Question

A parallel-plate capacitor has plates of dimensions 2.0 cm by 3.0 cm separated by a 1.0 mm thickness of paper. The relative dielectric constant of paper is 3.7. Find its capacitance.

• A. 20 pF
• B. 20 nF
• C. 200 pF
• D. 20 micro F

Hint:

The capacitance increases with the dielectric constant and the area of the plates and decreases with the separation between the plates.

Answer 1

The Correct Option is A

The capacitance of a parallel-plate capacitor is given by: \[ C = \frac{\varepsilon_r \varepsilon_0 A}{d} \] Where:
- \( \varepsilon_r \) is the relative dielectric constant,
- \( \varepsilon_0 \) is the permittivity of free space \( (8.85 \times 10^{-12} \, \text{F/m}) \),
- \( A \) is the area of the plates,
- \( d \) is the separation between the plates.
Substituting the values: \[ A = 2.0 \, \text{cm} \times 3.0 \, \text{cm} = 6.0 \times 10^{-4} \, \text{m}^2 \] \[ d = 1.0 \, \text{mm} = 1.0 \times 10^{-3} \, \text{m} \] \[ C = \frac{3.7 \times 8.85 \times 10^{-12} \times 6.0 \times 10^{-4}}{1.0 \times 10^{-3}} = 20 \, \text{pF} \]