Capacitance of two parallel plate capacitors

In summary, you found the capacitance of the parallel plate capacitor to be 8.85 x 10-11 Farads by taking the area of each plate and multiplying by the distance between the plates.
  • #1
Violagirl
114
0

Homework Statement



Two square metal plates with sides of length 0.1 m are separated in vacuum by 10-3 m. Find their capacitance.

Homework Equations



C = Q/V

For a parallel plate capacitor:

C = E0A/L


The Attempt at a Solution



If A is supposed to be the length of the metal plates with 0.1 m, since A has units of m2, I took A = (.1m)2 = .01 m2 and L = 10-3 m.

From there and plugging everything in, I get:

C = (8.85 x 10-12 C2/N m2 * .01m2/10-3 m

C = 8.85 x 10-11 Farads

If someone could confirm whether this is correct or not, I would appreciate it.
 
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  • #2
Violagirl said:
If A is supposed to be the length of the metal plates with 0.1 m, since A has units of m2, I took A = (.1m)2 = .01 m2 and L = 10-3 m.
A is the area of each plate. (The area of one face of a plate.)
 
  • #3
Right about A. What I meant was, of the two numerical values that I'm provided, which would be my A? Or how do I determine my A? I know then that my separation distance corresponds to L. So do I need to determine A by my length values in this situation?
 
  • #4
A is the area of the square plate. It's 0.1m per side, so A = (0.1m)² as you showed.
 
  • #5
Violagirl said:

Homework Statement



Two square metal plates with sides of length 0.1 m
This is the dimension of plate.
are separated in vacuum by 10-3 m. Find their capacitance.

This is the distance between the plate
Do not get confused by them. Dimensions of the plate has nothing to do with distance between the plates.

(Also the title is wrong there's only one parallel plate capacitor here, two plates make a capacitor.)
 
  • #6
Enigman said:
This is the dimension of plate.


This is the distance between the plate
Do not get confused by them. Dimensions of the plate has nothing to do with distance between the plates.

(Also the title is wrong there's only one parallel plate capacitor here, two plates make a capacitor.)

Oh shoot! I did not realize that. Do I need to redo this problem then and look at other equations to determine capacitance then?
 
  • #7
Er...no everything is correct; just show how you found area by the dimensions given.
 
  • #8
Oh got it! Will do, thanks so much for looking at this!
 

1. What is capacitance?

Capacitance is a measure of the ability of a capacitor to store electric charge. It is represented by the symbol C and is measured in farads (F).

2. How is capacitance of two parallel plate capacitors calculated?

The capacitance of two parallel plate capacitors can be calculated using the formula C = εA/d, where 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.

3. What factors affect the capacitance of two parallel plate capacitors?

The capacitance of two parallel plate capacitors is affected by the area of the plates, the distance between the plates, and the type of material between the plates (permittivity). It is also affected by the dielectric constant of the material between the plates.

4. How does the distance between the plates affect capacitance?

The capacitance of two parallel plate capacitors is inversely proportional to the distance between the plates. This means that as the distance between the plates increases, the capacitance decreases, and vice versa.

5. What is the significance of capacitance in electronic circuits?

Capacitance plays a crucial role in electronic circuits as it allows for the storage and release of electrical energy. It is used in a variety of applications such as filters, oscillators, and memory devices.

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