Determine the capacitance, in pF, of a paper-filled parallel-plate capacitor (dielectric constant=3.70) having a plate area of 5.40cm2 and plate separation of 590 µm. Use 8.85 x 10-12 F/m for εo, the permittivity of free space
To determine the capacitance of a parallel-plate capacitor, you can use the formula:
C = (ε₀ * εᵣ * A) / d
where:
C is the capacitance of the capacitor,
ε₀ is the permittivity of free space,
εᵣ is the relative permittivity (or dielectric constant) of the material in between the plates,
A is the plate area, and
d is the plate separation.
Let's substitute the given values into the formula:
ε₀ = 8.85 x 10^⁻12 F/m (given)
εᵣ = 3.70 (given)
A = 5.40 cm² = (5.40 x 10^⁻⁴) m² (convert cm² to m²)
d = 590 µm = (590 x 10^⁻⁶) m (convert µm to m)
Now, let's calculate the capacitance:
C = (8.85 x 10^⁻¹² F/m) * (3.70) * (5.40 x 10^⁻⁴ m²) / (590 x 10^⁻⁶ m)
First, multiply the numerical values together:
C = (8.85 x 3.70 x 5.40 x 10^⁻¹² x 10^⁻⁴) / (590 x 10^⁻⁶) F
Now, simplify the powers of 10:
C = (36.9978 x 10^⁻¹²) / (5.9 x 10^⁻⁶) F
Divide 36.9978 by 5.9:
C = 6.2716 x 10^⁻⁶ F
Finally, convert the capacitance to pF by multiplying by 10^12 (since 1 F = 10^12 pF):
C = 6.2716 x 10^⁻⁶ F * 10^12 pF / 1 F
C ≈ 6271.6 pF
Therefore, the capacitance of the paper-filled parallel-plate capacitor is approximately 6271.6 pF.