Physics 37.1 Gauss's Law Understood (23 of 29) Infinite Slab of Charge 1

In this video I will find the electric field E=? a distance from a slab (slab is a CONDUCTOR) of charge where the area charge density is given.

Next video in this series can be seen at:

[muhammedtalhayasar]

could you please explain difference between sheet and slab. at the and don't we have two surfaces and sholdn't we divide sigma/2 epsilon. and in serway university physic book it gives us measured very thin conductor sheet and in solution we didn't divide sigma by two. How could that possible. Is it related with being infinitely large ? ı don't think so. Thank you also. sincerely

[Mortgageman]

I am all kinds of confused after this one, why don’t we include a second circle? And what if our Gaussian surface runs through the entire slab? Then the maths would suggest that the electric field at every point of that field would be two times as strong even though we have the exact same situation

[danishsarfaraz]

why you don't take the area of circles above and below the slab, as due to symmetry same electric field will be at the bottom of the slab ...

[michaeltamajong]

The only conducting slab video that I see makes sense.

[leodeer]

why do you get the same answer as with a parallel plate capacitor but in this case you only have one side

[RiyaTomar-jdmw]

Prof. why didn't you make the Gaussian surface in such a way that it crosses the slab from the bottom also. Was it given already in the question? And if that would be the case then we would have considered electric field at the bottom also. Right?

[bebinashanty]

Sir why can't we take 2 times pi R square in the denominator . Isn't there any electric field lines passing through bottom of the cylinder

[CyberFenix]

Could you plese touch (no matter how superficially, no pun intended) on the Electric Field for a liquid (3D object)?

[helpmereachksubswithoutvi]

prof. won't the charge in the bottom of of the slab also affect the Electric field at that point, If yes why didn't we take the gaussian surface big enough so that it touches the bottom of the slab [Including the charges in the bottom]

[omaraboutaleb]

In your video where you mentioned the three conditions that need to be true to draw a gaussian surface, the third point is that the magnitude of the electric field must be constant on the surface. However, in this video, the bottom part of the surface has an electric field of zero but the top part does not.
Where is the mistake in my logic?