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For a circular surface, the surface area of the circle is cancelled by factors in the denominator of $E$, leaving behind only $Q/\epsilon_0$. But what about when we conside a cylinder? How so we arrive on the same expression

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    $\begingroup$ So your question essentially is "Why is Gauss's law valid for all geometries?"? $\endgroup$ Oct 1, 2019 at 14:44
  • $\begingroup$ Is this for say a point charge? That first result is easy to prove because the field lines are perpendicular to the spherical surface at every point, but try drawing the lines as compared to a cylindrical surface $\endgroup$
    – Triatticus
    Oct 1, 2019 at 17:39

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At the end of the day, flux is the number of field lines passing through a surface area. Imagine that the cylinder you mentioned is enclosed in a sphere, now the number of field lines passing through the cylinder should be equal to the field lines passing through the sphere(assuming there doesn't exist any other electric field). Therefore we can say that the flux through the cylinder is the same as the flux through the sphere. This is also true for any other closed surface because a sphere can enclose any closed surface.

$$\phi_{cylinder} = \phi_{sphere}$$

We also know the following result from your derivation of Gauss' Law for a sphere,

$$\phi_{sphere} = q_{enclosed}/\epsilon_0$$

$$\therefore \phi_{cylinder} = q_{enclosed}/\epsilon_0$$

Similarly, we can say that Gauss' Law works for any closed surface.

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The flux due to point charge is the same in both the sphere and the cylinder. However this is hard to show with ${\phi}={\int}E{\cdot}ds$ You would have to do some complex integration in the second case but the two are ultimately equal even though t=they do not look so. You can use the equation ${\phi}={q\over {\epsilon}}$ to easily verify that the two flux are actually equal.

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The derivation of Gauss's Law starts with a point charge (q) inside a closed surface of arbitrary shape. The E field is radial, dropping off as 1/r^2. Each tiny segment of area on the surface is represented by a vector, perpendicular to the surface, and pointed outward. The dot product in the expression E.dA gives a component of the vector dA which is parallel to E and represents the base of a small cone (in 3D) which has its apex at q. The area of this base increases as r^2. So the increase in dA matches the decrease in E, and the total area can be taken as a sphere of any size. Putting more charges anywhere inside the surface (of any shape) increases the flux proportionately. Gauss's Law is always true, but is generally used in situations with high symmetry.

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