Potential difference across the branch de is 6V. Net capacitance of de branch is 2.1µF So, q=CV ⇒q=2.1×6µC⇒q=12.6µC Potential across 3µF capacitance is V=
12.6
3
=4.2Volt Potential across 2 and 5 combination in paralle is 6−4.2=1.8V So, q′=(1.8)(5)=9µC