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E field between 2 parrallel conductors

  1. Mar 18, 2010 #1
    I’m trying to determine the how close 2 parallel infinitely long cylindrical conductors can get before they will arc together. Assume the conductors will be separated only by air, and I know that the Dielectric Strength of air is 3x10^6 V/m. So I figured that if I can come up with a formula to determine the E field I can find where the E field = the dielectric strength of air which should be the point at which they will begin to arc.

    Assuming the following...
    I have two parallel conductors with radius r, separated by distance d. Each conductor is carrying a time varying (AC) current I at 60 hz. The currents are flowing in opposite directions. The conductors also have a time varying (AC) voltage of V, but I don’t think this matters for determining E field magnitude.

    What I hope to come up with is a formula for the magnitude of E in terms of r, d, I, and V.

    Can anyone please help with the formula for E.

    P.s. This is not a homework question but if it would be moer apprpriate for that section I could post it there.
    Last edited: Mar 18, 2010
  2. jcsd
  3. Mar 18, 2010 #2
    Well, divide voltage by the distance between plates.
  4. Mar 18, 2010 #3
    sorry, my conductors are wires not plates

    But If I simply divide votlage I find that at 13200 volts the distance is only 4.4 mm and I find it really hard to believe that I can get a wire at 13kV that close to ground without it arcing
  5. Mar 18, 2010 #4
    You're right. Look up the equation for a long wire. You will have to approximate the wire with a finite radius and set that as your reference voltage. The e field from a small wire gets very large.
  6. Mar 18, 2010 #5
    Does anyone have have or know the equation for a long wire? I don't have any texts on this subject and I can't seem to find anything online.
  7. Mar 18, 2010 #6
    The two parallel cylindrical conductors form a transmission line of capacitance

    C = πε0[cosh-1(d/2r)]-1 Farads per meter

    See Smythe Static and Dynamic Electricity 3rd edition page 78. Because the signal propagation velocity is c=1/sqrt(ε0μ0) = 1/sqrt(LC) (where L is the inductance per unit length), L can be calculated. The transmission line impedance is Z = sqrt(L/C), so the ratio of voltage to current can be calculated. Any signal put on these two conductors will have this ratio of voltage to current, and it travels away from the source at the speed of light. You cannot put a voltage on these infinitely long lines without simultaneously supplying current, because any charge put on these two conductors travels away from the source at the velocity of light.

    There is a repulsive magnetic force between the two conductors, due to the I x B Lorentz force, and a simultaneous attractive Coulomb force due to the opposite charge Q=CV per unit length.

    The surface of each conductor is an equipotential, but the charge distribution is not azimuthally uniform. Thus the surface electric field is not azimuthally uniform. This can be solved for using the 2D electostatic equations in Smythe Chapter IV by applying the proper boundary conditions at the conductor and at a flat conducting midplane (plane of symmetry). (I will think about it).

    Bob S
  8. Mar 19, 2010 #7


    Staff: Mentor

    That seems to be the right order of magnitude. Spark plugs typically operate in the 12-25 kV range with a gap of about 1-2 mm.
  9. Mar 19, 2010 #8
    You may be able to get an approximate answer by using an infinite line of charge:

    [tex]E = 2k\frac{\lambda}{r}[/tex]

    So, you have two wires (where I put the other at radius d):

    [tex]E = 2k\lambda [\frac{1}{r} - \frac{1}{r - d}][/tex]

    And now you have to use your boundary conditions of the voltage at the surface of the wires.

    [tex]V=2k\lambda [ln(r_2) - ln(r_2 - d) - ln(r_1) + ln(r_1 - d)][/tex]

    keep in mind this would really only be a good approximation if the distance (d) is much larger than the radius of the wires. But its up to you to use.
    Last edited: Mar 19, 2010
  10. Mar 21, 2010 #9
    This is right sort of first-cut, but things are a bit more complex than this. For example the standard way to measure high voltage is to use two spherical electrodes spaced somewhat closer than the diameter. Standard measurements for such produce figures such as 17 kV for 5mm spacing with spheres of 5, 10, 20 etc Cm diameter or about 30 KV for 10mm spacing. If you go to larger and larger spheres the readings stay the same. If you go to smaller spheres or make the spacing much farther than the diameters the numbers begin to go all over the place.

    The reason for this has to do with the mechanisms of breakdown. A high voltage between two conductors creates an electric field there. This electric field depends upon the voltage and the geometry of the conductors. Where the radius of curvature of a conductor is small, the E field becomes very large. This large field is capable of pulling electrons out of the conductors creating a "pre-breakdown current". That current then becomes a precursor to actual breakdown which is an avalanche phenomena.

    In this case where conductors are cylindrical (we are ignoring "end" fields) the breakdown again depends on the spacing compared to the diameter of the cylinders. It's a sort of modification on the two spheres thing. If the spacing is close the results are very similar to using spheres. If you are using thin wires then fields are high and with large spacing values become erratic and hard to predict. High voltage breakdown is not a simple phenomena.
  11. Mar 21, 2010 #10


    Staff: Mentor

    Yes, that is all I intended. I think he was just surprised by how high the voltage was and how short the distance. Most people don't realize that "everyday" static discharges can be in the 100s of kV range and only jump a few mm.
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