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Parllel Plate Capcitor separation distance and voltage

  1. Dec 8, 2009 #1

    I'm studying for an exam tomorrow and am facing a question I can't for the life of me fathom.

    " A PP air capacitor is made by using to plates 16 cm squared, spaced 4.7 mm apart, connected to a 12V battery"

    Find Capacitance, charge on each plate, electric field between plated, energy stored in the capacitor. If the battery is disconnected and then the plated pulled apart to a separation of 9.4 mm repeat answers.

    I have no problem solving the first round of questions but when repeating with the new distance as expected capacitance halves.

    using C=Q/V i try and figure out Q (Q=CV)
    mathematically i expect Q to halve since Capacitance has also halved
    physically i know theres been no change in charge
    Since it is connected to a battery i can't justify in my mind the voltage increasing to 24V in order to harmonize my physical answer and my mathematical answer...
    I think im making mountains out of mole hills... i guess exam season will sometimes throw logic out the window.

    help please?
  2. jcsd
  3. Dec 8, 2009 #2


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    Gold Member

    If you disconnect the battery and keep the parallel plates unconnected, then the one thing you can be certain is that the charge on the plates stay the same. This is because there is no longer any circuit path for charge to migrate between the plates (unless the dielectric between the plates breaksdown but that is not a situation we usually consider and certainly won't happen here since separating the plates a larger distance can only decrease the electric field).

    So rework your equations under assumptions that you know the area, distance of separation, and charge on the plates.
  4. Dec 9, 2009 #3
    Note that pulling apart the plates adds energy since they are oppositely charged and attract each other. Considering work helps solve many physics problems.
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