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Faraday cage potential

  1. Jul 5, 2011 #1


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    I was wondering about the Faraday cage. When lightning strikes a car, the driver is not hurt since he/she is sitting in a Faraday cage. The charges cancel out and the net E field in the car is zero, though the potential inside the car should reach whatever is on the outside, meaning maybe hundreds of thousands of volts, which should kill the driver when eg. he grabs one door handle with one hand and the other handle with the other hand (potential difference)?.

    Maybe I answered my question here, there is no potential difference, since it is all constant voltage?

  2. jcsd
  3. Jul 5, 2011 #2


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    What do you mean? The potential between the ground/care and the incoming stroke of lightning is very very high, otherwise the lightning wouldn't be able to form. However, the current initially flows through a small area through the air that has been ionized. Once the current gets do your car, it now has a much easier path to travel and doesn't need ionized air to conduct it. The inside of your car is generally not completely bare metal, so the current would much rather travel around the frame and body than through the compartment and the passenger.

    I believe that the current and voltage of your car is about equal throught the entire body/frame, similar to how parallel circuits split their current and have the same voltage. If so, then there shouldn't be a potential difference between door handles.
  4. Jul 6, 2011 #3


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    So it is Kirchoffs voltage law - "the directed sum of the electrical potential differences (voltage) around any closed circuit is zero." I guess.
  5. Jul 6, 2011 #4
    Gauss bro, s'all about Gauss's Law.
  6. Jul 6, 2011 #5


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    With gauss you come to the conculsion that


    I find that that does not explain why you don't get electrocuted in the car. Though thinking of the car as a closed circuit does.
  7. Jul 6, 2011 #6
    Kirchoff may also be a way to do this, but I found that Gauss was easier to understand. The key is Q inside, think of the lighting bolt as a ton of excess electrons. When it strikes the car, all of the excess electrons remain on the outside of the car. Then applying Gauss's law, since Qinside = 0 ==> Vinside = 0.
  8. Jul 6, 2011 #7


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    It does not work like that. Think of how potential is defined, via doing work on a charge to home it in from infinity. If the potential of a lightning bolt is 100000V, then by Gauss, the [itex]V_{inside}[/itex] is 100000V.
  9. Jul 6, 2011 #8
    I'm sorry but I disagree. Treat the metal frame of the car as a Gaussian surface, a simpler example is the potential in a hollow conducting sphere. If you look at this example:

    3.2.3 Electric Potential Inside a Spherical Thin Shell Conductor, from

    http://www.physics.upenn.edu/courses/gladney/phys151/lectures/lecture_jan_31_2003.shtml [Broken]

    it's the same concept. But here is the short version:

    "Inside the sphere [or car frame in our case], the electric field is zero since any Gaussian surface we draw which is completely contained inside the sphere would contain no net charge."

    Hope this helps.
    Last edited by a moderator: May 5, 2017
  10. Jul 8, 2011 #9


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  11. Jul 8, 2011 #10
    whoops, you are right. That's my bad.
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