Answer:
a. 8 sheets of paper is needed between her plates to get the proper capacitance
b. Area of Aluminum Foil needed = 0.45m²
c. To keep a 1.0-nF, a larger area of Teflon is required.
Explanation:
a.
First, we need to calculate the distance between two plates.
This is given by
d = Kε0A/C
Where
K = 3
ε0 = Physical Constant = 8.854 * 10^-12 C²/Nm²
A = Area = 22 * 28 = 616cm² = 0.0616m²
C = 1.0-nF = 1 * 10^-12F
So, d = (3 * 8.854 * 10^-12 C²/Nm² * 0.0616) / (1 * 10^-12F)
d = 1.64 * 10^-3m
d = 1.64mm
Now, that the distance has been solved.
The Number of Sheets, N is given by
N = d/d,sheet where d, sheet =the sheet thickness = 0.2mm
N = 1.64/0.2
N = 8.2
N = 8 sheets --- Approximated
b.
Here, she's changed the diameter of the sheets to 12mm
Well make use of the formula in (a) above
Using d = Kε0A/C
Where
d = 12 * 10^-3m
Other constraints remain unchanged
Make A the subject of formula
A = dC/Kε0
A = (12 * 10^-3m * 1 * 10^-12F)/(3 * 8.854 * 10^-12 C²/Nm²)
A= 0.45m²
c. From (b) above
A ∝ 1/K
As the dielectric constant increase, the area decreases
The dielectric constant of a Teflon is 2.1
This means that if she used a Teflon instead, the area will be larger.
So, to keep a 1.0-nF, a larger area of Teflon is required.
