
Under the balanced condition, the capacitor will act as an open circuit.
The current (i1) through the 1Ω resistor is given by
i1==1Ω+2Ω10V310A
The current (i2) through the 6Ω resistor is given by
i2==6Ω+4Ω10V1A
From the above diagram, it can be written that
VA+i1×1−i2×6=VB⇒VA−VB=1×6−310×1=38V
Hence, the charge on the capacitor is given by
Q===C(VA−VB)150μF×38V400μC