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.