Electricity and magnetism
Proof (Advanced): 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)
Proof (Advanced): Field from infinite plate (part 2)
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- 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
- 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
- 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
- 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
- See you soon.
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