A capacitor with a capacitance of 100 is charged to a potential...

Since the battery is disconnected after charging, the charge on the capacitor remains constant.

Initial charge on the capacitor is $Q = CV$

When a dielectric of dielectric constant $K$ is inserted, the new capacitance becomes $C' = KC$.

The new potential difference $V'$ across the capacitor is given by:

$V' = \dfrac{Q}{C'} = \dfrac{CV}{KC} = \dfrac{V}{K}$

Substituting the given values $V = 50 \text{ V}$ and $K = 2.5$:

$V' = \dfrac{50}{2.5} = 20 \text{ V}$