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

[@Mr_mechEngineer]

The only conducting slab video that I see makes sense.

[@muhammedtalhayasar6687]

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

[@MichelvanBiezen]

It really come down to: 1) Does the electric field emanate on one side only or 2) Does the electric field emanate in both directions. In the second case the electric field is only half the strength.

[@danishsarfaraz6173]

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 ...

[@MichelvanBiezen]

Since the amount of charge in the circle is proportional to the size of the circle, we can use any shape and size circle to calculate the electric field. (It is uniform in density).

[@vazp3]

@@MichelvanBiezen As it should be with conductors :)

[@Mortgageman145]

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

[@MichelvanBiezen]

The Gassian surface is chosen to deterine the Electric field at a particular point.

[@beredaye]

Since it is a conductor, charges all gathered around on the two surface of the slab which are above and below surfaces. So question asked us the find the E on above that is why we didn't include a second circle that goes down.

[@omaraboutaleb3331]

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?

[@MichelvanBiezen]

The electric field is zero in the middle of the slab. The electric field will be the same on either side of the slab, since the slab is a conductor and half the charges will move to each of the sides of the slab.

[@Uygʻun-q2c]

thanks

[@RiyaTomar-jd3mw]

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?

[@MichelvanBiezen]

The choice for the Gaussian surface depends on how the charge is distributed and what you are trying to determine. In this case we are trying to determine the electric field on one side of the slab, which is a conductor and carries charge on both sides of the slab. Take a look at the other examples that show different situations.

[@RiyaTomar-jd3mw]

@@MichelvanBiezen ok :)

[@bebinashanty5545]

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

[@vazp3]

In this case the bottom of the cylinder is INSIDE the conducting slab. And in metals there is no net electric field inside the conductor. So the surface integral there vanishes.

[@leodeer9515]

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

[@MichelvanBiezen]

Because just like with a parallel plate capacitor, the electric field eminates from one side only and only depends on the charge on that one side.

[@CyberFenix000]

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

[@vazp3]

Great Video!