You all probably dont care at all but does anyone know of a trick to log back into an instagram account? I was dumb lost the login password. I would appreciate any tricks you can give me!
@Theodore Corbin I really appreciate your reply. I got to the site on google and Im trying it out now. Looks like it's gonna take a while so I will get back to you later with my results.
Dear Lasseviren1, I would just like to thank you personally for putting up all these videos on Gauss. You literally just saved my 1st year EM exam from being a total disaster. Couldn't have done it without you! Who knew Gauss' Law was so simple?
The electric field above the plane, points straight up. The electric field below the plane, points straight down. So there is no flux through the sides of the cylinder and therefore you do not count the area of the sides of the cylinder. You only count the top and bottom of the cylinder, where there is actually flux.
Thank you so much, i understood all underlying concepts because of your videos. I was really frustrated over gauss's surfaces, luckily i found your videos
When I write my midterm I'm gonna have to bring an empty pop can so I can tilt my test sheet sideways and make "choonk" sounds as I work out the problem.
If the slab of charge is of finite thickness but infinite length and width, then the E does not become weaker as you move away from it. When field lines do not diverge or converge, then the field is uniform. That is only for outside the slab. If you are wondering when we ever would have an infinite slab or plane of charge, the answer is never. But if you have a large slab or plane of charge and you are very near this slab or plane, it behaves very much like an infinite one.
Good video. BUT it needs to make clear that this only applies to an *infinite* sheet of charge. For a finite sheet, the field lines diverge. At large enough distances, a finite sheet of charge is 'seen' as a point charge. So the derivation and formula will not apply.
I dont understand why you need to calculate the flux out of both sides of the sheet to get the total e at a point above the sheet. Why cant you just calculate the flux of HALF a soda can, therefore getting sigma/epsilon instead of sigma/2epsilon ?
Because then you wouldn't be able to make use of the symmetry of the problem and would have to worry about actually evaluating the integral. The way he does it, you can assume the field is of constant magnitude over the whole area.
i don't see how at 3:38 he says "the net flux is the net charge enclosed over epsilon naught" -- isn't that only the case when there is spherical symmetry ?
Because the charge has to be enclosed in a Gaussian surface. The Electric flux disperses in two directions which is the reason you get 1/2 in the answer.
Something seems incorrect with your reasoning. If your reasoning were true, the size of the sheet would not matter, right? The real reason that the flux is constant is because the field lines are NOT just straight. They go in all directions. So as you get further from the sheet, the charge directly beneath you has a smaller effect, but the other charges have a greater effect because theta angle is smaller
+Jonathan Burns I disagree with your interpretation. If we model the problem as you want to, actually the other charges would have a greater effect as *close* as they are to the plate, not as far as they get. This would happen because in a single particle, E ~ x^-2, and the quadratic effect makes E rises much faster than sin(theta) decreases as theta approaches 0. www.wolframalpha.com/input/?i=sin(arctan(x))+*+1%2Fx%5E2,+x%3D0+to+x%3D10 This would imply that E is larger as close as we get to the plate, which is false.
I believe so if you have a flat sheet the field lines are all perpendicular to each other and on a sphere, the field lines are all perpendicular to the specific area dA. The field lines of the sphere are then diverging from each other as they go off to infinity, therefore weakening the electric field as it continues off into space.
What if the sigma, charge density is a function of x and y. E is not a constant at different positions so we cant pull E out of the integral what to doooo???
It doesn't really make sense to me. I don't get this, why is this gaussian surface good for the sheet plane. If we are going to calculate the electric field from the sheet, don't we have to consider the whole sheet?
What makes the field? The field lines. All field lines are perpendicular to the sheet. At the point where we are calculating the field we don't have to consider the other part of the paper as that has nothing to do with it. In reality we just need a very small gaussian cylinder.The field would still be the same.
