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I am reading Purcell's E&M book and in one of the example questions, it shows that there is no E field between outer and inner surface after a a point charge is located at an arbitrary position inside a neutral conducting shell. My question is the positive charge +q induces -q on the inner surface and +q on the outer surface. There is an E field drawn in RED going from +q(inside the shell) to -q(inner surface) and -q(inner surface) to +q(outer surface). I drew the fields on the left picture.

The author states that the E field inside the conducting material is ZERO. In the absence of an external electric field, how is the electric field from +q(outer surface) to -q(inner surface) getting canceled?Question

enter image description here

I believe the question was answered Electric field inside a conductor and induced charges But I was still confused after reading the explanation.

enter image description here

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In the absence of an external electric field, how is the electric field from +q(outer surface) to -q(inner surface) getting canceled?

It is getting cancelled by the field from charge q located inside the shell.

An equivalent statement is that the induced charges, +q(outer surface) to -q(inner surface), cancel the field produced by change q inside the conductor, i.e., between the inner and outer surfaces of the shell.

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  • $\begingroup$ Just a follow up question. How does the +q on the outer surface distribute uniformly while the -q on the inner surface distributes non-uniformly (clustered more towards the +q in the cavity)? $\endgroup$ – Rumman Nov 17 '18 at 19:24
  • $\begingroup$ @Rumman Here is a possible way to explain it. We agree that, after all charges get re-distributed, the field inside the conductor (between the outer and the inner surfaces) is zero. If so, the charges on the outer surface are not affected by the charges inside (there are no field lines getting through) and, therefore, they will get distributed evenly due to the symmetry of the sphere. $\endgroup$ – V.F. Nov 17 '18 at 19:45
  • $\begingroup$ Yes, Purcell uses the same logic! So, i understand that logic, and I agree. I added another picture and this is going from what you mentioned "An equivalent statement is that the induced charges, +q(outer surface) to -q(inner surface), cancel the field produced by change q inside the conductor, i.e., between the inner and outer surfaces of the shell." When I draw the fields out following your statement, I can't "visualize" how the +q charges would distribute each other uniformly on the outer surface. +q should follow the -q charges to cancel each E field from +q charge inside the shell $\endgroup$ – Rumman Nov 17 '18 at 20:08
  • $\begingroup$ I am curious if the ordering matters. Lets say, I have +q charge then I surround the +q charge with a shell of negative charge of -q. Then when I put the +q charge it distributes uniformly across the outer surface, and this is what you and purcell motivated to. $\endgroup$ – Rumman Nov 17 '18 at 20:10
  • $\begingroup$ 2nd ordering where I have a conducting shell, and I somehow manage to put a +q inside the hollow part (not centered) of the shell, and the inner and the outer surface follows to cancel the +q field produced inside the shell. Here, i don't catch your and Purcell's argument of uniform distribution on the outer surface. I guess this is my confusion. $\endgroup$ – Rumman Nov 17 '18 at 20:12
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There is no field inside a (non-driven) conductor because if there were a field, it would move free electrons around until there is no field. In this case the induced charges on the inner and outer surfaces act like the charges on the plates of a capacitor to produce a field which cancels that from the enclosed charge. At the point closest to the enclosed charge, the surface charge densities must be equal and at a maximum. (At the far point a minimum.) At other points the densities must vary to compensate for the tilt of the field from the inner charge. I don't believe that the charge density on the outer surface can be uniform. I suspect that the field outside the sphere would be the same if the sphere were removed.

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The relationship between the Charge inside a surface and the intensity of the field on said surface if given by (Gauss theorem):

$ \oint \vec{E}\cdot d\vec{A}= \frac{q_{\mathrm{enc}}}{\varepsilon_0} $

If $S$ is a closed surface located inside the shell, then:

$E . A = (+q) + (-q) = 0$,

therefore $ E = 0 $ (since A is non null)

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