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. $Q_{2} = \frac{Q}{R},\mspace{6mu} Q_{1} = Q - \frac{Q}{R}$
• B. $Q_{2} = \frac{Q}{4},\mspace{6mu} Q_{1} = Q - \frac{2Q}{3}$
• C. $Q_{2} = \frac{Q}{4},\mspace{6mu} Q_{1} = \frac{3Q}{4}$
• D. $Q_{1} = \frac{Q}{2},\mspace{6mu} Q_{2} = \frac{Q}{2}$

Answer 1

Answer: (D)

$Q_{1} = \frac{Q}{2},\mspace{6mu} Q_{2} = \frac{Q}{2}$

Answer 2

$Q_{1} + Q_{2} = Q$ ..... (i) and $F = k\frac{Q_{1}Q_{2}}{r^{2}}$ .....(ii)

From (i) and (ii) $F = \frac{kQ_{1}(Q - Q_{1})}{r^{2}}$

For F to be maximum

$\frac{dF}{dQ_{1}} = 0$⇒ $Q_{1} = Q_{2} = \frac{Q}{2}$