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Conducting Sphere

Here’s the figure.The thin ring is the conducting spherical shell. Now, for all Gaussian surfaces of radius less than R, the enclosed charge is Q only. So, shall we have a non-zero electric field at those points? Doesn’t electric field have to be 0 everywhere in the region enclosed by a conductor?

This second figure shows one such Gaussian surface I am talking about. 2nd figure

The blue line indicates the Gaussian surface. Will the electric field at points on the circumference of the blue Gaussian surface be 0 or non-zero?

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  • $\begingroup$ it has to be zero if "inside", not if "enclosed" $\endgroup$ – user126422 Jun 30 '17 at 12:51
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From the Feynman lectures on physics, volume II, 5.1

We return now to the problem of the hollow container—a conductor with a cavity. There is no field in the metal, but what about in the cavity? We shall show that if the cavity is empty then there are no fields in it, no matter what the shape of the conductor or the cavity.$$$$[...]$$$$ We have always said “inside an empty” cavity. If some charges are placed at some fixed locations in the cavity—as on an insulator or on a small conductor insulated from the main one—then there can be fields in the cavity. But then that is not an “empty” cavity.

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