Advanced proof of the formula for the electric field generated by a uniformly charged, infinite plate. Created by Sal Khan.
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- this is a really dumb question, but the parallel plates that aren't infinite are just an approximation of the infinite when charge and length of the plate are big enough compared to the distance right?(57 votes)
You are sort of correct. Khan actually mentions that at the end of the next video in this series: that for a finite plate where you are far away from the edge, this theory holds, so yes, you are correct. However, just remember that right near the edge, things fall apart.
Most theories are really good approximations; the only real values come when you test experimentally.
This isn't really a dumb question; I pondered in once myself, and most people will.(88 votes)
- At9:30, why is the area of the ring the circumference times dr? If dr were much larger (call it w), wouldn't the area be pi((r+w)^2-r^2) ? Like the difference between two circles.(24 votes)
- You are actually correct. If dr was larger, than it would be pi((r+w)^2-r^2) for the area. However, this is not the place to use that. We are assuming an infinitely small width between the two circles, we just assume that there is a central circle, right in the middle of the two circles, and we extend a length dr for the width. Because of the size, this average value is actually really accurate, so we stick to it, and that makes the math much more simple.
I hope that is not confusing; one of my teachers spent half a class teaching us why that is so, but his explanation was much better than mine.(18 votes)
- Around9:59, Sal says that the force generated by the ring is = kqq/d^2. However, how is the entire charge of the ring enclosed to one point? More specifically, how come he is treating the entire charge of the ring as enclosed to one point? Is he allowed to do that? Thanks.(14 votes)
- Since the ring is of uniform charge, and the charge density is sigma then Q (<-Charge of ring) = sigma (<-charge density) * (2 * pi * r * dr) (<-area of the ring). He is correct in doing this, but only in the case of a surface of uniform charge (The charge at every point of the surface is equal) and with the test charge (q) at the center of the ring so that all of the horizontal components of the electrostatic forces cancel, leaving only the vertical components.
Since forces are vector quantities and in the same direction, we can treat them as if they were all directly under the test charge and them sum them up to receive the total net force on the test charge from the ring.
Now if the ring was not of uniform charge and/or the test charge was not at the center of the ring, then we would have a much more difficult problem.
I hope this clears things up for you. ^^(12 votes)
- At9:18why is the width of the circle 'dr'; why does it share the same variable as the radius? Why not 'dx' or 'dw'? What allows us to use 'r' both times? Does it have to do with firstname.lastname@example.org's question? Thanks!(12 votes)
- because dr is just a small change in radius
since he's talking a bout a circle with radius r it seems fitting to use dr
and because in the next video he's going to integrate across r
if he used dw he would have to rename r to w to be consistent(1 vote)
- Doesn't this video assume that the magnetic field will repel the particle, instead of attracting it?(7 votes)
- Yes it does. It was mentioned early in the video that the the chanrge on the plate is positive and the particle is positive as well so the force would be repulsive.(11 votes)
- Why does Sal take an infinite plane? I find it quite hard to imagine.(5 votes)
- If you pick a plane of finite size, you would have to compute the net effect of whatever you are studying at that boundary and include it in your final answer. In much of physics, an effect (gravitational potential or electric field for examples) are defined as zero at infinity, so you can ignore that 'boundary' value.
