In the circuit shown, potential difference across the 4 μF capacitor is

Equivalent capacitance Ce=3×4+23+4+2μF=2 μF∴ Charge on 3μF capacitor, Q=CV=2×10 μC∴Potential difference across 3μF capacitor =2×103 volt = 6.66 volt∴ Potential difference across 4 μF capacitor =10−6.66 volt = 3.34 volt