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In this video, we're going to
study the electric field

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created by an infinite uniformly
charged plate.

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And why are we going
to do that?

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Well, one, because we'll learn
that the electric field is

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constant, which is neat by
itself, and then that's kind

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of an important thing to realize
later when we talk

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about parallel charged plates
and capacitors, because our

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physics book tells them that
the field is constant, but

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they never really prove it.

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So we will prove it here, and
the basis of all of that is to

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figure out what the electric
charge of an infinitely

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charged plate is.

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So let's take a side view of the
infinitely charged plate

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and get some intuition.

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Let's say that's the side view
of the plate-- and let's say

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that this plate has a charge
density of sigma.

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And what's charge density?

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It just says, well, that's
coulombs per area.

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Charge density is equal
to charge per area.

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That's all sigma is.

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So we're saying this has a
uniform charge density.

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So before we break into what may
be hard-core mathematics,

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and if you're watching this in
the calculus playlist, you

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might want to review some of
the electrostatics from the

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physics playlist, and
that'll probably be

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relatively easy for you.

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If you're watching this from the
physics playlist and you

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haven't done the calculus
playlist, you should not watch

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this video because you will
find it overwhelming.

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But anyway, let's proceed.

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So let's say that once again
this is my infinite so it goes

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off in every direction and it
even comes out of the video,

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where this is a side view.

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Let's say I have a point
charge up here Q.

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So let's think a little bit
about if I have a point--

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let's say I have an area
here on my plate.

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Let's think a little bit about
what the net effect of it is

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going to be on this
point charge.

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Well, first of all, let's say
that this point charge is at a

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height h above the field.

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Let me draw that.

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This is a height h, and let's
say this is the point directly

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below the point charge, and
let's say that this distance

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right here is r.

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So first of all, what is the
distance between this part of

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our plate and our
point charge?

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What is this distance that
I'll draw in magenta?

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What is this distance?

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Well, the Pythagorean theorem.

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This is a right triangle, so
it's the square root of this

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side squared plus this
side squared.

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So this is going to be
the square root of h

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squared plus r squared.

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So that's the distance between
this area and our test charge.

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Now, let's get a little
bit of intuition.

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So if this is a positive test
charge and if this plate is

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positively charged, the force
from just this area on the

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charge is going to be radially
outward from this area, so

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it's going to be-- let me do it
in another color because I

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don't want to-- it's going to
go in that direction, right?

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But since this is an infinite
plate in every direction,

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there's going to be another
point on this plate that's

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essentially on the other side of
this point over here where

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its net force, its net
electrostatic force on the

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point charge, is going
to be like that.

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And as you can see, since we
have a uniform charge density

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and the plate is symmetric in
every direction, the x or the

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horizontal components of the
force are going to cancel out.

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And so that's true for really
any point along this plate.

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Because if you pick any point
along it, and we're looking at

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a side view, but if we took a
top view, if that's the top

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view and, of course, the plate
goes off in every direction

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forever and that's kind of where
our point charge is, if

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we said, oh, well, you know,
there's this point on the

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plate and it's going to have
some y-component that's on

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this top view coming out of the
video, but it'll have some

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x-component, this point's
x-component effect

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will cancel it out.

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You can always find another
point on the plate that's

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symmetrically opposite whose
x-component of electrostatic

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force will cancel out
with the first one.

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So given that, that's just a
long-winded way of saying that

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the net force on this point
charge will only be upwards.

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I think it should make sense
to you that all of the

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x-components or the horizontal
components of the

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electrostatic force all cancel
out, because they're infinite

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points to either side
of this test charge.

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So with that out of the way,
what do we need to focus on?

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Well, we just need to focus
on the y-components of the

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electrostatic force.

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So what's the y-component?

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So let's say that this point
right here-- and I'll keep

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switching colors.

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Let's say that this point-- and
once again, this is a side

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view-- is exerting-- its field
at that point is e1, and it's

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going to be going in
that direction.

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What is its y-component?

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What is the component
in that direction?

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And, of course, it's
pushing outwards if

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they're both positive.

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So what is the y-component?

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What is that?

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Well, if we knew theta, if
we knew this angle, the

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y-component, or the upwards
component is going to be the

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electric field times
cosine of theta.

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Cosine is adjacent over
hypotenuse, so hypotenuse

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times cosine of theta is
equal to the adjacent.

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So if we wanted the vertical
or the y-component of the

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electric field, we would just
multiply the magnitude of the

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electric field times the
cosine of theta.

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So how do we figure out theta?

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Well, that theta is also the
same as this theta from our

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basic trigonometry.

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And so what's cosine of theta?

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Cosine is adjacent
over hypotenuse

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from SOHCAHTOA, right?

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Cosine of theta is equal to
adjacent over hypotenuse.

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So when we're looking at this
angle, which is the same as

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that one, what's adjacent
over hypotenuse?

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This is adjacent, that
is the hypotenuse.

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So what do we get?

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We get that the y-component of
the electric field due to just

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this little chunk of our plate,
the electric field in

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the y-component, let's just call
that sub 1 because this

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is just a little small
part of the plate.

