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I've read some answers on stack exchange for this question, but I don't find them particularly clear. I am convinced of the case for a sphere, since the area of a sphere is used in the derivation for the law.

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So maybe relating all other areas to the area of a sphere would provide an intuitive 'proof'?

I was uncertain, however, about approximating an infinitesimal curved surface to a flat surface.

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    $\begingroup$ The electric flux through an open in general surface $\,\rm S\,$ in empty space due to a point electric charge $\,q\,$ is $$ \Phi_{\rm S}\boldsymbol=\dfrac{\Theta}{4\pi}\dfrac{q}{\epsilon_{0}} $$ where $\,\Theta\,$ the solid angle by which the charge $\,q\,$ $''$ sees $''$ the surface $\,\rm S$. In the case of closed surfaces, like a sphere, we have a complete solid angle $\,\Theta\boldsymbol=4\pi\,$ so $$ \Phi_{\text{closed }\rm S}\boldsymbol=\dfrac{q}{\epsilon_{0}} $$ $\endgroup$
    – Frobenius
    Commented Sep 25, 2021 at 10:23

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Gauss law deals with electric flux and electric flux is related to number of field lines cutting or passing through a surface. And a charge create definite number of field lines, thus if any closed surface is enclosing the charge every field lines emmited or created by the charge will pass through the closed surface. So, whatever the closed surface be, the number of field lines passing through the surface will always be the same.

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In the picture you included, there are two distinct elements. One is Gauss’s law proper $$ \oint \vec E\cdot d\vec A=\frac{q}{\epsilon_0}\, , $$ which is always true and works for any closed surface.

The other part deals with using Gauss’s law to recover the field, i.e. the step $E=kq /r^2$. This is only possible in some circumstances where the surface and the field have high levels of symmetry.

In the example of your picture, imagine first the source charge is outside the Gaussian sphere. Then by Gauss’s law $\oint \vec E\cdot d\vec A=0$ since there is no charge inside the Gaussian surface. Of course, this does not imply there is no field. In other words, $\oint \vec E\cdot d\vec A=0$ does not imply $\vec E=0$.

Next imagine the Gaussian sphere is replaced by a long rectangle containing the charge. There is now charge inside the rectangular Gaussian rectangle but I can make the area of the rectangle whatever I want so it would seem I could get any field I want. In other words, $\oint \vec E\cdot d\vec A=q/\epsilon_0\ne E A$ always.

So while $\oint \vec E\cdot d\vec A=q/\epsilon_0$ is always true irrespective of the Gaussian surface, it is only when $\oint \vec E\cdot d\vec A=E A$ (which only holds for specific surfaces placed in certain ways around the source) that Gauss’s law is useful in recovering the field.

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This is because every shape can be changed into other more symmetric shape. We can convert it to sphere and thus derive Gauss' law.

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Gauss' Law is valid for any closed surface, provided that it meets the conditions for forming a "Gaussian surface".

However, most textbooks derive this equation using a sphere because it helps avoid a lot of the nasty integrals involved; a point charge has spherical symmetry, i.e., the magnitude of electric field experienced at every point of the sphere is equal as each area element of the sphere is at the same distance from the charge. Moreover, the as the field is pointing radially outwards, we'd always get θ = 0 (w.r.t. area element) which simplifies it a lot more for our use case.

Obviously, this cannot be generalised for all shapes, so the derivation is not as simple for another shape such as a torus or a cylinder. This is also why we try to pick the most symmetric Gaussian surface when applying this law, even though in theory we could have ANY closed surface and it would still work (imagine integrating over a horse-shaped Gaussian surface vs a simple sphere).

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