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Current in ideal wires -- circuits

  1. Jan 30, 2010 #1
    Current in "ideal" wires. - circuits

    I am trying to get a "complete" picture of what happens in circuits. I know how to solve typical physics circuit problems using kirchoff's laws and what not, but I am trying to get a "picture" of what is happening inside of the circuit itself in terms of the electric field in each part of the circuit. (the battery, the resistor and the wire)

    For simplicity sake, lets say the circuit is just a battery, some "ideal" wire (no resistance), and a resistor.

    Here is the picture in my head.. please correct me if I am wrong.

    So the battery has a uniform electric field pointing from the positive end to the negative end. This allows the battery to have a potential difference across it that increases the potential energy of the electrons.

    As you connect the battery to the circuit, an uniform electric field is created throughout the wire and resistor. In the "ideal" wire, the charges immediately position themselves to cancel out the electric field within the wire, like a conductor. Because the charges are already in motion though, they just continue in motion without an electric field pushing them. (do the charges move along the surface only?)

    When charges reach the resistor, there is now an electric field that "pushes" them through the resistor in such a way that no energy is lost in the electrons (the energy given to the electrons from the electric field is the same as the energy dissipated in the resistor), which allows them to keep the same steady current.

    The electrons then flow through another ideal wire, then back to the battery where the battery increases the electron's potential energy and the process repeats. Is this about right?
     
  2. jcsd
  3. Jul 7, 2015 #2
    While this is 5 years old, I'd like to know as well if this description of what happens is right ! This is also what I understand from my learning so far.

    However, one point is still mysterious to me : how does the potential difference across the resistor builds up? If electrons would "bulk up" in one side, that would indeed create an electric field between each ends of the resistor, thus a potential difference. But electrons can't actually bulk up. If they would do that, current couldn't stay constant across the circuit, I guess.

    So how does electric field inside a resistor builds up?
     
  4. Jul 7, 2015 #3
    You can not have an electric field on the interior of an ideal conductor. You can not have charge in the middle of an ideal conductor. The tangential electric field on the surface of an ideal conductor vanishes (= 0). The tangential magnetic field on the surface of an ideal conductor is equal to the surface current density on the conductor. Current does not flow in the middle of an ideal conductor. Current flows on the surface of an ideal conductor. Electromagnetic energy does not transmit through an ideal conductor; rather, the conductor guides the electromagnetic waves.

    You may have an electric field in a material that has finite conductivity. When you do, Ohm's Law is in effect: ##J = \sigma E## where ##J## is the current density, ##\sigma## is the conductivity of the material, and ##E## is the electric field. When an electric field is applied in the presence of an electron, a force acts on the electron in the opposite direction of the electric field. If the electron is free, it will accelerate freely and continuously. If the electron is within a crystalline material, the progress of the electron is impeded by collisions with the thermally excited crystalline lattice structure, and a constant average velocity (called drift velocity) is attained. This drift velocity is linearly related to the electric field by the mobility of the electron.

    ##v_d = -\mu_e E##

    The current density is linearly related to the velocity of the charge by the charge density:

    ##J = \rho_v v##

    So, ##J =-\rho_e \mu_e E##

    where ##\rho_e## is the free-electron charge density.

    So, ##\sigma = -\rho_e \mu_e## for metallic conductors.

    In semiconductors, in addition to free electrons, you also have holes.

    ##\sigma = -\rho_e \mu_e + \rho_h \mu_h##

    By doping semiconductor materials with very low levels of impurities, you can dramatically increase the conductivity of the semiconductor. This is how diodes and transistors are made.
     
  5. Jul 7, 2015 #4
    Thank you for answering.

    I do understand those principles (at least, I think o)), including the different forms of Ohm's law. The particular point which I can't digest is how exaclty can there be an E field inside a semiconductor plugged in a circuit (like a resistor)? While avoiding charges accumulation on one side of the semiconductor, so current can be constant along the circuit.

    Or in other words, where the potential difference across the resistor comes from? In the point of view that V = E*l, with l being the length of the resistor. Not in the point of view that electrons give away energy by colliding with the semiconductor atoms (producing heat). Where the E comes from within the resistor, how does it builds up?
     
  6. Jul 7, 2015 #5
    The E-field is generated by the battery. (How the battery generates an E-field is a very profound question, which I won't tackle right now, except to say that it has to do with the electrochemistry of the battery).

    The field generated by the battery is then guided by the conductors to the resistor. A perfect conductor is an equipotential surface. So, you have a battery generating an E-field. The ideal conductor that is touching the positive terminal of the battery is an equipotential surface. The ideal conductor in touch with the negative terminal of the battery is an equipotential surface. Now, between the two equipotential surfaces, you have an electric field. Now, you have a resistor that is there in the midst of that field. ##J = \sigma E## The electrons flow from the lower potential to the higher potential.
     
  7. Jul 7, 2015 #6
    Oh, right, that makes sense! This was a simple answer, sometimes I can't see the most obvious thing. Thank you. :)
     
  8. Jul 7, 2015 #7
    No problem. I regularly have to (re)think through ideas that at one time were very obvious to me.

    With fields, you just have to get used to imagining vector fields and equipotential surfaces all over the place. And then when one starts oscillating, then the disturbance causes oscillations in the fields that propagate outwards from the disturbance. And when you start sticking metal in various places in space, you get reflections and scattering, and currents, etc.
     
