Question

Total charge $Q$ is broken in two parts $Q_{1} \& Q_{2}$ 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. $Q_{2}=\frac{Q}{4}, Q_{1}=Q-\frac{2 Q}{R}$
• C. $Q_{2}=\frac{Q}{4}, Q_{1}=Q-\frac{3 Q}{4}$
• D. $Q_{2}=\frac{Q}{2}, Q_{1}=Q-\frac{Q}{2}$

Answer 1

Answer: (D)

$Q_{1}+Q_{2}=Q$ (i)
$Q_{2}=Q-Q_{1}$ (ii)
$F =\frac{ KQ _{1}\left( Q - Q _{1}\right)}{ R ^{2}}$ (iii)
For $F$ to be maximum $\frac{d F}{d Q_{1}}=0$
$\frac{1}{ R ^{2}}\left[ KQ -2 KQ _{1}\right]=0$
$\Rightarrow Q _{1}=\frac{ Q }{2}$
$Q_{2}=Q-\frac{Q}{2}=\frac{Q}{2}$