Space between the plates of a parallel plate capacitor of plate area 4 cm$$^2$$ and separation of (d) 1.77 mm, is filled with uniform dielectric materials with dielectric constants (3 and 5) as shown in the figure. Another capacitor of capacitance 7.5 pF is connected in parallel with it. The effective capacitance of this combination is _____ pF. (Given $$\epsilon_0 = 8.85 \times 10^{-12}$$ F/m)

$$C_0 = \frac{\varepsilon_0 A}{d} = \frac{8.85 \times 10^{-12} \times 4 \times 10^{-4}}{1.77 \times 10^{-3}}$$

$$C_0 = 5 \times 4 \times 10^{-13} \text{ F} = 20 \times 10^{-13} \text{ F} = 2 \text{ pF}$$

Top slab: $$C_1 = \frac{k_1 \varepsilon_0 A}{d/2} = 2k_1 C_0 = 2 \times 5 \times 2 \text{ pF} = 20 \text{ pF}$$

Bottom slab: $$C_2 = \frac{k_2 \varepsilon_0 A}{d/2} = 2k_2 C_0 = 2 \times 3 \times 2 \text{ pF} = 12 \text{ pF}$$

$$C_{\text{eq}} = \frac{C_1 C_2}{C_1 + C_2} = \frac{20 \times 12}{20 + 12} = \frac{240}{32} = 7.5 \text{ pF}$$

$$C_{\text{total}} = C_{\text{eq}} + 7.5 \text{ pF} = 7.5 + 7.5 = 15 \text{ pF}$$