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Lightning Rods

  1. Apr 28, 2009 #1
    What are the physics behind using a pointed lightning rod as opposed to a rounded one. I believe it has something to do with the electric field being concentrated to a point intensifying the field but can someone thoroughly explain it. Thanks!!!
     
  2. jcsd
  3. Apr 28, 2009 #2
    The point of the point (:blushing:) is to ionize the air and initiate a breakdown.
     
  4. Apr 28, 2009 #3
    The charge is more concentrated round the point and this sets up an intense electrical field which breaks down the insulation of the air.This results in charged particles moving between the point and the cloud this achieving some neutralisation.
     
  5. Apr 28, 2009 #4

    ZapperZ

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    It turns out that if one were to solve the field equations (i.e. Poisson's equation) whereby you put a sharp, pointing metal in, say, a uniform electric field, something called "field enhancement" occurs in the region of the sharp, pointy object. This is purely classical E&M solution out of Maxwell equation. So such sharp objects causes an enhancement of the electric field at its location. So the geometry of the object within such field is the cause.

    The resulting effect here is that there's a huge charge accumulation in that region. With the enhanced field, and the large charge accumulation, a phenomenon called "field emission" can occur (or at least, has a higher probability of occurring). This is where the charge carriers can quantum mechanically tunnel through the surface barrier. When this occurs, a number of possible mechanism can occur that can lead to the ionization of the surrounding air, and that can lead ultimately to an air breakdown/lightning/etc.

    Zz.
     
  6. Mar 28, 2010 #5
    Hey i am interested in something here

    what hapens if we place an irregular conductor in a uniform electric fields! Actualy whats the distribution of charges at the ends and at the middle and every where, whats the net electric field inside the conductor, and whats and how is the potential over the surface of that conductor?
     
  7. Mar 29, 2010 #6

    Born2bwire

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    The electric field inside the conductor will be zero, likewise the potential will be a constant zero. The charge distribution would depend on the geometry of the conductor.
     
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