Physics 37 Gauss's Law (15 of 16) Variable Charge Distribution: 'Infinite' Slab

In this video I will find the electric field inside and outside of an infinite slab of variable charge distribution.

[lichking]

Great vid and explanation. The concept finally clicked in my head and I understand it

[kaursingh]

sir - every lecture is best on the internet -sir , do u have complete series on electromagnetic theory of light ? do u have complete series on langragian mechanics ? --thank u sir

[MusicLove]

This helps me a lot to do my homework!! Thank you!!

[abhinandanverma]

Sir thanks you solved my doubt you are awesome ☺️

[richardrigling]

Good!
I'm somewhat befuddled by this. Let me try to describe my confusion.

Consider a square conducting plate of sides L and any thickness in an Electric field free area. Add one electron and it experiences no forces since there are no other E fields. Now add another electron. Each electron experiences a repulsive force from the other electron and in response they migrate eventually to opposite corners of the square. Add a 3rd electron and they will migrate under their mutual E fields in such a manner as to maximize the distance between each other and distribute themselves along the edge of the plate. With the 4th electron the charges each find a corner. It seems to me that adding more charges will continues this process where the charge tend to the corners and the center of the plate
In the case where the thickness of the plate is significant, the remains charge free. further, the charge density on the surface of the plate will not be uniform, being 0 at the center, higher along the edges (would the center of each edge also have 0 charge density?) and highest charge density at the corners?) and highest charge density near the corners.

If the plate has appreciable thickness (as in a cube). The first 2 electrons would migrate to opposite corners, the first 8 would each occupy a corner and so on. Also there would be no charge density inside the plate (or cube) since the charges would repel each other to some surface.

The end state is all charges settle into an equilibrium where the net E field from all other electrons is minimized.

Does this sound right? If so, what is the charge distribution for a very thin plate and for a cube?

[richardrigling]

This issue was raised by a physics teacher today and I tried to explain things to him. quickly got wrapped around the axle. His question was - What is the charge distribution of a conducting cube?

[Kolton]

Shouldn't you integrate form 0 to d/2 then multiply by 2 to account for both halves? The other half, even though it is also positive is still contributing to the E field on the positive side.

[griefinnub]

Great video!
By the way, why didn't you consider the Gaussian surface at x=0? Shouldn't integral of dA be 2L^2?

[fkafka]

You say it makes sense that E is not dependent on the distance from the slab because it is infinite in length and height. But when putting Gaussian surfaces around the spheres, the field varies with distance. Why does the infinite slab mean that the E will be felt at arbitrarily large distances away the same as just outside the slab?

[PinkPixelRabbit]

Why is the definite integral from 0 to d/2 but not from -d/2 to d/2?

[rogerramber]

sir, since you take the electric field inside the slab, dont you have to use the electric permittivity of the slab instead of the permittivity of vacuum?

[omarabu-khalaf]

Thank you very much, I am wondering about the electric field due to charges to the right of the plate with equation x=a, the electric field found at the end is rho*a^2/(2epsilon) is only due to the charges to the left of the plate x=a, don't we have to take into account the field generated by charges to the right of it?

[alieser]

When we are inside the stab, why do we say that electric field is only on the x direction? what prevents it from going sideways? I know the area vector and electric field vector should be parallel, but I don't understand why aren't there electric field going through the other faces of the cube.

[richardrigling]

Would there be any charge inside the plate? Any free electrons would migrate under the Electric field to the surface of the plate.

[jamilf]

Wouldnt integrating from 0 to d/2 be inside the slab?

[deniz-gunay]

You said the charge between -d/2 and 0 causes an electric field in the other direction but you didn't explain why is that. Actually I would ask you why didn't you take other side of the slab into account. But i have just noticed the reason is that both sides of slab cancel out each other since they have equal charge density on both sides. So, the left side of slab doesn't effect the electric field of the right side because charges on right side also creates an anti electric field in the left direction, which cancels out the electric field that is created by the left side.

If I am wrong please tell me.

[raoufharoune]

Hi sir, why did you take the integration fro 0 to D/2 just in the positive side .

[ersinsener]

you forget the L² at the left blue side . and i think it looks outside is actually inside solution ?? is it :))

[leefelix]

Hi sir,
If we remove the absolute sign on the charge distribution, i.e., the charge on the negative side would be negatively charged. Does that make the field outside of the slab become 0?

[hamzaejaz]

Is this always true for any slab? Even if the charge is uniformly distributed?