Consider the circuit shown in the given figure. What will be the...

In the steady state, the capacitor acts as an open circuit. Thus, no current flows through the middle branch containing the $5\ \mu\text{F}$ capacitor and the $4\ \Omega$ resistor.

The steady current $I$ flows only through the $2\ \Omega$ and $3\ \Omega$ resistors.

The equivalent resistance of the circuit is:
$R_{\text{eq}} = 2 + 3 = 5\ \Omega$

The current in the circuit is:
$I = \dfrac{10}{5} = 2\text{ A}$

The potential difference across the $3\ \Omega$ resistor is:
$V = I \times 3 = 2 \times 3 = 6\text{ V}$

Since the capacitor branch is in parallel with the $3\ \Omega$ resistor, the potential difference across the capacitor branch is also $6\text{ V}$. As there is no current in this branch, the voltage drop across the $4\ \Omega$ resistor is zero.

Therefore, the potential difference across the capacitor is $V_C = 6\text{ V}$.

The charge on the capacitor is:
$Q = C V_C = 5\ \mu\text{F} \times 6\text{ V} = 30\ \mu\text{C}$

Answer: $30\ \mu\text{C}$