A charge Q is divided into two parts Q_1 and | vedclass

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(D) Given $Q = Q_1 + Q_2$,so $Q_2 = Q - Q_1$.The electrostatic force between them is given by Coulomb's Law: $F = \frac{k Q_1 Q_2}{R^2} = \frac{k Q_1

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

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(D) Given $Q = Q_1 + Q_2$,so $Q_2 = Q - Q_1$.The electrostatic force between them is given by Coulomb's Law: $F = \frac{k Q_1 Q_2}{R^2} = \frac{k Q_1 (Q - Q_1)}{R^2} = \frac{k}{R^2} (Q Q_1 - Q_1^2)$.For the force to be maximum,the derivative of $F$ with respect to $Q_1$ must be zero: $\frac{dF}{dQ_1} = 0$.$\frac{d}{dQ_1} [\frac{k}{R^2} (Q Q_1 - Q_1^2)] = \frac{k}{R^2} (Q - 2Q_1) = 0$.Since $\frac{k}{R^2} 
eq 0$,we have $Q - 2Q_1 = 0$,which implies $Q_1 = \frac{Q}{2}$.Substituting this back,$Q_2 = Q - Q_1 = Q - \frac{Q}{2} = \frac{Q}{2}$.Thus,the force is maximum when $Q_1 = Q_2 = \frac{Q}{2}$.

$A$ charge $Q$ is divided into two parts $Q_1$ and $Q_2$. These charges are placed at a distance $R$. What will be the values of $Q_1$ and $Q_2$ for the maximum repulsive force between them?

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