Main content

## Electric potential energy, electric potential, and voltage

Current time:0:00Total duration:9:50

# Electric potential energy (part 2-- involves calculus)

## Video transcript

In the last video, we figured
out how much work or how much energy we have to put into a
particle to move it within a constant electric field. Let's see if we can do the same
thing now with a variable electric field, and actually,
then we could figure out the electric potential of one
position relative to another. So let's say we have
a point charge. It doesn't have to be a point
charge, but let's just say that there's some field being
generated by this charge that is a positive q1 coulombs. And if we wanted to draw field
lines-- and, of course, this is different than the example
with the infinite uniformly charged plate because this will
have a variable electric field, right? What is the electric field
of this charge? The electric field is going
to look like this. It's essentially the amount of
force it will exert on any particle, and it's always going
to be pointed outwards, because we assume that the test
particle in question is always a positive charge, so
a positive charge will be repelled from this
positive charge. That field is Coulomb's constant
times this charge q1 over the distance that we
are from this charge. So if I were to draw this
electric field close up, it's pretty strong. And then as we get further, it
gets weaker a little bit. It's always radially outward
from the charge. This a bit review for you. And this is a vector field that
I'm drawing, where I'm just randomly picking points and
showing the vector of the field at that point, that it's
pointed outward, and at any given radius away from the
circle, these vectors should be the same magnitude. I know mine aren't exactly,
but hopefully, you get the point, right? And the magnitude of these
vectors decrease with the square of the distance. So that's what the field is
going to look like if I were to draw a bunch of vector
field lines. And that makes sense, just
as a review, right? Because we know the force-- if
we had another test charge q2, we know that the force on that
test charge is just q2 times this, and that's just Coulomb's
Law, right? That the force on some other
particle q2 is equal to the electric field times q2, which
is equal to k q1 q2 over r squared, right? This is just Coulomb's Law, and
actually this comes from Coulomb's Law. So since we know that, let's
take some other positive particle out here, and
let's call that q2. I'll write it up here since I
already messed up down there. q2, and let's say this is a
positive charge, so it is going to be repelled
from this q1. And let's figure out how much
work does it take to push in this particle a certain
distance, right? Because the field is
pushing it outward. It takes work to
push it inward. So let's say we want
to push it in. Let's say it's at 10 meters. Let's say that this distance
right here-- let me draw a radial line-- let's say that
this distance right here is 10 meters, and I want to push this
particle in 5 meters, so it eventually gets right here. This is where I'm eventually
going to get it so then it's going to be 5 meters away. So how much work does it take
to move it 5 meters towards this charge? Well, the way you think about
it is the field keeps changing, right? But we can assume over a very,
very, very, very infinitely small distance, and let's call
that infinitely small distance dr, change in radius, and as
you can see, we're about to embark on some integral and
differential calculus. If you don't understand what any
of this is, you might want to review or learn the calculus
in the calculus playlist, but how much work does
it require to move this particle a very, very
small distance? Well, let's just assume over
this very, very, very small distance, that the electric
field is roughly constant, and so we can say that the very,
very small amount of work to move over that very, very small
distance is equal to Coulomb's constant q1 q2 over
r squared times dr. Now before we move on,
let's think about something for a second. Coulomb's Law tells us that this
is the outward force that this charge is exerting on
this particle or that the field is exerting on
this particle. The force that we have to apply
to move the particle from here to here has to
be an inward force. It has to be in the exact
opposite direction, so it has to be a negative. And why is that? Because we have to completely
offset the force of the field. Maybe if the particle was
already moving a little bit, then our force will keep it
from decelerating from the field, and if it wasn't already
moving, we would have to nudge it just an infinitely
small amount just to get it moving, and then our force would
completely offset the force of the field, and the
particle would neither accelerate nor decelerate. So this is the amount of work,
and I just want to explain that we want to put that
negative sign there because we going in the opposite direction
of the field. So how do we figure out the
total amount of work? We figured out the amount of
work to get it from here to here, and I even drew it much
bigger than it would be. These dr's, this is
an infinitely small change in radius. If we want to figure out the
total work, then we keep adding them up. We say, OK, what's the work to
go from here to here, then the work to go from there to there,
then the work to go from there to there, all the way
until we get to 5 meters away from this charge. And what we do when we take the
sum of these, we assume that it's an infinite sum of
infinitely small increments. And as you learned, that is
nothing but the integral, and so that is the total work is
equal to the integral. That's going to be a definite
integral because we're starting at this point. We're summing from-- our radius
is equal to 10 meters-- that's our starting point--
to radius equals 5 meters. That might be a little
unintuitive that we're starting at the higher value and
ending at the lower value, but that's what we're doing. We're pushing it inwards. And then we're taking the
integral of minus k q1 q2 over r squared dr. All of these are
constant terms up here, right? So we could take them out. So this is the same thing--
I don't want to run out of space-- as minus k q1 q2 times
the integral from 10 to 5 of 1 over r squared-- or to the
negative 2-- dr. And that equals minus k-- I'm running
out of space-- q1 q2. We take the antiderivative. We don't have to worry about
plus here because it's a definite integral. r to the negative 2, what's
the antiderivative? It's minus r to the
negative 1. Well, that minus r, the minus
on the minus r will just cancel with this. That becomes a plus r to the
negative 1, And you evaluate it at 5 and then subtract it
and evaluate it at 10. And then-- let me
just go up here. Actually, let me erase
some of this. Let me erase this up here. Valuable space to work on. So we said that the work
is equal to-- I'll just rewrite that. We had the minus out here, but
then we had a minus and we took the antiderivative and they
canceled out, so we have k q1 q2 times the antiderivative
evaluated at 5, so 1 over 5, right? r to the negative 1, so 1 over
r minus the antiderivative evaluated at 10 minus 1 over 10
is equal to-- well, 1 over 5, that's the same thing
as 2 over 10, right? So we have the work is equal
to k q1 q2 times 2 over 10 minus 1 over 10 is 1 over
10, so that equals k q1 q2 over 10. That's the work to take the
particle from here to here. And so similarly, we could say
that the potential energy of the particle at this-- the
potential difference of the particle at this point relative
to this point, that the potential difference here
is this much higher. It's going to be in joules
because that's the unit of energy or work or potential,
because potential is energy anyway. So the electric potential
difference between this point and this point: at this point,
it is this value higher. Let's do another example, and
actually, you might just find this interesting, just from
something to think about. The big difference between--
actually, let me just continue this into the next video because
I realize I'm already at 10 minutes. I'll see you soon.