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Energy stored in a capacitor works

  1. Apr 26, 2009 #1
    Im having trouble understanding how capacitors store energy. If a potential difference is applied to two parallel plates each with +-Q, it will set up polarisation charges on the dieletric between, which reduces the electric field inside the material. The potential difference between the plates is reduced and hence capacitance is increased.

    But in terms of the electric field how does it store the energy?
     
  2. jcsd
  3. Apr 26, 2009 #2

    tiny-tim

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    Hi hhhmortal! :smile:
    I don't see the problem…

    a capacitor stores energy because opposite charges are on opposite plates of the capacitor …

    they want to get together, but the absence of any connection stops them …

    that's potential energy, because if you join the plates (either the quick way, or more usually the long way round via a circuit), the charges will move, and will do work in doing so. :wink:
     
  4. Apr 26, 2009 #3
    I also have had trouble with detailed descriptions dielectrics in capacitors, but then I am not a materials person, and so it doesn't matter much to me.

    I will try to repeat the information of the previous reply in a different way.

    An ideal capacitor has no real materials, neither for the plates, nor the dielectric. One may make a "real" capacitor with two plates, perhaps using a metal, but even in this case there is nothing wrong with the dielectric material being absent. In this case, the dielectric is the vacuum. Notice that the dielectric constant of any "real" dielectric (capacitor) is always a ratio of its permittivity compared to the vacuum.

    Now, the vacuum is nothing, and so it has no polarization charges. The only charges "in the capacitor" are "on the plates". The entire electric field may be considered simply between these plates without further distraction.

    If you should find enlightenment regarding how the polarization charges in a non-ideal dielectric contribute to its relative permittivity (an easy answer, that is), I am curious.
     
  5. Apr 28, 2009 #4
    Ok yea. The part which I don't understand is if you the dielectric constant is greater inbetween the plates, the capacitance increases. so by looking at equation:

    CV = Q

    Does this mean V between the plates must stay constant at all times hence the charges increase and thus amount of energy that the capacitor stores increases?
     
  6. Apr 28, 2009 #5

    tiny-tim

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    i] If you use the same battery (so the voltage, V stays the same) to charge two capacitors, identical except for the dielectric, then the amount of charge changes, and so does the energy stored

    ii] but if you disconnect the battery, with the capacitor charged, and move a dielectric in between the plates, then obviously the charge stays the same, and the potential difference (the voltage) changes … since that also changes the energy, I assume you have to do some work to force the dielectric in there! :smile:
     
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