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Poynting vector in wire and capacitor

  1. Sep 15, 2010 #1

    For the scenario of wire or capacitor, the poynting vector is found to point inward radially, which suggests the energy is flowing into the wire from the air. Isn't this a bit peculiar because energy seems to be transferred from the battery along the wire.

    What i am speculating is the energy from the battery is flowing outward into the air, and somehow it goes back into the wire again.

    Does anybody has any clue on this?
  2. jcsd
  3. Sep 15, 2010 #2
    Yes. It points "outward" where work is being done by current (@ the battery), and points "inward" where work is being done on current (@ the resistive wire). This is assuming a steady state solution. If the energy in the EM field is changing in time, then the field can also act as a "source/sink" to the Poynting vector.

    In technical terms:

    [tex]\nabla \cdot S = -J\cdot E - du/dt[/tex]

    S=poynting vector, u = field energy density, etc

    Look up poynting theorem for more explanation.
  4. Sep 15, 2010 #3
    The Poynting vector doesn't always say where energy is flowing FROM and it doesn't always say where energy is flowing TO (there are many times you can make the argument though). What the (divergence theorem applied to the) Poynting vector always tells you is whether energy is flowing out of or in to a specified volume and how much.

    For simplicity, let's assume the wire to be cylindrical and carrying a uniform DC current parallel to the axis of the wire; this is a common text book example. Well, by Ohm's law, it's evident that the E-field is confined to the wire, i.e. it is precisely zero outside of the wire. So it follows that the Poynting vector is also zero outside of the wire; however, the Poynting vector has non-zero value on the surface of the wire. It is precisely over this surface, that you recognize the Poyting vector to be pointing into the wire (no pun intended). You then take the surface integral over the wire's surface and are able to derive the usual I2R law.

    Note that, at the surface of the wire, you recognize energy to be flowing IN TO the wire. However, you don't actually "know" where it is coming from because, if you step a little bit away from the surface of the wire, the Poynting vector immediately drops to zero.

    If you are still circumspect about this description, then consider one final thing. A peculiarity of (classical?) electromagnetism is that the theory doesn't actually say where energy is stored, only that the fields are able to propagate it. I'll refer you to the text books for the details. :smile:
  5. Sep 15, 2010 #4
    Ohms law does not tell you what is happening outside the wire at all. If you say the divergence of the poynting vector is negative within the wire, it makes no sense to say that is is precisely zero outside the wire.
  6. Sep 15, 2010 #5
    For power flowing in (actually along) a resistanceless wire, the external magnetic field is azimuthal, and the external electric field radial, so the Poynting vector (E x H) is parallel to the wire. The Poynting vector is completely outside the wire. The relative phase of E and H defines the direction of power flow.

    Bob S
  7. Sep 15, 2010 #6
    Yeah, you're right; the point I was trying to make is that the E-field should run parallel to the axis of the wire and be zero outside of it. I suppose I should have explained this is a highly idealized situation; essentially that of an very long wire.

    Again, with the highly idealized example, you have zero electric field outside the wire, which, in-turn, gives a zero Poynting vector. The point that I was trying to make with this is lumped in both the first and last paragraphs of Post #3. Specifically, that the "location" of energy is actually quite ambiguous.

    That last sentence is essentially my answer to the original question; however unsatisfying it may be.
  8. Sep 15, 2010 #7
    For every current in a wire, there is always a return current somewhere (another wire, ground plane, etc.). The electric field is actually the field between the two wires. It is always orthogonal to the wire and orthogonal to the azimuthal H field. Because the Poynting vector is the vector cross peoduct of E and H, it is parallel to the wire.

    For an ideal wire without resistance (i.e., with negligible skin depth, or superconducting), the entire Poynting vector is outside the wire.

    [added] For two parallel wires connected to an ideal capacitor, the ac impedance is 1/jωC, so in the frequency domain, the voltage between the two wires is 90 degrees out of phase with the current in the wires, so the Poynting vector is zero everywhere.

    Bob S
    Last edited: Sep 15, 2010
  9. Sep 15, 2010 #8

    Sorry about that, you must have posted Post #5 while I was posting Post #6; I didn't even notice your reply!

    In my previous post, when I spoke of a "wire" I assumed a resistive material, i.e. not a perfect conductor; depending on the value of the conductance, we may call this a lossy wire or a resistor. In that case, the Poynting vector indeed points into the wire.

    Now for the case of a perfect conductor, I'll have to agree with you. I'd just like to add, the fact that the Poynting vector flows parallel to a perfectly conducting wire indicates that you cannot form a Gaussian surface that allows for a either net inward or outward flux.

    Now what's really interesting is when you put together my example with your example; i.e. a perfectly-conducting wire connected to a resistive wire (with a power source somewhere to get current flowing of course). In terms of a circuit diagram, the perfect conductor would be the "connections" or "wires" and the resistive material would be the resistor.

    From what we've discussed, the Poynting vector must flow parallel to the perfectly-conducting wires, but then flow into the resistive component. By superposition, the Poynting vector will no longer be zero outside of the resistive component (due to the Poynting vector that appears because of the perfect conductors). In this picture, Poynting power flows around (but not in) the wires and into the resistor. I believe this is what the OP was originally getting at.
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