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Gauss's Law says that electric field inside an infinite hollow cylinder is zero. My question however is that an infinite hollow cylinder can be constructed by taking rings as element and the field produced by a ring within it is non zero. electric field inside a ring . Wouldn't this imply that there would exist a field inside an infinite hollow cylinder?

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Keep in mind that the video you linked only deals with the electric field within the plane of the ring. So while it is correct that the infinite cylinder can be treated as an infinite stack of rings, we also need to concern ourselves with how the electric field of a ring behaves out of the plane of the ring.

If we consider a positively charged ring, it has been shown that within the plane of the ring, for an axial distance less than the radius, the electric field is directed inward. I suspect that your first instinct was that as you go out of the plane of the ring, the radial component of the field remains pointed inward. And if this were so, when you added up the contributions of all the rings, you would get a net non-zero electric field directed inward.

Gauss's law implies that the field inside is zero, and therefore it implies that this intuition is false. At some distance above (or below) the plane of the ring, the radial component of the ring's electric field must switch direction from inward to outward. So when you integrate all the field contributions over an infinite stack of rings, the nearby rings with an inward directed radial field will be exactly balanced by the more distant ones with an outward directed radial field.

Someone somewhere has probably numerically calculated the field of a ring and mapped out the magnitude and direction. I'm not going to attempt to do that. I'm just going to argue that the direction change must occur.

Let's consider the field for a single positively charged ring of radius R. Let a be the distance from the axis of the ring and d be the distance from the plane of the ring. If we consider a position where a = R and d is some finite non-zero distance. Clearly at this point the radial component of the field must be directed outward, because all parts of the ring are below and to one side. Now keeping d fixed, we move a small but finite distance inward so a < R. We are moving through free space, so there can be no discontinuities in the electric field. The radial component can not immediately change from a finite outward directed field to a finite inward directed field. There must be some range of a where the radial component remains directed outward.

Now I haven't shown that for all a between 0 and R that there is some d beyond which the radial component changes from inward to outward. But I hope that this is enough to give you more confidence in the result from Gauss's law.

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  • $\begingroup$ Thanks, the apparent contradiction between gauss's law and the analogy of ring had risen because I had not considered that the field vector would flip it's direction at some d. $\endgroup$ Jun 21, 2019 at 16:46
  • $\begingroup$ The value of d(at which the field flips the direction) must tend to zero as move from a=0 to a=R? $\endgroup$ Jun 21, 2019 at 16:49
  • $\begingroup$ Yes, that's my expectation as well: d will tend to zero as a approaches R. I also expect that d will tend to infinity as a approaches zero. $\endgroup$
    – MacA
    Jun 24, 2019 at 17:55
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The important point to note here is that Gauss' law can be used to find the electric field of charge distributions that are within the Gaussian surface chosen not the fields coming from charge distributions outside. In the case of hollow cylinder electric field is from the charge distribution outside the Gaussian surface as your Gaussian surface is inside the cylinder(although they get cancelled in the ring analogy you mentioned).

In the case of ring analogy you mentioned, you haven't considered fields from the rings placed on the top and bottom which will cause the field to go to zero inside.

Hope this helps.

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  • $\begingroup$ The gauss's law relates flux to charge enclosed within the gaussian surface. The flux mentioned here is from all the charges (not only the ones inside the surface). $\endgroup$ Jun 20, 2019 at 1:29
  • $\begingroup$ Besides, in the analogy of the ring won't the field produced by charges above and below an elemental ring cancel out? So that only the field produced by the elemental ring (in the plane of the ring) is left? $\endgroup$ Jun 20, 2019 at 1:31
  • $\begingroup$ No the vertical components get cancelled out not the horizontal ones. You can try drawing it out. Or else gauss law would be wrong. $\endgroup$ Jun 20, 2019 at 6:14

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