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(D) Given that the total charge $Q$ is divided into two parts $Q_1$ and $Q_2$,we have $Q_1 + Q_2 = Q$,which implies $Q_2 = Q - Q_1$.The electrostatic

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

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(D) Given that the total charge $Q$ is divided into two parts $Q_1$ and $Q_2$,we have $Q_1 + Q_2 = Q$,which implies $Q_2 = Q - Q_1$.The electrostatic force of repulsion between them is given by Coulomb's law: $F = k \frac{Q_1 Q_2}{R^2}$.Substituting $Q_2$ in terms of $Q_1$: $F = \frac{k}{R^2} Q_1 (Q - Q_1) = \frac{k}{R^2} (Q Q_1 - Q_1^2)$.To find the maximum force,we differentiate $F$ with respect to $Q_1$ and set it to zero: $\frac{dF}{dQ_1} = \frac{k}{R^2} (Q - 2Q_1) = 0$.Solving for $Q_1$,we get $Q - 2Q_1 = 0$,which means $Q_1 = \frac{Q}{2}$.Since $Q_2 = Q - Q_1$,we find $Q_2 = Q - \frac{Q}{2} = \frac{Q}{2}$.Thus,the force is maximum when $Q_1 = Q_2 = \frac{Q}{2}$.

$A$ total charge $Q$ is broken into 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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