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 V battery, the 6Ω resistor, and the 2Ω resistor.
The total resistance of this loop is Req=2+6=8Ω.
The current flowing through the loop is I=82=0.25 A.
The potential difference across the parallel branches is equal to the voltage drop across the 2Ω resistor:
V=I×2=0.25×2=0.5 V
Since there is no current in the middle branch, the voltage drop across the 4Ω resistor is zero. Thus, the entire potential difference of 0.5 V appears across the capacitor.
Answer: 0.5
