Problem 21.79
The plates of a 3.2nF parallel-plate capacitor are each 0.25 m^2 in area.
Part A -
How far apart are the plates if there's air between them?
Express your answer using two significant figures.
=
Part B -
If the plates are separated by a Teflon sheet, how thick is the sheet?
Express your answer using two significant figures.
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html
To solve this problem, we need to use the formula for the capacitance of a parallel-plate capacitor:
C = (ε₀ * A) / d
where C is the capacitance, ε₀ is the permittivity of free space, A is the area of the plates, and d is the distance between the plates.
Given values:
C = 3.2 nF = 3.2 * 10^(-9) F
A = 0.25 m^2
Part A:
To find the distance between the plates when there's air between them, we can rearrange the formula as:
d = (ε₀ * A) / C
- First, we need to determine the value of ε₀, which is the permittivity of free space. It is a constant with a value of 8.85 x 10^(-12) F/m.
- Next, plug in the values into the formula:
d = (8.85 * 10^(-12) F/m * 0.25 m^2) / (3.2 * 10^(-9) F)
d ≈ 0.696 m
Therefore, the distance between the plates when there's air between them is approximately 0.696 meters.
Part B:
To find the thickness of the Teflon sheet when it is placed between the plates, we need to rearrange the formula again:
d = (ε * A) / C
- In this case, we need to find the value of ε, which is the permittivity of Teflon. The permittivity of Teflon is approximately 2.1 x 10^(-12) F/m.
- Plug in the values into the formula:
d = (2.1 * 10^(-12) F/m * 0.25 m^2) / (3.2 * 10^(-9) F)
d ≈ 0.0164 m
Therefore, the thickness of the Teflon sheet is approximately 0.0164 meters.