Question

A total change 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

Answer 1

$Q_2=Q-Q_1$

$F=\frac{k}{R^2}{Q}_1{Q}_2=\frac{k}{R^2}{Q}_1(Q-Q_1)$

$⟹F=\frac{k}{R^2}(Q_{1}Q-Q_1^2)$

Differentiating w.r.t $Q_1$ and equating to $0$ for maximum force-

$Q-2Q_1=0$

$⟹Q_1=\frac{Q}{2}$

Hence, maximum repulsion will be there when both charges are equal and have $Q_1=Q_2=\frac{Q}{2}$