An air-filled parallel plate capacitor has a capacitance of 37 pF.
(a) What is the separation of the plates if each plate has an area of 0.8 m2? (cm)
(b) If the region between the plates is filled with a material with κ = 5.0, what is the final capacitance? (pF)
It is a formula..
C=k eo A/d where k is dielectric constant (1 for air, 5.0 for part b)
eo=8.86E-12
To solve these problems, we will need to use the formula for capacitance of a parallel plate capacitor:
C = (κ * ε₀ * A) / d
Where:
C is the capacitance,
κ is the dielectric constant or relative permittivity of the material between the plates,
ε₀ is the permittivity of free space (approximately 8.85 x 10^-12 F/m),
A is the area of each plate, and
d is the separation or distance between the plates.
Now let's solve each part of the question:
(a) What is the separation of the plates if each plate has an area of 0.8 m2? (cm)
We are given the capacitance, C = 37 pF (pico farads) or 37 x 10^-12 F, and plate area, A = 0.8 m^2.
Using the formula, we can solve for the separation, d:
37 x 10^-12 = (κ * 8.85 x 10^-12 * 0.8) / d
Simplifying the equation:
d = (κ * 8.85 x 10^-12 * 0.8) / (37 x 10^-12)
d = (κ * 8.85 * 0.8) / 37
Since we weren't given the dielectric constant, κ, we assume that the region between the plates is filled with air, which has a dielectric constant of 1.0 (approximately).
Using this value of κ = 1.0:
d = (1.0 * 8.85 * 0.8) / 37
d = 0.191 cm
Therefore, the separation of the plates is approximately 0.191 cm.
(b) If the region between the plates is filled with a material with κ = 5.0, what is the final capacitance? (pF)
Given:
κ = 5.0
Area of each plate, A = 0.8 m^2
Using the same formula, we can solve for the final capacitance, C:
C = (κ * ε₀ * A) / d
Since we already found that d = 0.191 cm, we can substitute the given values:
C = (5.0 * 8.85 x 10^-12 * 0.8) / (0.191)
C ≈ 18.42 pF
Therefore, the final capacitance when the region between the plates is filled with a material with κ = 5.0 is approximately 18.42 pF.