The given circuit shows a balanced Wheatstone bridge. Now, the circuit becomes
In the branch AB, both capacitor are arranged in the series combination. Hence, its equivalent capacitance is given by ‌
1
CAB
=‌
1
20
+‌
1
20
=‌
2
20
=‌
1
10
CAB=10µF Similarly, in branch CD. CCD=10µF Now, CAB and CCD are connected in parallel combination. Hence, the equivalent capacitance of the circuit, Ceq=10+10=20µF We know that, charge Q=CeqV Q=20×10−6×150V Amount of charge stored Q=3×10−3C