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Halliday and Resnick present a system with a conducting shell enclosing a cavity with a negative charge located somewhere in the cavity, but off-center. The authors claim that the induced negative charge on the outer surface is equally distributed and has the field of a negative point charge because "the shell is spherical and because the skewed distribution of positive charge on the inner wall cannot produce an electric field in the shell to affect the distribution of charge on the outer wall."

My question is whether this logic is solid? I am not questioning whether the description of the outer charge distribution is correct, but rather WHY it is deduced to be this way. (my intuition is leading me to view the outer surface as an equipotential surface, that because it is symmetrical, should have a uniform field). However, I don't see how the author's description of the null field inside the material of the shell implies a foundational principle that charges on one side of a conductor do not contribute to a field on the other side. My understanding of the property of conductors is that the final arrangement of charges must have null field in the conductor, but to me, that doesn't necessitate that the inner induced charge can't contribute a component to the field at the surface.

Am I incorrect in stating that the electric field at some point on the outer surface or just beyond it is the superposition of the field components from ALL of the other charges in the system, including the inner induced charges? So the field contribution from the inner charges shouldn't just be dismissed? In other words, if some of the inner induced charge was to be magically removed, wouldn't this create a new initial condition that would be immediately felt by the outer charges, even before the internal charges of the metal have time to re-equilibrate?

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It is sound. Another way to see it is that the induced charges exactly cancel out the electric field of the inner charge. If you imagine getting rid of the bulk and the outer shell of the conductor, leaving you with only the inner induced charges, then the electric field would be zero everywhere outside the shell.

So, in principle, you're correct that the electric field is everywhere the sum of the electric fields due to all sources. But the components due to the induced inner charge and the source inside the cavity exactly cancel everywhere outside the inner shell.

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  • $\begingroup$ Thank you for your reply. I did see that the net charge is zero, but I understand that only to guarantee that the net flux through a Gaussian surface is zero, and not necessarily that the field everywhere has to be zero. As a counterexample, a dipole has no net charge and no net flux, but does produce an electric field. $\endgroup$ – lamplamp Jan 16 '18 at 5:54
  • $\begingroup$ I had also thought of a system of only the inner charges, but think that the zero net field in the bulk of the conductor is a consequence of the sum of the fields of the inner AND outer charges. I do see that the outer charges are uniformly distributed and don't contribute to the net field in the material. I don't see how the author's line of reasoning excludes an asymmetrical arrangement of charges on both the outer and inner surfaces. I believe the arrangement is as the author says, but not necessarily for the reason given. $\endgroup$ – lamplamp Jan 16 '18 at 5:57
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    $\begingroup$ @lamplamp It's not just that the net charge is zero that implies the field is zero everywhere outside, but also Poisson's equation $\nabla^2\phi=0$. It's pretty easy to show that since the inner shell is an equipotential, the electric field must be spherically symmetric. And the only spherically symmetric solution with zero flux through any surface is the one that is uniformly zero. $\endgroup$ – Chris Jan 16 '18 at 6:10
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    $\begingroup$ @lamplamp Essentially by making the surface an equipotential you remove any information about the positions of any charges you have inside: all you have left is the net charge. $\endgroup$ – Chris Jan 16 '18 at 6:12
  • $\begingroup$ Thanks for your follow up! I has suspected that Poisson's equation played a role. Purcell makes a pretty big deal of the concept of the uniqueness of solutions to Poisson's equation to justify that the field inside a conducting cavity is zero, even with an external field. I was just puzzled that Halliday and Resnick don't seem to address this. I recognize that it is a freshman-level book, but I would have thought that they could have acknowledged "additional ideas." $\endgroup$ – lamplamp Jan 16 '18 at 6:21

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