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Shouldn't the electric flux through a circular disc due to a point charge kept at some finite distance from it be zero as all field lines which enter it exit it also and hence the net would be zero?

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Your disk isn't a closed surface, so Gauss's law doesn't apply here. The idea of field lines entering and exiting the surface relies on the idea of your field line entering on one part of the surface and exiting on the other part of the surface. At a single point on a surface you don't say the field line is entering and exiting the surface. There will be a non-zero flux through the disk since at all points on the disk the field will have components in a single direction through that disk. The only way to make the flux $0$ through the disk is if the point charge and the disk were in the same plane.

This goes to show that the "entering and exiting" of field lines is a nice qualitative picture, but when you want to actually calculate flux you need to go back to the mathematical definition: $$\Phi=\int \mathbf E\cdot\text d\mathbf a$$

For your point charge example with your disk as your surface, this is obviously not $0$ since, as I mentioned earlier, all values of the integrand on the disk will have the same sign (positive or negative depending on the orientation of your surface and the charge in question).

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  • $\begingroup$ If all the field lines have one component in the same direction then what is wrong with saying that 'all field lines which enter the disc also exit it so the net flux would become zero?' $\endgroup$ – Schwarz Kugelblitz Jul 9 at 13:59
  • $\begingroup$ @HerambPodar Because that isn't how flux works. As pointed out in Tojrah's answer, this line of reasoning means that flux is always $0$. The "entering and exiting" picture of net flux relies on field lines entering one part of the surface and exiting another part of the surface $\endgroup$ – Aaron Stevens Jul 9 at 14:16
  • $\begingroup$ So that line of thought is only valid for closed surfaces? $\endgroup$ – Schwarz Kugelblitz Jul 9 at 14:18
  • $\begingroup$ ..(according to flux out -flux in) $\endgroup$ – Schwarz Kugelblitz Jul 9 at 14:19
  • $\begingroup$ @HerambPodar Not necessarily. What I am saying is that at a single point on a surface you will become mislead if you think of the field as both entering and exiting at that point. With that reasoning flux is always $0$. $\endgroup$ – Aaron Stevens Jul 9 at 14:23
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The two "surfaces" of disc are actually the "sides" of same surface. It is obvious,that field lines, entering from one side, exit from the other side of the surface. So, this would imply that flux is always zero!

Change your thinking perspective.

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  • $\begingroup$ Downvote. Consider a vector field $\:\mathbf{E}\left(\mathbf{x}\right)\in \mathbb{R}^{3},\mathbf{x}\in \mathbb{R}^{3}\:$ and a surface $\:\mathbf{a}\in \mathbb{R}^{3}\:$ not necessarily closed. Could you define mathematically what we call the flux $\:\Phi \in \mathbb{R}\:$ of this vector field through the surface $\:\mathbf{a}$ ??? Note that concepts like lines of the field, entering in, outgoing out are mathematically unacceptable to use in such a definition. If to the negative, see in the Aaron Stevens' answer. $\endgroup$ – Frobenius Jul 10 at 16:03

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