In figure is shown a system of four capacitors connected across a 10V battery. Charge that will flow from switch S when it is closed is :

When switch was open as shown in the figure 1, the equivalent capacitance across path PaQ is Ceq=2×3/(2+3)µF=1.5 µF. Similarly the equivalent capacitance across path PbQ is also Ceq=1.5 µF. When switch is closed the charge distribution across each capacitor is as shown in the figure 2. Thus taking potential at the negative terminal of capacitor to be zero and at the switch to be V0 We can write the charges across the capacitors as follows, For path PaQ, Q1=(10−V0)×2 µF.. (i) Q2=(10−V0)×3µF.. (ii) For Path PbQ Q1=V0×2 µF.. (iii) Q2=V0×3µF.. (iv) Using equation (iii) in equation (i) we get V0=5V From equation (iii) and (iv) we get charges Q1 Q2 as Q1=10µC Q2=15µC Thus charge flowing across switch is from point b towards point a The net charge flowing is 15 µC − 10 µC = 5 µC