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Question about potential energy in capacitors

  1. Oct 24, 2013 #1
    Hello,

    I'm studying capacitors in my freshman physics class and I'm not quite following my book here.

    My book states that we can calculate the potential energy U stored within a capacitor by calculating the work done to charge it. Let q and v be the charge and the potential difference, respectively, at an intermediate stage during charging.

    To transfer an additional element of charge dq, the work dW required is given by

    dW = v dq

    I'm not quite seeing how they got the above equation though. I'm aware that the work from point a to b divided by a unit charge gives the difference in voltage, but I'm not seeing how they can get the above equation from that and I'm clueless as to how else they'd get it from.

    Any clarification would be much appreciated.
     
  2. jcsd
  3. Oct 24, 2013 #2

    meBigGuy

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    Intuitively you probably already understand the concept expressed in that equation. For a given increment of work you can move a given increment of charge with a particular voltage. If you increase the voltage, you can move less charge for the same amount of work.

    If you then move on to unit analysis, you can see that work is joules, charge is coulombs, and volts can be expressed in many different ways, one of which is joules/coulomb (http://en.wikipedia.org/wiki/Volt). so just given the units you get V =J/C which can also be written J = V*C which is equiv. to W = V*Q or dW = vdQ.

    Your description of work from a to b breaks down to V = J/C
     
  4. Oct 24, 2013 #3

    tiny-tim

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    Hello Mangoes! :smile:

    difference in energy = work done (∆E = W)

    voltage (= electric potential energy difference) is defined as energy per charge (V = ∆E/Q, or ∆E = QV)

    this assumes the charge, Q, is constant: if it isn't (as in a capacitor), we have to write a differential equation:
    for a small increase dQ in charge, the small increase in energy is dE = VdQ​
    (strictly, V is changing slightly, so it's between VdQ and (V+dV)dQ, but the difference is dVdQ, a double-differential which we ignore)
     
  5. Oct 24, 2013 #4

    meBigGuy

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    which is a much better explanation!
     
  6. Oct 24, 2013 #5
    Ah, the answer was right in front of my face. It follows from the definition of voltage.

    Thanks a lot for the help!
     
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