A dielectric slab is introduced between the plates of an isolated...

For an isolated capacitor, the charge $Q$ on the plates remains constant.

The force between the plates of a capacitor can be calculated using the relation $F = -\dfrac{dU}{dx}$, where $U$ is the electrostatic potential energy stored in the capacitor and $x$ is the separation between the plates.

When a dielectric slab of thickness $t$ and dielectric constant $K$ is introduced between the plates, the capacitance becomes:
$C = \dfrac{\epsilon_0 A}{x - t + \dfrac{t}{K}}$

The energy stored in the capacitor is:
U = \dfrac{Q^2}{2C} = \dfrac{Q^2}{2\epsilon_0 A} \left( x - t +\dfrac{t}{K} \right)

Differentiating $U$ with respect to $x$ to find the force:
$F =$-\dfrac{dU}{dx} = -\dfrac{Q^2}{2\epsilon_0 A}

The magnitude of the force is $|F| =$\dfrac{Q^2}{2\epsilon_0 A}, which is exactly the same as the force between the plates in the absence of the dielectric slab.

Alternatively, the electric field produced by one plate at the location of the other is $E =$\dfrac{Q}{2A\epsilon_0}. The polarized dielectric slab produces zero net electric field outside its volume. Thus, the field experienced by the charge on the other plate remains unaffected, and the force $F = QE$ remains unchanged.