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(D) The electrostatic force between two charges $Q_1$ and $Q_2$ separated by distance $R$ is given by $F = k \frac{Q_1 Q_2}{R^2}$.Since $Q_1 + Q_2 = Q

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The electrostatic force between two charges $Q_1$ and $Q_2$ separated by distance $R$ is given by $F = k \frac{Q_1 Q_2}{R^2}$.Since $Q_1 + Q_2 = Q$,we can write $Q_2 = Q - Q_1$.Substituting this into the force equation: $F = \frac{k}{R^2} (Q_1 Q - Q_1^2)$.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 implies $Q_1 = \frac{Q}{2}$.Since $Q_2 = Q - Q_1$,we find $Q_2 = Q - \frac{Q}{2} = \frac{Q}{2}$.Thus,the force is maximum when $Q_1 = Q_2 = \frac{Q}{2}$.

$A$ charge $Q$ is divided into two parts $Q_1$ and $Q_2$. For a given distance $R$,the force between them is maximum when:

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