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What causes the potential (voltage) in a conductor to rise?

  1. Sep 4, 2009 #1
    I noticed a couple of similar threads, but I thought it might be better to start a new one for clarity.

    Imagine a simple DC circuit with a static (time invariant) source of potential and a closed circuit with a simple conductor. If you were to trace the potential from the negative terminal to the positive terminal you would expect a linear gradient in potential between the two terminals.

    Now, take any point on the conductor. What determines the potential at this point? In electrostatics, electric potential at a point in space is determined by the grouping of static charges and distance from those static charges.

    What determines the potential inside a conductor? Do the electrons rearrange themselves within the conductor (e.g. bunching up closest to the negative terminal)? Does this occur at the speed of light?
  2. jcsd
  3. Sep 4, 2009 #2


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    Are you familiar with Ohm's Law? The potential at any point in the circuit is determined by the current and the resistance between the point and the source.
  4. Sep 4, 2009 #3
    Yes, Ohms law is a familiar abstraction. I'm looking for a more indepth explanation of what actually creates potential in a conductor.

    To clarify I know that the charges have a drift velocity and I guess the potential drop is proportional to the drift velocity and rate of collisions within the conductor, but I can't figure out what "potential drop" really means. What physical parameter within the conductor is being reduced?
  5. Sep 4, 2009 #4


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    So you understand the "abstraction" that is Ohm's law and are comfortable with the relationship of current, resistance, and potential? The actual mechanics are not as important as the fundamental relationship.
  6. Sep 4, 2009 #5
    Yes, I understand Ohm's law. The underlying mechanics are what I'm curious about here.
  7. Sep 5, 2009 #6
    In a battery, there are two terminals, one which is more negatively charged than the other. Since there's more charge around that terminal, the charge is resisted more at that terminal. Therefore there is a difference in electric potential between both the positive and negative terminals.

    As a Coulomb of charge moves towards the positive terminal, it loses electric potential energy. Since electric potential is location-dependent, you can conclude that the voltage will be a certain value depending on where you measure it.

    Hope I helped.
  8. Sep 5, 2009 #7
    what does it mean for the charge to lose electric potential? it isn't losing kinetic energy as the drift velocity is constant through the conductor.
  9. Sep 5, 2009 #8
    Assuming that you have ideal ohmic contacts connecting the metal and the conductor, the only "source" of voltage drop in the system is the contacts. Think about the 1D Poisson's equation. It is second-order, and requires two boundary conditions to solve. One condition just sets the potential gauge. The other sets the drop in the system.

    Mathematically, it's exactly equivalent to a simple capacitor. In both cases, the potential drops linearly across a region of space, and the physical reason is the voltage (charge buildup) on the metal plates. So, perhaps the better question is why the presence of a current doesn't distort the constant field profile in the case of a conductor. If a "bump" was somehow able to form in the field, electrons would either pile up or deplete from that region, until they exactly canceled out the effect of the bump.
  10. Sep 5, 2009 #9


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    It is losing kinetic energy, that is why the drift velocity is constant. If the charges did not lose any kinetic energy then they would keep speeding up since the potential applies a force on the charges. The electrons will collide with the large atoms in the conductor's lattice, losing kinetic energy and causing vibrations that give rise to heat. This is the ohmic heating that occurs due to the conductor's resistance. But this is different from what you first asked, the electric potential is set at the terminals of the voltage sources, what it is at other points in the circuit with respect to your reference points is different. We care about the potential difference.
  11. Sep 5, 2009 #10
    That makes sense but at the same time it doesn't. Electrons move slowly in the conductor, and they have very little mass, which makes it hard to believe that the PE of the electrons is turning into KE.

    The charges have potential energy and lose it as they pass through a resistance. Yes the current does carry KE, but it's so small that it's negligible.
  12. Sep 6, 2009 #11
    Okay, that seems fairly logical. I guess this means that locally electrons are constantly accelerating and decelerating, and the aggregate effect is a constant drift velocity. So can we say that the accelerating potential across the circuit (from negative to positive terminal) is uniform.

