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I checked the website for a similar problem but although there are other questions with a similar title, I couldn't find any question which directly deals with my problem. This question might be trivial but for some reason I have difficulty understanding it.

Imagine a uniformly charged ring. A picture is given below for clarity. We want to find the electric field in an arbitrary point inside the circular ring (point $P$). The primed coordinates are source coordinates and the others are field coordinates. Just to make it clear, I am not talking about analyzing a 3D object in a 2D plane. Imagine the whole universe is 2D and we want to solve this electrostatics problem in that universe.

enter image description here

Based on symmetry it is clear that the field must be along the radial direction. Using Coulomb's law we can integrate over the contributions of infinitesimal elements. Assume that the charge density is $\lambda$ and it is positive and $r'=R$ is the radius of the ring. This problem is actually solved here, and the answer is:

$E = \frac{\lambda}{4\pi \epsilon_0 R}\int_0^{2\pi} \frac{cos\theta'-a}{(1+a^2-2acos\theta')^{3/2}} d\theta'$

Here, $a=\frac{r}{R}$ and the results for the numerical solution of the integral is given in the linked video. Alternatively, applying Gauss' law for an arbitrary Gaussian surface containing $P$ (not necessarily close to the boundary of the surface) gives:

$\oint E.da = \frac{Q_{enc}}{\epsilon_0} \rightarrow E=0$

Here, we have assumed that the electric field is uniform inside the Gaussian surface (the red circle) and this assumption is wrong, therefore the result is not reliable. Some people (for example here) claim that this is the correct result but I don't think it is, because as I showed, direct calculation using Coulomb's law gives a different result.

As a workaround, we can look at a Gaussian surface very very close to the point we are interested in (make the red circle smaller so the point $P$ is very close to it's surface). Doing so, we don't need to assume that the field is uniform in order to take the surface integral. As stated before, due to symmetry the electric field will be along the radial direction (which is perpendicular to the Gaussian surface and makes the calculation of the integral the same as before) and all points with the radius $r$ from the origin will have an electric field of same magnitude along their respective radial directions.

The result for the electric field is again zero which doesn't agree with the initial result. Moreover, since the direction of the field along all the points just below the surface is the same, I can't understand how do we get zero as the result of the integral. If all the arrows are pointing to either inside or outside the surface (depending on the sign of the charge) due to the symmetry, how can we have a zero flux?!

Since the Gauss law is fundamental I must be applying it wrong or I am assuming a symmetry that isn't there. What am I doing wrong?

There are other problems with different geometries that I have a similar problem with. For instance, the first part of the problem 4 in this document (PhD qualifying exam in Yale) solutions to which is available here, deals with the electric field due to a rotating, uniformly charged and infinitely long cylinder. The cylinder is made up of infinite number of rings with non-vanishing radial fields (as explained earlier). As you can see in the solutions document, using Gauss law as I did in my problem, they conclude that the field inside the cylinder is zero.

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  • $\begingroup$ The fact that the flux is zero through a Gaussian surface does not mean that the field is zero everywhere on the surface. What is your Gaussian surface? $\endgroup$
    – nasu
    Commented Jan 26, 2022 at 23:41
  • $\begingroup$ I know. That's why I tried to explain the situation in length. The Gaussian surface is a circle as explained in the text. $\endgroup$
    – Ali Pedram
    Commented Jan 26, 2022 at 23:43
  • $\begingroup$ A circle is not a Gaussian surface. It is not a closed urface. You need a real, closed, 3D surface and it will go out of the plane of the circle. $\endgroup$
    – nasu
    Commented Jan 27, 2022 at 0:10
  • $\begingroup$ The problem is in 2D. Is electrostatics mathematically impossible in 2D? Also there is a similar problem of infinite cylinder which is explained in the final paragraph if you don't want to deal with 2D (which is the problem in the first place). The infinite cylinder problem can basically be built by putting infinite rings side by side. The Solutions suggest that the field zero. However, I don't understand how because each element has a non-vanishing radial component in 2D. $\endgroup$
    – Ali Pedram
    Commented Jan 27, 2022 at 0:19
  • $\begingroup$ It obviously does not work as you try to use it. That circle has no flux through it but not because Gauss; law. Just because the field is parallel to the surface. So there is nothing to say about the field from this. It looks very simple, no further consideration can add to this. Gauss' law comes from the field going like 1/r^2 for a pont charge, How would the field go like for "electrostatics in 2D"? $\endgroup$
    – nasu
    Commented Jan 27, 2022 at 0:26

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I do not believe Gauss law can help you here. To apply Gauss law you need a symmetry both in geometry and in the magnitude of the electric fields. You can only draw the conclusion that the electric field is zero if the electric field at the surface (more correctly $\vec{E} \cdot \vec{A}$ , where $\vec{A}$ is the surface vector, is constant everywhere at the Gauss surface, which is not true in this situation.

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  • $\begingroup$ Gauss law is a statement which relates the flux of the electric field with the source charges. If there are useful symmetries one can deduce the strength of the electric field using it. However, people use Gauss law for situations with non-uniform electric fields all the time. I have updated my question so now you can see another example. $\endgroup$
    – Ali Pedram
    Commented Jan 26, 2022 at 23:01
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    $\begingroup$ In the problem you sent me the field is zero inside the cylinder, not outside. In the ring situation you are looking outside the ring. If you have a random surface outside the ring, then you will only learn that the total flux is zero and that's it. The electric field is not constant on the surface to be able to go further and say that the electric field is zero. In the example you gave me, also the electric field was constant at the surface. $\endgroup$ Commented Jan 26, 2022 at 23:08

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