Yes - if you are calculating field outside the charged plate - it would still be independent of distance above the plate. For calculation inside, lasseviren has another video here
really ,, i dont understand why magnetic field due to infinite sheet or plane dont change when we walking far from sheet ? . knowing that E due to point charge radially outwards change inversely with distance ... i wanna understand it .
Consider the infinite sheet to be a wall of infinite height and length. No matter how far you get away, it will always be a wall, and never reduce to a point like a defined shape would. or using trig, if you are standing at theta and sine is infinitely large, is there any value for cosine that will make sin approach zero? if sine is infinite, and cosine goes to infinity, sin is still infinite. So there is no way to reduce the efield of an infinite sheet to a point charge, but there is a way to define it over a given area, and apply that knowledge over the whole.
The pop can sounds-- 10/10
You all probably dont care at all but does anyone know of a trick to log back into an instagram account?
I was dumb lost the login password. I would appreciate any tricks you can give me!
@Maxwell Onyx instablaster =)
@Theodore Corbin I really appreciate your reply. I got to the site on google and Im trying it out now.
Looks like it's gonna take a while so I will get back to you later with my results.
Hii james
Dear Lasseviren1,
I would just like to thank you personally for putting up all these videos on Gauss. You literally just saved my 1st year EM exam from being a total disaster. Couldn't have done it without you! Who knew Gauss' Law was so simple?
I recommend more sound effects! Choonk!
CHOONK
The electric field above the plane, points straight up. The electric field below the plane, points straight down. So there is no flux through the sides of the cylinder and therefore you do not count the area of the sides of the cylinder. You only count the top and bottom of the cylinder, where there is actually flux.
I've learned so much about choonk from these youtubes
I wish my teacher would just send us videos like these instead of wasting our time
Thank you for these videos! They really helped me!
AlexBond4ever same
1:30 that's some exquisite sound effects
Thank you so much, i understood all underlying concepts because of your videos. I was really frustrated over gauss's surfaces, luckily i found your videos
This was very helpful thank you. Nice sound effects too
He teaches well and also a fantastic artist.
I had a question, and your video answered. Thank you.
When I write my midterm I'm gonna have to bring an empty pop can so I can tilt my test sheet sideways and make "choonk" sounds as I work out the problem.
If the slab of charge is of finite thickness but infinite length and width, then the E does not become weaker as you move away from it. When field lines do not diverge or converge, then the field is uniform. That is only for outside the slab. If you are wondering when we ever would have an infinite slab or plane of charge, the answer is never. But if you have a large slab or plane of charge and you are very near this slab or plane, it behaves very much like an infinite one.
Thank you for this video! I had a similar problem like this for my homework but no one can explain this to me as well as you can
Good video. BUT it needs to make clear that this only applies to an *infinite* sheet of charge. For a finite sheet, the field lines diverge. At large enough distances, a finite sheet of charge is 'seen' as a point charge. So the derivation and formula will not apply.
thankyou i didnt know that
Nice! "Pop can"
"choonk"! What a great video
Above a slab of charge does the electric field become weaker the further you are from it? Or is it the same as an infinite plane of charge?
at 1:31, i prefer to imagine a beer can
Is it sigma/2eps on both sides sumed up or once on top and once to the bottom ?
Love the sound effects
I dont understand why you need to calculate the flux out of both sides of the sheet to get the total e at a point above the sheet. Why cant you just calculate the flux of HALF a soda can, therefore getting sigma/epsilon instead of sigma/2epsilon ?
im confused over the same thing you are confused about
Choonk!
Why does the bottom have to be at the same distance ? What would it change if it was 2y for example ?
Why are A1 and A3 added together? They're vector quantities with opposite directions, so why don't they subtract from each other?
openskies11 but flux is scalar
Haha omg I imagine him sitting in his kitchen making a UA-cam video from his phone teaching physics w nothing but copy paper and a sharpie
Because then you wouldn't be able to make use of the symmetry of the problem and would have to worry about actually evaluating the integral. The way he does it, you can assume the field is of constant magnitude over the whole area.
i don't see how at 3:38 he says "the net flux is the net charge enclosed over epsilon naught" -- isn't that only the case when there is spherical symmetry ?
