The circuit shown in the figure, there are two parallel plate capacitors each of capacitance $C$. The switch $s_1$ pressed first to fully charge the capacitor $C_1$ and then released. The switch $S_2$ is then pressed to charge capacitor $C_2$. After some time, $S_2$ is released and then $S_3$ is pressed. After some time

After pressing $S_1$ charge on upper plate of $C_1$ is $+2C{V}_0$. After pressing $S_2$ this charge equally distributes in two capacitors. Therefore charge an upper plates of both capacitors will be $+C{V}_0$.

When $S_2$ is released and $S_3$ is pressed, charge on upper plate of $C_1$ remains unchanged $\left(=+C{V}_0\right)$ but charge on upper plate of $C_2$ is according to new battery $\left(=-C{V}_0\right).$