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Assume there is a thin conducting shell with a fixed point charge in the center, Q. Scattered outside the shell (not on its surface) there are more fixed point charges.

If I would like to know the electric field inside the shell at point P, do I need to know the magnitude of the point charges outside of the shell?

If yes: why are these relevant? Gauss' law says that flux (and electric field) is proportional to only enclosed charge, so a spherical Gaussian surface centered around Q through P would only include the internal point charge.

If no: why can't I draw a spherical Gaussian surface centered around a non-central point charge through P (intersecting the shell) such that the enclosed charge depends also on point charges outside the shell?

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No, they're not relevant, but your arguments don't really show that.

The problem with your statements is that Gauss's law gives you total flux through the surface, not the specific flux (or electric field) at any point on the surface. You have to be able to use other information or symmetry to then simplify down and find the strength at a particular point.

In your second example of a surface that contains some of the external charges, the field strength through the portion of the surface outside the conducing sphere may be very different from portion inside. You have the information on their sum, but not their individual contributions.

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  • $\begingroup$ Isn't a complete conducting shell a Faraday Cage? $\endgroup$
    – Bob D
    Feb 27, 2019 at 21:21
  • $\begingroup$ You can probably consider it so. Is that relevant? $\endgroup$
    – BowlOfRed
    Feb 27, 2019 at 21:54
  • $\begingroup$ Maybe not. But doesn't a Faraday Cage prevent external fields due to external charges from going into the cage and internal fields due to internal charges going out of the cage? $\endgroup$
    – Bob D
    Feb 28, 2019 at 0:08
  • $\begingroup$ The first one is true, the second one is not. But that seems to be a separate question. $\endgroup$
    – BowlOfRed
    Feb 28, 2019 at 0:22
  • $\begingroup$ Hmm. I always thought a Faraday Cage works both ways. It shields equipment within from EM generated from equipment outside and equipment outside from EM generated from equipment within. I.e., there is a field within due solely to the charge within, but there is no flux across the shell. Correct? $\endgroup$
    – Bob D
    Feb 28, 2019 at 15:16

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