In the following diagram, the charge and potential difference across 8 μF capacitance will be respectively
214 μC, 27 V
Given circuit can be redrawn as follows, because capacitors, 9μF, 9μF and 7μF are short circuited. So they can be removed.
V1+V2=40V and V1V2=3618=2
Hence V1=803V and V2=403V
Charge on 8μF capacitor =8×803=213.3 μF≈214μF