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Homework Help: Parallel Plate Capacitor and Battery

  1. May 30, 2014 #1
    1. The problem statement, all variables and given/known data
    Two parallel plates, each having area A = 3558 cm2 are connected to the terminals of a battery of voltage Vb = 6 V as shown. The plates are separated by a distance d = 0.54 cm.

    1) What is Q the charge on the top plate? 3.5*10^-9 C
    2) What is U, the energy stored in this capacitor? 1.05 * 10^-8 J
    3) The battery is now disconnected from the plates and the separation of the plates is doubled ( = 1.08 cm). What is the energy stored in this new capacitor?
    4) What is E, the magnitude of the electric field in the region between plates?
    5) Compare V, the magnitude of the new potential difference across the plates to Vb, the voltage of the battery. V > Vb
    6)Two uncharged parallel plates are now connected to the initial pair of plates as shown. How will the electric field, E, and potential difference across the plates, V, change, if at all? Both E and V will decrease?

    2. Relevant equations
    U = (1/2)*Epsilon-naught*E^2

    3. The attempt at a solution
    So the problems I'm having are 3 and 4. For three, even if the battery is disconnected, the voltage will remain the same as well as the charges, and therefore the capacitance. Once the distance is doubled, the charge will remain the same, and the capacitance, meaning the voltage needs to be doubled to maintain C and Q at the same values. Therefore, the new U should equal (1/2)*Epsilon-naught*E^2 = (1/2)*Epsilon-naught*(V/d)^2

    Plugging in V=12volts, d= 0.0054m and the standard 8.854*10^-12, I get 2.18*10^-7J which is incorrect.

    As for question 5, wouldn't doing 12V/0.0054m give E? It isn't obviously, however I'm having trouble understanding why.

    Thank you
  2. jcsd
  3. May 30, 2014 #2
    You're using the wrong formula for U. There is a formula that relates U, Q, and V. Do you know it? How did you answer question 2?

    For question 5. Didn't the distance double to 1.08 cm?
  4. May 31, 2014 #3
    Thanks for the answer,

    Yes it did double to 1.08 cm. It was a typo on my part.

    In terms of question 2, I really have no idea. I played around with some equations and subbed them into each other and ended up with the correct answer. I can have a quick look at my notes when I get back home, however I can't recally the formula for U off the top of my head at the moment. Can I get a refresher?
  5. May 31, 2014 #4


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    Homework Helper

    Your formula for U is the energy density, energy per unit volume. You need the whole energy of the capacitor.

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