Two identical metal plates are given positive charges Q₁ and Q₂ (< Q₁) respectively. If they are now brought close together to form a parallel plate capacitor with capacitance C, the potential difference between them is :

(Q₁ – Q₂) / 2C

Electric field with the plates $\vec{E}=\vec{E}_{Q_1}+\vec{E}_{Q_2}$

$E =E_1-E_2$

$=\frac{Q_1}{2 A \varepsilon_0}-\frac{Q_2}{2 A \varepsilon_0}$

E = $\frac{Q_1-Q_2}{2 A \varepsilon_0}$

∴ Potential difference between the plates

$V_A-V_B =E . d=\left(\frac{Q_1-Q_2}{2 A \varepsilon_0}\right) d$

$=\frac{Q_1-Q_2}{2\left(\frac{A \varepsilon_0}{d}\right)}$

$=\frac{Q_1-Q_2}{2 C}$