Q. 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) $Q_{1} = \frac{Q}{2},\ Q_{2} = \frac{Q}{2}$
Explanation
(d) $Q_{1} + Q_{2} = Q$ …(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 $\frac{dF}{dQ_{1}} = 0 \Rightarrow Q_{1} = Q_{2} = \frac{Q}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.