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What seems to incredibly bother me is why Gauss's law applies to any shape of a closed surface. Moreover, the fact that the electric flux is proportional to the enclosed charge is by many sources simply proven by using a point particle that is enclosed by a sphere. Hence the electric flux is proportional to the enclosed charge for any closed surface, which to me isn't obvious by just using a specific proof that includes merely a sphere.
Furthermore I have watched various videos on youtube, including a lecture given by Walter lewin, and visited numerous websites, however, all sources didn't resolve my confusion. Lastly, why does Gauss's law also work for any collection of randomly distributed charges? Many sources state the any collection of charges can be thought of a collection of separate point charges, and since the electric fields of point charges should be added vectorially, they can also be be considered to be one total charge that generates one net electric field. Does this imply that the closed surface integral, used in Gauss's law, can be divided into separate closed surface integrals? Like so:

$$ \int\sum\ \vec E\bullet d\vec A = \int\ \vec E1\bullet d\vec A +\int\ \vec E2\bullet d\vec A + \ldots +\int\ \vec Ei\bullet d\vec A$$ Where every seperate line integral includes the electric field of a single point charge.

Now my mathematical toolkit is relatively limited, thus, please do not utilize complicated mathematical equations that originate from divergence, differential equations, etc.

Thank you in advance.

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  • $\begingroup$ It isn't clear to me how a law like Gauss's law can be "proved". Where do you want to start from? It's quite simple to show that Gauss's law holds for any configuration of charges if you believe Coulomb's law and the principle of superposition. So from where would you like such a proof to start? Also, remember that while the law is true always, it isn't always useful to calculate the field: that only happens when the charge configuration has some special symmetries. $\endgroup$ – Philip Oct 12 '20 at 9:46
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If you are convinced by the sphere argument, then consider for any irregular surface a spherical surface within it that contains entirely your charge. Then within a "sector" of certain solid angle, the field lines that pass through the intersection of this "sector" and your irregular surface are exactly those that pass through the projection of the irregular surface onto the sphere, and so the total number must be exactly the same.

If you want a more quantitative (read: rigorous) solution, I will have to direct you to the divergence theorem, there is just no way around it. To be frank, if you don't want to learn vector calculus, you will probably just have to accept some results for what they are.

Also, yes, the closed surface integral of the sum can be expressed as the sum of the integrals.

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The integrals are not “line integrals”; they are an expression of flux (field x area). If you divide the enclosed volume into many very small solid angles, the dot product means you can use the component of the area at the outer end of the solid angle (represented by a vector pointing out) which is parallel to the field (and the radius). That area increases with the square of the radius and the field decreases with the square of the radius. That means that the flux depends only on the charge and the size of the solid angle; and not the radius or the slant of the actual surface. The integral involves summing the solid angles (the area of a sphere of any radius divided by the radius).

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  • $\begingroup$ Thanks, mostly clear, but how is it possible that a small area of dA has the shape of sphere and not just a flat surface? Why can we assume that dA adopts the form of the surface of a sphere? Furthermore, could it be related to the fact that the electric force is a conservative force, similar to gravity? So regardless of the slant of dA from a given shape, if you integrate over the entire surface only the parallelle components of dA constitute to the total flux, which is the same as simply using a sphere as shape? Maybe not as conceptual and mathematical, but more like a compromise? $\endgroup$ – Jelle 3.0 Oct 12 '20 at 19:07
  • $\begingroup$ For a very small dA there is very little difference between a flat or curved surface. Once you are comfortable with saying that the flux is proportional to the solid angle, then the problem becomes how to sum all the solid angles. This can be done with a sphere of any size. $\endgroup$ – R.W. Bird Oct 13 '20 at 18:31

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