A total charge Q is broken into two parts Q_1 | vedclass

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(D) Let the two parts of the charge be $Q_1$ and $Q_2$ such that $Q_1 + Q_2 = Q$. Thus,$Q_2 = Q - Q_1$. The electrostatic force of repulsion between t

Vedclass · Accepted Answer

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

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(D) Let the two parts of the charge be $Q_1$ and $Q_2$ such that $Q_1 + Q_2 = Q$. Thus,$Q_2 = Q - Q_1$. The electrostatic force of repulsion between them is given by $F = \frac{k Q_1 Q_2}{R^2}$. Substituting $Q_2$,we get $F = \frac{k Q_1 (Q - Q_1)}{R^2} = \frac{k}{R^2} (Q Q_1 - Q_1^2)$. For the force $F$ to be maximum,the derivative of $F$ with respect to $Q_1$ must be zero: $\frac{dF}{dQ_1} = 0$. $\frac{d}{dQ_1} [\frac{k}{R^2} (Q Q_1 - Q_1^2)] = \frac{k}{R^2} (Q - 2Q_1) = 0$. This implies $Q - 2Q_1 = 0$,or $Q_1 = \frac{Q}{2}$. Since $Q_1 + Q_2 = Q$,we find $Q_2 = Q - \frac{Q}{2} = \frac{Q}{2}$. Therefore,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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