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In Griffiths' Introduction to Electrodynamics, at the Electrostatics chapter, in particular, in the conductors section, he says this after the stating that within an empty cavity surrounded by a conductor material, the electric field is 0:

"(...) This is why you are relatively safe inside a metal car during a thunderstorm - you may get cooked, if lightning strikes, but you will not be electrocuted".

So, in a conductor, if I bring (for example) a positive charge q outside the conductor, induced negative charge will appear at its surface, negating the positive charge, killing off the field of q for points inside the conductor. My question is: how fast does such process occurs in order to negate the charge of a lightning strike, "protecting" me inside the cavity (interior of the car)?

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  • $\begingroup$ "in order to negate the charge of a lightning strike" - I'm not sure what that means. Will you elaborate? I'm under the impression that lighting is essentially a current. $\endgroup$ – Alfred Centauri Jun 25 '18 at 1:39
  • $\begingroup$ What i'm picturing is the lightning strike as a "source of charge" that will spread around the car. What i'm really trying to picture is how the car can protect me of such lightning strike, as if it were a faraday cage. $\endgroup$ – RicardoP Jun 25 '18 at 2:39
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The car dos not "repel" the lightning strike, as you seem to be picturing. Yes, it may actually separate the charges in the car's metal surface. But, say, the car has 5 positive particles and 5 negative. Even though they might be separated in different parts of the surface, the resulting charge in the car is still zero. Therefore, it cannot "repel" the lightning strike.

What does happen is that the electrons in the lightning stick to the car's surface, because the metal in it is a good conductor. That's why you're safe in the interior of the car, because the electric charges of the lightning strike cannot reach you.

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You are forgetting how lightning strikes are formed. When the electric difference potential, between a charged cloud and the ground, is greater than the dielectric constant of the air, a current goes down from the negative pole (Cloud) to the positive pole (Earth). So, what he is trying to say is that the current will travel along the conductor and because the electric potential is zero, there cannot be any charge on the internal surface of the conductor.

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If a container has a conducting surface it forms a Faraday cage.. There is a related answer here.

A car, if it is not plastic, is a metalic enclosure on rubber wheels. To be struck by lightning, ionizations paths between the charges on the ground and the opposite in the clouds form , and current flows discharging the energy. See this video of a car hit by lightning . Obviously the energy was not discharged inside the car, which as a Faraday cage did not offer inside the least resistance path for the current to flow to the ground.

So, in a conductor, if I bring (for example) a positive charge q outside the conductor, induced negative charge will appear at its surface,

yes,

negating the positive charge,

this is confusing.

What happens is that the surface of the conductor will become negative .The field of the positive charges stay within the conductor material ( or flow to the ground if grounded) and cannot penetrate inside a metallic enclosure, a Faraday cage, and thus no current can be induced inside the car for the lighting current to flow.

It is Gaus' law that does not allow field lines within a conducting enclosure. See "inside a sphere of charge"

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    $\begingroup$ Note that video is a movie stunt and the apparent lightning is lens flare from the pyrotechnics. There are people watching from the roofs and the spectators that rush out are all lined up along the buildings waiting for their que. $\endgroup$ – C. Towne Springer Jun 25 '18 at 7:45
  • $\begingroup$ @C.TowneSpringer ok, I see it is very popular on the net and the locations is changing. I will replace it with an experiment. $\endgroup$ – anna v Jun 25 '18 at 9:14
  • $\begingroup$ Lightning can behave in extremely odd ways. While you'll be safe with a 1MV discharge, or if a high voltage line drops on your car (as long as you don't try to step out...) real lightning is a different animal. It can even travel through windows, destroying the glass first. $\endgroup$ – hdhondt Jun 25 '18 at 10:22
  • $\begingroup$ @hdhondt walls and windows do not make a Faraday cage. There is even ball lightning. youtube.com/watch?v=D62KzIc-R-4 The question is about conductors. $\endgroup$ – anna v Jun 25 '18 at 10:41
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The reason a person in a car does not get hit by a lightning strike is because the roof and walls of the car present a path of less resistance than the interior of the car.

For the same reason, a person could also be protected from a direct strike by a nearby tall conductive object - not a Faraday cage, but it would not be as reliable as a car and the person would have much greater exposure to light, heat and EM field produced by the current of the strike.

Your question is about the exposure of a person in a car to the EM field. It should be relatively low, but not zero, since a car has windows. The effect of the finite speed of the electrons cancelling the field should be negligible.

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protected by Qmechanic Jun 25 '18 at 7:09

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