^2}{r^2}=\frac{9}{16}\frac{kQ^2}{r^2}$
$F_1=\frac{9F}{16}$