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Q43) In the upper part of circuit, 2 and 2 μF are in series.
Equivalent capacitance = 2/2 = 1 μF
This 1μF is parallel to 1μF.
Equivalent Capacitance = 2 μF
Total charge = Q = CV
Q = 2 (2) = 4 μC
Capacitance of upper capacitors (1μF) is equal to capacitance of lower capacitor(1μF).
Hence charge is divided equally between upper and lower capacitors.
So charge in lower capacior = 2 μC
Charge in any of the upper capacitor = 2 μC
So answer is (4)