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

In steady state, the capacitor acts as an open circuit, meaning no current flows through the middle branch.

The circuit simplifies to a single loop containing the \(2 \text{ V}\) battery, the \(6 \Omega\) resistor, and the \(2 \Omega\) resistor.

The total resistance of this loop is \(R_{\text{eq}} = 2 + 6 = 8 \Omega\).

The current flowing through the loop is \(I = \dfrac{2}{8} = 0.25 \text{ A}\).

The potential difference across the parallel branches is equal to the voltage drop across the \(2 \Omega\) resistor:
\(V = I \times 2 = 0.25 \times 2 = 0.5 \text{ V}\)

Since there is no current in the middle branch, the voltage drop across the \(4 \Omega\) resistor is zero. Thus, the entire potential difference of \(0.5 \text{ V}\) appears across the capacitor.

Answer: \(0.5\)