Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

What is voltage gradient good for? (Read post first)

  1. Sep 10, 2009 #1
    There is an equation for the electric field E=V/d. This tells me the change in voltage per distance.

    Lets say I have a 1-meter wire and a 1-volt battery, so the electric field would be 1V/m. What is the significance of this in the circuit?
     
  2. jcsd
  3. Sep 10, 2009 #2
    From a broad perspective, if you know the electrical properties of the wire, and the transport limit it will operate at 1Volts, the emerging view in the electronic transport community is that electric field does not have any significance, and all limits (Diffusive, Quasi-Ballistic, Ballistic) can in principle be treated by a more general framework.

    But since you are a beginner, and we are talking about BIG things, you can assume (while being aware of it) that you wire (since it's 1 meter) will operate in the diffusive (Ohmic) transport regime.

    In this regime, current flow can be divided up into two distinct parts (Drift and diffusion), a feat that cannot be achieved in quantum transport.

    Traditionally, drift current is caused by the electric field you are describing. And if carrier densities are uniform (no doping gradient etc...) in your conductor you can write the current as:

    [tex]I_{total}=I_{drift}= q \mu_n E n_s[/tex]

    mu_n is the mobility [cm^2/ V.s]
    E is the electric field [V/cm]
    ns is the electron density per unit length [1/cm]
    and q is the electron charge [C]

    But as I said, with the breakthrouhgs in the theory of mesoscopic conductors people found other ways to describe current flow without necessarily talking about electric fields.

    This became inevitable (rahter than a convenience) because current flow when there's NO SCATTERING (i.e, operation in the ballistic limit) cannot be explained by Ohm's Law - which assumes there's enough scattering in the conductor that we could talk about a mobility expression. Mobility is proportional to the time it takes before a carrier is scattered back from its original direction, and in the ballistic limit this time becomes "infinite"... Because there's no scattering.

    Then Ohm's Law fails to describe anything that operates close to this limit... Why electric field needs to be discarded in the process is a subtle issue and requires some few hours of lectures to convey.

    nanohub.org is an excellent place to start.
     
  4. Sep 11, 2009 #3
    I understand the E-field is unimportant, but I'm talking about the voltage drop per meter. If I already know the overall potential difference across the circuit, what is the importance in knowing how many volts there are per meter?
     
  5. Sep 11, 2009 #4

    berkeman

    User Avatar

    Staff: Mentor

    If you have a 1-meter wire, the battery will be short circuited. You mean a string of resistors of reasonable resistance, 1 meter long, right? Does that change your question?
     
  6. Sep 11, 2009 #5

    berkeman

    User Avatar

    Staff: Mentor

    V/m generally only comes up in E&M field values. Not in actual closed circuits.
     
  7. Sep 11, 2009 #6
    As I said, there's no importance in knowing that by itself. Even drift-diffusion equations (in the Ohmic limit) can be expressed in terms of a single voltage gradient...

    The electric field by itself doesn't say anything. In the case of pn junctions for instance, even in equilibrium you have enormous electric fields, but no net current across the whole device...

    The notion that the electric field "sweeps" all electrons is wrong. Only the states that are around the energy difference your voltage source creates are important. This might be too much for you at the moment, but there's no simple answer.

    I strongly suggest checking nanohub.org again! (specifically nanoelectronics/ supriyo datta)
     
  8. Sep 11, 2009 #7

    berkeman

    User Avatar

    Staff: Mentor

    sokrates, please stop posting like this. It is not addressing the OP's question and level of understanding.
     
  9. Sep 11, 2009 #8
    Never connect an ordinary low resistance wire to a battery,this will result in a short circuit and can be dangerous as well as expensive( you can zap your battery)
    High resistance wires are connected to batteries in circuits called potentiometers.By knowing the volts per metre you can measure the voltage of other supplies.Potentiometers have several other uses but are bulky and time consuming to use.You would use one as a voltmeter only when you wanted a really accurate value of voltage i.e. a more accurate value than an ordinary voltmeter would give.
     
  10. Sep 11, 2009 #9

    Dale

    Staff: Mentor

    I disagree most emphatically with the statements that the E-field is not important, it is the force on a unit charge. E=V/d is only correct for uniform E-fields, in general [itex]E=-\nabla V[/itex].
     
  11. Sep 11, 2009 #10
    Yes E is extremely important for example in charged particle dynamic applications such as X ray tubes and in dielectric breakdown of insulators
     
  12. Sep 11, 2009 #11
    Basically, yes.
     
  13. Sep 11, 2009 #12
    I would like to make it clear; I know the 1 meter wire would zap the battery. I discovered that when I was 6 years old.
     
  14. Sep 11, 2009 #13
    It depends on the resistance of the wire.I'm not sure what you mean with your question so can you be more specific please?
     
  15. Sep 11, 2009 #14
    My question is: is the E-field around a circuit meaningful? Or is it just an effect created by the moving charges?
     
  16. Sep 11, 2009 #15
    Hello user,there would be a potential gradient whether the charges are moving or not,for example there is a potential gradient across two separated parallel metal plates of opposite charge.With this example we can use the potential gradient to do calculations such as finding the electric force on an electron between the plates.
    As far as wires are concerned the concept of potential gradient is meaningful and useful when doing calculations with potential dividers and when using potentiometers as measuring instruments.There are probably other areas where the concept is useful but I cannot think of any at the moment.
     
  17. Sep 12, 2009 #16

    Dale

    Staff: Mentor

    It is always meaningful, it always means the force on a unit charge at that location. It is important any time that the force on a charge is important.
     
  18. Sep 12, 2009 #17

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Yes, the E-field is meaningful. But not particularly useful for calculating currents and voltages in circuits.
     
  19. Sep 12, 2009 #18
    I pulled out my 1956 issue of the ARRL handbook and looked up the resistance of 40 Ga copper wire (1069 ohms per 1000 ft), or roughly 3.3 ohms per meter. So the current for a 1 volt battery would be about 300 milliamps. not too much for most batteries.
    Bob S
     
  20. Sep 12, 2009 #19
    I can see that...But what practical use is of knowing the force on a charge in the circuit? Doesn't seem very practical.
     
  21. Sep 13, 2009 #20

    Dale

    Staff: Mentor

    True, that is why I only said that it is important any time the force on a charge is important.
    As Redbelly said, if your interest is only in solving circuit equations then you won't need it. However there are many cases where it (or its generalization) becomes important:
    writing the Lagrangian (most important)
    determining the force on a capacitor
    determining force in a mechanical device (e.g. motor)
    finding how hot a resistor will get
    determining the speed of an electron in an x-ray tube
    calculating the average force in the Drude model
    etc.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: What is voltage gradient good for? (Read post first)
  1. Reading up on voltage (Replies: 15)

Loading...