  9. Jul 7, 2015 #8
    BTW, that means the E field exists between the ideal conductors (but not inside them) all the way from the battery to the resistor right? Any chance you'd have some "grass seed" picture of a E field in a simple circuit?
     
  10. Jul 7, 2015 #9
    Here you can get a visual of the electric field between two charges.

    You can also find the expression for the E-field of an infinitely long conductor. When you have to parallel infinitely long conductors - one that is positive charged and one that is negatively charged, the E-fields in the vicinity superimpose on each other so that the fields are strong and oriented away from the positively charged conductor and towards the negatively charged conductor in between the conductors.
     

    Attached Files:

  11. Jul 7, 2015 #10
    Thanks for the details. I have more questions relative to resistors. Hope that's not too much to ask!

    So, first : when electrons move along a resistor, they give away energy by colliding with atoms, ok. Thus, the potential drop. BUT, are electrons also losing potential due to the distance travelled? I mean : V = E*l, when E is constant. For instance, If a charge of -1 Coulomb move a distance 1 m closer to a charge +1 Coulomb, it loses some potential energy, right? Therefore, is the net potential drop in a resistor due to both collisions AND to distance travelled?

    And secondly, since the current must stay constant along the circuit, it must also stay constant within the resistor. BUT, the difference of potential across the resistor means the electrons accelerate in it (there is a constant E field within the resistor). In the same time, collisions must slow them down. SO, do the electrons travel at constant drift velocity within the resistor because the collisions exactly cancel out the speed gained by the acceleration due to the E field? Thus, keeping the current constant even within the resistor. If so, why is that? Why does it cancels it out exactly?

    I hop what I am writing is understandable, I'm sorry if it looks confused. Not easy to explain clearly.
     
  12. Jul 7, 2015 #11
    This is not so clear to me.

    In the presence of an electric field, a negatively charged electron is subjected to an electric force that is opposite to the direction of the field.

    If the electron was completely free, it would continuously accelerate - gaining speed and energy.

    In a crystalline material, the electron speeds up to a certain drift velocity and then pretty much stays at that drift velocity. That drift velocity is proportional to the electric field, which is to say that the drift velocity of an electron is proportional to the electric force acting on it.

    So, in this sense, a resistor is something like the electrical analogy of a mechanical damper. The speed of the electron is proportional to the force acting on it, due to the impedance of the crystalline material.

    In good conductors, a drift velocity of a few inches a second is sufficient to produce a noticeable increase in temperature that would melt the wire if the heat is not removed by conduction, convection, or radiation.

    Additionally, the conductance of a conductor (and the mobility of an electron in that conductor) increases as temperature increases. Thus, at higher temperatures, there is a greater crystalline lattice vibration, more impeded electron progress for a given electric field strength, lower drift velocity, lower mobility, lower conductivity, and higher resistivity.

    An electron (with its negative charge) is drawn from a lower potential to a higher potential like you and I are drawn toward the center of the earth. Imagine falling. Now, imagine falling through a bunch of branches of a tree that are relatively dense (but not so dense as to stop your fall, only dense enough to slow you down). The assumption that the E-field is constant at all places within a resistor is generally false. Simply, the electron's electric potential energy at any particular location depends on its charge and the strength of the E-field at that location. Similarly, your gravitational potential energy depends on your mass and the strength of the gravitational field wherever you are. When an electron moves closer to a positive charge, it does lose potential energy, even while it climbed to a higher voltage. This is due to the difference of signs between negative charges and positive charges. But the change of potential energy of the electron is due to its change of position in an electric field - not due to collisions in the lattice structure. Falling through branches may slow you down (decreasing your kinetic energy), but it doesn't impact your gravitational potential energy.
     
  13. Jul 7, 2015 #12
    I think your entire series of posts is misguided. The only source of all the electric fields in a circuit with a steady current is the distribution of charges in the circuit.
    If the wires guide the field they do so by rearranging electric charges within themselves to make the field inside the wires zero, and the field inside the resistor just high enough to the same potential as the battery across it.

    To get an electric field in the resistor, but no field or a very small field in the wires there have to be electric charges on both ends of the resistor. At the place where the two materials touch, the divergence of the electric field is non-zero, so there must be a charge density. (gauss' law)
    I don't know why DoubleD thinks there is a problem with this.
     
  14. Jul 7, 2015 #13
    I don't see any problem with what you are saying here, but I also don't see that I have said anything contrary to what you are saying here either.
     
  15. Jul 7, 2015 #14
    The main problem is "The field generated by the battery is then guided by the conductors to the resistor". This in replay to a question of "where does the potential difference across the resistor come from". If you don't mention that charges will rearrange, your statement becomes very vague and misterious. How does the conductor guide the fields?
     
