We learned several videos ago that if I had an infinite
uniformly charged plane-- let me draw one right here, and I
won't draw it infinite and I'll tell you why in a
second-- that if we had an infinite uniformly charged
plane, and let's say this one's positive, that the
electric field generated by it is constant.
Those are the field lines.
They should all be the same size.
And the strength of the field, or the magnitude of the field,
is equal to 2 times Coulomb's constant times pi times the
charge density of the plate.
So if this is infinite-- so what was charge density?
We defined it when we proved that this truly is a uniform
electric field, but what is charge density?
Charge density is just the total amount of charge divided
by the area, or charge per area.
Well, if we have an infinite plane, the area's going to be
infinite, and so if this is a constant number, this is also
going to be infinite, so it's kind of hard to work with.
But what we also know is that when we have a non-infinite
plane that has some finite area, that near the center of
it and fairly close to it, it approximates an infinite
uniformly charged plane.
So with that said, let's see if we can figure out some of
the properties of the voltage and how the voltage relates to
the charge.
If we were to have two parallel-- let me draw it
before I say it, because I think saying it'll just
confuse it.
So let's say I have two plates, that plate-- and then
I'll do this one in a different color-- and I have
that plate, and let's say they're the same size and they
both have area A.
Let's say that I place plus Q worth of charges here.
So this is plus Q so this is positively charged, right?
I could draw a bunch of charges here.
Let's say this is minus Q, so this is negatively charged.
So what's the electric field going to look like
between these two?
Well, it's essentially going to be the combination of the
electric field generated by this plate on top of the
electric field generated by this plate.
And they're both going to be constant close to the center,
assuming that they're reasonably-- and let's say
that they're d apart.
Assuming that d isn't too big, near the center we're going to
have a constant electric field.
For example, this green one is going to be generating-- its
field lines are going to look something like this.
Near the center, it's constant.
These are meant to look constant.
Near the center, it'll look like that, and it'll start to
bulge out when you get to the edges.
Once again, near the center, it's constant.
That one should've been at an angle.
They start to bulge out, and it'll look
something like that.
And similarly, this purple plate will generate a constant
electric field, and since it's negatively charged, the field
lines will be going towards it, not away from it, so its
field lines are going to look something like this.
Near the center, they'll be constant, and its field lines
are going to look something like that.
As you can see, they're going to be of the same magnitude
and in the same direction, and they also
will bulge out there.
So the big picture is that you just kind of have twice the
electric field as you would have if you just had one of
these plates.
So let's say we're operating near the center of these,
where we have roughly a constant electric field and
see if we can figure out the relationship between the
voltage across these two plates and the area, and maybe
the distance between the two plates.
So we know that the electric field generated by any one of
these charge plates-- I'll do it in the blue of this color.
So for the bottom plate right here, what is the electric
field generated?
It's 2K pi times sigma.
Sigma is just the total charge divided by the area.
So Q/A, right?
And we know that the total electric field generated by
this one is going to be essentially the same thing.
I mean, we could say it's a minus, a negative, because
it's going towards it, but it's
essentially the same thing.
Because we see that they overlap just drawing the field
line, so the electric field from that one, and we know
that they go in the same direction.
If this was somehow-- well, this is negative, so the field
lines go towards it.
So plus 2K pi and this is Q/A.
We could have said minus and then had a minus Q/A, but we
know that they go into same direction, so we know that
they're going to be additive.
And so we know that the total electric field is
going to be 4K pi Q/A.
So now we know the exact strength of
the electric field.
Let's see if we can figure out the voltage difference between
this point and this point.
What was voltage difference, just as a review?
Well, voltage difference is the electrical potential
energy per charge if the charge was here versus here.
So how much more potential energy per coulomb is there
for a charge to be here relative to here?
So another way to view it is a charge here, a positive charge
here, because by default we're always assuming a positive
charge when we talk about positive numbers and the
direction of the field lines or what the positive
charge would do.
So by default, a positive charge here really wants to go
up to this negative plate, although we later learned that
most of the movement in electronics and electricity,
it's actually the negative charge that's moving.
It's the electrons moving.
But let's say we did have a positive ion
or a positive charge.
The voltage is a measure of if any charge is here, how badly
does it want to move to this point if it has a way to move?
If we have air here or if we have a vacuum here, it might
be difficult or impossible for it to move up here.
