Question

What charges (in $\mu C$) will flow through section $B$ of the circuit in the direction as shown in figure when switch $S$ is closed?

Answer 1

When switch is opened, 2 and $3\mu F$ capacitors are in series. So,
$C_{eq}=\frac{2\times 3}{5}=\frac{6}{5}\mu F$
Hence, charge on each capacitor is
$q=CV=\frac{6}{5}\times 90=108\mu C$
When switch S is open, let $q_1$ and $q_2$ be charges on the two capacitors as in figure (b). So,
$q_1=2\times 30=60\mu C$
$q_2=3\times 60=180\mu C$
Let charge $q_B$ goes to the upper plate $3\mu F$ capacitor and lower plate of $2\mu F$ capacitor. Initially, both the plates have charge $+q-q=0$. Finally, they have charges $q_2-q_1$. So
$q_2-q_1=q_B+0$
or $q_B=q_2-q_1=180-60=120\mu C$
Alternatively:
After closing the switch, charge flowing out of $2\mu F$ capacitor is
$\Delta q_1=108-60=48\mu C$
charge flown inot $3\mu F$ capacitor is
$\Delta q_2=180-108=72\mu C$
So the net charge
$\Delta q=\Delta q_1+\Delta q_2=48+72=120\mu C$