1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Parallel plate air capacitor

  1. Feb 11, 2012 #1
    A parallel-plate air capacitor is made by using two plates 16 cm square, spaced 3.7 mm apart. It is connected to a 12-V battery. What is the capacitance? What is the charge on each plate? What is the electric field between the plates? What is the energy stored in the capacitor?



    What I believe to be relevant equations:
    C=KC_0=Kε_0*A/d
    Q=CV
    E=Q/(ε_0*A)
    U=.5QV=.5CV^2
    K for air is1.00059




    3. The attempt at a solution:
    Really I'm just stuck on the first part. I tried C=KC_0=Kε_0*A/d, and put in C=(1.00059*8.854*10^-12*.0016/.0037)
     
  2. jcsd
  3. Feb 11, 2012 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Yes, evaluate.

    ehild
     
  4. Feb 11, 2012 #3

    gneill

    User Avatar

    Staff: Mentor

    Be careful with the area of the plates, particularly in converting to square meters.

    Is "16 cm square" the same as 16 cm2, or does it mean (16 cm) x (16 cm)?
     
  5. Feb 11, 2012 #4
    That's the exact question word-for-word, so I can't tell either. The answer I got from doing the calculation was 3.83pF, which is wrong.
     
  6. Feb 11, 2012 #5

    gneill

    User Avatar

    Staff: Mentor

    Suggest you try the other interpretation.
     
  7. Feb 11, 2012 #6
    The other option also resulted in a wrong answer. 6.13*10^-11F using an area of (.16m)^2. I've 3 tries left.
     
  8. Feb 11, 2012 #7

    gneill

    User Avatar

    Staff: Mentor

    Does the system want the answer in any particular units? Significant figures?
     
  9. Feb 11, 2012 #8
    It asks for Farads, and to put everything as XXX*10^Z. I have done so. I've used Mastering Physics all last year, so I know how to deal with the inputs.

    Is there some special integral or another equation that I could try?

    P.S. The way I'm doing the problem is exactly like the example in the book.
     
  10. Feb 11, 2012 #9

    gneill

    User Avatar

    Staff: Mentor

    The equation is correct, and it appears to me that you've obtained a correct answer. If the method of data entry is also correct then it is possible that the machine's answer key is wrong. This has been known to happen from time to time. You should report it to your course instructor.

    You might want to see if the capacitance value obtained returns correct results for other parts of the problem.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook