AIIMS 2019 Physics Capacitors MCQ Question

To find the area of the capacitor plates, we use the formula for capacitance $C = \frac{\varepsilon_0 \varepsilon_r A}{d}$, where $\varepsilon_0$ is the permittivity of free space ($8.85 \times 10^{-12} \, \text{F/m}$), $\varepsilon_r$ is the relative permittivity (dielectric constant), $A$ is the area of the plates, and $d$ is the separation between the plates. Given that the dielectric strength is 30 MV/m and the potential difference is 30 V, we can calculate the maximum plate separation $d$ as $d = \frac{V}{E} = \frac{30 \, \text{V}}{30 \times 10^6 \, \text{V/m}} = 1 \times 10^{-6} \, \text{m}$. Substituting $C = 15 \times 10^{-9} \, \text{F}$, $\varepsilon_r = 2.5$, and $d = 1 \times 10^{-6} \, \text{m}$ into the capacitance formula allows us to solve for $A$, yielding approximately $6.7 \times 10^{-4} \, \text{m}^2$,