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

Initial charge on the capacitor is given by $Q = CV$.

Substituting the given values, $Q = 100 \mu\text{F} \times 50 \text{ V} = 5000 \mu\text{C} = 5 \text{ mC}$.

When the battery is disconnected, the charge on the capacitor plates remains constant.

The induced charge $Q_i$ on the surface of the dielectric is given by the formula Q_i = Q \left( 1 -\dfrac{1}{K} \right).

Substituting $Q = 5 \text{ mC}$ and $K = 2.5$, we get Q_i = 5 \left( 1 -\dfrac{1}{2.5} \right) =5 \left( 1 -\dfrac{2}{5} \right) =$5 \times$\dfrac{3}{5} =$3 \text{ mC}$.