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If the electric field inside an infinitely long charged cylinder is non-zero except origin, how can be the inward flux zero?

https://www.youtube.com/watch?v=BF3wEV4tWq8

If we look at this video, we would see that a charged ring has a non-zero E-field inside it except for the origin. A cylinder can be constructed by infinitely many rings, and their vertical components would just cancel out each other to give us back the same formula.

E-field inside a ring

So if we place inside a surface S (let's say another cylinder), I can't understand how can the net flux into the S be zero.

Gaussian law tells us that the flux passing through a surface is exactly the sum of charges in it. Flux has to be zero because we have no charges inside S. But I am not so sure about it. The argument for the Gaussian law in the case of external charges is:

External charge

Here the field lines entering the sphere leave them, making the total flux zero. But in our charged cylinder, we see that field lines end in the origin, where the electric field is zero by symmetry, they don't exit the cylinder. So it doesn't seem plausible to me that we will have zero net flux. Please help me understand what is going on here. Thanks!

İbrahim İpek
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  • One possibility is that you are ignoring the fact that you need to have a closed surface through which to calculate the flux. So, you have to imagine, for instance, a sphere centered at the origin, and the electric field will be pointing out of the surface at certain parts of the sphere. – march May 04 '23 at 02:23

2 Answers2

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A cylinder can be constructed by infinitely many rings, and their vertical components would just cancel out each other to give us back the same formula.

This is incorrect. The field inside a cylinder of charge is zero.

Consider a cylindrical pillbox inside the infinite cylinder of charge. By symmetry, the E field through the ends must be 0, and also by symmetry the field through the side must be radial and uniform magnitude. Since there is no charge inside, then by Gauss’ law the net flux must be zero. The only uniform magnitude field which can give zero net flux is zero.

Your assumption that we get the same formula out is incorrect because you are neglecting the radial components of the other rings. While it is true that the vertical components cancel out, the radial components of the other rings cancel out the radial component of the in-plane ring.

Dale
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What you show in the first figure is not the field inside a cylinder (which is zero). The field of the ring is not limited to the plane of the ring. When you add the contributions of the rings to make a cylinder you have to add the whole field produced by each ring and not just the field which is in the plane of each ring. This is not an easy job. First you will have to calculate the field of the ring in the whole space.
In introductory physics texts only the field along the axis going through the center, if not just in the center, is treated as an example. For of-axis points the integrals require a lot more work and are left for advanced treatements. But this does not mean that the field does not exist outside the plane or off-axis.

nasu
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