BITSAT 2020 Paper

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