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How Does a Faraday Cage Work?

  1. Apr 21, 2007 #1
    I've looked this up extensively on the web but nothing seems to answer this question satisfactorily. For example, here's Wikipedia's answer.

    I try thinking of it considering the properties of a conductor in electrostatic equilibrium. The charge on a conductor in equilibrium resides on the surface and the electric field is zero inside. Okay...

    But once lightning hits a metal cage with a person inside, isn't the cage momentarily NOT in electrostatic equilibrium? Isn't the cage not in electrostatic equilibrium until the charges are done redistributing themselves, and in this period of time, won't the person inside be fried?
  2. jcsd
  3. Apr 21, 2007 #2


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    i think you need to think about Gauss's Law and what the component parallel to the cage shell of any E field is.
  4. Apr 21, 2007 #3
    Those are words that make no sense to me. I know Gauss's Law, but what does it have to do with lightning striking a cage and the person inside not being hurt?
  5. Apr 21, 2007 #4
    This is true, but lightning is a very different phenomenon than electromagnetic radiation. Lightning is an electrostatic discharge, and it will deposit a charge on a conductor, resulting in an electric current. EM radiation requires an electric field to pass through the conductor, which will quickly rearrange its charges to cancel out the field.
  6. Apr 21, 2007 #5
    So why doesn't the lightning hurt the person inside?
  7. Apr 21, 2007 #6
    Why does lightning even hurt people? Why does it kill people?
  8. Apr 21, 2007 #7
    Last edited by a moderator: Apr 21, 2007
  9. Apr 21, 2007 #8


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    it has something to do with the fact that there can't be any electric fields inside the cage. whether or not there is a nasty E-field just outside of the cage.
  10. Apr 21, 2007 #9
    When the lightning strikes the metal cage, why isn't there an electric field on the inside? The cage isn't in electrostatic equilibrium anymore, it just got some charge added to it and caused a current to flow.
  11. Apr 22, 2007 #10


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    This is correct, but the deviation from equilibrium comes about from the finite conductivity of the cage. There is a relaxation time associated to a conductor (rho/epsilon) which is the time constant which governs the redistribution of charge in a conductor which is not in electrostatic equilibrium. For copper, it is of the order of 10^(-19) seconds, which means that electrostatic equilibrium is re-established on that timescale (see for instance section 4.3.5 in Lorrain, Corson and Lorrain).
  12. Apr 22, 2007 #11
    Maybe it would be appropriate if we decide if we are talking about a Faraday cage made of ideal conductor (zero resistivity) or of real metal like copper.
    The thunderbolt implies a large current. If the cage where made of perfect conductor, there will be no difference of potential between the hit point and the exit point. If the cage if made of normal metal there will be an difference of voltage due to the current going through the metal resistance. This voltage difference will be measurable from the inside of the cage. Will it be big or not depends on the current and the resistance of the metal.
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