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Suppose we had a hollow conducting sphere with a net charge q on it. There is no charge in the cavity; the conductor itself has a charge q. The idea is that this net charge would reside on the 'surface', since the conductor has free charges otherwise to make the net field inside the meat of the conductor zero.

  1. My question is with regard to what 'surface' means. Does 'surface' refer to the interface between the conductor and air? If so, why isn't there charge residing on the inner surface of the hollow sphere? Has this something to do with uniqueness theorems?

  2. Related question: If there is an external charge q outside an uncharged hollow conductor, why is there no induced charge on the inner 'surface'?

  3. Also related: How is it that information of charge inside a cavity is known outside, but that outside is completely unknown inside? In a sense, aren't both regions of air the same and separated by the conductor only? Better put: when there is a charge inside the cavity, the inner surface charge distribution cancels the field in the conductor due to the cavity charge, and residual charge q sits uniformly on the outer surface, thus 'revealing' the presence of q to the outside. Why does the same not happen to a charge outside?

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  • $\begingroup$ What do you mean by a hollow sphere, is it have thickness? $\endgroup$
    – asd.123
    Oct 5, 2020 at 14:59
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    $\begingroup$ @Butane yes, it does have thickness. So think of a solid conducting sphere of radius say, 1m, and a concentric solid sphere of radius 0.5m removed from the first sphere. $\endgroup$ Oct 6, 2020 at 4:47
  • $\begingroup$ Regarding "knowledge": that an ideal conductive shell can screen the structure of external fields completely requires an infinite number of degrees of freedom. Imagine a shell which has only two charges (one positive, one negative) that can move freely: in an external electrical field the only "response" of this system is a dipole moment. Four balanced charges can cause a dipole and a quadrupole moment etc. but there will always be a residual field on the inside. Only an infinite number of charges can zero out the external field, but even they can't remove a net total charge. $\endgroup$ Apr 28, 2023 at 17:28

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Initially, the meaning of the surface for your question is the upper boundary of the sphere which is equivalent to the sphere shell with a thickness infinitesimally small (mathematically, goes to zero). When it comes to the problem of why charges are not residing at the inner surface or in between the layers is, I need to explain how the charges are defined. Charges are defined as points so if you consider the cardinality of the surface of the sphere you will see that it is capable of accomodating uncountably infinite amount of charge. If you have any suspicion about injecting charges to the initially charged surface, a sphere shell can carry all the charges on only the (mathematical) surface. Also, the charges distribute themselves in a fashion that minimizes their potential energy, so if you consider Coulomb's law, for some distribution some charges can reside in the interior of the material but this condition doesn't minimize the potential energy. Applying the same condition for your second question, you will easily see that under an electric field charges distribute themselves (for the spherical shell) symmetrically, and using the minimization of potential energy there will be no charge interior of the material. In addition, if you put a charged sphere under an electric field the distribution for the final state requires a detailed analysis and calculation to carry and I think this may help you to understand the charge distribution from a more general perspective.

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  • $\begingroup$ The OP did not ask about the charge inside the material. $\endgroup$
    – nasu
    Apr 28, 2023 at 15:16
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A net charge on the inside surface of a hollow sphere will contradict Gauss's law. You can take gaussian surfaces completely inside the conductive material. The flux is zero so the net charge on the inside surface must be zero. But is is not just this. The fact that the circulation of the (static) electric field on a closed path is zero prevent the existence of separated charges on the inner surface. So not just net chage is zero, there is no polarization (separation of charges). This conclusion is independent of the existence (or nonexistence) of an external field.

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  • $\begingroup$ I think the problem is that we can have a net charge inside a conducting shell, which will also cause mirror charges on the inner surface. What is counterintuitive is that the conducting shell will not screen the total charge that are on its inside. They will appear as a total charge on the outside. What will not appear on the outside are the details of the internal fields. I just fell for this myself in a comment on another post, even though I have decades of practical experience with Faraday cages and I know in my sleep what they can and can not do. $\endgroup$ Apr 28, 2023 at 17:07
  • $\begingroup$ What do you mean by "inside a conducting shell"? In the hole? $\endgroup$
    – nasu
    Apr 28, 2023 at 18:35
  • $\begingroup$ Yes, in the free volume on the inside. A freely moving charge won't stay there, of course. It will drift towards the wall. What is counterintuitive is that it doesn't matter where that charge is. On the outside it always looks like it's sitting on the surface. It's easier to do the following: take a single charge, then add a pair of external screening charges. What's the total field visible from "infinity"? It's that of the total charge plus a dipole moment. Add another pair of screening charges: now it's a single charge plus a dipole plus a quadrupole etc.. We never get rid of total charge. $\endgroup$ Apr 28, 2023 at 18:42
  • $\begingroup$ So, why do you say that this is a problem? What problem? $\endgroup$
    – nasu
    Apr 28, 2023 at 19:42
  • $\begingroup$ It is not a physics problem. The problem is how to give people the correct mental model. $\endgroup$ Apr 29, 2023 at 0:31

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