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Proof: Field from infinite plate (part 2)

Video transcript
So where I left off, we had this infinite plate. It's just an infinite plane, and it's a charged plate with a charge density sigma. And what we did is we said, OK, well, we're taking this point up here that's h units above the surface of our charge plate, and we wanted to figure out the electric field at that point, generated by a ring of radius r essentially centered at the base of where that point is above. We want to figure out what is the electric field generated by this ring at that point? And we figured out that the electric field was this, and then because we made a symmetry argument in the last video, we only care about the y-component. Because we figured out that at the electric field generated from any point, the x-components cancel out, because if we have a point here, it'll have some x-component. The field's x-component might be in that direction to the right, but then you have another point out here, and its x-component will just cancel it out. So we only care about the y-component. So at the end, we meticulously calculated what the y-component of the electric field generated by the ring is, at h units above the surface. So with that out of the way, let's see if we can sum up a bunch of rings going from radius infinity to radius zero and figure out the total y-component. Or essentially the total electric field, because we realize that all the x's cancel out anyway, the total electric field at that point, h units above the surface of the plane. So let me erase a lot of this just so I can free it up for some hard-core math. And this is pretty much all calculus at this point. So let me erase all of this. Watch the previous video if you forgot how it was derived. Let me even erase that because I think I will need a lot of space. There you go. OK, so let me redraw a little bit just so we never forget what we're doing here because that happens. So that's my plane that goes off in every direction. I have my point above the plane where we're trying to figure out the electric field. And we've come to the conclusion that the field is going to point upward, so we only care about the y-component. It's h units above the surface, and we're figuring out the electric field generated by a ring around this point of radius r. And what's the y-component of that electric field? We figured out it was this. So now what we're going to do is take the integral. So the total electric field from the plate is going to be the integral from-- that's a really ugly-looking integral-- a radius of zero to a radius of infinity. So we're going to take a sum of all of the rings, starting with a radius of zero all the way to the ring that has a radius of infinity, because it's an infinite plane so we're figuring out the impact of the entire plane. So we're going to take the sum of every ring, so the field generated by every ring, and this is the field generated by each of the rings. Let me do it in a different color. This light blue is getting a little monotonous. Kh 2pi sigma r dr over h squared plus r squared to the 3/2. Now, let's simplify this a little bit. Let's take some constants out of it just so this looks like a slightly simpler equation. So this equals the integral from zero to-- So let's take the K-- I'm going to leave the 2 there. You'll see why in a second, but I'm going to take all the other constants out that we're not integrating across. So it's equal to Kh pi sigma times the integral from zero to infinity of what is this? So what did I leave in there? I left a 2r, so we could rewrite this as-- well, actually, I'm running out of space. 2r dr over h squared plus r squared to the 3/2, or we could think of it as the negative 3/2, right? So what is the antiderivative of here? Well, this is essentially the reverse chain rule, right? I could make a substitution here, if you're more comfortable using the substitution rule, but you might be able to eyeball this at this point. We could make the substitution that u is equal to-- if we just want to figure out the antiderivative of this-- if u is equal to h squared plus r squared-- h is just a constant, right-- then du is just equal to-- I mean, the du dr-- this is a constant, so it equals 2r, or we could say du is equal to 2r dr. And so if we're trying to take the antiderivative of 2r dr over h squared plus r squared to the 3/2, this is the exact same thing as taking the antiderivative with this substitution. 2r dr, we just showed right here, that's the same thing as du, right? So that's du over-- and then this is just u, right? H squared plus r squared is u. We do that by definition. So u to the 3/2, which is equal to the antiderivative of-- we could write this as u to the minus 3/2 du. And now that's easy. This is just kind of reverse the exponent rule. So that equals minus 2u to the minus 1/2, and we can confirm, right? If we take the derivative of this, minus 1/2 times minus 2 is 1, and then subtract 1 from here, we get minus 3/2. And then we could add plus c, but since we're eventually going to do a definite integral, the c's all cancel out. Or we could say that this is equal to-- since we made that substitution-- minus 2 over-- minus 1/2, that's the same thing as over the square root of h squared plus r squared, right? So all of the stuff I did in magenta was just to figure out the antiderivative of this, and we figured it out to be this: minus 2 over the square root of h squared plus r squared. So with that out of the way, let's continue evaluating our definite integral. So this expression simplifies to-- this is a marathon problem, but satisfying-- K-- let's get all the constants-- Kh pi sigma-- we can even take this minus 2 out-- times minus 2, and all of that, and we're going to evaluate the definite integral at the two boundaries-- 1 over the square root of h squared plus r squared evaluated at infinity minus it evaluated at 0, right? Well, what does this expression equal? What is 1 over the square root of h squared plus infinity, right? What happens when we evaluate r at infinity? Well, the square root of infinity is still infinity, and 1 over infinity is 0, so this expression right here just becomes 0. When you evaluate it at infinity, this becomes 0 minus this expression evaluated at 0. So what happens when it's at 0? When r squared is 0, we get 1 over the square root of h squared, right? So let's write it all out. This becomes minus 2Kh pi sigma times 0 minus 1 over the square root of h squared. Well this equals minus 2Kh pi sigma times-- well, 1 over the square root of h squared, that's just 1 over h, right? And there's a minus times minus 1 over h. Well, this minus and that minus cancel out. And then this h and this 1 over h should cancel out. And all we're left with, after doing all of that work, and I'll do it in a bright color because we've done a lot of work to get here, is 2K pi sigma. So let's see it at a lot of levels. First of all, what did we even do here? We might have gotten lost in the math. This is the net electric, the total electric field, at a point at height h above this infinite plate that has a uniform charge, and the charge density is sigma. But notice, this is the electric field at that point, but there's no h in here. So it essentially is telling us that the strength of the field is in no way dependent on how high above the field we are, which tells us this is going to be a constant field. We can be anywhere above the plate and the charge will be the same. The only thing-- oh, sorry, not the charge. The field will be the same, and if we have a test charge, the force would be the same. And the only thing that the strength of the field or the strength of the exerted electrostatic force is dependent on, is the charge density, right? This is Coulomb's constant, pi is pi, 2pi, and I think it's kind of cool that it involves pi, but that's something else to ponder. But all that matters is the charge density. So hopefully, you found that reasonably satisfying, and the big thing that we learned here is that if I have an infinite uniformly charged plate, the field-- and I'm some distance h above that field-- above that plate, it doesn't matter what that h is. I could be here, I could be here, I could be here. At all of those points, the field has the exact same strength, or the net electrostatic force on a test charge at those points has the exact same strength, and that's kind of a neat thing. And now if you do believe everything that occurred in the last two videos, you can now believe that there are such things as uniform electric fields and they occur between parallel plates, especially far away from the boundaries. See you soon.