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Why is the electric field constant in a wire?

  1. Feb 4, 2007 #1
    When deriving Ohms law from basic principles (at an introductory physics level) the electric field is always assumed to be constant in a wire (like the electric field between two capacitor plates that are close together). However, a wire may be very long and the distance between the postive and negative terminals can be very far away, yet the electric field is constant. I think I can answer this based on atoms (electrons and protons in a metal), but how is this done classically (or what is your atomic explanation)?
     
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  3. Feb 4, 2007 #2

    ZapperZ

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    Er... the electric field is simply the gradient of the potential. Since you are applying a constant potential at the ends of the wire, and the wire's length is constant, this means that the potential gradient is constant. Thus, the electric field is a constant.

    Zz.
     
  4. Feb 4, 2007 #3
    I don't see how holding a constant potential between two surfaces forces the electric field to be constant. Conside a capacitor with two concentric spherical shells held to a constant potential V. The electric field is the same as for a point charge (i.e., inversely proportional to r^2) between the two shells.

    What is the physical mechanism that causes the electric field to remain a constant instead of dropping off as 1/r^2? It has something to do with the physical wire.
     
  5. Feb 4, 2007 #4

    ZapperZ

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    Your example here is different because the field density is different.

    Try solving the E-field for an infinite flat plane of charge. You'll see that, since the field is all perpendicular to the plane, it doesn't "diverge" out, keeping the field density to be constant throughout. You get a constant E-field in that case.

    If you have a finite source such as a spherical or line charge, then this is no longer true.

    In the case of a wire, ALL the fields are confined to the wire, so one can consider that the field density is uniform throughout the wire. So it is a valid approximation to consider that the E-field is constant throughout the conductor.

    Zz.
     
  6. Feb 4, 2007 #5
    Yea, I know how it works for 1-D and you are mostly right in that the field is constrained to the wire. There is a field outside the wire, but it decays as 1/r^2, so we can ignore it. The question is why does the field stay in the wire as a constant.

    Let me tell what I think it is. If the wire were a perfect conductor everything would be at the same potential, so there would be no current, so part of the wire cannot be a perfect conductor. The wire has atoms which has + and - charges. As a electron is pulled by one battery terminal (the + end) it creates a + charge in an adjacent atom (for simplicity I looking at the wire a string of atoms and a real wire is a tube of atoms). The + charge pulls another electron etc., so electrons are dragged down the wire by the local adjacent electrical forces which are approximate the same since the metal atoms are about the same distance apart.

    If the above description is correct, then how can one explain 'classically' why the electric field is constant in a wire.
     
  7. Feb 4, 2007 #6

    ZapperZ

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    The explanation is not correct. If you want to dive THAT deeply into it, then you might want to look up a solid state physics text and figure out what is meant by a "conduction band" and the semi-classical Drude model. I suggest you don't.

    Why is the field only within the conductor? It is because the potential difference is applied only to the ends of the conductor, not to the ends of the volume surrounding the conductor. It doesn't mean that the latter can't be done. It means that in the question that you are asking for a typical conduction, the potential different is applied to the conductor only.

    I think I am done with this one...

    Zz.
     
  8. Feb 4, 2007 #7
    Well thanks anyway. My explanation was purposely oversimplied just to get the idea (yes I know about conduction bands etc.) I thought there should be an explanation of why the electric field is constant from first principles (i.e., by just discussing the charges in the battery and wire and the electric fields created by said charges.) I know the field is pretty much constant or else circuit theory wouldn't work, I just don't see how one can simply state the electric field is constant in a wire other than it works. It seems you need to understand what is going on in the wire.
     
  9. Feb 5, 2007 #8
    Maybe it would be helpful to consider what would happen if the E-field were not constant along the length of the wire.

    Imagine you could divide up the wire into many many infinitesimal slices along its length. Each slice is "dz" meters long. The voltage drop "dV" across a slice at position z will be given by:

    dV(z) = -E(z)*dz

    Notice that we have not assumed that the E-field is constant across the length of the wire, just constant enough across the thin slice dz.

    Now if E can vary along the length of the wire, the voltage drop dV across any slice will vary across the length of the wire as well. This would be a very curious result --- why would the current flowing through the wire cause a different voltage drop in a slice at one end of the wire compared to the other end? (We're assuming the wire is of uniform cross section and material...) What would happen if you were to disconnect the wire and switch the ends connected to the generator, reversing the direction of current?

