Problem 21.79


The plates of a 3.2 nF 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.

To find the answers to Part A and Part B, we need to use the formula for capacitance:

C = ε₀ * (A / d)

Where:
C is the capacitance
ε₀ is the permittivity of free space
A is the area of the plates
d is the distance between the plates

Let's start by solving Part A:

Given:
Capacitance, C = 3.2 nF (nanofarads)
Area, A = 0.25 m^2

We know that the permittivity of free space, ε₀, is approximately 8.85 x 10^-12 F/m.

Rearranging the formula, we get:

d = A / (C * ε₀)

Plugging in the values:

d = 0.25 m^2 / (3.2 nF * 8.85 x 10^-12 F/m)

To get the result in meters between the plates, we need to convert 3.2 nF to F by multiplying by 10^-9:

d = 0.25 m^2 / (3.2 x 10^-9 F * 8.85 x 10^-12 F/m)

Simplifying:

d = 0.25 m^2 / (2.82 x 10^-20 F^2/m)

d = (0.25 / 2.82) m

d ≈ 0.089 m

Therefore, the distance between the plates with air between them (Part A) is approximately 0.089 meters.

Now, let's move on to Part B:

Given:
Capacitance, C = 3.2 nF (nanofarads)
Area, A = 0.25 m^2

This time, the plates are separated by a Teflon sheet. We need to find the thickness of the sheet (d).

We can use the same formula as before:

d = A / (C * ε₀)

Plugging in the values:

d = 0.25 m^2 / (3.2 nF * 8.85 x 10^-12 F/m)

Again, converting 3.2 nF to F:

d = 0.25 m^2 / (3.2 x 10^-9 F * 8.85 x 10^-12 F/m)

Simplifying:

d = 0.25 m^2 / (2.82 x 10^-20 F^2/m)

d = (0.25 / 2.82) m

d ≈ 0.089 m

Therefore, the thickness of the Teflon sheet (Part B) is also approximately 0.089 meters.

Bear in mind that the value of d in Part B is the same as the value obtained in Part A because the thickness of the Teflon sheet is considered as part of the distance between the plates.