Answer
During charging, the magnitude of the charge that passes through the meter is $~~+\frac{CV}{3}$
Work Step by Step
In part (a), we found that the potential difference across each capacitor is $\frac{V}{3}$
We can find the magnitude of the charge on each plate:
$q = (C)(\frac{V}{3})$
$q = \frac{CV}{3}$
Since the right side of the battery is the negative side, the right plate of each capacitor will have a negative charge, and the left plate of each capacitor will have a positive charge.
The charge on the left plate of each capacitor is $~~+\frac{CV}{3}$
During charging, the magnitude of the charge that passes through the meter will be equal in magnitude of the final charge on the left plate of the capacitor on the far left of the circuit.
During charging, the magnitude of the charge that passes through the meter is $~~+\frac{CV}{3}$.
