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1. Electric Charges and Fields
medium
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
${Q_1} = \frac{Q}{2},\;{Q_2} = \frac{Q}{2}$
Solution
(d) ${Q_1} + {Q_2} = Q$ ….. $(i)$ and $F = k\frac{{{Q_1}{Q_2}}}{{{r^2}}}$ …..$(ii)$
From $(i)$ and $(ii)$ $F = \frac{{k{Q_1}(Q – {Q_1})}}{{{r^2}}}$
For $F$ to be maximum $\frac{{dF}}{{d{Q_1}}} = 0$ $==>$ ${Q_1} = {Q_2} = \frac{Q}{2}$
Standard 12
Physics
