Let the two parts be $q$ and $(4-q)$

The total charge is fixed at $4\,\mu C$.

Write the electrostatic force between them

Since the distance $r$ between the two charges is constant, Coulomb's law gives $F = k\frac{q(4-q)}{r^2}$

Here, $k$ and $r^2$ are constants, so to maximize $F$, we need to maximize $q(4-q)$

Maximize the product

$q(4-q) = 4q - q^2$

This is a downward-opening parabola. Its maximum occurs at $\frac{d}{dq}(4q-q^2)=4-2q=0$ $2q=4 \Rightarrow q=2$

Hence the other charge is also $4-q=2$

Therefore

The force is maximum when the charge is divided equally: $2\,\mu C \text{ and } 2\,\mu C$

Check options

• A: $1$ and $3$ $\Rightarrow$ product $=3$
• B: $2$ and $2$ $\Rightarrow$ product $=4$ ✅ maximum
• C: $0$ and $4$ $\Rightarrow$ product $=0$
• D: $1.5$ and $2.5$ $\Rightarrow$ product $=3.75$

So the maximum force occurs for Option B.