Question

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

• A. $\frac{16F}{9}$
• B. $\frac{4F}{3}$
• C. F
• D. $\frac{9F}{16}$

Answer 1

Answer: (D)

$F=\frac{k{Q}^2}{r^2}$
If 25% of charge of A transferred to B then
$q_A=Q-\frac{Q}{4}=\frac{3Q}{4}$ and $q_B=-Q+\frac{Q}{4}=\frac{-3Q}{4}$
$F_1=\frac{k{q}_A{q}_B}{r^2}$
$F_1=\frac{k{\left(\frac{3Q}{4}\right)}^2}{r^2}$
$F_1=\frac{9}{16}\frac{kQ}{r^2}$
$F_1=\frac{9F}{16}$