Question

A total charge $Q$ is broken in two parts $Q_1$ and $Q_2$ , and they are placed at a distance $R$ from each other. The maximum force of repulsion between them will occur when

• A. $Q_2=\frac{Q}{R},Q_1=Q-\frac{Q}{R}$
• B. $Q_2=\frac{Q}{3},Q_1=\frac{2Q}{3}$
• C. $Q_2=\frac{Q}{4},Q=\frac{3Q}{4}$
• D. $Q_1=\frac{Q}{2},Q_2=\frac{Q}{2}$

Answer 1

The correct option is D $Q_1=\frac{Q}{2},Q_2=\frac{Q}{2}$

$Q_1+Q_2=Q...\left(i\right)$
and $F=k\frac{Q_1{Q}_2}{r^2}....\left(ii\right)$
From (i) and (ii) $F=\frac{k{Q}_1(Q-Q_1)}{r^2}$
For $F$ to be maximum, $\frac{dF}{d{Q}_1}=0$
$\Rightarrow Q-2Q_1=0$
$\Rightarrow Q_1=Q_2=\frac{Q}{2}$