1. Write the electrostatic force between the two parts

If the total charge $Q$ is divided into two parts: $q \quad \text{and} \quad (Q-q),$ and they are placed at a fixed distance $r$ apart, then by Coulomb’s law the repulsive force is $F = k\frac{q(Q-q)}{r^2},$ where $k$ and $r$ are constants.

So, to maximize $F$, we only need to maximize $f(q)=q(Q-q)=Qq-q^2.$

2. Maximize the expression

We have $f(q)=Qq-q^2.$ Differentiate with respect to $q$: $\frac{df}{dq}=Q-2q.$ For maximum force, $\frac{df}{dq}=0 \implies Q-2q=0 \implies Q=2q.$

3. Check that it is maximum

Second derivative: $\frac{d^2f}{dq^2}=-2<0,$ so this gives a maximum.

4. Interpretation

Thus the force is maximum when the total charge is divided equally: $q = Q-q \implies q=\frac{Q}{2}.$ Hence, $Q=2q.$

5. Match with options

• A: $Q=2q$ ✅
• B: $Q=4q$ ❌
• C: $Q=3q$ ❌
• D: $Q=\dfrac{q}{2}$ ❌

Therefore, the correct option is A.