Conductors & Fields: Feynman's Argument Explained

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atavistic
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I had been reading Feynman lectures , and in it he has shown an argument which proves that E field inside an empty cavity of a conductor is zero.OK. Now he says a similar argument can be used to show that if there is some charge in a cavity of a conductor than the field outside must be zero. Electrostatic shielding works both ways. Doubt: But then if we consider a gaussian surface containing the conductor , then the net charge is not zero => integral(E.da) is non zero, but E is zero. HOW?
 
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Feynman made a mistake. See the comments by Thorne at the bottom of http://www.feynmanlectures.info/flp_errata.html.
 
How does grounded-ness preserve the argument?
 
If the conductor is grounded, an amount of charge equal and opposite to the amount in the cavity can come into the conductor and the net charge inside a Gaussian surface containing the conductor would be zero. Of course this doesn't necessarily mean E must be zero, but it turns out that under electrostatic conditions the charges always rearrange themselves such that it is.
 
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