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Proof: Field from infinite plate (part 2)

We see that the infinite, uniformly charged plate generates a constant electric field (independent of the height above the plate). Created by Sal Khan.

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  • male robot donald style avatar for user Blade
    Sal says that no matter at what height the test point is the strength of the field remains same. If this is so then the force exerted by the field should remain constant. But as far as i know the magnitude of force keeps changing as we keep on changing the distance(i.e height in this case). Please help me. I'm so confused.
    (19 votes)
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    • blobby green style avatar for user Ash
      Imagine a bunch of strings connecting each point on the plate to your test point. When you're really close to the plate, the repulsive force is extremely strong since you're very near the plate points directly underneath you, but you're at such a shallow angle to the rest of the plate's points that their y-components make relatively little impact. The further away those points are, the shallower the angle relative to you. As you move a little bit away, the distance between you and the points directly below you increases, but the angle between you and every other point on the plate gets steeper. The y-components of those now-steeper angles gets larger, and they compensate for the increased distance. The further away you move from the plate, the more you increase the influence of ever more-distant points on the plate.
      (87 votes)
  • blobby green style avatar for user Khashon Haselrig
    Does this mean if you had 2 infinite and identically charged plates, a test charge would feel an equal force from either plate even if it was 1mm from one plate and 100km from the other? And if this can be extended to gravity does that mean you would be weightless regardless of your distance from equally dense infinite plates?
    (12 votes)
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  • leaf green style avatar for user Veronica Schmidt Teron
    How did he decide that the charge of the ring, Qr=2*(π)r*d*r(sigma)? I know that 2*(pr)*r is circumference. How did he figure d*r=width? What is d, distance? Why doesn't he just use the area of the circle Area = π × r^2 equation to solve for Qr? Please let me know if you are able to explain and where I can figure this out. Thank you! :-)
    (5 votes)
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    • leaf green style avatar for user Mark Zwald
      I didn't watch the video, but I'm guessing he is integrating the area of concentric rings to derive the E field from an infinite plate. In that case the area of a ring with radius r and width dr would be the length of the ring (ie. the circumference) times the width. So the incremental ring area is
      dA = 2*π*r*dr
      where the incremental charge on the ring is the ring area times charge per unit area (σ), so dQ = σ*dA
      and the total charge Q is the integral of σ*dA from r = 0 to infinity.
      (7 votes)
  • leaf green style avatar for user Kris Kalavantavanich
    The field is infinitely large; therefore, wouldn't the force applied to the test charge be infinitely strong as well?
    (7 votes)
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  • leaf green style avatar for user Ruben
    lets say that there is a positive charged particle above the infinite plate. Will the particle move up infinitely because of the force produced by the constant electric field?
    (1 vote)
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  • male robot hal style avatar for user Konain Ali
    if the test charge is at infinite height from the infinitely charged plate then would there be an electric field from the plate ??
    (3 votes)
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  • blobby green style avatar for user ittisaf07
    we know when we do definite integrals, we fine the area under a curve between two points. in that sense, what does this actually mean? how does summing up all the forces relate to finding the area of a function between two points .. i just cant relate these two things..
    (3 votes)
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    • blobby green style avatar for user Teacher Mackenzie (UK)
      good question

      One way I find it useful to make sense of this in a simple way is to think of the units of the area as being equal to the units of the y axis (F) multiplied by the units of the x axis (r). Just as you would calculate area of any other surface.

      So, now you are not just adding up all the forces; your integral is giving you y times x. In this case the amount of work done (Fr). (Think about it as each ,very thin column as having an area equal to F times delta r)

      The area under a F/x curve is the work done by the force in moving from one point to the other. and is given by the intergal of the function F=f(x) between two points (or posiitons)

      You can use the hookes law line F against extension to confirm this idea (energy stored in a spring = area under the curve = 1/2 Fx)
      (2 votes)
  • aqualine seed style avatar for user avithebest
    Intuitively, there should be some point where the particle would be out of the electric field, if we go high enough, but the math says otherwise? Would this be the case if the plate was not infinite, but just very large?
    (2 votes)
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    • spunky sam orange style avatar for user Willy McAllister
      It's challenging to use your intuition in the presence of a non-intuitive infinite plate. If it really is infinite, then the field never falls off, no matter how far away you get.

      If the plane is not infinite then yes, the field strength will fall off as you move away. If you get really far away, the plane gets really small and starts to look like a point charge for all practical purposes. Somewhere in between the plane can be considered large or small or whatever intermediate size you want. The math in this problem works for those planes, too. The only difference is you have to modify the limits on the integrals as you sweep the hoop radius from 0 to some non-infinite intermediate value.
      (4 votes)
  • blobby green style avatar for user j6453h4h335h
    At the very end, there is still the σ left. Isn't σ = Q / A which is σ = Q / (2*pi*r*dr)? Isn't this "constant" σ not really a constant because there is still a 'dr' in there? Why is it valid to not substitute the values of σ back into the Q / A form?
    (3 votes)
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  • leafers seedling style avatar for user reso123
    A charge of 17.7 *10^4 C is distributed uniformly over a large sheet of area 200 sq m. Calculate the electric field intensity at a distance of 20 cm from it in air.

    i calculated the answer using the formula
    E = σ/ 2ε ( ie .surface charge density/ permittivity of the medium), but the answer in my book is twice of my answer. please guide me..
    (2 votes)
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    • leafers seedling style avatar for user Suvadip Mandal
      Perhaps the question in the book was 'determine the electric field at a point 20 cm above a conducting sheet ,not a non-conducting sheet.For a non-conducting sheet the answer will be E = σ/ 2ε, but for a conducting sheet, the answer will be just the twice of that , i.e. E = σ/ ε. The video shows the electric field for a non-conducting sheet not for a conducting sheet.
      (3 votes)

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.