A total charge Q is broken in two parts Q_1 a | vedclass

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Question

$\mathrm{Q}_{1}+\mathrm{Q}_{2}=\mathrm{Q}$          ..........$(i)$  and

${\rm{F}} =  - k\frac{{{{\rm{Q}}_1}{{\r

Vedclass · Accepted Answer

${Q_1} = \frac{Q}{2},{Q_2} = \frac{Q}{2}$

Vedclass · Answer

$\mathrm{Q}_{1}+\mathrm{Q}_{2}=\mathrm{Q}$          ..........$(i)$  and

${\rm{F}} =  - k\frac{{{{\rm{Q}}_1}{{\rm{Q}}_2}}}{{{{\rm{r}}^2}}}$             .........$(ii)$

For $\mathrm{F}$ to be maximum $\frac{\mathrm{dE}}{\mathrm{dQ}_{1}}=0 \Rightarrow \mathrm{Q}_{1}=\mathrm{Q}_{2}=\frac{\mathrm{Q}}{2}$

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

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