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Electric Field Propagation

  1. Mar 26, 2007 #1
    Can anyone explain how an electric field propagates from one point to another?

    For example if i touch two 6ft wires to the ends of a 5V battery, i'll get 5V across the end of the wires after a propagation delay. How does the electric field (voltage) of the battery make it's way down to the ends?

    also, if my understand is correct, there are no additional electrons being added, so are the electrons just under a greater force now?

    If electric field is a measure of distance, do i get a different potential if i hold the wires 1 ft apart as opposed to 3ft apart?

  2. jcsd
  3. Mar 26, 2007 #2


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    I think you are mixing up the concepts of field, force and potential.
    Try looking this over to see if it clears anything up.

    But to answer the question of how the field propogates. Maybe this visual representation will help. The force of the field is dependant on the amount of charge, and the position of the charges. So when you add the battery to the system of the wires you cause a shuffling of the charge in the wire. But the shuffling is not initially equal because the force by the battery varies with distance. But this first shuffling puts some of the charge in a new position, which means a new set of forces, which means more shuffling. Repeat (or propogate) until all the forces balance and the system is stable.

    This is the mathematical model used to describe it:
  4. Mar 26, 2007 #3
    es1, thanks for the reply! ...

    there's still some gaps in my understanding ...

    say you have a 5V battery ... on the +ve terminal we have an excess of postive charge and on the -ve terminal we have a negative charge. in essence the battery acts as only a pump and does not provide any extra charge to the circuit.

    now when i add the battery to my two wires , when you say 'shuffling of charge' do you mean that the positive charge from the battery makes it's way to the end of the wire by pulling in an electrons and thereby creating another another positive charge? if would start a "chain reaction" until the reaches the end...if this were the case, then we would only have the voltage on the end of the wire and no where else ... i guess i understood your reply wrong.

    also, what would be the difference between a 5V line and a 3V line. do we just have more positive charge along the wire since we have more positive charge on the battery?
  5. Mar 26, 2007 #4


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    When I say 'shuffling of charge' I mean the new net force on the charge gives it an acceleration, and thus a change in position. Not very precise, I admit. :)

    Why do you think there would only be net charge at the end of the wires?

    If you draw this case I think you will find the net force on all the charges is not zero, so there will be a force on some, which means there will be charge movement. So this is not the stable state.

    To really understand how the charge moves and reaches the stable state you will have to study the telegrapher's equation. I am not sure if the usual water analogy works here (based on your description of a voltage source as a pump). Maybe if the pipe was only half full of water or something, then it might work, maybe.
  6. Mar 26, 2007 #5


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    This can give you a visual representation of what you want.
    Set Rsoure equal to Zo, and make them small. Set Rload to the largest value. Then hit start. The applet will now sim the situation you describe. Pay close attention the the voltage vs position plot in the lower left.
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