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Two point charges $q$ and $—q$ are separated by the distance $2L$. Find the flux of the electric field strength vector across a circle of radius $R$.
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Now, I don't understand the answer given by the book regarding this question. I mean, suppose instead of a disk, we have a Gaussian cylinder with radius $R$ between both charges. Now, with $2L$ being the length of the cylinder, let both ends length tends to zero, so that in the limit, the cylinder and the disk are the same. Second by Gauss' Law, the flux through a surface that does not cover a charge should be zero, shouldn't it?

But, apparently, the answer is not zero. Why? Where is the error on my thought?

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  • $\begingroup$ I think if you consider the simpler case of a uniform electric field directed along the axis of a cylinder vs. along the normal of a disk, you can see why the cylinder to disk limit does not work. $\endgroup$ – d_b May 7 at 3:48
  • $\begingroup$ Is ans $(qK4π) - (lq4πK/√[R^2+l^2])$ $\endgroup$ – Jay May 7 at 4:45
  • $\begingroup$ The question is for a circle, not a disk. It is just a geometric area, not a material object. $\endgroup$ – Bill Watts May 7 at 6:30
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    $\begingroup$ Very good question. I stuck for a moment. $\endgroup$ – Frobenius May 7 at 6:41
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    $\begingroup$ I agree with frobenius, this shouldn't be closed, I will edit to make it more general $\endgroup$ – Buraian May 7 at 9:22
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enter image description here

The limit of the cylinder as its length $\,2L\,$ approaches zero is two disks one on the other but with opposite normal orientations $\,\mathbf n_{\texttt{left}}\,$ and $\,\mathbf n_{\texttt{right}}\,$ as shown in above Figure-01. You must choose exclusively one of the normals to have one oriented surface (disk).

Hint :

In general the flux through an oriented open or closed surface $\:\mathrm{S}\:$ due to a point charge $\:Q\:$ is
\begin{equation} \Phi_{\mathrm{S}}=\dfrac{\Theta}{4\pi}\dfrac{Q}{\epsilon_{0}} \tag{01} \end{equation} where $\:\Theta\:$ the solid angle by which the charge $\:Q\:$ $''$sees$''$ the surface.

$\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=$

Related : (1) Electric flux through an infinite plane due to point charge (2) What is the electric field flux through the base of a cube from a point charge infinitesimally close to a vertex? (3) Flux through side of a cube

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A cylinder consists of two disks, with opposite orientation, and a lateral side. By symmetry, the flux through the lateral side is zero. Since the two disks have opposite orientations, again by symmetry, the sum of the flux through them is also zero. This is consistent with what you would expect from Gauss' Law, since the cylinder contains zero charge.

However, what you have here is quite different from a cylinder, since a disk is not a Gaussian surface, meaning that you cannot use Gauss' Law*. Instead you need to calculate the electric field at each point on the disk using Coulomb's Law, and then integrate. If you think about which way the electric field points at each point on the disk, it should become intuitively clear why the flux is non-zero. The key thing that distinguishes this from the cylinder, is that here, you have a single disk, whereas the cylinder has two disks whose flux cancels out because of their opposite orientations.


*In principle, there is a way to indirectly use Gauss' Law by finding the solid angle of the cone with the disk as its base and apex at one of the point charges, and then finding the flux through that solid angle.

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I don't really have a very rigorous answer but I think I may be able to say the 'right' way to think about it. The problem is that the shrinking the cylinder action is not really 'continuous' in sense that what happens in the limit is not what happens when you literally drop height $h=0$.

For a concrete example of the above thought, if $f(z)$ denotes the flux through the cylinder as a function of cylinder's height, then:

$$ f(0) \neq \lim_{z \to 0} f(z)$$

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