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It is equal to the electric
field generally, the magnitude

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of the electric field from this
point, times cosine of

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theta, which equals the electric
field times the

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adjacent-- times height-- over
the hypotenuse-- over the

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square root of h squared
plus r squared.

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Fair enough.

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So now let's see if we can
figure out what the magnitude

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of the electric field is, and
then we can put it back into

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this and we'll figure out the
y-component from this point.

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And actually, we're not just
going to figure out the

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electric field just from that
point, we're going to figure

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out the electric field from a
ring that's surrounding this.

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So let me give you a little bit
of perspective or draw it

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with a little bit
of perspective.

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So this is my infinite
plate again.

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I'll draw it in yellow
again since I

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originally drew it in yellow.

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This is my infinite plate.

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It goes in every direction.

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And then I have my charge
floating above this plate

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someplace at height of h.

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And this point here, this could
have been right here

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maybe, but what I'm going to do
is I'm going to draw a ring

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that's of an equal radius around
this point right here.

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So this is r.

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Let's draw a ring, because all
of these points are going to

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be the same distance from
our test charge, right?

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They all are exactly like this
one point that I drew here.

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You could almost view this as
a cross-section of this ring

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that I'm drawing.

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So let's figure out what the
y-component of the electric

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force from this ring is
on our point charge.

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So to do that, we just have to
figure out the area of this

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ring, multiply it times our
charge density, and we'll have

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the total charge from that
ring, and then we can use

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Coulomb's Law to figure out its
force or the field at that

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point, and then we could use
this formula, which we just

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figured out, to figure
out the y-component.

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I know it's involved, but it'll
all be worth it, because

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you'll know that we have a
constant electric field.

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So let's do that.

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So first of all, Coulomb's Law
tells us-- well, first of all,

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let's figure out the charge
from this ring.

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So Q of the ring,
it equals what?

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It equals the circumference
of the ring times the

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width of the ring.

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So let's say the circumference
is 2 pi r, and let's say it's

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a really skinny ring.

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It's really skinny.

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It's dr. Infinitesimally
skinny.

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So it's width is dr. So that's
the area of the ring, and so

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what's its charge going to be?

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It's area times the charge
density, so times sigma.

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That is the charge
of the ring.

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And then what is the electric
field generated by the ring at

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this point here where
our test charge is?

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Well, Coulomb's Law tells us
that the force generated by

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the ring is going to be equal
to Coulomb's constant times

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the charge of the ring times our
test charge divided by the

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distance squared, right?

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Well, what's the distance
between really any point on

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the ring and our test charge?

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Well, this could be one of the
points on the ring and this

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could be another one, right?

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And this is like a
cross-section.

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So the distance at any point,
this distance right here, is

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once again by the Pythagorean
theorem because

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this is also r.

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This distance is the
square root of h

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squared plus r squared.

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It's the same thing as that.

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So it's the distance squared
and that's equal to k times

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the charge in the ring times
our test charge divided by

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distance squared.

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Well, distance is the square
root of h squared plus r

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squared, so if we square
that, it just becomes h

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squared plus r squared.

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And if we want to know the
electric field created by that

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ring, the electric field is
just the force per test

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charge, so if we divide both
sides by Q, we learned that

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the electric field of the ring
is equal to Coulomb's constant

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times the charge in the
ring divided by h

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squared plus r squared.

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And now what is the
y-component of the

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charge in the ring?

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Well, it's going to
be this, right?

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What we just figured out is the
magnitude of essentially

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this vector, right?

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But we want its y-component,
because all of the

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x-components just cancel out,
so it's going to be times

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cosine of theta, and we figured
out that cosine of

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theta is essentially this, so
we multiply it times that.

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So the field from the ring in
the y-direction is going to be

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equal to its magnitude times
cosine of theta, which we

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figured out was h over
the square root of h

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squared plus r squared.

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We could simplify this
a little bit.

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The denominator becomes what?

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h squared plus r squared
to the 3/2 power.

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And what's the numerator?

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Let's see, we have kh and then
the charge in the ring, which

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we solved up here.

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So that's 2 pi sigma r-- make
sure I didn't lose anything--

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dr. So we have just calculated
the y-component, the vertical

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component, of the electric
field at h

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units above the plate.

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And not from the entire plate,
just the electric field

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generated by a ring of radius r
from the base of where we're

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taking this height.

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And so I've already gone 12
minutes into this video, and

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00:12:53,160 --> 00:12:55,130
just to give you a break and
myself a break, I will

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00:12:55,130 --> 00:12:56,230
continue in the next.

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00:12:56,230 --> 00:12:58,110
But you can imagine what
we're going to do now.

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00:12:58,110 --> 00:13:03,680
We just figured out the electric
field created by just

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00:13:03,680 --> 00:13:05,320
this ring, right?

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00:13:05,320 --> 00:13:08,425
So now we can integrate across
the entire plane.

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00:13:08,425 --> 00:13:11,010


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00:13:11,010 --> 00:13:14,040
We can solve all the rings of
radius infinity all the way

250
00:13:14,040 --> 00:13:17,170
down to zero, and that'll give
us the sum of all of the

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00:13:17,170 --> 00:13:19,880
electric fields and essentially
the net electric

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00:13:19,880 --> 00:13:24,460
field h units above the
surface of the plate.

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See you in the next video.

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