Because GAUSS'S THEOREM based on only enclosed surfaces. When r is greater then a and less then b then we are with Gaussian surface passing through Gaussian surface of radius r,in this time outer surface will be in effective.
For r>c the electric field should be Q/4(pie)(epsilon)(r^2) since the charge on the inner shell is zero, negative charge(-Q) will get induced on the inner surface of the outer shell. But , since the outer shell is given no charge initially therefore net charge on the outer shell is zero and for this the outer surface of outer shell must be equal to positive charge Q. Hence electric field at a point r>c is as I have mentioned above. If we go by your logic, you are denying the law of conservation of charges. Please let me know, If you disagree with my argument.
Amar Munkhjargal Thank you for clearing that up. I couldn't figure out why he ignored the charge, but assuming it was grounded it makes total sense. Thank-you
I was just about to ask the same question as April Pun - shouldn't the negative charge in the outer sphere increase the magnitude of the electric field for a
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?
I know Im a year late, but anyways, because what you consider in the equation is only Q that is enclosed in Gaussian Surface (Q enc.), there will only be +q enclosed by the g.s. when r is a
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
I'm confused by this as well but I agree with the original poster. In the video, he grounds only the conductive shell which surely means that it then has a net charge of zero. The sphere isn't grounded and retains the charge, Q. so the total charge enclosed in a gaussian sphere of r > c should be Q and the electric field outside the conductive sphere should be non-zero. If not, where am I going wrong?
If the shell were grounded, wouldn't that mean that it would collect an opposite charge Q- due to induction from the inner sphere? In that case the two charges cancel out and the field outside disappears.
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!
I can't understand why the electric field at th-cam.com/video/WQ6adZYd9_0/w-d-xo.html is 0, and why it isn't E*4π*a^2= o*(4/3)π*r^3. Isn't Q spread evenly in the sphere volume? Then why a conducting sphere has to have its charge Q=0?
If we try to calculate the field at a point just past a, will it be different than the field at a point just before b? Because the radius will differ, so would we get a different answer for 2 points in that same area?
this is the hardest thing ever, I understand what you are saying but I'm having so much trouble understanding the problems I'm working on. I have a horrible teacher who doesn't explain anything!!
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?
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.
Although your result is right there seems to be a problem with the argumentation. Correct me if I'm wrong. The total flux through a surface can be zero but that doesn't say anything about the field locally. An example of that would be a dipole: its total charge is zero, as is the flux of E through a surface enclosing the dipole, but the field created by the dipole is not.
Katia Checkwicz it is clear it is normal to the surface but what i am wondering is how can I determine the direction( i.e.pointing inwards or outwards)? or just arbitrary as long as it is consistent in a system?
It makes sense to point the area vector outward, since you start with an area of 0 and work your way outward. If the vector pointed inward, this would indicate a decrease in area, which would make more sense in the case of potential, but not electric field.
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.
+Victoria Yepez r is an arbitrary point, a certain distance (radius) from the center of the spheres, used to determine the presence of an electric field in this particular setup.
Bakıyorum 2 ay önce sormuşsun. Umarım final dönemi olduğu için geç olmamıştır. Eğer "pi*r^2" değil miydi diye düşündüysen o çemberin alanı. Şuan anlatılan kürenin alanı.
içeriye bir tane farzı misal gaussian küre çiziyorsun, o kürenin alanı kalkülüste belirtildiği üzere r^2*sin(theta)*d(theta)*d(phi) nin integrali oluyor, integrali alınca 4pi r^2 geliyor. daha ayrıntılı bilgi için küresel alan integraline bakabilirsin
I don’t see how it would change other than a decimal going back and forth on the squeezing of the square root of ten that is timed and then divided into a concreted number.
No I meant that in the following video after this you explain that the E outside the concentric sphere isn't 0 but in this video you say that E=0 on the outside of the concentric sphere.
The electric field is zero inside the small sphere because the metal it is made of is a conductor. If the sphere was not a conductor, then the electric field would not be zero.
![“อาร์ต พศุตม์” เกือบเอี่ยวทุกคดีดัง เอ๊ะไว้ก่อน ไม่มีอะไรได้มาง่ายๆ | แฉ 16 ต.ค. 67 [2/3] | GMM25](http://i.ytimg.com/vi/0Gx-0-kUBv8/mqdefault.jpg)
Even 14 years later your videos are helping me
So essentially between concentric shells where a < r < b, the electric field is only due to the shell with radius a?
+Charles Goh I think that statement is correct, since the outer shell is grounded all of te E field is essentially due to the smaller charged sphere
Thanks! but for a
Because GAUSS'S THEOREM based on only enclosed surfaces. When r is greater then a and less then b then we are with Gaussian surface passing through Gaussian surface of radius r,in this time outer surface will be in effective.
