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Energy stored on a capacitor

  1. Mar 23, 2010 #1
    Hello,

    I don't understand how my general physics textbook describes how energy is stored in a capacitor.

    Can someone please explain this? One of the details I don't understand is who comes first? Is it the potential difference that creates charge accumulation,or is it charge accumulation that creates potential difference?

    Once this charge is stored as "potential energy" what actually happens to it?

    I guess once the capacitor is discharged it is turned into kinetic energy? What about before the capacitor is discharged? Charge will hang around on the plates? :P

    How is a capacitor discharged?

    Many thanks.
     
  2. jcsd
  3. Mar 23, 2010 #2
    There is a pretty thorough discussion here:
    http://en.wikipedia.org/wiki/Capacitor

    A charged capacitor has electrical potential energy stored in its electric field equal to the work required to charge it. This energy can be recovered when the capacitor is allowed to discharge...by letting the stored electrons circulate in a complete circuit. You can touch a wire to the charged plates and see a spark...with almost no resistance, the electrons run around almost instantaneously in a sudden burst. The charge can also heat a resistor (heat energy) or turn an electric motor (rotational mechanical energy) for example. Electricity is neat because it can be used for a number of different energy applications.

    To charge a capacitor, you can imagne an external agent pulling electrons from the positive plate and pushing them onto the negative plate....an external battery can do this, for example. It takes work (energy) to move all the electrons around and to push them into proximity on the negative plate...that takes voltage (electrical potential). You need potential (V) and charge (q) simultaneously and can see this from the relationship q = CV........and note that q = it...so a current i (flow of electrons) must generally occur over time (t) while a potential pushes electrons around....


    It turns out the work required is W = 1/2CV2
     
  4. Mar 23, 2010 #3
    Thank you for trying to help.

    I still don't understand how the plates are charged. I also don't understand if this charge creates or requires a potential energy?
     
  5. Mar 23, 2010 #4
    Ok well I was reading through another textbook (Halliday Resnick Walker) and things are a little clearer now. This is part of what I was looking for:
    startquote:
    Although the total charge on the capacitor is zero (because there is as much excess positive charge on one conductor as there is excess negative charge on the other), it is common practice to refer to the magnitude of the charge on either conductor as “the charge on the capacitor.” endquote

    It would follow that the charge on the conductors create potential energy because of their position relative to each other. Am I correct?
     
  6. Mar 23, 2010 #5
    yes, the two capacitor plates will face each other in close proximity so that the positive and negative charges are attracted to each other. However, between the plates exists something nonconductive (maybe air) so the charges aren't able to get from plate A to plate B. That doesn't stop the charges from being attracted to each other though. It is this attraction that allows the charge to be stored on the plates. As you would expect the closer the plates are to each other the stronger the potential will be. To release the energy simply connect the two plates with a conductor (wire, etc) and the opposite charges will rush to meet up with each other.
     
  7. Mar 23, 2010 #6
    You will need to do some thinking in addition to just reading....q = CV = it....think about what those simple expressions represent...can you explain them to yourself???

    thats why the electron deficit on the positive plate equals the electron surplus on the other plate....

    You should should look up terms like current flow, electron mobility, and similar, like electron drift speed....

    it takes some time when first exposed to all this because there is a lot going on...later it will become quite obvious....be patient and stick with it...
     
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