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.