Conductors & Fields: Feynman's Argument Explained

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The discussion centers on Feynman's argument regarding the electric field (E) inside a conductor's cavity, asserting it is zero, and extends this to suggest that the field outside the conductor must also be zero if there is charge within the cavity. A participant raises a doubt about this conclusion, noting that a Gaussian surface around the conductor would show a non-zero net charge, leading to a non-zero integral of E. The resolution lies in the concept of grounding, which allows charge to flow into the conductor, balancing the charge in the cavity and resulting in zero net charge within the Gaussian surface. This charge rearrangement under electrostatic conditions ensures that E remains zero both inside and outside the conductor. The discussion highlights the nuances of electrostatic shielding and grounding in relation to electric fields.
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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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