The plates of an air capacitor have an area of 150 cm² and are 6.0 mm apart…

Step 1. The effective gap. $d - t + \dfrac{t}{K} = 6.0\ \text{mm} - 4.0\ \text{mm} + 1.33\ \text{mm} = 3.33\ \text{mm}$.

Step 2. The capacitance with the dielectric slab. $C = \dfrac{\varepsilon_0 A}{3.33 \times 10^{-3}\ \text{m}} = \dfrac{8.85 \times 10^{-12} \times 1.5 \times 10^{-2}}{3.33 \times 10^{-3}} = 3.98 \times 10^{-11}\ \text{F}$, that is $39.8$ pF.

Step 3. The capacitance with the metal slab. The gap is $6.0\ \text{mm} - 4.0\ \text{mm} = 2.0\ \text{mm}$, so $C = \dfrac{8.85 \times 10^{-12} \times 1.5 \times 10^{-2}}{2.0 \times 10^{-3}} = 66.4$ pF.

Step 4. The fields. In the air $E_0 = \dfrac{V}{d - t + t/K} = \dfrac{20\ \text{V}}{3.33 \times 10^{-3}\ \text{m}} = 6.0\ \text{kV/m}$. In the slab $E = \dfrac{E_0}{K} = 2.0$ kV/m.

Step 5. The voltage across the slab. $E \times t = 2.0 \times 10^{3}\ \text{V/m} \times 4.0 \times 10^{-3}\ \text{m} = 8.0\ \text{V}$.