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)?
Best Answer
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.
yes,
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"