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: \[ A = 2.0 \, \text{cm} \times 3.0 \, \text{cm} = 6.0 \times 10^{-4} \, \text{m}^2 \] \[ d = 1.0 \, \text{mm} = 1.0 \times 10^{-3} \, \text{m} \] \[ C = \frac{3.7 \times 8.85 \times 10^{-12} \times 6.0 \times 10^{-4}}{1.0 \times 10^{-3}} = 20 \, \text{pF} \]

Answer 2

The capacitance of a parallel-plate capacitor is set by its area, its plate separation, and the dielectric between the plates. Working out the size of each answer choice against the physical dimensions given tells us which one is realistic.

1. 20 pF: Using \( C = \frac{\varepsilon_r \varepsilon_0 A}{d} \) with \( A = 6.0 \times 10^{-4} \, \text{m}^2 \), \( d = 1.0 \times 10^{-3} \, \text{m} \), and \( \varepsilon_r = 3.7 \), we get \( C = \frac{3.7 \times 8.85 \times 10^{-12} \times 6.0 \times 10^{-4}}{1.0 \times 10^{-3}} \approx 2.0 \times 10^{-11} \, \text{F} = 20 \, \text{pF} \), which matches this option exactly.
2. 20 nF: This is \( 1000 \) times larger than \( 20 \, \text{pF} \). For the capacitance to reach the nanofarad range with these small plates and a millimetre-scale gap, the dielectric constant would need to be in the thousands, far above the given value of \( 3.7 \), so this option is too large.
3. 200 pF: This is \( 10 \) times larger than the computed value. Such a jump would need either ten times the plate area or a dielectric constant of about \( 37 \), neither of which matches the numbers given in the question.
4. 20 microF: This is roughly a million times larger than the computed value. A capacitance in the microfarad range from plates this small and a gap this wide is not physically realistic without an enormously higher dielectric constant, so this option can be ruled out immediately.

Only the direct computation using the given area, separation, and dielectric constant lands in the picofarad range, matching one specific option.

Therefore, the correct answer is 20 pF.