You may also study with DHANANJAY PHYSICS CLASSES
He Solves this such that THE OUTER SPHERE IS GROUNDED good to know if you're new to this
In electric I guess you would ground the last one or you’d have to keep calculating?
actually, the E_r
Same goes for E_b
For r>c the electric field should be Q/4(pie)(epsilon)(r^2) since the charge on the inner shell is zero, negative charge(-Q) will get induced on the inner surface of the outer shell. But , since the outer shell is given no charge initially therefore net charge on the outer shell is zero and for this the outer surface of outer shell must be equal to positive charge Q. Hence electric field at a point r>c is as I have mentioned above.
If we go by your logic, you are denying the law of conservation of charges.
Please let me know, If you disagree with my argument.
thats what i thought as well
I think he ignored the electric field outside the outer shell because he assumed the outer shell is grounded.
The outer shell is grounded. Any charge on the outer shell goes away. Therefore, the law of conservation of charges is has not been disobeyed.
Amar Munkhjargal Thank you for clearing that up. I couldn't figure out why he ignored the charge, but assuming it was grounded it makes total sense. Thank-you
Divye Bhutani Ya I'm pretty sure you're right. I think he just messed up because he never said it was grounded.
This is the first time i come closer to so many genius professors in Physics.Thanksssssssssssssssssssssssssssssssssssssssssssssssssssss............
I was just about to ask the same question as April Pun - shouldn't the negative charge in the outer sphere increase the magnitude of the electric field for a
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?
Wow. I just want to say thank you so much for these videos, they're really saving my life for my Physics midterm. You're the man!
"Snuck in there and looked at a dA." 10/10
I have a final tomorrow and I trust these videos insead of my actual notes so thanks for your efforts :)
I did the same thing and I did just well on the exam lol.
did you pass
@@MultiFilip12 i think i did !
I am a Brazilian guy and I liked this classes. He is so calm and explain the subjects so good.
so,,, you don't consider -q when it r is a
I know Im a year late, but anyways, because what you consider in the equation is only Q that is enclosed in Gaussian Surface (Q enc.), there will only be +q enclosed by the g.s. when r is a
What do you mean with "because this is inside the metal"? when you are talking about the electric field inside the sphere?
What if instead of a
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
I'm confused by this as well but I agree with the original poster. In the video, he grounds only the conductive shell which surely means that it then has a net charge of zero. The sphere isn't grounded and retains the charge, Q. so the total charge enclosed in a gaussian sphere of r > c should be Q and the electric field outside the conductive sphere should be non-zero. If not, where am I going wrong?
If the shell were grounded, wouldn't that mean that it would collect an opposite charge Q- due to induction from the inner sphere? In that case the two charges cancel out and the field outside disappears.
How is the charge enclosed equal to 0 in the c < r example?
+Jordan Singer Outside has charge of -Q, inside has charge of +Q. The charges cancel to get 0.
What if the concentric sphere is an insulater what happens in r>c and r
It would gradually increase imo
OK thanks
You need to integrate the charge density out to an arbitrary radius to find the enclosed charge, then apply gauss's law like usual
If C is a conductor there should be charge at both B
Exam in 10 hours, you're my hero.
Catherine Hannah did you pass it ?
How was your exam 😐😁😁
So when r < a, would E still be zero if the sphere in the middle has +q charge uniformly distributed throughout the volume?
+William Mitchel that's what i was thinking. Field inside a shell is zero. But inside a charged solid sphere is (kQx)/R^3. So how does he say zero?
The field inside the smallest shell. Nope. You wouldn’t want to start another space trip on that one. Count shells. Lol
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!
Fantastically explained. Love from 🇮🇳
I can't understand why the electric field at th-cam.com/video/WQ6adZYd9_0/w-d-xo.html is 0, and why it isn't E*4π*a^2= o*(4/3)π*r^3. Isn't Q spread evenly in the sphere volume? Then why a conducting sphere has to have its charge Q=0?
Who needs to work those stupid C problems when this guy can just gift me his knowledge.
If we try to calculate the field at a point just past a, will it be different than the field at a point just before b? Because the radius will differ, so would we get a different answer for 2 points in that same area?
If the outer surface is not grounded, then what is E on outer surface?
Draw a spherical Gauss. Surf. just outside the outer sphere. The qen is Q+ -Q+Q so E=(1/4pi eps.naught)Q/r^2.
this is the hardest thing ever, I understand what you are saying but I'm having so much trouble understanding the problems I'm working on. I have a horrible teacher who doesn't explain anything!!
thanks so much! this helped me understand the base concepts for another problem i was having difficulty with. Your help is truly appreciated!
Taking AP Physics C in a few hours. I thank you for these videos.
whats the electric field at r = a and r= b
So, if the shell was not grounded, the electric field would be equal to Q/(epsilon*4*pi*r^2)?
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?
so does the hollow sphere has net -q charge?
Yes. The negative charge on the inside wall of the hollow sphere is the same magnitude as the positive charge of the inner sphere.
