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

Answer 1

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

From conservation of charge $Q_1+Q_2=Q$ ... (i)

Electrostatic force
$F=k\frac{Q_1{Q}_2}{r^2}$ and ... (ii)
From (i) and (ii)

$F=\frac{k{Q}_1(Q-Q_1)}{r^2}$
For $F,$ to be maximum $\frac{dF}{d{Q}_1}=0$
$Q-2Q_1=0$
$\Rightarrow Q_1=\frac{Q}{2}$
$\Rightarrow Q_1=Q_2=\frac{Q}{2}$