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Capacitance of two parallel plate capacitors

  1. Oct 2, 2013 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations

    C = Q/V

    For a parallel plate capacitor:

    C = E0A/L


    3. 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.
     
  2. jcsd
  3. Oct 2, 2013 #2

    NascentOxygen

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    Staff: Mentor

    A is the area of each plate. (The area of one face of a plate.)
     
  4. Oct 2, 2013 #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?
     
  5. Oct 2, 2013 #4

    NascentOxygen

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    Staff: Mentor

    A is the area of the square plate. It's 0.1m per side, so A = (0.1m)² as you showed.
     
  6. Oct 2, 2013 #5
    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.)
     
  7. Oct 2, 2013 #6
    Oh shoot! I did not realize that. Do I need to redo this problem then and look at other equations to determine capacitance then?
     
  8. Oct 2, 2013 #7
    Er...no everything is correct; just show how you found area by the dimensions given.
     
  9. Oct 2, 2013 #8
    Oh got it! Will do, thanks so much for looking at this!
     
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