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What is Voltage?

  1. Apr 16, 2015 #1


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    I'm currently taking an EM course. We're doing AC circuits, but I'm having a hard time understanding how voltage is defined.

    In the electrostatic context I understand how the voltage is defined as the line integral of E over a path, and I get how that line integral has nice path-independent properties because E has no curl. But then our course introduces magnetism, and circuits. The sort of circuits we're looking at have inductors, or lots of self-inductance, so even magneto-statics goes out the window.

    When I ask about how voltage is defined in this more complicated context I get lots of answers like "voltage is still just the integral of E * dl", but I'm not happy with that because it's path-dependent, and therefore meaningless unless a convention is chosen for taking the path (which I've never seen done), or unless an argument is given for why that detail isn't important (which I've also never seen done).

    I also sometimes get answers like "voltage is just the difference in potential energy divided by charge," but I'm also not happy with that because people giving this answer uniformly fail to answer the follow-up question of "what is potential energy?" You can't define electric potential energy in usual way, for the reasons I just explained in the last paragraph. And anyway there are some choices of gauge for which the [itex]\phi[/itex] in [itex]\vec E = -\nabla\phi - d{\vec A}/dt[/itex] is declared by fiat to be [itex]\phi=0[/itex] everywhere and at all times (which has interesting consequences for how [itex]\vec A[/itex] has to be defined, but nonetheless renders the concept of the scalar potential null and void).

    Lastly, I get a lot of people saying that I'm overcomplicating things. But all I'm asking is for some definition of this pervasively used concept of "voltage" that doesn't have big gaping holes in it. I don't think that's too much to ask for, and I think it's worth nitpicking about. If there are assumptions or approximations going on that render my nitpicking extraneous, then I want to know exactly what those assumptions or approximations are.

    Also, to the admins: I wasn't sure where to post this question. If I'm in the wrong section then feel free to move me.
    Last edited: Apr 16, 2015
  2. jcsd
  3. Apr 16, 2015 #2


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    No, it's not path-dependent. Say you have a coil with some electric field induced in the turns. You may, during integration of the total voltage-difference over the coil, choose a path through the turns and get a result, or you may choose a path that jumps from one turn, through an airgap, to another turn (missing one or more turns). You will get the same result because in the airgap there will be an electric field (induced by the voltage-difference between one turn to the other), and when you integrate through this airgap, you will exactly get a voltage-difference that matches the voltage-difference integrated through the missing turns. So you may choose the path as you like, making loops, jumps, whatever: You will get the same result.

    ( I don't know if I have understood your confusion correctly here? )
    Last edited: Apr 16, 2015
  4. Apr 16, 2015 #3
    You are right that in AC circuits and generally circuits with time varying current the line integral of E depends on the path because CurlE is not zero, however you do the so called "quasi static" approximation that CurlE=0 , which essentially means that the time varying of E is solely due to a time varying scalar potential [itex]\phi[/itex].

    It should be clear how this approximation works in the case of a capacitor (where the time varying E between its plates is mainly due to the time varying charge accumulation in its plates which creates a time varying scalar potential, that is for frequencies of no more than a few Ghz).

    In the case of inductor, the time varying vector potential A and the associated non conservative electric field [itex]E_{\vec{A}}=-\frac{d\vec{A}}{dt}[/itex] initiate the process of creating surface charges in the surface of the wire that the inductor is made, which surface charge result in a conservative electric field [itex]E_{\phi}=-\nabla\phi[/itex] such that the total electric field [itex]E_{\phi}+E_{\vec{A}}\approx 0[/itex] is almost zero in the region inside the wire of the inductor. So you can take as E just the conservative constituent [itex]E_{\phi}[/itex] and define as [itex]V=\int E_{\phi}dr=-\int E_{\vec{A}}dr[/itex].
    Last edited: Apr 16, 2015
  5. Apr 16, 2015 #4


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