Solution:

When a total charge \(Q_{\text{total}} = 2 \mu C\) is given to one plate, the effective charge stored on the capacitor plates is \(Q = \frac{Q_{\text{total}}}{2} = \frac{2 \mu C}{2} = 1 \mu C\). Given capacitance \(C = 1 \mu F\). Potential difference \(V = \frac{Q}{C} = \frac{1 \mu C}{1 \mu F} = 1 \text{ V}\).