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Free electrons in conductors

  1. Apr 24, 2015 #1
    when the external electrical potential difference is applied across the conductor ,electrons flow in the direction opposite to the current.Whether the number of free electrons in the conductor decrease?
    I think ,No.As electrons return ,because they move in closed loop.
     
  2. jcsd
  3. Apr 24, 2015 #2
    your question is not clear, can you please restate it?
     
  4. Apr 24, 2015 #3
    Electrons, like fluid, flow from an area of high concentration to low concentration. This is why we have electron current (essentially DC- to DC+ in a DC circuit) and conventional current (DC+ to DC-). You don't actually lose electrons per se; the force acting them is reduced as they perform work.

    It would also help if you could clarify between whether you're referring to AC or DC. If it's AC the electrons continually move back and forth according to the electromagnetic force acting on the conductor, in DC the electrons will move from the high-concentration area to the lower-concentration area until equilibrium (zero voltage) is reached.
     
  5. Apr 24, 2015 #4
    I have stated the question as it was given in my textbook.
     
  6. Apr 24, 2015 #5
    Then IMHO the textbook needs to be re-written. "electrons flow in the direction opposite to the current" is extremely misleading. Current is itself the movement of electrons. I think what the textbook may be driving it as the issue of conventional vs. electron current flow.
     
  7. Apr 24, 2015 #6
    Yes.
     
  8. Apr 24, 2015 #7
    Am I right.
     
  9. Apr 24, 2015 #8
    I honestly don't know because I'm not sure what exactly the question is. Is the direction of electron current the opposite of conventional current in a DC circuit? yes.
     
  10. Apr 24, 2015 #9

    ZapperZ

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    By convention, the direction of current is opposite to the direction of motion of electrons.

    Haven't you connected an ammeter to a circuit? What does an ammeter measure? And do you think if you sit there and look at the measured current long enough, the current drops over time, i.e the "free electrons in the conductor decreases"?

    If this is a question from your textbook, it should have been done in the HW/Coursework forum, not here.

    Zz.
     
  11. Apr 24, 2015 #10
    All these information is not given .It was just a conceptual doubt.
     
  12. Apr 24, 2015 #11

    ZapperZ

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    I thought they were out of your texbook?

    Besides, haven't you done simple basic circuit already? Or haven't you looked at the light bulb in your house after you turned it on? Do you see it diminishing in brightness over a short period of time, which would be consistent with the circuit losing electrons?

    What I'm trying to teach you here is to look around you and see if you can find the answer yourself. Just because physics is taught in a classroom in school, it doesn't mean that it has no connection to reality or the world around you. You need to ask yourself "If it is true that the circuit is losing electrons, then....... " (i) you should expect the current to drop over time and (ii) things connected to it will start behaving in such a way as to exhibit this drop in current over time, and (ii) my lightbulb should start getting noticeably dimmer!

    Zz.
     
  13. Apr 24, 2015 #12
    No.
     
  14. Apr 24, 2015 #13

    ZapperZ

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    Great!
     
  15. Apr 24, 2015 #14
    Thanks.
    Is it correct?
     
  16. Apr 24, 2015 #15
    In as much as the question seems to be asking about losing electrons, yes. Like I said though the question isn't worded real well; in particular it doesn't take into account the circuit you're working with.

    Let's use Zz's example of a light bulb's brightness. No, you will not see the bulb in your house get dim as you leave the light on because there's always a power source (the main feed to your home). However, if you take the same bulb and put it in a DC circuit, you WILL notice it steadily getting dimmer over time, with how much time being dependent on P=VI, where P=power in watts, V=voltage, and I=current flow (amperage) and then calculated against the battery's rating. Eventually the bulb will go out. However, this isn't because any electrons have been "lost", it's because there is no longer a force to move them. Voltage is a potential difference between two points. If you attach a DC power source (battery) to a load, the difference in electron concentration will cause the electrons to flow from a high concentration to low concentration. This will continue until they reach a point of equilibrium throughout the circuit, ie zero voltage. At that point there is no longer any force to move the electrons. They are still there; they're just not moving!

    AC brings a different set of variables but the basic concept is the same; as long as there is a force to move the electrons you will have electric current until the force is removed. As such the electrons are not "lost" so much as they are moved through the conductor, but they remain in the circuit itself.
     
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