But maybe if we were to connect a wire where the
charges could freely conduct, then it will move.
And the voltage is just kind of how badly
does it want to move?
You could almost view it as electrical pressure.
And maybe I'll do a whole video on trying to get an
intuitive understanding of voltage, because that really
is probably the most important thing to get an intuitive
understanding of, if you ever want to study electrical
engineering or whatever.
But anyway, back to the problem.
We know that the combined electric field is this, right?
It goes upwards in that direction.
So what is the electric potential at this point
relative to this point or the potential difference
from here to here?
Well, that's the amount of energy per charge it would
take to move a positive charge from here to there, right?
Remember, electric potential energy is the amount of work
necessary to move a charge from there to there, and then
the voltage is how much to do it per charge.
Let me write that down.
So the work necessary to move a charge from there to there--
let's say a 1-coulomb charge, it will be 1 coulomb times the
electric field, because we're always going to have to be
going against the electric field.
So we have to apply an equal and opposite force.
So the force that is going to be the electric field-- so
far, this just generates this force. coulomb times electric
field, charge times electric field, tells us the force on
the charge, right?
That's force, and then we have to
multiply that times distance.
Force times distance.
So we see the work necessary is going to be the electric
field times d joules-- the J is joules-- and so what is the
voltage difference or the electric potential difference
between this point and this point?
Let's call that point a.
Let's call that point b.
So Va minus Vb, which is the voltage difference, that's
essentially the electric potential energy difference
divided by the charge.
Or, per charge.
Well, here, the charge was just 1, so we can just divide
by 1, and we see that it is equal to the electric field
times the distance.
And the units are going to be joules because we divided both
sides by charge joules per coulomb, or volts, right?
That's just the units.
So what does that equal?
So the voltage difference-- so we can say change in voltage.
The voltage difference is equal to the electric field,
which we know is constant 4K pi Q over A times distance.
Or we could rewrite this.
Let's see if we could write Q as a function of V.
So if we just do a little bit of algebraic manipulation, we
can get Q is equal to what?
We would essentially divide both sides by 4 pi Kd and
multiply both sides by A, so we would get A
over 4K pi d voltage.
And why is this interesting?
Why did I go through all of this work to get this
relationship?
Well, what it shows you, if you look at this, if we assume
that the area of the plates aren't changing-- that's a
constant; this is definitely a constant-- and if we assume
that the distance between the plates don't change, what we
see is that there's a proportional difference
between the voltage and the amount of the combined charge
in the plates.
And that's interesting because, before doing this,
maybe voltage is somehow proportional to the square of
the charges or to the square root, but now we know that
it's directly proportionally.
And actually, this term right here has a name, and it is
called capacitance.
And so another way of rewriting this, if we divide
both sides by voltage, we get Q/V is equal to 1 over 4K pi
area over distance.
And so what it essentially says is that the amount of
energy that-- well, actually, I don't want to
go into that yet.
But for a given configuration, and the configuration is
defined by the area of the plates and the distance-- for
a given configuration, if I know the amount of charge that
I put onto the plates, if I did a minus Q here and a plus
Q here, I know the voltage across the
plates or vice versa.
If I know the voltage across the plates and I know its
configuration, I know how much charge there is, and this is
called capacitance, and the unit for capacitance
is called the Farad.
And if you become an electrical engineer or even
take a couple of electrical engineering courses, you'll
become very familiar with this.
And one other thing to point out; this term right here,
just so you know a little bit of terminology.
This term right here.
This 1 over 4K pi, this is often called epsilon nought,
or just epsilon, and that's called the permittivity of
free space or permittivity of the vacuum.
And maybe in a future lecture or a future video, I'll talk
more about why it's called that.
But anyway, I'm already well over the time limit.
So I just wanted to give you a sense of, one, that you can
calculate the voltage across what we call, in
this case, a capacitor.
It has capacitance.
That voltage, you can kind of view it as
the electric pressure.
How bad does the charge here want to move here?
And if you put a wire here, you'll learn in a second-- not
in a second, in several videos-- that
that charge will flow.
Or actually the negative charges will flow this way and
generate current.
And we'll do that when we start learning a little bit
more about electricity.
For any given configuration, it has a corresponding
capacitance, and then given that capacitance, if I put
some amount of charge, I can figure out the voltage, or if
I know there's some voltage, I can figure out the charge.
Anyway, I will see you in the next video.