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Current density in a conductor at DC

  1. May 30, 2013 #1
    Why is there a uniform current density in a conductor at DC. It’s my understanding that generally in a conductor, a sphere for instance, in a static situation, all charge is at the surface of the conductor because it the charges repel each other. So in a cylindrical conductive wire even though there is an electric field that is causing the charges to flow longitudinally with respect to the wire, why don't the electric fields of the charges themselves repel each other and cause the current to be confined to the surface of the conductor?
     
  2. jcsd
  3. May 30, 2013 #2
    A conductor like copper has a large amount of movable negative charge carriers (electrons) and nearly the same amount of positive charges (copper ions). Any excess or deficit of electrons will be at the surface, but the material will always be filled a much larger amount of electrons and copper ions, and this will result in a current wherever there's an electric field.
     
  4. May 30, 2013 #3

    Jano L.

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    Very good question.

    The short answer may be that in electrostatics, only the superfluous charges get to the surface. The negative electrons that cancel the positive nuclei are still inside the metal. They form negative charge density, which is cancelled by the positive charge density of the nuclei. In statics, these densities are uniform.

    These distributed electrons later form the DC current. The electrons do repel each other, but this is balanced by the positive nuclei, so the electrons are not pushed out to the surface, but can stay distributed in the volume.
     
  5. Jun 3, 2013 #4
    Thanks for the explaination. That brings up another question though. In Electromagnetics we often used the concept of a line of charge to find an electric field and to derrive the E&M field in a cable. To find the E&M fields in the cable it seems we treated the positive conductor as positive line of charge and the negitive cable as a negitive line of charge. I'd assumed at the time that a wire with a current would appear as a line of charge, but thinking about your guys' explaination about current density this doesn't seem to be the case. I gather from your explainations that there is an equal positive and negitive charge in the conductors, so as to be a net zero charge. So as viewed from outside of a wire, no matter what the current flowing through it, it would not appear as a line of charge? Is that correct?
     
  6. Jun 3, 2013 #5

    Jano L.

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    Was your example about the coaxial cable?

    Anyway, when DC current flows through wire, the wire will be charged on its surface, in order to produce electric field inside the wire that pushes the charges inside. The charge density will not be uniform though, but will decrease as one advances from the end of higher potential to the end of lower potential.
     
  7. Jun 4, 2013 #6
    In my EM book, Wentorth, he used the example of a transmission line. The book states "...a test charge placed a couple of centimeters from an elevated transmission line will see what appears to be an infinite length line (of charge)." I'm confused why this would appear to be a line of charge if there is an equal amount of positive and negative charges inside the transmission line or conductor. It would seem to me that it would appear as an infinite line of current.

    Also i was a bit confused about your explanation about how the current density at the surface of the conductor would be greater. By surface do you mean where the potential is applied?

    I've attached a diagram similar to one in my EM text book. It depicts a conductor with a voltage applied to both ends. According to the book the E field in the conductor is E = V[itex]_{ab}[/itex]/L . Because current density, J = σE , I would expect the current density to be constant through the length of the conductor, which I would expect from circuit theory as well.

    Sorry for the lengthy questions. I took EM quite a while ago. It seems like I spent a lot of time doing vector calculus, and just now trying to ferret out what it all means.
     

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