The charge on two plates of a parallel plate capacitor of capacitance C are 3Q and –Q, respectively. The potential of the capacitor is [None of the plates of capacitor is grounded].

electric field due to positive plate is $\frac{3Q}{A2{\epsilon }_0}$ and due to negative plate $\frac{Q}{2{\epsilon }_{0}A}$ and both are in the same direction  so net electric field between the plates is $\frac{4Q}{2{\epsilon }_{0}A}$

potential V = Ed = $\frac{2Qd}{{\epsilon }_{0}A}$=$\frac{2Q}{C}$                               $\left(\frac{{\epsilon }_{0}A}{d}= C\right)$