For calculating field outside a charged plane conductor, a gaussian surface of Cylinder is considered. I have been said that we consider cylinder because the circle in its upper end is in equal distance to the charges on the circle of its lower end. This way electric field intensity on the upper surface is equal at all points of that area.

I am particularly confused over why couldn't there be a cube instead of the cylinder. Is there a mistake or misunderstanding with how I am understanding choice of gaussian surface?

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    $\begingroup$ Yes, you could also use a cube. In fact, Gauss' Law works for any closed surface. It's just that some surfaces lead to simpler calculations than others. $\endgroup$ – WAH May 8 at 10:29
  • $\begingroup$ @WAH is correct. You can use a gaussian surface as a cube. You put a pillbox on an infinite conducting plane of surface density $\sigma$. You can put any surface you want but your surface integral of $\vec{E}\cdot d\vec{a}$ will be painstakingly difficult to evaluate as there might be no symmetry involved for you to deduce some things and put the $\vec{E}$ outside of the integral $\endgroup$ – Dominik Car May 8 at 10:36
  • $\begingroup$ Gauss' law holds for any and all closed orientable surfaces. We choose surfaces to match the symmetry of a source in order to use Gauss' law to compute E in cases of high symmetry. This may be a source of confusion. $\endgroup$ – ggcg May 8 at 10:59

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