How does a Faraday Cage block electric fields?

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solzonmars
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Why does a Faraday Cage work? (or more generally, why does the inside of a conducting shell have no electric field if there are charges placed outside it?)

I understand that this is the result of polarisation, but why does the polarisation happen to exactly cancel out the field? Could there be a way to prove this with Gauss' Law?
 
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Ah, so it doesn't actually cancel out fully. I was intending to use a Faraday cage as an example of a charged conductor (in the Falstad stimulation I had actually used a spherical shell). Any idea why a spherical shell would cancel out the field though?
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solzonmars said:
why does the polarisation happen to exactly cancel out the field?
If it isn't balancing then charge will flow until it is. A Faraday cage is metallic (a good conductor) to allow charges to move as necessary.
 
@sophiecentaur I understand what you mean for a solid conducting object, but for a shell does the electric field in the centre cavity actually affect the electrons in the conductor?
 
solzonmars said:
@sophiecentaur I understand what you mean for a solid conducting object, but for a shell does the electric field in the centre cavity actually affect the electrons in the conductor?
It does. If there is charge inside the shell, an E field exists within the shell. Opposite charge (which is electrons moving either towards the inside surface or away from it depending on the sign of the inside charge) will move to the inside surface of the shell due to electrostatic induction. The opposite charge will be equal in magnitude (total coulombs) to the internal charge so that the net E field within the metal is everywhere zero. And charge equal in magnitude and sign to the internal charge will accumulate on the outside of the shell.
 
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Some engineers spend a lot of time trying to prevent high frequency electric fields escaping from products like computers and smart TV. Using a Faraday cage is one tool in their armoury.