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Electric potential energy

  1. Nov 29, 2008 #1
    all my question is listed in the file below

    thanks for your help
     

    Attached Files:

  2. jcsd
  3. Nov 29, 2008 #2

    Doc Al

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    Staff: Mentor

    1/2 QV is the energy stored in the capacitor. The other half of the energy was dissipated during charging as heat or radiation.
     
  4. Nov 30, 2008 #3

    clem

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    Q and V for a capacitor are proportional to each other by Q=CV so the simple E=QV does not apply. The change in energy is dU=Vdq=CVdV.
    Then [tex]U=\int_0^VCV'dV'=(1/2)CV^2=(1/2)QV.[/tex]
     
    Last edited: Nov 30, 2008
  5. Nov 30, 2008 #4

    Doc Al

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    Clem is, of course, correct. Perhaps I misinterpreted your question (or just answered a different question!) about charging a capacitor with a battery. The energy imparted to the charge by the battery is QV, but the energy stored in the capacitor only half that. Sorry about that!
     
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