Q160 / 180 · COMEDK 2024 (Evening Shift) · Physics
A parallel plate capacitor having a dielectric constant 5 and dielectric strength $10^6 \mathrm{~V} \mathrm{~m}^{-1}$ is to be designed with voltage rating of $2 \mathrm{~kV}$. The field should never exceed $10 \%$ of its dielectric strength. To have the capacitance of $60 \mathrm{~pF}$ the minimum area of the plates should be
- A$27.1 \times 10^{-4} \mathrm{~m}^2$
- B$2.71 \times 10^{-4} \mathrm{~m}^2$
- C$27.1 \times 10^{-2} \mathrm{~m}^2$
- Correct answerD$2.7 \times 10^{-2} \mathrm{~m}^2$
Solution
Working electric field: $E = 10\%$ of $10^6 = 10^5$ V m$^{-1}$
Plate separation: $d = \dfrac{V}{E} = \dfrac{2000}{10^5} = 2 \times 10^{-2}$ m
$C = \dfrac{K\epsilon_0 A}{d} \Rightarrow A = \dfrac{Cd}{K\epsilon_0} = \dfrac{60 \times 10^{-12} \times 2 \times 10^{-2}}{5 \times 8.85 \times 10^{-12}}$
$A = \dfrac{120 \times 10^{-14}}{44.25 \times 10^{-12}} = \dfrac{1.2 \times 10^{-12}}{44.25 \times 10^{-12}} \approx 2.7 \times 10^{-2}$ m$^2$
Plate separation: $d = \dfrac{V}{E} = \dfrac{2000}{10^5} = 2 \times 10^{-2}$ m
$C = \dfrac{K\epsilon_0 A}{d} \Rightarrow A = \dfrac{Cd}{K\epsilon_0} = \dfrac{60 \times 10^{-12} \times 2 \times 10^{-2}}{5 \times 8.85 \times 10^{-12}}$
$A = \dfrac{120 \times 10^{-14}}{44.25 \times 10^{-12}} = \dfrac{1.2 \times 10^{-12}}{44.25 \times 10^{-12}} \approx 2.7 \times 10^{-2}$ m$^2$
Hence, the correct answer is option: 'D'.