Gauss's Law and Concentric Spherical Shells

How do you calculate r?

I have a final tomorrow and I trust these videos insead of my actual notes so thanks for your efforts :)

Thanks for the videos. You are the best.

So essentially between concentric shells where a < r < b, the electric field is only due to the shell with radius a?

thank you sooooooo much! you are a live saver!

So when r < a, would E still be zero if the sphere in the middle has +q charge uniformly distributed throughout the volume?

How is the charge enclosed equal to 0 in the c < r example?

Who needs to work those stupid C problems when this guy can just gift me his knowledge.

I'm heavily concerned on your proof of r>c for the field being zero. There is a positive surface charge and therefore a field emitted from that, you even do this exact problem in your part II video with a twist on variables.

My result for E in r>c:

E=k(Q/r^2)

(Equal in magnitude and direction to the field between the materials.)

Proof:
E(total)=E(a<r<b)+E(inside cond. surface)+E(outside cond. surface)+E(inside cond.)
where
E(outside cond. surface)= -E(outside cond. surface)
and
E(inside cond.)=0

Therefore
E(total)=E(outside cond.)=E(a<r<b)

E(a<r<b) Proof:
(using CSI)
EA=Q/ε0
where
A=4πr^2 (surface area)

Therefore
E=Q/(ε04πr^2) = k(Q/r^2)

I think you should do a video explaining how to suspend the first sphere inside the second sphere. (P.S. Great Explanations!)

so,,, you don't consider -q when it r is a<r<b?

"Snuck in there and looked at a dA." 10/10

I have my quiz later, thanks to this ❤️

beyler alan neden 4piR^2 oluyor?

you are hopeless

It is very good video for learning electric field. I got the answer from your video. Thank for your good video.

And what about the electric field at point r=b? Is we go by Gauss law, would we take total enclosed charge 0 or +q. as the charge is on the inner surface, I thought that -q would be included and that would make the net enclosed charge 0. But my book says otherwise.

sorry but expiation is unclear and all over the place

So, if the shell was not grounded, the electric field would be equal to Q/(epsilon*4*pi*r^2)?

Nice explain

whats the electric field at r = a and r= b

r is.... ...greater than c..... ahhaahah

You sound like a guy I could have a beer or smoke a cigar with

nice

Where is r but?

actually, the E_r<a is zero only if the metal is a spherical shell.

bilkent phys 102ciler doluşun.

He Solves this such that THE OUTER SPHERE IS GROUNDED good to know if you're new to this

You are so amazing. Thank you so much. You have no idea how much this saved me.

pff i wish i watch thisi before exam

Vizede aynısı çıktı haberiniz olsun

allah razı olsun

If the outer surface is not grounded, then what is E on outer surface?

How is that last part right? induced charge on outer sphere means q on outermost surface, so gaussian surface surrounding entire things has Q_enc = +q, not 0?

Que :
A hollow charged conductor has a tiny hole cut into its surface.
Show that the electric field in the hole is (σ/2ε0) nˆ , where nˆ is the
unit vector in the outward normal direction, and σ is the surface
charge density near the hole.

thanks

ffs i wish i saw this before my exam. Great Explanation

The serial of describing the concepts is incorrect.

Fantastically explained. Love from 🇮🇳

so does the hollow sphere has net -q charge?

why does the negative charge stick with outer hollow metal? If it sticks with the inner one, the b>r>a portion would have q_en = 0 thus 0 electric fields. Why is this not the case?

what happens if it is at b (the inner surface of shell)

Thankyou. It simplifies the meaning of the equations.

W

Are they both spherical shells? The drawing looks like it's a solid inside a shell and I was extremely confused. Then I read the description and now I understand, there isn't any interaction betweeen the two on the inside I can just plug and chug yahh!

Even 14 years later your videos are helping me