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I have a rather general question about electric field lines. If we have a hollow and neutral spherical shell and we place somewhere inside it a charge QA=+q, the electric field outside of the spherical shell will be that of a point charge +q located at the center of the sphere (with the sphere being absent).

The solution to this problem depended on Gauss's law which said that there is a charge Q(inner)=-q on the inner surface of the spherical shell (which has infinitesimal thickness) and a charge +q on the outer surface of the shell. They said that the Q(Out)=+q is spread uniformly over the outer surface because the electric field lines of QA and Q(inner) cannot pass through the conductor (which is basically the shell with its small thickness) therefore no force acts on Q(out). As they've explained, electric field lines cannot pass through the conductor for then E will either form a closed loop (cannot happen cause curl E=0) or it will pass from the inner surface to the outer and there will be potential difference which is a contradiction (Conductors have equipotential surfaces).

My question is, if we have N parallel infinite parallel sheets with surface charge density Sigma(i) i=1,2,...,N such that Sheet (1) is placed at the origin (it's normal is in the y direction) and all the other sheets are placed next and parallel to it in the +y direction. If I want to calculate the electric field on the left of Sheet(1) (in the -y area), do I have to find the final surface charge on the left inner plate of Sheet(1) (which has an infinitesimal thickness)? Or can I just pretend that the electric field lines pass through the plates and sum over all the electric fields produced by the sheets individually (Sigma/(2*epsilon0))*N (this method is used in a couple of exercises).

PS: In the spherical shell question, if I look only at a system which has a hollow spherical shell with Q(inner) on its surface and QA Somewhere inside it, how is Q(inner) spread on the surface? Will it be spread in such a way that E is 0 outside?

Thanks in advance for any answers, and sorry for the long essay!

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Quite a lot of questions, but I think they can all be answered by understanding the main argument: 'electric field lines can not pass through a conductor'. I will illustrate with the spherical shell-question.

The statement that electric field lines cannot pass through a conductor, is simply wrong. The reason that your flashlight works, is that the battery produces an electric field in the wiring, which causes free electrons to move and convert their kinetic energy into light and heat in the bulb.

However, when there is a static equilibrium (no moving charges), it is true that the electric field inside a conductor must be zero. If the electric field wouldn't be zero inside a conductor, the free electrons undergo a force and move (rearrange) untill they dont feel a force anymore. If the electrons don't feel a force, the electric field must be zero. This doesn't apply to insulators, since there, the charged particles can't move. (The same reasoning holds for 'why is the angle between electric field and a conductor always 90 degrees').

Now, the rearranging means that (in the case of a positive charge inside), the free electrons will go to the inner side of the shell, so that within the shell, the total electric field is zero: $$\vec{E} = \vec{E}_{q,inside}+\vec{E}_{free \ electrons}=\vec{0}$$ This of course leads to an induced charge on the outer side of the shell, which creates an electric field outside the shell:

$$\vec{E} = \vec{E}_{q,inside}+\vec{E}_{free \ electrons}+\vec{E}_{induced}=\vec{E}_{induced}$$

But, as the induced charge is equal to q, you can act as if the shell isn't there when calculating the field outside the shell (just calculate $E_q$).

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  • $\begingroup$ Thanks for the answer! So the induced +q doesn't feel any force acting on it? And if we forget, for a second, about the induced +q, is the field E=0 outside due to the -q and the original q inside the shell? And what about the infinite sheets problem? $\endgroup$ – Dylan132 Jun 22 '15 at 17:46
  • $\begingroup$ The last line is the key to this answer: "you can act as if the shell isn't there" as far as computing the field everywhere else is concerned. $\endgroup$ – Floris Jun 22 '15 at 18:43

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