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

Initial charge on the capacitor is given by $Q_0 = C_0 V_0$

Substituting the given values:

$Q_0 = 100 \mu\text{F} \times 50 \text{ V} = 5000 \mu\text{C} = 5 \text{ mC}$

Since the battery is disconnected, the charge remains constant. When a dielectric of constant $K = 2.5$ is inserted, the new capacitance becomes $C = K C_0$

The new potential difference is $V = \dfrac{Q_0}{C} = \dfrac{Q_0}{K C_0} = \dfrac{V_0}{K}$

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

The charge $Q'$ required to produce this new potential difference in the absence of the dielectric slab (i.e., with capacitance $C_0$) is:

$Q' = C_0 V$

$Q' = 100 \mu\text{F} \times 20 \text{ V} = 2000 \mu\text{C} = 2 \text{ mC}$