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Electric field inside a resistor

  1. Apr 4, 2004 #1
    A resistor (20 ohm) is made of a very thin piece of metal wire, length = 3mm and diameter = .1mm. Given that it has a potential of 8 volts and .4 amps of current running through it, what is the electric field inside the wire?

    I know there's no electrostatic equilibrium here, so I can't just take the easy answer and say there is no electric field internal to a cylinder. It seems I must use the equation that relates voltage to electric field:

    V = integral(E * ds)

    But, as this seems to be based upon Gauss's law (which I haven't the firmest hold) I hate to assume that I can use the equation when the charge is in motion (current).
     
  2. jcsd
  3. Apr 4, 2004 #2

    Doc Al

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    Staff: Mentor

    Your equation is almost right: ΔV = - ∫ E ds. This is not based on Gauss's law, but on the relationship between field and potential.

    In your example, the magnitude of the field is E = ΔV/L.
     
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