Why does Gauss's law seem violated when a charged cavity shields external fields?

[atavistic]

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 zerK. 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?

 

[atyy]

Feynman made a mistake. See the comments by Thorne at the bottom of http://www.feynmanlectures.info/flp_errata.html.

 

[atavistic]

How does grounded-ness preserve the argument?

 

[dx]

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.