A 100 pF capacitor is charged to a potential difference of 24 V ....

The initial charge on the first capacitor is given by $Q = C_1 V_1$.

Substituting the given values, $Q = 100 \times 24 = 2400 \text{ pC}$.

When this capacitor is connected to an uncharged capacitor of capacitance $C_2 = 20 \text{ pF}$, the total charge remains conserved and is shared between them until they reach a common potential $V$.

The equivalent capacitance of the parallel combination is $C_{eq} = C_1 + C_2 = 100 + 20 = 120 \text{ pF}$.

The common potential difference $V$ across both capacitors is given by $V = \dfrac{Q}{C_{eq}}$.

Substituting the values, $V = \dfrac{2400}{120} = 20 \text{ V}$.

Therefore, the new potential difference across the $100 \text{ pF}$ capacitor is $20 \text{ V}$.