Finding the electric field a distance z above axis of a square ex 2616

Griffith's 2-4

Finding the electric field a distance z above axis of a square

[wandering-willow]

did you integrate only on the edges? what about the charge spread on the surface?

[pt-au-hg]

In finding the limits for alpha when x = a/2 it's stated that alpha is Arctan (a/2); what happen to the K in the denominator. The definition for tangent of alpha was define as: tan(x/K) not just tan(x) so when a/2 is plugged back in to find the limits for alpha it would seem to have to include the K also?

[sunilmandal]

arctan(a/2K) only mistaken. method is safe!

[BernardJacobs-dj]

If a (=one side of the square) goes to infinity, the final result goes to 0. How to reconcile this result with the formula of the electric field calculated in a point z above an infinite plane, being 2ϵ0σ?

[cytyy]

how do we check the limits? we have the same solution but finding its limit when z >> a is where I find it hard.

[carlsanderson]

Just one criticism, we know that we only care about the Ez portion of the field, but you drop the z^ unit vector. It would've been useful to keep that to make the video easier to follow. Otherwise, thank you, I understand this much better for watching your video.

[praveshkumar]

how u take the limits from 0 to a/2 think it would be from 0 to a/2k...

[BrownPen_BluePen]

do these thin plates have electric field vectors on the top and bottom?

[atiqchep]

You are great dear .... You have done a great job ... Love you ...!

[kemalyigitantepuzumu]

I wonder one thing. You have said K^2 instead of z^2+((a^2)/4) at 9:24, but the square of both sides of K = z + (a/2) does not equal to the answer which you have found in the solution. If it were, it was supposed to be K^2 = z^2 + az + ((a^2)/4). Could you explain this bro since I am so confused :d

[jayeshhemani]

can you explain how you take alpha in 1st figure where theta has been shown ?

[candoga]

watch him with x2 speed 😅 but it was a nice and clear solution though :)

[zarifsjs]

hey, i have a question.. You integrated from 0 to a/2 .. then shouldn't you multiply by 8 ?

[eng.eletrica.concursos]

mistake only limits on sin alpha.

it is arctan(-a/2*sqrt(a^2/4+z^2)) and arctan(a/2*sqrt(a^2/4+z^2))

[shrutidubey]

How a/2 ? a/4 is coming distance btwn O to center of side length a

[rutvishah]

what if there is a rectangle instead of a square?