A parallel plate capacitor has plate area 40 cm$$^2$$ and plates separation 2 mm. The space between the plates is filled with a dielectric medium of a thickness 1 mm and dielectric constant 5. The capacitance of the system is:

Parallel plate capacitor: plate area $$A = 40 \text{ cm}^2 = 40 \times 10^{-4} \text{ m}^2$$, plate separation $$d = 2 \text{ mm}$$.

Dielectric: thickness $$d_1 = 1 \text{ mm}$$, dielectric constant $$K = 5$$. Remaining air gap: $$d_2 = 1 \text{ mm}$$.

This is equivalent to two capacitors in series:

$$C_1 = \frac{K\varepsilon_0 A}{d_1} = \frac{5 \times \varepsilon_0 \times 40 \times 10^{-4}}{1 \times 10^{-3}} = 20\varepsilon_0 \text{ F}$$

$$C_2 = \frac{\varepsilon_0 A}{d_2} = \frac{\varepsilon_0 \times 40 \times 10^{-4}}{1 \times 10^{-3}} = 4\varepsilon_0 \text{ F}$$

For series combination:

$$\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{20\varepsilon_0} + \frac{1}{4\varepsilon_0} = \frac{1 + 5}{20\varepsilon_0} = \frac{6}{20\varepsilon_0} = \frac{3}{10\varepsilon_0}$$

$$C = \frac{10\varepsilon_0}{3} \text{ F}$$

The answer is Option C: $$\frac{10}{3}\varepsilon_0$$ F.