    The way to reconcile these problems is if we force E to be constant along the length of the wire. This means that the voltage drop across each slice will be constant, and independent of where along the length of the wire you are. After all, looking at each slice of wire, all you see is a current "I" going in, a current "I" coming out, and a thin slice of metal with a certain cross section. The voltage drop across this slice must look the same as the voltage drop across any other slice, because the slices are all the same.

    This symmetry leads to the conclusion that

    -dV/dz = constant = E

    Hope this helps.
     
  10. Feb 6, 2007 #9
    Your explanation is clear and accurate, but let me explanation the difficulty I am having understanding why the electric field is constant in a wire.

    Textbooks generally describe the parallel plate capacity electric field (when end effects can be neglected) and then describe the wires electric field somehow relating them or they say most metals obey J=(sigma)E (where sigma is the conductivity) for steady state and then derive the low-level (no quantum) physics of conductivity. However, both start with the electric field being constant as steady state, plus J=(sigma)E implies E is constant for a wire with uniform volume characteristics. So, as you have stated "I" going in and "I" being the same requires a constant E field.

    What I am trying to do is get a mental picture, involving charges at one end and oppositely charged charges at the other end with a piece of wire in between using Coulomb's law and some basic assumptions about the material in the wire (classically). Since Coulombs law descreases as 1/r^2 and a wire can be very long, the wire itself by have local fields that cause E to be a constant in the wire. Here is a picture of what I am talking about.

    http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

    Go to Electric circuits (near top left) and then choose conductors (far left, middle). The picture explains how a field applied to one end can propagate through the material by way of local fields. The thing that I don't get from this picture is why isn't the electric field independent of the length of the wire. In a circuit the electric field diminishes as the wire is lengthened (E = V/L). Its kind of wierd that if the wire is made longer the drift velocity shrinks, but otherwise we'd have conservation of energy violated and we don't want that.

    Thanks again.
     
  11. Feb 6, 2007 #10

    ZapperZ

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    But see, this is where you're applying the wrong expression for the geometry. Coulomb's law does NOT indicate that all forces and E-field vary as 1/r^2. That's just not right. The example you gave in which for a parallel place capacitor that has no edge effect, E-field does NOT vary with the distance! You get this by solving Coulomb's law!

    The 1/r^2 factor is geometry dependent. If you have a finite bound charge, then at a sufficiently far distance away ( a far field region), then it looks like a point charge and thus, the 1/r^2 dependence (this is where the joke where one approximate a cow as a sphere comes in). However, when the field is sufficiently uniform, as when neglects edge effects, then one does not get such 1/r^2 dependence anymore! This is what you get in the parallel plate capacitor. Applying Gauss's law to such a situation (which is nothing more than Coulomb's law in disguise), you do NOT get the same field dependence. The E-field is uniform and does not "diverge", causing a uniform "field density" throughout. It has no dependence on length!

    This is what I've been trying to equate in a conductor. The field lines are uniform and do not "leak" out, so it maintain the same field density. In fluid mechanics, this is equivalent to a laminar flow. It doesn't matter how long the pipe is. As long as the laminar flow is maintained in a uniform pipe, then the rate of flow remains constant.

    Zz.
     
  12. Feb 6, 2007 #11
    I can see the E field constant for a parallel plate capacitor with no problem because the plates are close together and we assume infinite (effectivly) extent in and plane. I also understand that the field lines are uniform in a uniform wire. However, I cannot get a mental picture where the only the charges at each end of the wire cause the electric field to be constant in the wire as that is the way it is usually explained. I know it is true, but if I drew a picture with charges and a wire, I wouldn't be able to explain why the electric field from the collection of point charges at one end of wire do not diminish with distance (give a fixed length wire and a constant potential between them). It seems the charges in the wire have to be responding to the field at the ends to keep the field from diminishing.

    Thanks for being patient. I'm just trying to get a simple picture of what is going on.
     
  13. Feb 6, 2007 #12

    ZapperZ

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    No, it has nothing to do with the plates being closed together. If you solve the problem, you'll notice that there is no dependence on the distance from the plates, nor the separation between the plates. All we care about is that the edges are infinitely far away.

