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Capacitance, Charge and Voltage (P.D.)

  1. Sep 5, 2013 #1
    1. One question I'm having a lot of trouble solving is this: The plates of a parallel plate air capacitor have an area of 100cm2. The capacitor has a capacitance of 3.5nF and is connected to a 12V battery. It is fully charged. Calculate the:

    a) separation of the plates
    b) magnitude of electric field intensity between the plates
    c) the charge stored on the capacitor plates

    ε0 is equal to 8.854x10-12/m

    2. I used C = (ε0εr (1 in this case)A)/(d) and E = V/d and Q = CV

    3. The area is 0.01m2, and using the first equation, along with the given capacitance, I found the distance to be 2.53x10-5m. Using the second equation I found the magnitude to be 474308N. Then I used the last equation to find the charge to be 4.2x10-8.

    Now, the real difficulty... the circuit is broken, and an insulating material filling the whole distance between the two plates is inserted. It has a dielectric constant of 3.5. This material covers the same area of the plates, plus a border of 2cm.

    This border doesn't affect the area, since the first formula deals with the common area, in this case 0.01m2. I was asked for the new capacitance, for which I multiplied the one I was given (3.5nF) by the dielectric constant - for a new capacitance of 1.22x10-8. So far, so good.

    However, my stumbling block was when I was asked to find the change in the potential difference between the two plates. How do I find that?

    Thanks in advance, any help is extremely appreciated!
  2. jcsd
  3. Sep 5, 2013 #2
  4. Sep 5, 2013 #3


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    Staff: Mentor

    Incomplete units.
    Except for your units. You WILL lose marks if your units are not correct or not given.
    What capacitor formula do you have that relates capacitance, charge, and potential difference? What do you know about the charge on the "new" capacitor?
  5. Sep 5, 2013 #4
    So the charge doesn't change between the two parts? Hence I can use Q/C = V? If yes, then the voltage is 3.44V.

    Also, sorry about the units. I had been going at this question for an hour, and I was extremely frustrated. Thanks a lot guys! I really appreciate your help!
  6. Sep 5, 2013 #5


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    Staff: Mentor

    Yup. The charge can't change if the capacitor is not connected to an external circuit.
    Glad to help!
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