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If you refer to Gauss's law, it states the electric flux through any closed surface is proportionate to the enclosed electric charge.

The electric flux density is the same as the electric field intensity. A Gaussian surface is a closed, three dimensional surface (there's no holes in it).

Here's an example to help clarify what this is and is not saying. Suppose I have a clear glass ball. The "light charge" inside the ball is zero because there is no light source inside the ball. If I put the ball in the sunlight, light will go into one side, and out the other (ignore any sort of prism effect, etc. just don't think too hard about this example!). This ball still does not have an internal "light charge" because the light flowing into the ball is equivalent to the light flowing out (the "light density" through the surface sums to zero, or the line integral of the light density = 0 for this surface).

If I put a light source inside the ball, the line integral of "light density" leaving the ball would be proportional to the "light charge" inside the ball; in other words the line integral tells you what is enclosed by the Gaussian surface (my fictitious light source, but not the sun). Even if I put it in the sunlight again, the line integral will remove the "light charge" due to the sun and I will be left with only my internal light source.

In both these instances, absolutely nothing is being stated about the "light density" / "light intensity" inside the ball. For both instances, there is a light intensity INSIDE the ball, even though the "light charge" inside is non zero in only one case.

Relating to the question, this means if you have a Gaussian surface (such as a sphere), and it has/does not have an enclosed electric charge, you can have an electric field through the sphere - the fact this field is there tells you nothing about the internal charge of the Gaussian surface until you perform the line integral to measure what's coming in and what's going out.

So, what I'm stating is the question is not true - the electric field is not necessarily zero inside a Gaussian surface, even if the surface does not contain an electrically charged particle. This should be easily seen by taking the typical point charge example: You have a point charge, and you draw the Gaussian surface around it. The point charge radiates electric field lines in all directions away from itself. If you move the Gaussian surface to the left until the point charge is no longer enclosed in it, you will see the radiating electric field lines due to this point charge still go into and out of the surface (so there is an electric field due to the point charge inside the surface), but the point charge is no longer enclosed by the surface (so the line integral sums to zero).

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12y ago
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11y ago

I think there is no charge distribution in side a ring

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Q: Why is electric field inside a ring is zero?
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