Answer
$q_2 = 0$
Work Step by Step
In part (a), we found that $q_1 = 480~\mu C$
In part (c), we found that $q_2 = 480~\mu C$
If plates with opposite signs are wired together, we can find the net charge on two plates that are wired together:
$q_{net} = +480~\mu C-480~\mu C = 0$
The capacitors are wired in parallel, so the potential difference across each capacitor is equal.
Since the net charge on the plates that are wired together is 0, the charge on all four plates must be 0.
Therefore:
$q_2 = 0$
