Question

A parallel plate capacitor having a dielectric constant 5 and dielectric strength $10^6 \, \text{V/m}$ is to be designed with a voltage rating of 2 kV. The field should never exceed 10% of its dielectric strength. To have the capacitance of 60 pF, the minimum area of the plates should be

• A. \( 2.71 \times 10^{-4} \, \text{m}^2 \)
• B. \( 2.7 \times 10^{-2} \, \text{m}^2 \)
• C. \( 2.71 \times 10^{-4} \, \text{m}^2 \)
• D. \( 2.7 \times 10^{-2} \, \text{m}^2 \)

Hint:

In capacitor problems, remember that the dielectric strength limits the electric field, and use the formula \( C = \kappa \epsilon_0 \frac{A}{d} \) to relate capacitance to the plate area and separation.

Answer 1

The Correct Option is B

We are given: - Dielectric constant \( \kappa = 5 \), - Dielectric strength \( E_{\text{max}} = 10^6 \, \text{V/m} \), 
- Voltage rating \( V = 2 \, \text{kV} = 2000 \, \text{V} \), 
- Capacitance \( C = 60 \, \text{pF} = 60 \times 10^{-12} \, \text{F} \). 
The maximum electric field the capacitor can handle is 10% of its dielectric strength, so the maximum electric field is: \[ E_{\text{max, design}} = 0.1 \times 10^6 = 10^5 \, \text{V/m}. \] Now, using the relationship between the voltage, electric field, and separation between the plates for a parallel plate capacitor: \[ E = \frac{V}{d} \] where \( d \) is the separation between the plates. Rearranging for \( d \): \[ d = \frac{V}{E} = \frac{2000}{10^5} = 0.02 \, \text{m}. \] The capacitance \( C \) of a parallel plate capacitor is given by: \[ C = \kappa \epsilon_0 \frac{A}{d} \] where: - \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \) is the permittivity of free space, - \( \kappa = 5 \) is the dielectric constant, - \( A \) is the area of the plates, and - \( d = 0.02 \, \text{m} \) is the separation between the plates. Substituting the given values into the formula: \[ 60 \times 10^{-12} = 5 \times 8.85 \times 10^{-12} \times \frac{A}{0.02} \] Simplifying: \[ 60 \times 10^{-12} = 44.25 \times 10^{-12} \times \frac{A}{0.02} \] \[ A = \frac{60 \times 10^{-12} \times 0.02}{44.25 \times 10^{-12}} = 2.7 \times 10^{-2} \, \text{m}^2. \] 
Thus, the minimum area of the plates required is \( 2.7 \times 10^{-2 \, \text{m}^2} \).