Question

Than how much charge (in $\mu C$) will flow through the switch after it is closed?

Answer 1

When S is open, potential across each $3\mu F$ capacitors is $V_3=\frac{6}{3+6}\times 360=240V$
and potential across each $6\mu F$ capacitors is $V_6=\frac{3}{3+6}\times 360=120V$
Charge on each $3\mu F$ capacitors is $q_3=3\times {10}^{-6}\times 240=720\mu C$ and Charge on each $6\mu F$ capacitors is $q_6=6\times {10}^{-6}\times 120=720\mu C$
When S is closed, potential across each capacitors is $V=\frac{9}{9+9}\times 360=180V$
Charge on each $3\mu F$ capacitors is $q_3^{\prime }=3\times {10}^{-6}\times 180=540\mu C$ and
Charge on each $6\mu F$ capacitors is $q_6^{\prime }=6\times {10}^{-6}\times 180=1080\mu C$.
After closing the switch charge flow from c to d is $\Delta q=(q_6^{\prime }-q_6)+(q_3-q_3^{\prime })=(1080-720)+(720-540)=540\mu C$