So think of it as 'choose a plane whose size is large enough that we can use infinity as the 'far' distance (for example, we could calculate (or integrate) from zero radius to infinity). If the effect has dropped to zero at infinity, that term becomes simpler to calculate - or may drop out altogether.(8 votes)
- At ~9:30, why didn't we take πr^2, since there are charged particles everywhere on the ring?(5 votes)
- We have to consider the point charges on the ring. So the total charge isn't the area because the "ring" is assumed to be hollow. So the charge is :-
(circumference) * (thickness) * (charge density)
Watch the next part and Sal will integrate this equation and obtain a general formula for this situation.(6 votes)
- He saying i need to learn about calculus first, can anybody tell me from where should i start learning it, i mean there are 3 different calculus topics ;Differential, multivariable, integral.Or any other type?(2 votes)
- For this specific video, you'd want to learn basic integration techniques, though learning most of differential and integral calculus (single-variable) would be a good idea so as to establish a thorough knowledgebase.(5 votes)
- So, a test charge could be light-years away and still feel the same force?(4 votes)
- If the plate were infinite, yes. Infinite is big even compared to light-years. But there's no such thing as an infinite plate, right?(2 votes)
- Why do you calculate the charge from the whole area, if youre only gonna use the points on the edge of the circle?(3 votes)
- since it is a ring and ring does not has a symmetry except at the ends of diameter, therefore he has taken two small charges 'dq' and then integrate it(2 votes)
In this video, we're going to study the electric field created by an infinite uniformly charged plate. And why are we going to do that? Well, one, because we'll learn that the electric field is constant, which is neat by itself, and then that's kind of an important thing to realize later when we talk about parallel charged plates and capacitors, because our physics book tells them that the field is constant, but they never really prove it. So we will prove it here, and the basis of all of that is to figure out what the electric charge of an infinitely charged plate is. So let's take a side view of the infinitely charged plate and get some intuition. Let's say that's the side view of the plate-- and let's say that this plate has a charge density of sigma. And what's charge density? It just says, well, that's coulombs per area. Charge density is equal to charge per area. That's all sigma is. So we're saying this has a uniform charge density. So before we break into what may be hard-core mathematics, and if you're watching this in the calculus playlist, you might want to review some of the electrostatics from the physics playlist, and that'll probably be relatively easy for you. If you're watching this from the physics playlist and you haven't done the calculus playlist, you should not watch this video because you will find it overwhelming. But anyway, let's proceed. So let's say that once again this is my infinite so it goes off in every direction and it even comes out of the video, where this is a side view. Let's say I have a point charge up here Q. So let's think a little bit about if I have a point-- let's say I have an area here on my plate. Let's think a little bit about what the net effect of it is going to be on this point charge. Well, first of all, let's say that this point charge is at a height h above the field. Let me draw that. This is a height h, and let's say this is the point directly below the point charge, and let's say that this distance right here is r. So first of all, what is the distance between this part of our plate and our point charge? What is this distance that I'll draw in magenta? What is this distance? Well, the Pythagorean theorem. This is a right triangle, so it's the square root of this side squared plus this side squared. So this is going to be the square root of h squared plus r squared. So that's the distance between this area and our test charge. Now, let's get a little bit of intuition. So if this is a positive test charge and if this plate is positively charged, the force from just this area on the charge is going to be radially outward from this area, so it's going to be-- let me do it in another color because I don't want to-- it's going to go in that direction, right? But since this is an infinite plate in every direction, there's going to be another point on this plate that's essentially on the other side of this point over here where its net force, its net electrostatic force on the point charge, is going to be like that. And as you can see, since we have a uniform charge density and the plate is symmetric in every direction, the x or the horizontal components of the force are going to cancel out. And so that's true for really any point along this plate. Because if you pick any point along it, and we're looking at a side view, but if we took a top view, if that's the top view and, of course, the plate goes off in every direction forever and that's kind of where our point charge is, if we said, oh, well, you know, there's this point on the plate and it's going to have some y-component that's on this top view coming out of the video, but it'll have some x-component, this point's x-component effect will cancel it out. You can always find another point on the plate that's symmetrically opposite whose x-component of electrostatic force will cancel out with the first one. So given that, that's just a long-winded way of saying that the net force on this point charge will only be upwards. I think it should make sense to you that all of the x-components or the horizontal components of the electrostatic force all cancel out, because they're infinite points to either side of this test charge. So with that out of the way, what do we need to focus on? Well, we just need to focus on the y-components of the electrostatic force. So what's the y-component? So let's say that this point right here-- and I'll keep switching colors. Let's say that this point-- and once again, this is a side view-- is exerting-- its field at that point is e1, and it's going to be going in that direction. What is its y-component? What is the component in that direction? And, of course, it's pushing outwards if they're both positive. So what