The electric field is parallel to the normal coming from the surface, not perpendicular as stated, which is why it is cos(0). If it was perpendicular it would be cos(90) which would be equal zero. There should be a note stating the correction I think.
Hello professor, i think the concept of flux needs few more videos for better understanding as it is probably going to be 1 question of most of people here's midterms.
Sir I have a question: what happens if a sphere is not hollow? is there still electric flux inside the sphere body? If so, how would you calculate the total electric flux and the electric potential?
Q inside the Gaussian surface equals the net charge inside the surface. (when the charge is positive, it just means that there are more positive charges than negative charges).
It depends on the geometry and the context. If you have a point charge and you draw a spherical Gaussian surface around the charge, then it doesn't matter how large the radius of the Gaussian surface is, the flux through the surface will be the same.
The field associates with each coordinate (x,y,z) some force F due to the electric force, and this is the electric field. Electric flux is how much field is "flowing" out of a body (or through a surface).
Michel van Biezen For a constant electric field, and also in a situation in which the electric field is parallel to the area vectors of the surface, the definition of electric flux reduces to the product of electric field and surface area. But if we look at what a surface integral of a field gives you, that is a flow, or a flux. So, electric flux is the "flow" of the electric field through a surface.
+andrewskyworker Electric flux is kinda like the amount of electric field lines that passes through a "surface" or area. While electric field is the vector field generated by charges/objects ......
+Rahul Tiwari If you know the strength of the field and the direction of the field and the surface then you can simply do a dot product. Flux = E x A x cos (angle between the normal of the surface and the direction of the field). Gauss theorem is used to find the electric field.
this is a very funny comment because approximately 99.9% of comments under this channel are positive, and there are too many to count about how he is a very good teacher. You go right ahead and keep reading your textbook.
@@surry99 yup but take a look at the playlists. Majority of playlists on this channel require no knowledge of the subject. so videos about the basics, as u said things that are written in print, are taught here. And I appreciate that especially because I dont have a good textbook to use. Cheers.
![Lp. Сердце Вселенной #60 РОЖДЕНИЕ ЛОЛОЛОШКИ [Финал] • Майнкрафт](http://i.ytimg.com/vi/YoR0pAV9FVQ/mqdefault.jpg)
He misspoke but wrote it correctly: cosine of 0 is 1, not 0.
Thank you sir..you are a great teacher..the best teachers really are invested in their students understanding the material..excellent job..
Thank you. That is what we are trying to do. 🙂
Really... you are one of my best teacher from now...
Dear Professor thank you sooo much for your incredible physics videos, you make everything much understandable, easy and fun.
Thank you so much for all these videos. I am very grateful. They have been a great help to me!
The electric field is parallel to the normal coming from the surface, not perpendicular as stated, which is why it is cos(0). If it was perpendicular it would be cos(90) which would be equal zero. There should be a note stating the correction I think.
0:53 quite confused about the cos(0) equals 0
That was a misspeak.
I am literally watching every single topic to study for the finals, which might be 100% of my grade.
You are my Hero.
and became mine as well
Why aren't you my teacher. First thing I understood all semester!
Hello professor, i think the concept of flux needs few more videos for better understanding as it is probably going to be 1 question of most of people here's midterms.
Great work sir
Sir I have a question: what happens if a sphere is not hollow? is there still electric flux inside the sphere body? If so, how would you calculate the total electric flux and the electric potential?
There are examples like that in the playlist.
Sir,does Q(ins) includes the electrons present over the gaussian surface?
Q inside the Gaussian surface equals the net charge inside the surface. (when the charge is positive, it just means that there are more positive charges than negative charges).
Thank you Sir! You are the best
sir d formula is same for regular and irregular surfaces?
You need to be able to draw the Gaussian surface so that the electric field direction at the Gaussian surface is perpendicular to the surface.
Is 4pir^2 only for spheres?
The area of a cirle is pi r^2 The surface area of a sphere is 4 pi r^2
So if the flux doesn't depend on the radius because Φ=Qin/ε0 -- but if you were to look at Φ=E•A , then the radius would affect the flux?
It depends on the geometry and the context. If you have a point charge and you draw a spherical Gaussian surface around the charge, then it doesn't matter how large the radius of the Gaussian surface is, the flux through the surface will be the same.
How about electrostatic force in cavity ?
cos(0) is one on zero as was stated
It is clear he didn't mean it. If he had meant it the flux would be zero.
what is different between electric flux and electric field?
Andrew,
Electric flux = Electric field x area
The field associates with each coordinate (x,y,z) some force F due to the electric force, and this is the electric field. Electric flux is how much field is "flowing" out of a body (or through a surface).
Juxtaroberto
Electric flux does not "flow".
It is simply the product of the electric field and the surface area.
Michel van Biezen For a constant electric field, and also in a situation in which the electric field is parallel to the area vectors of the surface, the definition of electric flux reduces to the product of electric field and surface area. But if we look at what a surface integral of a field gives you, that is a flow, or a flux. So, electric flux is the "flow" of the electric field through a surface.
+andrewskyworker Electric flux is kinda like the amount of electric field lines that passes through a "surface" or area. While electric field is the vector field generated by charges/objects ......
sir y we take the Gaussian surface??? is it must???
For some situations the technique of the Gaussian surface makes it a lot easier to find the electric field.
Sir how E flux will be Zero
and all the surface will appear zero
Are you referring to the flux inside the conductor?
cos(0) is not 0, it's 1.
correct
Thanks my man, thats the most import part of this video.
XD
Sir is it possible to calculate flux through an open surface without using gauss theorem
+Rahul Tiwari
If you know the strength of the field and the direction of the field and the surface then you can simply do a dot product.
Flux = E x A x cos (angle between the normal of the surface and the direction of the field). Gauss theorem is used to find the electric field.
+Michel van Biezen thanks sir
sir if the outer surface contains + charge and and the inner surface contains + charge their should be a reputation between them right?? 💭
repulsion
Yes, it is the repulsion that causes the (extra) positive charge to move as far away from each other as possible.
sir , the electric field exert in the center of the charged spherical conductor ??
Michel dayı you are the best all of them rest
one not zero
sideways spaceship = flux
cos(0) = 1
Be my Professor plis
👍
Why bother doing a video like this? It can be found in any textbook. He provides absolutely no insight about Gauss's Law.
this is a very funny comment because approximately 99.9% of comments under this channel are positive, and there are too many to count about how he is a very good teacher. You go right ahead and keep reading your textbook.
@@Dina-he1uc Do you understand what the word "insight" means?
@@surry99 yup but take a look at the playlists. Majority of playlists on this channel require no knowledge of the subject. so videos about the basics, as u said things that are written in print, are taught here. And I appreciate that especially because I dont have a good textbook to use. Cheers.
@@Dina-he1uc I am glad you find them helpful.
That is the most important part.