A parallel-plate capacitor with a capacitance of 1.2 imes 10^ -3...

When charges $Q_1$ and $Q_2$ are given to the two plates of a parallel-plate capacitor, the charge that appears on the inner facing surfaces is $q = \dfrac{Q_1 - Q_2}{2}$.

Given $Q_1 = 2.0 \times 10^{-8} \text{ C}$ and $Q_2 = -1.0 \times 10^{-8} \text{ C}$.

Substituting the values:

$q = \dfrac{2.0 \times 10^{-8} - (-1.0 \times 10^{-8})}{2}$

$q = \dfrac{3.0 \times 10^{-8}}{2} = 1.5 \times 10^{-8} \text{ C}$

The potential difference $V$ developed between the plates is given by $V = \dfrac{q}{C}$.

Given $C = 1.2 \times 10^{-3} \text{ F}$.

$V = \dfrac{1.5 \times 10^{-8}}{1.2 \times 10^{-3}}$

$V = 1.25 \times 10^{-5} \text{ V}$