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.
You are so amazing. Thank you so much. You have no idea how much this saved me.
It is very good video for learning electric field. I got the answer from your video. Thank for your good video.
Although your result is right there seems to be a problem with the argumentation. Correct me if I'm wrong. The total flux through a surface can be zero but that doesn't say anything about the field locally. An example of that would be a dipole: its total charge is zero, as is the flux of E through a surface enclosing the dipole, but the field created by the dipole is not.
ffs i wish i saw this before my exam. Great Explanation
You rocks! My mother tonge is spanish but i actually understand you more than my physics professor :)
Cheers for posting this! Been struggling to get my head round Gaussian surfaces for a while now...
Daniel Burns do you still remember how to do it ?
wait so what happens if you don't ground the outer sphere?
lmfaooooo right?
how can I determine the direction of the area vector?
Simon Ngai it's normal to the surface.
Katia Checkwicz it is clear it is normal to the surface but what i am wondering is how can I determine the direction( i.e.pointing inwards or outwards)? or just arbitrary as long as it is consistent in a system?
It makes sense to point the area vector outward, since you start with an area of 0 and work your way outward. If the vector pointed inward, this would indicate a decrease in area, which would make more sense in the case of potential, but not electric field.
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.
Owing to this video, I would be able to solve a problem on a capacitor. Thanks a lot.
what happens if it is at b (the inner surface of shell)
Great videos greetings from turkey
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How do you calculate r?
+Victoria Yepez
r is an arbitrary point, a certain distance (radius) from the center of the spheres, used to determine the presence of an electric field in this particular setup.
U look like Shanola Hampton
9:53 you mean... one more youtube?
lmfaooooooo only the real know..... (ie. the ones who've been binge watching this man's videos)
Thankyou. It simplifies the meaning of the equations.
Thank you so much! This video really helped me make sense of my notes.
The serial of describing the concepts is incorrect.
this is good video. it so clear explanation and i can understand that explain.
Thanks for the videos. You are the best.
I have my quiz later, thanks to this ❤️
Where is r but?
nice question
beyler alan neden 4piR^2 oluyor?
Bakıyorum 2 ay önce sormuşsun. Umarım final dönemi olduğu için geç olmamıştır. Eğer "pi*r^2" değil miydi diye düşündüysen o çemberin alanı. Şuan anlatılan kürenin alanı.
içeriye bir tane farzı misal gaussian küre çiziyorsun, o kürenin alanı kalkülüste belirtildiği üzere r^2*sin(theta)*d(theta)*d(phi) nin integrali oluyor, integrali alınca 4pi r^2 geliyor. daha ayrıntılı bilgi için küresel alan integraline bakabilirsin
this just gave me 15 points on my exam.
bilkent phys 102ciler doluşun.
watch these vids, got 96 on physics midterm at 1 of the top universities in the world (its in canada).
@stephenxluo Yes i agree i was well confused did he make a mistake in this one with that then.
Anybody else have a midterm in a few hours?
Nice explain
You sound like a guy I could have a beer or smoke a cigar with
Thanks for the TH-cam.
I think you should do a video explaining how to suspend the first sphere inside the second sphere. (P.S. Great Explanations!)
I don’t see how it would change other than a decimal going back and forth on the squeezing of the square root of ten that is timed and then divided into a concreted number.
pff i wish i watch thisi before exam
thank you sooooooo much! you are a live saver!
No I meant that in the following video after this you explain that the E outside the concentric sphere isn't 0 but in this video you say that E=0 on the outside of the concentric sphere.
E=0 on the outside of the sphere? i just got mind fkd.
lmaooooooo, it's because he grounded it
Hello I am DHANANJAY upadhyay from India. I am a teacher of PHYSICS. You can also search my channel DHANANJAY PHYSICS CLASSES
Very clear. Thank you.
Really thank you
Amazing, thank you!
THANK YOU!!!
THANKS!
Bu güzel videoyu türkçe dublaj yapacak ,birisi olsa ?
bunu biri yapsa geri kalan 3294729 videoyu kim yapacak, türkçe kaynak bulmak çok zor internette, en iyisi ingilizce öğrenmek
It doesn't affect the electric field because it is not enclosed by the Gaussian surface.
Thank you
i love these videos!! :]
thanks
so helpful thank you!
nope, its a final exam :/
Midterm is next week...
Finaller basladı aq
sorry but expiation is unclear and all over the place
nice
allah razı olsun
The electric field is zero inside the small sphere because the metal it is made of is a conductor. If the sphere was not a conductor, then the electric field would not be zero.
Vizede aynısı çıktı haberiniz olsun
Benim de haftaya vizem :D Bau da okuyorum
@@necatiakpnar5677 :D kolay gelsin
W
we are horrible students
loool me too
r is.... ...greater than c..... ahhaahah
too slow..