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This confuses me.

  1. Feb 16, 2012 #1
    Since the beginning, seeing the first definition of electric potential difference, ΔV = ΔU/q, I have had somewhat a harder time understanding it, or more so, utilizing it. While I have begun to understand it better now, I am still quite confused as to what it means in a circuit with an emf device.

    Take for example a simple single loop circuit with 1 battery with potential difference V and 1 resistor with resistance R. Assume that wires connecting the circuit are of negligible resistance. Why is it that the electric potential taken over any one side of the circuit is constant, or its difference is zero? That is, the potential difference has to be taken over the terminals or over the resistor to obtain any potential difference.

    The reason this doesn't make sense to me is that if ΔV = ΔU/q = -∫E ds and there is an electric field setup in the wire causing current flow, then as each electron flows, it surely does lose potential, even when there is no resistance because there is both electric field E and displacement ds.

    I find it easier to grasp this when thinking of a similar circuit but instead with a capacitor. Here it makes sense to me as once the capacitor is charged, there is no longer flow and the build up of charge on the plates of the capacitor causes a electric field over a distance of their separation which is indeed ΔV. When you have current flow, it just confuses me though.
     
  2. jcsd
  3. Feb 16, 2012 #2

    gneill

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    Staff: Mentor

    If the wires are truly without resistance then they are equipotential objects --- there would be no potential difference from end to end. And yet, current flows in such wires!

    Consider a mechanical analogy where objects are sliding on a horizontal frictionless surface. They continue to move even though no change in height is occurring. The surface may be at "ground" height, or high on a plateau (high potential) without affecting the constant motion.

    In a wire at some uniform potential where there is an 'entrance' for charge carriers at the same potential at one end and access to a lower potential 'exit' for charges at the other, then charges attracted to that lower potential leave a 'vacuum' that will be filled by other charges desiring to fill out the charge density in as uniform a manner as possible (like charges repel, so the charges fill up the available area like a uniform gas). When the average motion is established (after any initial impulse response has calmed down), then the wire maintains an essentially uniform potential from end to end and throughout, with a constant flow of charge carriers that doesn't require any potential difference to maintain it.
     
  4. Feb 16, 2012 #3
    So essentially what you are saying is that once current reaches some fixed uniform manor, its motion state is left unchanged there after and thus there is no net electric field and no potential difference?
     
  5. Feb 17, 2012 #4

    gneill

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    Staff: Mentor

    Yes, that's about it (for a zero resistance conductor).
     
  6. Feb 17, 2012 #5
    Ok, thanks for the help!
     
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