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Whys is product rule, not quocient rule ?

Wonderful content, thank you for making it.

Hello, at 13:35 when you say that, by geometry, the x-y components cancel. I was unconvinced but after carrying out the integral I found that in the d(phi) integral portion, the integral for the x-component from 0 to 2pi of cos(phi)d(phi) is zero since the integral is sin and sin of 2pi and 0 is 0 For the y-component the integral from 0 to 2pi of sin(phi)d(phi) is zero since the integral is cos and cos(2pi)-cos(0) is 1-1=0 Those values are multiplied to their respective double integral so the entire integral vanishes, leaving the z-component Now my question is, does it actually cancel because of geometry or does it cancel because each integral evaluates to zero? I don't particularly like saying "they cancel" since you can't add and subtract the y-hat and x-hat components. Is saying that they cancel just a kind of short-hand way of saying that the integrals evaluate to zero? Or, do they truly cancel? Thank you!

isn't the seperation vector supposed to be r-r'?

I also want to mention that you can get to the (1-1/2 R²/z²) by using the Taylor expansion for (1+R²/z²)^(1/2) at R=0

But, in da, the correct aren t rsen(teta)? Cause r2sen(teta) is used in dv

cos(npi/2) equals 1 for n=4, am I missing something???

I think it should be (slanted r = zr^ - rr'^) instead of zz^ -rr^

I don’t think you can necessarily say what the effects on the outer surface are when the third charge is brought near. You assumed that the electric field still points radially outward outside the conductor but this is not the case with newly introduced asymmetry

You didnt show why its true that any regular polygon w equal charges at its vertices will exert a net force of 0 in the center. Why not use Columbs law and geometry to prove that statement????

Im not convinced the net force in c) is 0

probably only vid which i found knows what they are doing , great one !!!!!

what did i just watch

excellent video, i'm not gonna lie I also used wolfram to solve the integral and looked at the steps. I must say the way the integral was evaluated was beautiful. Let a = root 2 z tan u.......etc

th-cam.com/video/WayntWTn2E4/w-d-xo.html&lc=UgybzenIIUSvTshDe_h4AaABAg

Thank you! I wonder why some solutions on the internet say if Z>>R it would be like a point charge, not 0.

small correction:The last step it should be 1/2ln2. Great video!

❤

P is electrostatic PRESSURE not potential. Potential is U, or Fd, while electrostatic pressure is Force per area

Thank you!

Thank you, I love that isn't that easy on the beginning but at the end it's so simple, amazing :D

Thanks 🎉

i also dont understand how you got the area to be dr dphi can you please explain

Thanks ❤

great video btw

why we took the gradient in cylindrical coords?

does not comply with charge conservation

don't get how you got that R<z = (z-R) and R>z = (R-z) ?

great explanation! thank you!

Thank you ❤

good to revise some old knowledge while learning electrostatic

Thank you so much for explaining where the values of s and s’ came from!

Debería ser a^2, b^2 no?

my goat

Gauss's law also works well for this problem.

Good explanation. Thank you

Thank you for using Griffith's notation to solve this problem as I was also following his notation. Thanks!

How you put tan inverse pls explain about this integral formula

I think for part a) the net force can also be nonzero. I don't get why the electric field from the surface charge has to cancel.

man just put the hardest part of the problem in a calculator 😒

Thank you

You are such a great teacher. You should continue your youtube journey and start re-uploading the videos

Thankyou

Great thanks.

The "r"-bitrary. Haha loved it. Thanks so much for your videos! I've been reviewing for my final and everything is clicking properly with your walk through videos!!

I'm a bit puzzled by your choice of what the scalar function and vector function are at 4:31. You state that we've got the freedom to define them "however we want", but had you chosen e^(-lambda*r) to be the vector function and (lambda/r + 1/lambda^2) the scalar function, you would have missed the delta function in the final density expression. Surely, a mathematical result cannot depend on our freedom of choice. A more riguros justification of your choice would be needed.

Excellent

The audio is low and I can't use the video without earphones.😢

nice asmr

Audio is very low