Three capacitors

when switch is open then as given charge =Q=1/2(c.v) ,since both capacitors are in series now if switch is closed then charge will be=2/3(c.v) now comparing it with previous case it will be 4/3(Q) v is supply voltage

Initially, without the switch, the relation would be $\frac{C}{2} = \frac{Q}{V}$ , because capacitors add in capacitance in reciprocals over a series. However, with the closing of the switch, we can factor all three capacitors into play. First, we see that the two capacitors in series add to 2C, because in parallel, they add. Then, we add the conjoined capacitance to the third capacitor on the leftmost side by adding their reciprocals and then taking the reciprocal of that, which is $\frac{1}{(\frac{1}{2C} + \frac{1}{C})} = \frac{1}{\frac{3}{2C}} = \frac{2C}{3}$ . From here, we can see that the change from $\frac{C}{2}$ to $\frac{2C}{3}$ is by multiplying $\frac{4}{3}$ to $\frac{Q}{V}$ . Since voltage remains constant due to Kirchoff's laws, the answer therefore would be $\boxed{\frac{4Q}{3}}$