    What this is implying is that in the region where the E-fields are uniform and do not diverge (i.e they maintain the field density), variation in distance from the source is irrelevant. This is the main point of everything that I've been trying to tell you.

    Zz.
     
  14. Feb 6, 2007 #13
    But the reason we can assume the edges are infinitely far apart is because the plates are close together. If we have finite plates then they have to be close together for the field to be approximately constant. I know you know this and we are saying the same thing. However, the wire is not a parallel plate capacitor, but its electric field acts like on. I can physically see why the parallel plate capacitor field is constant, I cannot see why the wire's field is constant, except by other assumptions (e.g., that J = (sigma)E which is an experimentally observed fact). I aiming for a physical picture using charges and fields produced by those charges to explain why the field in a wire is constant when a potential is applied at each end. It seems a picture using just charges and fields from those charges should be able to be drawn, but I cannot do it.
     
  15. Feb 6, 2007 #14

    ZapperZ

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    This is because the field lines do not leak out!

    You do the same thing in magnetic fields where instead of an air core solenoid, you use an iron core solenoid. The iron core confines the magnetic field lines tightly within itself rather than have it stay out in the air.

    The same thing occurs in a conductor. When you attach the two ends to a potential difference, the field lines are confined to only within the conductor and nowhere else! When you have such a condition, then the field lines density remains constant, the same condition you would have for an infinite plane charge!

    I know for a fact that I've mentioned this already early on in this thread.

    Zz.
     
  16. Feb 6, 2007 #15

    vanesch

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    I think I see your problem (I vaguely remember having had similar worries when I was studying the relationship between Maxwell EM and electric circuits - but the human mind is such that, once one understands a subject, one doesn't remember anymore why one didn't understand it before :smile: ).

    The whole point is, as Zz points out, that "the E-field cannot leak out". Now, that in itself is a non-evident element, but it is also part of "ohm's law".

    Imagine, as you do, two "concentrated balls of opposite charge" far away, in a dielectric. The field lines will radiate outward, and swarm all over the place. Now, imagine again these two balls of charge, but this time, say, in seawater. No difference! Except of course that a current is established, and the current density will be proportional to the E-field: Ohm's law. In fact, there will be a tiny difference with the previous situation, and that's because of the tiny amount of charge that is actually traveling, will slightly change the E-field. But as this amount of "traveling charge" is very tiny as compared to the amount of "reservoir charge", this will only influence the E-field so much.

    But next, consider variations in conductivity over space (different amounts of salinity). We will now have that certain pathways will have higher current densities, and others lower. The distribution of "traveling charge" will be depending on these variations, and in fact, whenever the E-field will go from a "high conductivity" to a "low conductivity" zone, there will be some accumulation of 'traveling charge'... until the E-field is modified so that a steady-state flow of charge is again established.
    So, charges will accumulate until the E-field is modified such, that by Ohm's law, the charge flux is the same in the "high conductance" and the "low conductance" region of the E-field flux tube.

    In the extreme case of a cylindrical conductor in a VERY low-conducting environment, charges will accumulate on its surface SUCH THAT the flux tube of charge flow will have uniform flux density (simply because if that isn't the case, charges will accumulate where it changes!). Now, as it is assumed that the only place where the flux tube can go, is the conductor itself, charges will place themselves such, UNTIL j is constant along the conductor. And hence, by Ohm's law, E is constant along that conductor. Any deviation from this pattern will make charges accumulate, and modify the E-field, until this is the case.
     
  17. Feb 6, 2007 #16
    Thanks for the very clear description. It clearly explains why J is constant and if one uses Ohms law, E is constant. I'll try to explain the problem I am having. In each derivation that I have seen one either starts with Ohm's law (J=(sigma)E) and shows that E is a constant or they assume E is a constant and derives Ohm's law. I understand both these arguments, although yours is the best I've seen. I also understand why E is a constant at the surface of the battery terminals (by using Gauss' law or by summing the vector contributing in the direction perpendicular with the surface). One thing I do not understand is how the wire forces the electric field to stay inside, for as you bend the wire, the field bends. There must be something about the wire material that influences the external field supplied by the battery terminals. One the one hand the electric field is treated as an external field acting within the wire and on other the hand, if I bend just the wire, the electric field changes direction. How is the external electric field from the battery terminals being influenced by the material makeup of the wire. I believe that is what I am trying to figure out.

