NEET 2020 Physics Question 166 Solution

For a parallel plate capacitor, capacitance with air (vacuum) is $$C_0 = \frac{\varepsilon_0 A}{d}$$. When the space between plates is filled with a dielectric medium of absolute permittivity $$\varepsilon$$, the capacitance becomes: $$C = \frac{\varepsilon A}{d}$$.

The relative permittivity (dielectric constant $$K$$) is the ratio of capacitances: $$K = \varepsilon_r = \frac{C}{C_0} = \frac{30\ \mu\text{F}}{6\ \mu\text{F}} = 5$$.

$$\varepsilon = K \varepsilon_0 = 5 \times (8.85 \times 10^{-12}\text{ C}^2\text{N}^{-1}\text{m}^{-2}) = 44.25 \times 10^{-12}\text{ C}^2\text{N}^{-1}\text{m}^{-2}$$.

$$44.25 \times 10^{-12} = 0.4425 \times 10^{-10} \approx 0.44 \times 10^{-10}\text{ C}^2\text{N}^{-1}\text{m}^{-2}$$. Matches Option 3.