What charges will flow through section B of the circuit (in μC) in the direction shown when switch S is closed.

When switch is opened, 2 and 3 μF capacitors are in series.So, Ceq=2×35=65μFHence, charge on each capacitor, q=CV=65×90=108μCWhen switch S is closed, let q1 and q2 be charge on the two capacitors as shown in Fig. (b). So, q1=2×30=60μC q2=3×60=180μCFrom figure it is clear that, after closing the switch the charge flowing out of 2 μF capacitor, Δq1=108−60=48μCAnd the charge flown into 3 μF capacitor, Δq2=180−108=72μC.So, the net charge flown through the switch, Δq=Δq1+Δq2=48+72=120μC