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Electric field and current in a conductor

  1. Oct 2, 2008 #1
    [My understanding is that in a conductor which is conducting a current there is and electric field present also. If the electric field can be found by E= - [tex]\frac{dV}{dl}[/tex], what is E in a conductor. On a conductor in a DC circuit the potential is the same everwhere on that conductor, so I beleive - [tex]\frac{dV}{dl}[/tex] would be 0? So is the equation E= - [tex]\frac{dV}{dl}[/tex] only valid for noncunductors?

    I'm only in my second month of Phys II so I'm still pretty new to this concept. If anybody can clear this up I'd really appreciate it.
     
  2. jcsd
  3. Oct 3, 2008 #2

    Doc Al

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    No. The potential is the same throughout a conductor in electrostatic equilibrium, not when there's a current through it as in a DC circuit.
     
  4. Oct 4, 2008 #3
    Ok, so I guess thats where I'm getting confused.
    If you look at the .jpg example circuit I've attached I can explain my question (its done with MS paint so its not great):

    Anyway, you can see it represents a DC circuit with voltage(potential) readings taken at several points. At points A,B, and C along the conductor there is no difference in voltage. So does E= -[tex]\frac{dV}{dl}[/tex] =0.
    I guess it doesn't, but then my question is how is there a change in voltage(potential) along the conductor, but my voltage readings show no difference.

    I know I'm missing something in my idea of potential, electric fields, current, etc but I'm not sure what it is.
    Thanks to anyone who can help enlighten me. :smile:
     

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  5. Oct 4, 2008 #4

    Doc Al

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    Ah, my bad. Now I see your point. Since the conductors have almost no resistance, they require only a very small electric field to drive the current through them. Compared to the resistors, they have zero voltage drop across them.

    Sorry for the confusion.
     
  6. Oct 4, 2008 #5

    atyy

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    Yes, there should be no potential difference between points A and C. In circuit theory, a conductor is always "super" with no resistance, so current can flow through it without a potential difference. There is a potential difference only when current flows across a "circuit element" like a resistor, capacitor or inductor.

    This is just a simplification to make life easier. In reality, copper wires are not "super" conductors with no resistance, they just have a very small resistance. Even in some cases where the wire has significant resistance distributed along its entire length, it may be a good approximation to model it as a combination of a "super" conductor and a resistor in series.

    Edit: I see Doc Al already posted an answer that is the same as mine, but shorter and sweeter.
     
  7. Oct 6, 2008 #6
    Thanks guys, that helps clear things up quite a bit.

    If you don't mind me beating a dead horse though, how does current flow thorugh a perfect conductor, like a superconductor with zero resistance? With zero resistance, there is no difference in potential and therefor no E field, correct? So what causes the current to flow?

    I'm still really new to this conceptual physics so I'm trying to sqare it away with what I already know, or what think I know about electronics. I'm not even halfway through the semester and I've already had to discard many of the assumtions i've made based on several years as an electronics tech. I'm glad theres a place like this to get answers.
     
  8. Oct 6, 2008 #7

    f95toli

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    Even with superconductors there is a potential difference somewhere in the circuit, remember that the superconductor is connected to "something" and that something has a resistance and all the voltage drop is across that resistance.
    I.e. while you can't really voltage bias a superconducting circuit there is nothing preventing you from using current bias.

    If you wan't to understand what goes on in the superconductor itself you can't use the usual equations, they are simply not valid. However, you can modify the usual equations slightly and use what is known as London theory (the London equations) to quite accurately predict what happens in superconducting circuits at the macroscopic level.
     
  9. Oct 6, 2008 #8

    atyy

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    Just to make sure, the "super" conductor I talked about only exists in circuit theory as an excellent approximation for a real conductor like copper wire which has small, not zero resistance. f95toli is talking about superconductors which exist in real life, really have zero resistance, and which are not made of copper wire.
     
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