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Electric Field distribution around two-conductor cable

  1. Mar 23, 2016 #1
    What is electric field distribution around two conductor cable connected to DC power supply? Assume power supply is isolated and not grounded. Assume cable is straight.

    Case 1: no current runs through the cable. One wire is positive, another – negative. Negative or zero? If we think in terms of charge: Positive wire has positive charge in it, electric field distribution is radially outwards. What charge is in negative wire? This will determine total field distribution. If there is not charge in negative wire total field distribution will be that from positive wire only - radially outward. If there are negative charges in negative wire the total distribution form the cable will be like from dipole – on positive wire side field will be outwards, on negative wire side field will be inwards. Which field distribution is correct?

    Case 2: load is attached to cable and current flows. Will the Electric field distribution changes?
  2. jcsd
  3. Mar 24, 2016 #2


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    Case 1. The dipole case applies, because the field lines from each conductor have no where else to go to.
    Case 2. Nothing changes once the steady state is reached. But at switch-on, for a brief moment, an electric field will arise acting along the wire. This field is the one which accelerates the electrons up to their steady state velocity.
  4. Mar 24, 2016 #3
    1) In my opinion, at first an electric charge-one positive and the other negative-will be installed in each conductor according to voltage and capacitance Q=Cap*V.

    The electric field will be E(x)=Q/eps/(2*pi)/x where eps=permittivity x=distance from conductor centerline.

    No field will be in inner part of conductor since E=ro*J and J[current density] =0.

    2) If a current will flow through conductors the potential difference will change with voltage drop and Q will change. In conductor inner part will be an electric field according the current density J and resistivity[ro].
  5. Mar 25, 2016 #4

    The electric field will be E(x)=Q/eps/(2*pi)/length/x [from both sides]. One has to add vectorially the electrical fields from both charges-positive and negative-in any point.
  6. Mar 25, 2016 #5


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    All agreed, but if there is a volt drop in a wire, there will also be a field component acting along the wire.
  7. Apr 2, 2016 #6
    Case 1: Dipole distribution is correct

    Case 2:Electric field distribution changes when load current flows. When there is no load current and the circuit is open, surface charge distribution on the conductors connected to DC source is such that there is no electric field inside the conductor. Hence the charges are in equilibrium. If we assume a uniform cylindrical conductor, this means a uniform surface charge distribution.
    When load is connected, surface charges are realigned such that there is a net electric field inside the conductor which helps the moving charges to overcome the resistance offered by conductor material. Hence the field distribution is also changed accordingly
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