is the y-component? What is that? Well, if we knew theta, if we knew this angle, the y-component, or the upwards component is going to be the electric field times cosine of theta. Cosine is adjacent over hypotenuse, so hypotenuse times cosine of theta is equal to the adjacent. So if we wanted the vertical or the y-component of the electric field, we would just multiply the magnitude of the electric field times the cosine of theta. So how do we figure out theta? Well, that theta is also the same as this theta from our basic trigonometry. And so what's cosine of theta? Cosine is adjacent over hypotenuse from SOHCAHTOA, right? Cosine of theta is equal to adjacent over hypotenuse. So when we're looking at this angle, which is the same as that one, what's adjacent over hypotenuse? This is adjacent, that is the hypotenuse. So what do we get? We get that the y-component of the electric field due to just this little chunk of our plate, the electric field in the y-component, let's just call that sub 1 because this is just a little small part of the plate. It is equal to the electric field generally, the magnitude of the electric field from this point, times cosine of theta, which equals the electric field times the adjacent-- times height-- over the hypotenuse-- over the square root of h squared plus r squared. Fair enough. So now let's see if we can figure out what the magnitude of the electric field is, and then we can put it back into this and we'll figure out the y-component from this point. And actually, we're not just going to figure out the electric field just from that point, we're going to figure out the electric field from a ring that's surrounding this. So let me give you a little bit of perspective or draw it with a little bit of perspective. So this is my infinite plate again. I'll draw it in yellow again since I originally drew it in yellow. This is my infinite plate. It goes in every direction. And then I have my charge floating above this plate someplace at height of h. And this point here, this could have been right here maybe, but what I'm going to do is I'm going to draw a ring that's of an equal radius around this point right here. So this is r. Let's draw a ring, because all of these points are going to be the same distance from our test charge, right? They all are exactly like this one point that I drew here. You could almost view this as a cross-section of this ring that I'm drawing. So let's figure out what the y-component of the electric force from this ring is on our point charge. So to do that, we just have to figure out the area of this ring, multiply it times our charge density, and we'll have the total charge from that ring, and then we can use Coulomb's Law to figure out its force or the field at that point, and then we could use this formula, which we just figured out, to figure out the y-component. I know it's involved, but it'll all be worth it, because you'll know that we have a constant electric field. So let's do that. So first of all, Coulomb's Law tells us-- well, first of all, let's figure out the charge from this ring. So Q of the ring, it equals what? It equals the circumference of the ring times the width of the ring. So let's say the circumference is 2 pi r, and let's say it's a really skinny ring. It's really skinny. It's dr. Infinitesimally skinny. So it's width is dr. So that's the area of the ring, and so what's its charge going to be? It's area times the charge density, so times sigma. That is the charge of the ring. And then what is the electric field generated by the ring at this point here where our test charge is? Well, Coulomb's Law tells us that the force generated by the ring is going to be equal to Coulomb's constant times the charge of the ring times our test charge divided by the distance squared, right? Well, what's the distance between really any point on the ring and our test charge? Well, this could be one of the points on the ring and this could be another one, right? And this is like a cross-section. So the distance at any point, this distance right here, is once again by the Pythagorean theorem because this is also r. This distance is the square root of h squared plus r squared. It's the same thing as that. So it's the distance squared and that's equal to k times the charge in the ring times our test charge divided by distance squared. Well, distance is the square root of h squared plus r squared, so if we square that, it just becomes h squared plus r squared. And if we want to know the electric field created by that ring, the electric field is just the force per test charge, so if we divide both sides by Q, we learned that the electric field of the ring is equal to Coulomb's constant times the charge in the ring divided by h squared plus r squared. And now what is the y-component of the charge in the ring? Well, it's going to be this, right? What we just figured out is the magnitude of essentially this vector, right? But we want its y-component, because all of the x-components just cancel out, so it's going to be times cosine of theta, and we figured out that cosine of theta is essentially this, so we multiply it times that. So the field from the ring in the y-direction is going to be equal to its magnitude times cosine of theta, which we figured out was h over the square root of h squared plus r squared. We could simplify this a little bit. The denominator becomes what? h squared plus r squared to the 3/2 power. And what's the numerator? Let's see, we have kh and then the charge in the ring, which we solved up here. So that's 2 pi sigma r-- make sure I didn't lose anything-- dr. So we have just calculated the y-component, the vertical component, of the electric field at h units above the plate. And not from the entire plate, just the electric field generated by a ring of radius r from the base of where we're taking this height. And so I've already gone 12 minutes into this video, and just to give you a break and myself a break, I will continue in the next. But you can imagine what we're going to do now. We just figured out the electric field created by just this ring, right? So now we can integrate across the entire plane. We can solve all the rings of radius infinity all the way down to zero, and that'll give us the sum of all of the electric fields and essentially the net electric field h units above the surface of the plate. See you in the next video.