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Energy of Capacitor

  1. Sep 13, 2005 #1
    My physics class is not a calculus based class; therefore, the formulas you see here will not be calculus. My problem states:

    A parallel-plate capacitor has 3.56 cm^2 plates that are separated by 6.08 mm with air between them. The permittivity of a vacuum is 8.85419e-12 C^2/Nm^2. If a 10.4 V battery is connected to this capacitor, how much energy dies it store. Answer in units of J.

    Capacitance between parallel plates:
    C = (epsilon * A)/d (epsilon is the permittivity of a vacuum.)

    Once you have capacitance, you can figure the stored energy using:
    U = 1/2CV^2

    A = 3.56 cm^2 = 3.56e-4 m^2
    d = 6.08 mm = .00608 m
    Epsilon = 8.85419e-12 C^2/Nm^2
    V = 10.4

    Plugging the above numbers into the formula:
    U = (1/2)((epsilon * A)/d)(V^2)

    I get 9.9811e-15 J, but the answer is wrong. Am I missing something here or just flat out working the problem wrong?
  2. jcsd
  3. Sep 14, 2005 #2


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    Homework Helper

    I get 2.80e-11 J. Everything you said looks correct except for the final answer, so you probably typed an error into your calculator.
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