A charge of 20 is deposited on the positive plate of an isolated...

Let the charge given to the first plate be $Q_1 = 20 \mu\text{C}$ and the charge on the second plate be $Q_2 = 0$.

When charges $Q_1$ and $Q_2$ are given to the plates of an isolated capacitor, the charge appearing on the inner facing surfaces of the plates is $q = \dfrac{Q_1 - Q_2}{2}$.

Substituting the given values:

$q = \dfrac{20 - 0}{2} = 10 \mu\text{C}$

The potential difference developed between the plates depends only on the inner surface charges and is given by $V = \dfrac{q}{C}$.

$V = \dfrac{10 \mu\text{C}}{10 \mu\text{F}} = 1 \text{ V}$