# Calculate Capacitance of Flat Plate Capacitor

• phys-lexic
In summary, the problem is to find the capacitance of a parallel plate capacitor using two flat plates of aluminum, a sheet of paper, a sheet of glass, and a slab of paraffin, with given dimensions and dielectric constants. The calculated capacitance values for each material are 5.885E-7 farads for paper, 5.885E-8 farads for glass, and 3.363E-9 farads for paraffin. The question also asks for the largest and smallest capacitance values in nF and pF units, with the smallest being 3363 pF for paraffin and the largest being 588.5 nF for paper. The formula used to calculate capacit
phys-lexic
the problem is:

To make a parallel plate capacitor, you have available two flat plates of aluminum (area = 190 cm2), a sheet of paper (thickness = 0.10 mm, κ = 3.5), a sheet of glass (thickness = 2.0 mm, κ. = 7.0), and a slab of paraffin (thickness = 10.0 mm, κ = 2.0). Find each's capacitance.

my answers are:

5.885E-7 farads for paper
5.885E-8 farads for glass
3.363E-9 farads for paraffin

...now the question asks to enter the largest capacitance possible, of the three, in nF and the smallest in pF...

-smallest= 3363 pF (paraffin)
-largest= 588.5 nF (paper)

am i right with my values and just converting wrong or am i totally off... i converted all the values to meters before plugging them into the equation below:

*****here is the equation i am using*****

(capacitance) = ((epsilon not)x(A)x(k)) / (d=thickness?)

where epsilon not is 8.85E-12, A is area of plates, k is the dielectric constant, and d is the distance between the plates (thickness).

Yes, your values and conversions are correct. The equation you are using is the correct formula for calculating capacitance for a parallel plate capacitor. The only thing I would suggest is to use the correct units for capacitance, which is Farads (F), instead of the scientific notation you used (E-7 and E-8). Other than that, your calculations are correct.

## 1. How do you calculate the capacitance of a flat plate capacitor?

The capacitance of a flat plate capacitor can be calculated by using the formula C = εA/d, where C is the capacitance, ε is the permittivity of the material between the plates, A is the area of the plates, and d is the distance between the plates.

## 2. What is the unit of capacitance?

The unit of capacitance is farad (F), named after the English physicist Michael Faraday.

## 3. What factors affect the capacitance of a flat plate capacitor?

The capacitance of a flat plate capacitor is affected by the area of the plates, the distance between the plates, and the dielectric material between the plates. It is also affected by the number of plates and the material used for the plates.

## 4. Can the capacitance of a flat plate capacitor be increased?

Yes, the capacitance of a flat plate capacitor can be increased by increasing the area of the plates, decreasing the distance between the plates, or using a material with a higher permittivity between the plates.

## 5. What is the use of calculating the capacitance of a flat plate capacitor?

Calculating the capacitance of a flat plate capacitor is important in designing and understanding electronic circuits. It helps in determining the amount of charge that can be stored in the capacitor and its ability to store and release energy. It is also useful in calculating the time constant of a circuit and determining the frequency response of a circuit.

### Similar threads

• Introductory Physics Homework Help
Replies
5
Views
8K
• Introductory Physics Homework Help
Replies
11
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
2K
• Introductory Physics Homework Help
Replies
11
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
7K
• Introductory Physics Homework Help
Replies
2
Views
3K
• Electromagnetism
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
10
Views
2K