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Electric Breakdown

  1. Dec 26, 2005 #1
    Hi,
    I have recently been reading through / self learning some material on Electrostatics and I came across this question.

    The minimum electric field that can be supported by dry air at atmospheric pressure is about 10^5 volts/m. What is the maximum potential difference to the earth for a conduction sphere of radius 10cmin air? (Take the distance of the sphere to the earth to infinite).

    I can't quite figure out how to approach it. I have tried placing a charge q inside the sphere and then calculating the value of q supposing the electric field on the surface is 10^5 v/m and from that calculating the potential at the surface. However this answer is incorrect and I am out of ideas. Any help would be much appreciated.

    Regards,
    Lucas
     
  2. jcsd
  3. Dec 28, 2005 #2

    CarlB

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    Take the earth's surface to be a conducting plane. You probably were recently exposed to the question of what the electric field of a point particle looks like in the neighborhood of a perfectly conducting plane.

    Your answer should come in terms of volts per meter. That is, even though you're assuming the sphere is an infinite distance away from the earth, the potential is still going to depend on that distance. The question seems to be worded in a kind of messy manner.

    Carl
     
  4. Dec 28, 2005 #3

    CarlB

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    When you get your answer, you will have to find the place where the electric field is greatest, as that is where the breakdown will occur.

    As a matter of intuition, I will share with you the well known fact that electric fields tend to be greatest where the conductors are sharpest. Since the globe (a sphere) is sharper than the earth (a plane), that is where the breakdown will occur.

    If you've got much time arc welding, you will find that it is easier to direct the arc to the edge of a piece of a work than it is to get it to heat up the opposite of an edge. For example, it is easier to weld to the outside edges of a box than the inside edges. This is at least partly due to the fact that the electric field is highest on those outside edges (which are "sharp") while the electric field is lowest on those inside edges (which are even duller than a flat sheet).

    As a youth, I climbed up a mountain and got a tour of a lightening research lab. They had a pole on top of the mountain that they hoped would get struck by lightening. To keep it highly charged, they had to prevent the air from breaking down near it. That meant that they had to use a large sphere instead of a small one.

    A small sphere will break the air down in an effect that the sailors called "St. Elmo's Fire". When they first built their pole, they miscalculated how much electric field they would get and used a sphere that was too small. So instead of getting struck by lightening, it constantly hissed with St. Elmo's fire.

    The sphere they ended up choosing was about a yard in diameter. Underneath it was a chamber with conducting walls (called a "Faraday cage") from which they could safely watch the pole get struck by lightening. When this happened, the electricity was conducted to a ground made of wires staked into the ground over a large region of the mountain. There was a very small electrical resistance between the Faraday cage (which supported the pole) and the mountain, and this resistor would get a voltage on it due to the current of the lightening strike. They measured the voltage using equipment.

    For someone in the Faraday cage, I would think that getting so close to a lightening strike would be exciting enough that I'd keep a fresh change of underwear convenient.

    What I mean to be saying by all this is that when you get your answer, it's going to depend on the diamater of that ball. Just like the lightening observatory needed a certain size ball for a given electric field, so you will be doing the reverse problem. That is, given the ball, you will be calculating the electric field.

    Go to it.

    Carl
     
    Last edited: Dec 28, 2005
  5. Dec 28, 2005 #4
    I understand that electrical breakdown will occur at the sphere ie. where the electric field is greatest and where the radius of curvature is least. However i am unable to calculate at what potential the sphere will breakdown. I tried
    Unfortunately this method failed and I am unsure why?

    Working through the problem logically I image that as the conducting sphere is placed in the field it will develop a potential difference to the earth, equal on all parts of the sphere and in a field of 10^5 v/m the potential developed will be enough to ionize the air and form corona, ie. breakout. This is where I am stuck; I do not know how to calculate the potential that exists on the surface of the sphere?

    Much appreciative,
    Lucas
     
  6. Dec 29, 2005 #5

    Tide

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    Perhaps if you showed us some details we could help figure out what the problem is.

    You seem to be on the right track. You want to set the field at the surface of the sphere to [itex]E_0 = Q/R^2[/itex] (fill in the appropriate constants to accomodate your specific choice of units!). Use that value of Q to find the potential [itex]V_0 = Q/R[/itex].
     
  7. Dec 29, 2005 #6
    Thanks to all for your help!

    I was able to get the correct answer using the methods outlined in my first post. It seems that my calculator was using a different a approximation for the permittivity of free space. :mad:
     
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