A total charge Q is broken in two parts Q₁ and Q₂ and they are placed at a distance R from each

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1. Electric Charges and Fields

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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

$\mathrm{Q}_{1}+\mathrm{Q}_{2}=\mathrm{Q}$ ……….$(i)$ and

${\rm{F}} = – k\frac{{{{\rm{Q}}_1}{{\rm{Q}}_2}}}{{{{\rm{r}}^2}}}$ ………$(ii)$

For $\mathrm{F}$ to be maximum $\frac{\mathrm{dE}}{\mathrm{dQ}_{1}}=0 \Rightarrow \mathrm{Q}_{1}=\mathrm{Q}_{2}=\frac{\mathrm{Q}}{2}$

Standard 12

Physics

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