    Thanks for you help.
     
  18. Feb 6, 2007 #17
    Ahhh, I see now. This is a much deeper question.

    Let's think of what's going on in the conductor. Remember, the definition of a conductor is a material with a lot of available free (not bound) charges to move around. Now again, think of your wire, and assume that the Electric field was somehow disturbed such that it is not constant along the length of the wire. Looking at some slice of the wire, you would have the E-field at one end greater than the E-field at the other end. This would mean that you have current at one end greater than the other end of the slice. (Note this assumption doesn't necessarily assume Ohm's law, but it does assume that the current is an increasing function of E-field. Considering that the force on an electron will be proportional to the E-field, this seems like a reasonable assumption. )

    Anyway, conservation of charge would demand that either positive or negative charge is accumulating inside the slice. What would the effect of this be? These charges would generate E-fields such as to balance out the original non-constant E-field in that section of wire.

    We can think of the time scale that such an effect would happen. Classical EM theory defines a "dielectric relaxation time" as the ratio of a material's dielectric permittivity to its conductivity. For a good conductor like copper, this time is astonishingly quick, on the order of 1e-19 seconds! For a crappy conductor, like seawater, it is still blazingly fast: on the order of a nanosecond!

    So I think what is happening is as you twist and jiggle the wire, the charges inside it are very very quickly redistributing such that the equilibrium (constant) E-field is reestablished.

    I don't think I'm saying anything new in this thread, just restating what has been said already. Neat question, and I hope you find this helpful...
     
  19. Feb 7, 2007 #18
    The only reason I'm trying to get a physical picture, in terms of charges moving around and fields is because all of DC circuits is built around Ohms law and this constancy of E within a wire. Once you say that, a lot of physics comes out of it and we know Ohm's law describes these things pretty well.

    As I mentioned, in the derivations I have seen the E field within the wire is stated to be external as far as the charged particles within the wire, but as you pointed out, their fields must be doing something keep the current constant (which is the same thing as keeping E constant).

    Thanks
     
  20. Feb 7, 2007 #19

    vanesch

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    One cannot really "derive" ohm's law (unless one goes deep into the quantum physics of solid state): it is a PROPERTY of certain materials. That property is such that, an amount of material (conductor) being exposed to an E-field, generates a flow of charge. If you want to have a simplistic model, consider the Drude model. However, there are several problems with the Drude model, which is more appropriate for the transport of charges in a gas than for the transport of charges in a solid conductor. Nevertheless, it illustrates a RESPONSE of a material to the exposure to an electric field. Ohm's law is just such a kind of response. You can't really *derive* it in macroscopic, classical electromagnetism. Some materials don't behave at all according to Ohm's law.

    Charges ! As the wire can transport charges, it can put them in such a way that these charges CHANGE the electric field. As mdelisio also points out, you get charge accumulation whenever the flux of charge density changes, by conservation of charge. It is a bit like with a traffic jam: if you have a wide road with 5000 vehicles/hr, and all of these go onto a small road with only 200 vehicles/hr, the vehicles will accumulate (jam) where the transition of the two roads occurs, and the traffic jam will increase until, on the big road, there's only 200 vehicles/hr driving because of the jam. (this analogy is of limited use, mind you). But charges do what cars don't: when they accumulate, they CHANGE the E-field! They diminish the E-field "from where they are coming", and they increase the E-field "where they ought to go". So it is as if the accumulation of cars doesn't only decrease the flow of vehicles on the big road, but also increases the flow of vehicles on the small road! (this is where the analogy breaks down).

    Now, what is an EXTREME change in charge flux, is when it tries to cross the surface of the conductor: it simply can't. So all charge that wants to flow "outside" of the conductor is accumulated on its surface! And this in such a way, that the E-field is modified UNTIL it is entirely parallel to the conductor surface (because only then, no charge wants to "flow outside" and hence gets accumulated more at the surface).

    EDIT: when I talk about Ohm's law not derivable, I mean: J = sigma.E. Of course, ONCE this is accepted, one can derive the "macroscopic Ohm's law" which is I = 1/R . V
     
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