Because the charge has to be enclosed in a Gaussian surface. The Electric flux disperses in two directions which is the reason you get 1/2 in the answer.
Something seems incorrect with your reasoning. If your reasoning were true, the size of the sheet would not matter, right?
The real reason that the flux is constant is because the field lines are NOT just straight. They go in all directions. So as you get further from the sheet, the charge directly beneath you has a smaller effect, but the other charges have a greater effect because theta angle is smaller
+Jonathan Burns I may be months late to this, but I was also under the impression that this was why E remained constant.
+Jonathan Burns
I disagree with your interpretation. If we model the problem as you want to, actually the other charges would have a greater effect as *close* as they are to the plate, not as far as they get. This would happen because in a single particle, E ~ x^-2, and the quadratic effect makes E rises much faster than sin(theta) decreases as theta approaches 0.
www.wolframalpha.com/input/?i=sin(arctan(x))+*+1%2Fx%5E2,+x%3D0+to+x%3D10
This would imply that E is larger as close as we get to the plate, which is false.
So the electrical field (E) is always perpendiculiar to the surface when it leaves it?
I believe so if you have a flat sheet the field lines are all perpendicular to each other and on a sphere, the field lines are all perpendicular to the specific area dA. The field lines of the sphere are then diverging from each other as they go off to infinity, therefore weakening the electric field as it continues off into space.
Wait a minute, why did you add up A1 and A3? Will not the electric fields cancel each other?
Stay ahhh idk A little late but flux is scalar not vector, therefore it won’t cancel.
@@tonyf1163 good answer
Thanks for the lovely explanation...... CHOONK
Thank you great Sir.
Thank you very much!
if i can press like twice i would be more than happy to like it twice. thank you:)
Why is integral of dA = (A1 +A3)?
I really want to understand this in it's entirety.
yeah me too
What if the sigma, charge density is a function of x and y. E is not a constant at different positions so we cant pull E out of the integral what to doooo???
It doesn't really make sense to me. I don't get this, why is this gaussian surface good for the sheet plane. If we are going to calculate the electric field from the sheet, don't we have to consider the whole sheet?
What makes the field? The field lines. All field lines are perpendicular to the sheet. At the point where we are calculating the field we don't have to consider the other part of the paper as that has nothing to do with it.
In reality we just need a very small gaussian cylinder.The field would still be the same.
good work thanks:)
excellent
Thank You
this problem uses the word plane, but would this be the same if the problem says find Efield above charge plate ?
Yes - if you are calculating field outside the charged plate - it would still be independent of distance above the plate. For calculation inside, lasseviren has another video here
why not put a Gaussian cube or a cuboid ?
Irrelevant, in any case the Area of the plane cancels with the top and bottom area to give you the Electric field...
Thanks.
can the gaussian surface be cube shape? o.o just curious
Yeah in this case it doesn't matter, since the area is being expressed as A anyway (not 4 pi r^2.)
wanna know that too..
really ,, i dont understand why magnetic field due to infinite sheet or plane dont change when we walking far from sheet ? . knowing that E due to point charge radially outwards change inversely with distance ... i wanna understand it .
Consider the infinite sheet to be a wall of infinite height and length. No matter how far you get away, it will always be a wall, and never reduce to a point like a defined shape would. or using trig, if you are standing at theta and sine is infinitely large, is there any value for cosine that will make sin approach zero? if sine is infinite, and cosine goes to infinity, sin is still infinite. So there is no way to reduce the efield of an infinite sheet to a point charge, but there is a way to define it over a given area, and apply that knowledge over the whole.
Interesting
I have this question here: "Using Gauss law, obtain electric field created by a charge uniformly distributed over a plane"
Did he say pop can? hahaha
Who’s here for physics c homework
Poggers
ua-cam.com/video/0rRWQlGleNM/v-deo.htmlm32s best part
Thank you very much!
CHOONK