Answer
a) $\frac{2V}{3}$
b) $\frac{-2CV}{3}$
c) $\frac{2CV}{3}$
Work Step by Step
(a) We know that the voltage across $ C_1$ is $\frac{2V}{3}$. The two capacitors have the same charge, and hence the smaller capacitor will have the larger voltage.
(b) The equivalent capacitance of $ C_1$ and $ C_2$ in series is $\frac{2C}{3}$. Hence, the charge on the equivalent capacitance is $ Q=C_{eq}V=\frac{2CV}{3}$ and the charge on the right plate of $ C_2$ is $\frac{-2CV}{3}$.
(c) We know that the magnitude of the net charge that flowed through the ammeter is actually the magnitude of the charge on each plate of the capacitor, $\frac{2CV}{3}$.
