A capacitor is rolled from two strips of aluminum separated by two strips of paraffin-coated paper. Each strip is 5.40cm wide. The foil is 7.00μm thick, and the paper is 50.0μm thick. What length, in cm, should the strips be if a capacitance of 24.0nF is desired? Use 3.7 for the dielectric constant κ of the coated paper and 8.85 x 10-12 F/m for εo, the permittivity of free space.

To find the length of the strips for the desired capacitance, we can use the formula for the capacitance of a parallel plate capacitor:

C = (κ * ε₀ * A) / d

where C is the capacitance, κ is the dielectric constant, ε₀ is the permittivity of free space, A is the area of the capacitor plates, and d is the distance between the plates.

In this case, we know the capacitance (C = 24.0 nF), the dielectric constant of the coated paper (κ = 3.7), and the thicknesses of the aluminum foil (7.00 μm) and the paper (50.0 μm). We need to find the length of the strips, which corresponds to the area of the plates (A).

The width of each strip is given as 5.40 cm, and we need to find the length (L) in cm.

We can calculate the area of the plates as:

A = 2 * (width of strip) * length = 2 * (5.40 cm) * L

Substituting the given values into the formula for capacitance, we get:

24.0 nF = (3.7 * 8.85 x 10^(-12) F/m * (2 * 5.40 cm * L)) / (7.00 μm + 50.0 μm)

Let's simplify the equation step by step:

- Convert the capacitance to farads by multiplying by 10^(-9):
C = 24.0 nF = 24.0 x 10^(-9) F

- Convert the thicknesses of the aluminum foil and the paper to meters:
7.00 μm = 7.00 x 10^(-6) m
50.0 μm = 50.0 x 10^(-6) m

Now we can substitute the values into the equation:

24.0 x 10^(-9) = (3.7 * 8.85 x 10^(-12) * (2 * 5.40 x 10^(-2) * L)) / (7.00 x 10^(-6) + 50.0 x 10^(-6))

Now, to solve for L, let's isolate it on one side of the equation:

L = (24.0 x 10^(-9) * (7.00 x 10^(-6) + 50.0 x 10^(-6))) / (3.7 * 8.85 x 10^(-12) * (2 * 5.40 x 10^(-2)))

Calculating the expression on the right-hand side will give us the length L in centimeters.

To find the length of the strips required for a desired capacitance of 24.0nF, we can use the formula for the capacitance of a parallel plate capacitor:

C = (κ * ε0 * A) / d

Where:
C is the capacitance,
κ is the dielectric constant,
ε0 is the permittivity of free space,
A is the area of the capacitor plates,
and d is the distance between the plates.

Let's calculate step-by-step:

Step 1: Calculate the area of the plates.
The area of each plate is given by:
A = width * length

Given:
Width = 5.40cm

Let's assign "L" as the length of the strips.
So, the area of each plate is:
A = (5.40cm) * L

Step 2: Calculate the thickness of the capacitor.
The total thickness of each plate is given by:
d = foil thickness + paper thickness

Given:
Foil thickness = 7.00μm
Paper thickness = 50.0μm

So, the total thickness is:
d = 7.00μm + 50.0μm

Step 3: Calculate the value of κ:
Given:
κ (dielectric constant) = 3.7

Step 4: Convert units and calculate the length:
To use the formula, we need to convert the units to meters:
Foil thickness = 7.00μm = 7.00 x (10^-6)m
Paper thickness = 50.0μm = 50.0 x (10^-6)m

Using the formula for capacitance:
24.0nF = (3.7 * 8.85 x 10^-12 F/m * A) / (7.00 x 10^-6 m + 50.0 x 10^-6 m)

Simplifying the equation and substituting the known values:
24.0nF = (3.7 * 8.85 x 10^-12 F/m * 5.40cm * L) / (7.00 x 10^-6 m + 50.0 x 10^-6 m)

Now, let's solve for L:

L = (24.0nF * 7.00 x 10^-6 m + 50.0 x 10^-6 m) / (3.7 * 8.85 x 10^-12 F/m * 5.40cm)

Calculating this expression will give you the length, in cm, required for the desired capacitance of 24.0nF.