  16. Jul 7, 2015 #15

    Dale

    Staff: Mentor

  17. Jul 8, 2015 #16
    Right, the electron being negative, it travels to higher potential, sorry. I think that the image of falling through branches made me understand a little more :

    - If I'm falling and there is no branches, my initial gravitational potential energy is converted to kinetic energy, and then to heat and sound when I touch the ground.
    - If I'm falling through branches, my initial gravitational potential energy is converted to kinetic energy as well, but this time some of it is given to branches (in the form of kinetic energy for them, and a little heat and sound), and I actually touch the ground with less energy than in the first case.

    In both cases gravitational potential have decreased of the same amount. The only difference is the remaining energy I have when I've passed the branches. For the electrons now :

    - In the case of a short circuit, with an ideal wire connected directly to each terminal of the battery, potential energy difference is turned into kinetic energy of the electrons, they accelerate until they reach the positive terminal, and the battery discharge very fast.
    - If we place a resistor in the circuit, potential energy difference of the electrons is turned into kinetic energy, but then some of this energy is given to the resistor in terms of heat. Once electrons have passed the resistor, the remaining kinetic energy of an electron is a little less than in the first case.

    Again in both cases, potential have decreased the same amount. The only difference is the remaining energy of electrons when they have passed the resistor.

    Is this explanation correct? If yes, then I think I do understand the potential drop, which has actually nothing to do with the energy converted to heat in the resistor. It's just that the charges move from one potential to another. Kinetic energy is lost during the journey, but that actually doesn't affects the potential drop.

    Regarding the E field within the resistor, are you sure is not constant? I thought it was, due to the fact that we could approximate each end of the resistor has a plane of charge, like a capacitor. With inside field E = charge density of an end / epsilon_0, using Gauss's law.

    Isn't the "macroscopic" version of Ohm's law works only because E is constant ?

    If E is constant, then : V = E*l

    Then J = σ*E = σ*V / l and J = I / A

    So V = l*J / σ = l*I / (σ*A) = R*I with R = l / (σ*A)

    => V = IR

    With l being the resistor length, A its cross-sectionnal area, I the current, J the current density, and σ the charge density at an end of the resistor.

    And finally, I still have troubles to understand that there is a constant drift velocity in the resistor, like if the collision within it would slow down the electrons just enough to cancel the acceleration of the E field.

    Even a simple circuit does not seem that simple to fully understand. o)

    Thanks for the clarification. Fortunately however that is what I understood when @EM_Guy answered. The battery creates potential, and when connected to the ideal wires, the charges in those wires rearrange themselves to cancel any E field within them, so each wire is an equipotential surface, thus the potential difference between each end of the resistor.

    Thanks as well. I have tried to read it, but it seems to require some knowledge I don't have yet, regarding magnetic field which I haven't studied yet. I'll give it another try later.
     
    Last edited: Jul 8, 2015
  18. Jul 8, 2015 #17

    Dale

    Staff: Mentor

    I think that if you do not know about magnetic fields then you cannot understand energy transfer in anything more than a cursory fashion. In other words, you should probably just learn the circuit rules at this point (understanding and accepting that they are useful simplifications) and not worry about the details until you have covered Maxwell's equations and Poynting's theorem. Anything that you can learn about the details without those will require un-learning in the future.
     
  19. Jul 9, 2015 #18
    Seems to be a wise advice. I'll just do that then. Thanks! :)
     
  20. Jul 9, 2015 #19
    DoobleD, I agree with Dale. Learn the circuits rules. Then, if you have the ambition, learn Maxwell's equations and the physics of electromagnetism. There is not a 1:1 analogy between every electrical interaction and mechanics.

    Also, when you fall and you hit the ground, very little of the energy will be transferred by heat and sound. Rather the ground will be somewhat deformed, and your body may be significantly deformed - or very possibly broken.

    I have never really thought about electrons having kinetic energy. I'm sure that moving electrons have kinetic energy. And the drift velocity of the electrons is proportional to the current. But I have never really considered any kind of circuits, RF, or electromagnetic problem that involves currents in terms of the kinetic energy of the electrons. If you want to delve into this, you can't be too careful in your analysis. Better I think to stick to the laws of electromagnetics.
     
  21. Jul 9, 2015 #20
    The answer to this gets into electromagnetic propagation. The point that I was making is that the E-field within a perfect conductor is 0, but that the conductor becomes an equipotential surface by guiding waves. Say you have a battery, a switch, and a transmission line. You flip the switch connecting the transmission line to the battery terminals. Now, a pulse propagates down the transmission line, and the conductor is guiding the pulse. When it gets to the end of the transmission line - there will probably be reflections. After a certain amount of time, we reach a steady state. Each conductor becomes an equipotential surface. At the boundary of the conductor, the E-field is normal to the conductor.

    You say that the charges rearrange themselves. This is true, but it doesn't happen spontaneously. When that switch is flipped, the E-field at the beginning of the conductor suddenly changes. This change of an E-field produces a change in the magnetic field.

    ##\bigtriangledown \times H = J + \frac{\partial D}{\partial t}##

    The tangential magnetic field on the surface of the conductor is equal to the current density on the surface of the conductor.

    ##\hat{n} \times H = J_s##

    So, yes, the charges rearrange themselves due to the changes of the fields at the surface of the conductor.
     
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