    If so, what causes the potential at say halfway along the circuit to be half of the potential difference between the two terminals. If the drift velocity is constant then what property of the electrons or conductor sets the potential at this point?
    Last edited: Sep 6, 2009
  13. Sep 6, 2009 #12
    In free space, the potential at an arbitrary point seems to be determined by spherical geometry (the steradian function 1/4pir^2). In a dielectric, potential distributes itself through polarisation. How does potential distribute itself in a conductor?
  14. Sep 6, 2009 #13


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    This is the Drude model: http://en.wikipedia.org/wiki/Drude_model and is still an accepted model for electron conduction in metals, I just sat in on a talk yesterday that made use of it. The electrons move slowly because of the constant collisions with the lattice. Otherwise, the electrons would move very fast because, as you stated, they have very little mass and would experience a force all along the length of the conductor from the applied fields.
  15. Sep 6, 2009 #14


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    Yeah, spot on. The main thing to realize is that the potential difference is set by the sources, all other objects in a circuit are passive (ok, so there are active things like transistors and so on, but let's restrict ourselves to simple LRC circuits). user111_23 did hit upon this indirectly, but the actual amount of kinetic energy lost by the electrons in a conductor is very small. Because they have very little mass, they lose little amounts of kinetic energy to keep a constant slow drift velocity. But like I said, this is something separate from what you asked. Because they lose so little kinetic energy, the potential difference along a wire is almost zero, there is little energy lost. But when the charges meet a circuit element, say a resistor, then a large amount of energy will be expended (purposely as in the case of the resistor). It is this drop in energy that gives rise to the change in the potential difference across the circuit when referenced from a common point.
  16. Sep 7, 2009 #15
    The way you make it sound is as though the potential at a point is a intrinsic property of the space there that is determined by the electrons and the conductors. But that isn't right, potential is a relative thing; the change in potential just means that between this point and the battery terminal, electrons are gaining/losing energy. Nothing special is happening to cause a "change" in the potential.

    On a side note, if we had a perfect conductor that electrons could travel through unimpeded, if the electrons are continually gaining energy and going faster does that mean that they would eventually pile up at the negative terminal of the battery? Or would the current keep getting larger? Never thought of that before, but I guess it's moot because it's totally idealized.

    And I have a question of my own, when electrons come to a resistor, the heat energy comes from their kinetic energy right? So they slow down; but electrons don't pile up at one end of a resistor. Does the battery somehow know to work harder to pump them through? Or is it like someone previously said, if there was a "bump" in the electron flow the repulsion between electrons would automatically sort it out?
  17. Sep 7, 2009 #16
    You make it sound as if the potential drop across a resistor isn't a measurable quantity.

    I think the point he's getting at is this:

    Say we have a simple circuit consisting of a battery and a resistor connected by two wires.

    Why is the measurable potential drop negligible in the wires when compared to the resistor, and how can this be explained in terms of the Drude model?
  18. Sep 8, 2009 #17
    Because the resistance is much greater at the resistor as compared to the wires.
  19. Sep 8, 2009 #18
    PhaseShifter, perhaps I didn't think through what I said enough, but measuring potential across two points is definitely possible, I just meant we can't just look at a single point inside a circuit and decide how the potential has changed. Voltmeters can do this by making a new circuit connected to the reference point, but just by looking the some arbitrary point, what could you observe that would tell you about a change in potential?

    If the resistance is greater, then in terms of the Drude model one could describe this by saying the mean free time between collisions is shorter. But if the electrons still move at their constant drift velocity after the pass the resistor, what could you observe to notice the drop in potential? I'm not sure.
  20. Sep 8, 2009 #19
    When electrical energy is converted into other types of energy, there is a voltage drop. Such as a light bulb, you know there is a drop because the energy of the electrons creates friction in the filament to produce light and heat energy.
  21. Sep 8, 2009 #20
    I can see that mathematically--In the case of direct current, if the resistor has a lower charge density (or shorter mean free path, as you point out, or is simply narrower diameter wire) then the potential gradient must increase to compensate in order to provide a steady-state current.

    I'm just trying to figure out a physical mechanism for this to take place--not in terms of bulk quantities like "resistance" and "current", but in terms of variables used in the Drude model. (acceleration due to electric field, distance between collisions, and carrier charge density)

    How do you get from "Electrons collide with the lattice more frequently in this region of space" to "the field accelerating them is proportionally stronger", for instance?
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