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## Electric field

# 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.