AIIMS 2019 Physics Capacitance MCQ Question

To find the area of the plates of a capacitor, we use the formula $C = \frac{{\varepsilon_0 \varepsilon_r A}}{{d}}$, where $C$ is the capacitance, $\varepsilon_0$ is the permittivity of free space ($8.85 \times 10^{-12} \, \text{F/m}$), $\varepsilon_r$ is the relative permittivity (2.4), $A$ is the area, and $d$ is the separation between the plates. Given the dielectric strength of 20 MV/m and a potential difference (P.D.) of 20 V, we can find $d$ as $d = \frac{V}{E} = \frac{20 \, \text{V}}{20 \times 10^6 \, \text{V/m}} = 1 \times 10^{-6} \, \text{m}$. Substituting these values into the capacitance formula and solving for $A$ yields approximately $4.2 \times 10^{-4} \, \text{m}^2$, which corresponds to option B. Other options are incorrect as they do not satisfy the capacitance equation with the given parameters.