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Energy in capacitor

  1. Jun 27, 2008 #1
    If the charge on one side of a capacitor is +Q and on the other side it is -Q, what's the energy in the capacitor?

    I know that for one charge it is: (1/2)(Q^2/C)
    so for two charges is it just Q^2/C? or do I even need to take into account both charges?
  2. jcsd
  3. Jun 27, 2008 #2
    Hmm, I don't like just "giving" answers :-)

    But maybe a hint... I assume this is a parallel plate capacitor (just for imagery sake, doesn't necessarily have to be). If you have:



    What is the electric field between them? What work would be required to move against this electric field?
  4. Jun 27, 2008 #3
    From the definition of voltage as the energy per unit charge, one might expect that the energy stored on this ideal capacitor would be just QV. That is, all the work done on the charge in moving it from one plate to the other would appear as energy stored. But in fact, the expression above shows that just half of that work appears as energy stored in the capacitor. For a finite resistance, one can show that half of the energy supplied by the battery for the charging of the capacitor is dissipated as heat in the resistor, regardless of the size of the resistor.

  5. Jun 27, 2008 #4


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    Hi pooka,

    No, I don't believe that statement is correct. If you go back to the beginning of the capacitor discussion in your textbook (when they introduce the definition C=Q/V), what does it say about the variable Q, and how it is related to the charges on the plates? And what assumption do they make about the charges on the plates?
  6. Jun 27, 2008 #5


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    The formula is all right, the statement before the formula is what I was referring to. The words right before the formula in the original post ("I know that for one charge it is:") are not correct.

    That's why I suggested looking at the assumptions being made when that formula was derived, especially in regards to the charges on both plates. When they derived that formula, they did not assume that one plate had a charge and the other one did not. What did they assume?

    EDIT: I thought I was responding to the original poster, so had to change the wording.
    Last edited: Jun 27, 2008
  7. Jun 27, 2008 #6
    Oh I see! They assumed that both plate had charges so the equation actually takes into account both +Q and -Q, so the energy in the capacitor would not be Q^2/C.
  8. Jun 27, 2008 #7
    Okay thanks guys! I understand now.
  9. Jun 27, 2008 #8
    They were messing with you about what the true charge in the capacitor was by making them +Q and -Q relative to some zero, and then using the same symbol (Q) in the formula for energy in a capacitor.

    Try using lower case (+q) and (-q), so that the charge difference between the plates is Q=(+q) - (-q) =2q

    .. I think :)
  10. Jun 27, 2008 #9


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    Hi GTrax,

    That does not look right to me. If one plate has 15 microCoulombs of charge, the other will have -15 microCoulombs. Then the Q in the equation needs to be 15 microCoulombs (not 30 microCoulombs like your last equation suggests.)

    Q is the magnitude of the charge on each of the plates.
  11. Jun 27, 2008 #10
    Sure - I agree - and I can get confused about this too.

    In my first scenario, I imagined one plate forcibly grounded, its potential being zero, and we do the work add 15 microcoulombs of electrons to the other plate, giving it a negative voltage. I am thinking the charge in that capacitor is 15 microcoulombs, and 15 uC worth of electrons would have to be moved to ground to discharge it.

    In my second scenario, I imagined to start with the capacitor plates isolated
    You can do work to force 15 microcoulombs of negative charge (electrons) onto one, relative to the zero volts earth. You can do more work to extract another 15 microcoulmbs from the other plate, (depleting it of electrons), making it have a positive charge.

    Each plate will have a voltage, say +v2 and -v1, relative to earth, the value depending on the self-capacitance of the plate body. The voltage difference between the plates v2 - (-v1) will depend on the plate spacing, and the dielectric between them. It took two loads of work to get them charged!

    This is hard to reconcile when one realizes only 15 microcoulombs of electrons need to be moved between the plates to leave the capacitor discharged. The initial putting of the electrons on one plate could have been done by removing them from the other, thus arriving at the starting scenario with only one load of work.

    Sorry for the temporary misunderstanding
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