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 \(\mathrm{F}\). If \(25 \%\) charge of \(A\) is transferred to \(B\), then force between the charges becomes:
(1) \(\mathrm{F}\)
(2) \(9 \mathrm{~F} / 16\)
(3) \(16 \mathrm{~F} / 9\)
(4) \(4 \mathrm{~F} / 3\)

Answer 1

The correct option is (2) 9F/16.
Explanation:
$$
\begin{aligned}
&+Q \stackrel{A}{\longrightarrow} \longrightarrow \stackrel{B}{r}-Q \\
&F=\frac{k Q^{2}}{r^{2}}
\end{aligned}
$$
If \(25 \%\) of charge of \(A\) transferred to \(B\) then
$$
\begin{aligned}
&q_{A}=Q-\frac{Q}{4}=\frac{3 Q}{4} \\
&\text { and } q_{B}=-Q+\frac{Q}{4}=\frac{-3 Q}{4} \\
&q_{A} \bullet r \\
&F_{1}=\frac{k q_{A} q_{B}}{r^{2}}
\end{aligned}
$$
$$
\begin{aligned}
&F_{1}=\frac{k\left(\frac{3 Q}{4}\right)^{2}}{r^{2}} \\
&F_{1}=\frac{9}{16} \frac{k Q}{r^{2}} \\
&F_{1}=\frac{9 F}{16}
\end{aligned}
$$