Physics 37 Gauss's Law (7 of 16) Capacitor Plates

In this video I will find the electric field of capacitor plates.

[MichelvanBiezen]

Buyisiewe,
My suggestion is to pause the video at times.
I try to find the optimum speed for most viewers which means that some will find them too fast while others find them too slow. The best way to watch them if you find them to fast is to pause the video here and there until the ideas sink in.

[alonsosch]

And from the video final expression E = q/(E_0 * A), you can get the capacitance value:

q = CV => C =q / V
V = E * d = q * d / (E_0 * A),

C = (E_0 * A) / d

[keysky_]

Sir you brought light to my thirty minutes of struggle with this formula!!!

[jasonheard]

Your videos are exceptionally helpful, and easy to follow. Thank you.

[ΜΕΓΑΚΛΗΣΑΣΤΕΡΗΣ]

If we draw the Gaussian surface in such a way to include the "inside" surface of the plate then there will be a Qin and a field E should exist on the outer side of the plate as well !! Couldn't it be better if we say that E = 0 outside the plates because the field of one plate cancels out the corresponding field of the other plate (as being equal in magnitude but in opposite direction)?

[nandhannatarajan]

Why isn't the electric field of the left plate added to that of the right?

[hassanalthuwaini]

really amazing ... way better explaining than my prof, i wish if you were teaching me.

[valeriereid]

Excellent example. Thank you very much.

[sondossalah]

Why we calculate the Qin as the ratio of the area but in the previous video mutiply Q with the ratio of volume ? << why we ignore the area of the side sylinder and calculte A as the area of the top only!

[abdohajar]

+Michel Van Biezen
Sir if I may ask doesn't the negative charge on the other plate have a electric field as well and should be calculated. Then by the law of superpositions we just add both electric fields . My point being why did we neglect the electric field of the other plate. Many thanks sir

[rajmonibasumatary]

Question: if you do that on the other plate, you find the same value of E, but why don't they add each other and give us twice the value in between the plates?

[nomann]

but the negetive plate has also electric field of its own. it should add up

[physicsem]

1:50 Why do we have to multiply Q by the ratio of the area Gaussian surface and the area of the plate? I've never seen that done before, normally I'd just get EA = (Q) / (epsilon knot), and then divide by A to get E = (Q) / ((A) (epsilon knot)). It's the same result but I wanna understand the logical reasoning for why we are multiplying by that ratio in this method.

[alvarozerkowski]

Well Professor, you lost me when you divided A by Ac. For the first time, I will really have to search elsewhere for capacitors. I will try your playlist of videos on the subject! :)

[raspberryboss]

This boi is feeding me depression medecine

[AndyU]

No electric field on the left side of the capacitor? I'm not buying that, unless the plates are infinitely long.

[valeriatorres]

God bless u sir, you're like an angel haha :) thank you

[FaizanAhWani]

The plates of the capacitor are conducting in nature. What i don't understand is that, why does not the charge spread equally on both faces of the plate of the capacitor?

[Bob-ylpm]

Charge is stored in the dielectric! (not the plates)

[Jose-obpc]

Why the magnitude of the eletric field is not multiplied by 2? considering that there is a positive and negative plate, that is what my logic would tell me to be correct