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Suppose there is an off-center charge inside a copper conducting shell. We know that we can use the method of images to calculate the charge distribution on the inner surface. We also know that the charge distribution on the outer surface is uniform so as not to cause any unwanted electric fields inside the conductor. However, it is claimed that the electric field outside the conductor is only determined by the charge on the outside of the conductor such that it is as if there was a charge at the center of the conductor. However, I don't see why the inner charges (both placed and induced) should be ignored when calculating the field at some random point outside the sphere. There were meant to cancel exactly inside the conductor but nowhere else.

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One way to explain why the field outside a conducting shell is completely defined by the charge distribution on the outer surface of the shell is to consider that the field inside the conductor is zero.

This means that there are no electric field lines going from the inside to the outside of the shell and, therefore, the field outside the shell due to the induced charges on the outer surface should be identical to the field due to the identically distributed "placed" charges.

In other words, once the re-distribution of charges is completed and the field inside the walls of the conducting shell becomes zero, the internal charges stop contributing to the outside field - their further contribution is "blocked" by the conductor.

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This is the case because besides the electric field produced by point and surface charges which look like it might be possible to influence the other side of a conductor there is also the differential form of Maxwell's equations which are local. Given that the electric field at the outer boundary of the conductor is that of a point charge at the center of the sphere and that there are no charges outside of the conductor, Laplace's equation only allows one solution, i.e. the internal charges cannot contribute anywhere outside of the conductor.

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  • $\begingroup$ This assumes none of the charges are entangled. $\endgroup$ – Alex Roberts Jul 12 '18 at 18:48

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