Under steady state condition the potential difference across the capacitor in the circuit is________  V.

Because the middle branch contains a $$2\ \mu\text{F}$$ capacitor, no steady-state current flows through it ($$I_{\text{middle}} = 0$$).

The potential drop across the $$4\ \Omega$$ resistor is $$0\text{ V}$$.

The active current loop consists only of the outer boundary: the $$2\text{ V}$$ battery, the bottom $$6\ \Omega$$ resistor, and the top $$2\ \Omega$$ resistor.

$$R_{\text{total}} = 2\ \Omega + 6\ \Omega = 8\ \Omega$$

$$I = \frac{V_{\text{battery}}}{R_{\text{total}}} = \frac{2\text{ V}}{8\ \Omega} = 0.25\text{ A}$$

$$V_{\text{nodes}} = I \times R_{\text{top}} = 0.25\text{ A} \times 2\ \Omega = 0.5\text{ V}$$

$$V_{\text{nodes}} = V_C + V_{4\Omega}$$

$$V_C = V_{\text{nodes}} = 0.5\text{ V}$$