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Let's review a little bit of
what we had learned many, many

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videos ago about gravitational
potential energy and then see

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if we can draw the analogy,
which is actually very strong,

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to electrical potential
energy.

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So what do we know about

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gravitational potential energy?

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If we said this was the surface
of the Earth-- we

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don't have to be on Earth, but
it makes visualization easy.

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We could be anywhere that has
gravity, and the potential

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energy would be due to the
gravitational field of that

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particular mass, but let's
say this is the

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surface of the Earth.

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We learned that if we have some
mass m up here and that

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the gravitational field at this
area-- or at least the

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gravitational acceleration-- is
g, or 9.8 meters per second

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squared, and it is h-- we could
say, I guess, meters,

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but we could use any units.

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Let's say it is h meters above
the ground, that the

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gravitational potential energy
of this object at that point

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is equal to the mass times the
acceleration of gravity times

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the height, or you could view
it as the force of gravity,

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the magnitude of the
force of gravity.

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You know, it's a vector, but
we can say the magnitude of

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the vector times height.

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And so what is potential
energy?

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Well, we know that if something
has potential energy

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and if nothing is stopping it
and we just let go, that

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energy, at least with
gravitational potential

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energy, the object will start
accelerating downwards, and a

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lot of that potential energy,
and eventually all of it, will

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be converted to kinetic
energy.

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So potential energy is energy
that is being stored by an

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object's situation or kind of
this notional energy that an

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object has by virtue
of where it is.

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So in order for something to
have this notional energy,

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some energy must have
been put into it.

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And as we learned with
gravitational potential

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energy, you could view
gravitational potential energy

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as the work necessary to move
an object to that position.

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Now, if we're talking about work
to move something into

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that position, or whatever, we
always have to think about,

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well, move it from where?

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Well, when we talk about
gravitational potential

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energy, we're talking about
moving it from the surface of

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the Earth, right?

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And so how much work is required
to move that same

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mass-- let's say it was here at
first-- to move it from a

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height of zero to
a height of h?

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Well, the whole time, the
Earth, or the force of

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gravity, is going to
be F sub g, right?

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So essentially, if I'm pulling
it or pushing it upwards, I'm

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going to have to have-- and
let's say at a constant

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velocity-- I'm going to have to
have an equal and opposite

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force to its weight
to pull it up.

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Otherwise, it would accelerate
downwards.

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I'd have to do a little bit more
just to get it moving, to

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accelerate it however much, but
then once I get it just

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accelerating, essentially I
would have to apply an upward

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force, which is equivalent to
the downward force of gravity,

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and I would do it for a
distance of h, right?

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What is work?

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Work is just force
times distance.

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Force times distance, and it
has to be force in the

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direction of the distance.

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So what's the work necessary
to get this mass up here?

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Well, the work is equal to the
force of gravity times height,

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so it's equal to the

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gravitational potential energy.

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Now this is an interesting
thing.

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Notice we picked the reference
point as the surface of the

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Earth, but we could
have picked any

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arbitrary reference point.

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We could have said, well, from
10 meters below the surface of

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the Earth, which could have been
down here, or we could

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have actually said, you know,
from a platform that's 5

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meters above the Earth.

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So it actually turns out, when
you think of it that way, that

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potential energy of any form,
but especially gravitational

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potential energy-- and we'll
see electrical potential

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energy-- it's always in
reference to some other point,

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so it's really a change in
potential energy that matters.

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And I know when we studied
potential energy, it seemed

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like there was kind of an
absolute potential energy, but

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that's because we always assume
that the potential

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energy of something is zero the
surface of the Earth and

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that we want to know the
potential energy relative to

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the surface of the Earth, so it
would be kind of, you know,

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how much work does it take to
take something from the

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surface of the Earth
to that height?

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But really, we should be saying,
well, the potential

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energy of gravity-- like this
statement shouldn't be, you

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know, this is just the absolute

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potential energy of gravity.

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We should say this is the
potential energy of gravity

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relative to the surface of the
Earth is equal to the work

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necessary to move something, to
move that same mass, from

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the surface of the Earth to
its current position.

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We could have defined some other
term that is not really

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used, but we could have said
potential energy of gravity

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relative to minus 5 meters
below the surface of the

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Earth, and that would be the
work necessary to move

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something from minus 5 meters
to its current height.

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And, of course, that
might matter.

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What if we cut up a hole and
we want to see what is the

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kinetic energy here?

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Well, then that potential
energy would matter.

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Anyway, so I just wanted to do
this review of potential

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energy because now it'll make
the jump to electrical

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potential energy all that
easier, because you'll

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actually see it's pretty
much the same thing.

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It's just the source of the
field and the source of the

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potential is something
different.

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So electrical potential energy,
just actually we know

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that gravitational fields are
not constant, we can assume

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they're constant maybe near the
surface of the Earth and

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all that, but we also know that
electrical fields aren't

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constant, and actually they
have very similar formulas.

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But just for the simplicity of
explaining it, let's assume a

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constant electric field.

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And if you don't believe me that
one can be constructed,

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you should watch my videos that
involve a reasonable bit

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of calculus that show that a
uniform electric field can be

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generated by an infinite
uniformly charged plane.

