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In the last video, we figured
out how much work or how much

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energy we have to put into a
particle to move it within a

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constant electric field.

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Let's see if we can do the same
thing now with a variable

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electric field, and actually,
then we could figure out the

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electric potential of one
position relative to another.

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So let's say we have
a point charge.

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It doesn't have to be a point
charge, but let's just say

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that there's some field being
generated by this charge that

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is a positive q1 coulombs.

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And if we wanted to draw field
lines-- and, of course, this

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is different than the example
with the infinite uniformly

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charged plate because this will
have a variable electric

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field, right?

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What is the electric field
of this charge?

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The electric field is going
to look like this.

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It's essentially the amount of
force it will exert on any

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particle, and it's always going
to be pointed outwards,

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because we assume that the test
particle in question is

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always a positive charge, so
a positive charge will be

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repelled from this
positive charge.

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That field is Coulomb's constant
times this charge q1

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over the distance that we
are from this charge.

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So if I were to draw this
electric field close up, it's

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pretty strong.

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And then as we get further, it
gets weaker a little bit.

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It's always radially outward
from the charge.

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This a bit review for you.

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And this is a vector field that
I'm drawing, where I'm

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just randomly picking points and
showing the vector of the

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field at that point, that it's
pointed outward, and at any

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given radius away from the
circle, these vectors should

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be the same magnitude.

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I know mine aren't exactly,
but hopefully, you get the

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point, right?

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And the magnitude of these
vectors decrease with the

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square of the distance.

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So that's what the field is
going to look like if I were

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to draw a bunch of vector
field lines.

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And that makes sense, just
as a review, right?

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Because we know the force-- if
we had another test charge q2,

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we know that the force on that
test charge is just q2 times

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this, and that's just Coulomb's
Law, right?

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That the force on some other
particle q2 is equal to the

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electric field times q2, which
is equal to k q1 q2 over r

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squared, right?

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This is just Coulomb's Law, and
actually this comes from

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Coulomb's Law.

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So since we know that, let's
take some other positive

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particle out here, and
let's call that q2.

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I'll write it up here since I
already messed up down there.

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q2, and let's say this is a
positive charge, so it is

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going to be repelled
from this q1.

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And let's figure out how much
work does it take to push in

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this particle a certain
distance, right?

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Because the field is
pushing it outward.

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It takes work to
push it inward.

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So let's say we want
to push it in.

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Let's say it's at 10 meters.

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Let's say that this distance
right here-- let me draw a

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radial line-- let's say that
this distance right here is 10

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meters, and I want to push this
particle in 5 meters, so

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it eventually gets right here.

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This is where I'm eventually
going to get it so then it's

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going to be 5 meters away.

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So how much work does it take
to move it 5 meters towards

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this charge?

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Well, the way you think about
it is the field keeps

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changing, right?

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But we can assume over a very,
very, very, very infinitely

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small distance, and let's call
that infinitely small distance

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dr, change in radius, and as
you can see, we're about to

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embark on some integral and
differential calculus.

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If you don't understand what any
of this is, you might want

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to review or learn the calculus
in the calculus

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playlist, but how much work does
it require to move this

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particle a very, very
small distance?

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Well, let's just assume over
this very, very, very small

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distance, that the electric
field is roughly constant, and

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so we can say that the very,
very small amount of work to

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move over that very, very small
distance is equal to

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Coulomb's constant q1 q2 over
r squared times dr.

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Now before we move on,
let's think about

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something for a second.

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Coulomb's Law tells us that this
is the outward force that

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this charge is exerting on
this particle or that the

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field is exerting on
this particle.

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The force that we have to apply
to move the particle

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from here to here has to
be an inward force.

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It has to be in the exact
opposite direction, so it has

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to be a negative.

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And why is that?

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Because we have to completely
offset the force of the field.

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Maybe if the particle was
already moving a little bit,

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then our force will keep it
from decelerating from the

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field, and if it wasn't already
moving, we would have

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to nudge it just an infinitely
small amount just to get it

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moving, and then our force would
completely offset the

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force of the field, and the
particle would neither

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accelerate nor decelerate.

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So this is the amount of work,
and I just want to explain

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that we want to put that
negative sign there because we

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going in the opposite direction
of the field.

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So how do we figure out the
total amount of work?

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We figured out the amount of
work to get it from here to

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here, and I even drew it much
bigger than it would be.

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These dr's, this is
an infinitely

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small change in radius.

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If we want to figure out the
total work, then we keep

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adding them up.

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We say, OK, what's the work to
go from here to here, then the

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work to go from there to there,
then the work to go

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from there to there, all the way
until we get to 5 meters

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away from this charge.

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And what we do when we take the
sum of these, we assume

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that it's an infinite sum of
infinitely small increments.

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And as you learned, that is
nothing but the integral, and

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so that is the total work is
equal to the integral.

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That's going to be a definite
integral because we're

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starting at this point.

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We're summing from-- our radius
is equal to 10 meters--

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that's our starting point--
to radius equals 5 meters.

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That might be a little
unintuitive that we're

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starting at the higher value and
ending at the lower value,

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but that's what we're doing.

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We're pushing it inwards.

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And then we're taking the
integral of minus k q1 q2 over

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r squared dr. All of these are
constant terms up here, right?

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So we could take them out.

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So this is the same thing--
I don't want to run out of

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space-- as minus k q1 q2 times
the integral from 10 to 5 of 1

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over r squared-- or to the
negative 2-- dr. And that

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equals minus k-- I'm running
out of space-- q1 q2.

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We take the antiderivative.

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We don't have to worry about
plus here because it's a

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definite integral.

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r to the negative 2, what's
the antiderivative?

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It's minus r to the
negative 1.

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Well, that minus r, the minus
on the minus r will just

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cancel with this.

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That becomes a plus r to the
negative 1, And you evaluate

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it at 5 and then subtract it
and evaluate it at 10.

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And then-- let me
just go up here.

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Actually, let me erase
some of this.

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Let me erase this up here.

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Valuable space to work on.

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So we said that the work
is equal to-- I'll

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just rewrite that.

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We had the minus out here, but
then we had a minus and we

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took the antiderivative and they
canceled out, so we have

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k q1 q2 times the antiderivative
evaluated at 5,

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so 1 over 5, right?

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r to the negative 1, so 1 over
r minus the antiderivative

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evaluated at 10 minus 1 over 10
is equal to-- well, 1 over

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5, that's the same thing
as 2 over 10, right?

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So we have the work is equal
to k q1 q2 times 2 over 10

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minus 1 over 10 is 1 over
10, so that equals

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k q1 q2 over 10.

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That's the work to take the
particle from here to here.

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And so similarly, we could say
that the potential energy of

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the particle at this-- the
potential difference of the

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particle at this point relative
to this point, that

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the potential difference here
is this much higher.

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It's going to be in joules
because that's the unit of

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energy or work or potential,
because

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potential is energy anyway.

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So the electric potential
difference between this point

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and this point: at this point,
it is this value higher.

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Let's do another example, and
actually, you might just find

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this interesting, just from
something to think about.

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The big difference between--
actually, let me just continue

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this into the next video because
I realize I'm already

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at 10 minutes.

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I'll see you soon.

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