The plates of parallel plate capacitor are given charges +4Q and – 2Q. The capacitor is then connected across are uncharged capacitor of same capacitance as first one. The final potential difference between the plates of the first capacitor is

Steady state before connection are with second capacitor.

After connection with second capacitor the charge on inner facing surface are shared. Finally two capacitors will have same potential difference across them with magnitude of charge on each of the facing surfaces as $\frac{3\mathrm{Q}}{2}$and potential difference $\frac{3\mathrm{Q}}{2\mathrm{C}}$