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Let's say this is the side view
of an infinite uniformly

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charged plane and let's
say that this

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is positively charged.

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Of course, you can never get
a proper side view of an

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infinite plane, because you
can never kind of cut it,

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because it's infinite in every
direction, but let's say that

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this one is and this
is the side view.

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So first of all, let's think
about its electric field.

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It's electric field is going to
point upward, and how do we

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know it points upward?

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Because the electric field is
essentially what is-- and this

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is just a convention.

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What would a positive charge
do in the field?

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Well, if this plate is positive,
a positive charge,

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we're going to want to
get away from it.

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So we know the electric field
points upward and we know that

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it's constant, that if these
were field vectors, that

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they're going to be the same
size, no matter how far away

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we get from the source
of the field.

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And I'm just going to pick
a number for the

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strength of the field.

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We actually proved in those
fancy videos that I made on

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the uniform electric field of an
infinite, uniformly charged

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plane that we actually proved
how you could calculate it.

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But let's just say that this
electric field is equal to 5

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newtons per coulomb.

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That's actually quite strong,
but it makes the math easy.

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So my question to you is how
much work does it take to take

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a positive point charge-- let
me pick a different color.

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Let's say this is the
starting position.

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It's a positive 2 coulombs.

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Once again, that's a massive
point charge, but

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we want easy numbers.

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How much work does it take it to
move that 2-coulomb charge

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3 meters within this field?

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How much work?

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So we're going to start here
and we're going to move it

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down towards the plate 3
meters, and it's ending

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position is going to be
right here, right?

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That's when it's done.

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How much work does that take?

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Well, what is the force of
the field right here?

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What is the force exerted on
this 2-coulomb charge?

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Well, electric field is just
force per charge, right?

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So if you want to know the force
of the field at that

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point-- let me draw that
in a different color.

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The force of the field acting on
it, so let's say the field

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force, or the force of the
field, actually, is going to

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be equal to 5 newtons per
coulomb times 2 coulombs,

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which is equal to 10 newtons.

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We know it's going to be upward,
because this is a

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positive charge, and this is a
positively charged infinite

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plate, so we know this is an
upward force of 10 newtons.

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So in order to get this charge,
to pull it down or to

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push it down here, we
essentially have to exert a

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force of 10 newtons
downwards, right?

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Exert a force of 10 newtons in
the direction of the movement.

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And, of course, just like we did
with gravity, we have to

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maybe do a little bit more than
that just to accelerate

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it a little bit just so you have
some net downward force,

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but once you do, you just have
to completely balance the

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upward force.

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So just for our purposes, you
have a 10-newton force

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downward and you apply that
force for a distance of 3

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meters, the work that you put to
take this 2-coulomb charge

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from here to here, the work
is going to be equal to 10

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newtons-- that's the force--
times 3 meters.

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So the work is going to equal
30 newton-meters, which is

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equal to 30 joules.

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A joule is just a
newton-meter.

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And so we can now say since it
took us 30 joules of energy to

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move this charge from here to
here, that within this uniform

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electric field, the potential
energy of the charge here is

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relative to the charge here.

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You always have to pick a point
relative to where the

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potential is, so the electrical
potential energy

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here relative to here and this
is electrical potential

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energy, and you could say P2
relative to P1-- I'm using my

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made-up notation, but that gives
you a sense of what it

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is-- is equal to 30 joules.

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And how could that help us?

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Well, if we also knew the mass--
let's say that this

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charge had some mass.

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We would know that if we let go
of this object, by the time

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it got here, that 30 joules
would be-- essentially

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assuming that none of it got
transmitted to heat or

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00:11:08,600 --> 00:11:11,180
resistance or whatever-- we know
that all of it would be

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00:11:11,180 --> 00:11:13,200
kinetic energy at this point.

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So actually, we could
work it out.

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Let's say that this does have
a mass of 1 kilogram and we

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were to just let go
of it, right?

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We used some force to bring it
down here, and then we let go.

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So we know that the electric
field is going to accelerate

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00:11:34,755 --> 00:11:36,320
it upwards, right?

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It's going to exert an upward
force of 5 newtons per

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coulomb, and the thing's going
to keep [COUGHS]--

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excuse me-- keep accelerating
until it gets

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to this point, right?

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00:11:47,260 --> 00:11:49,290
What's its velocity going
to be at that point?

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Well, all of this electrical
potential energy is going to

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be converted to kinetic
energy.

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So essentially, we have 30
joules is going to be equal to

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1/2 mv squared, right?

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00:12:01,040 --> 00:12:05,450
We know the mass, I said, is 1,
so we get 60 is equal to v

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squared, so the velocity is the
square root of 60, so it's

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7 point something, something,
something meters per second.

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So if I just pull that charge
down, and it has a mass of 1

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kilogram, and I let go, it's
just going to accelerate and

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be going pretty fast once
it gets to this point.

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Anyway, I'm 12 minutes into this
video, so I will continue

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in the next, but hopefully, that
gives you a sense of what

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electrical potential energy
is, and really, it's no

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different than gravitational
potential energy.

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It's just the source of the
field is different.

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See you soon.

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