### Video Transcript

The dielectric constant of Teflon
is 2.1. A Teflon-filled, parallel-plate
capacitor has a plate area of 50.0 centimeters squared. The spacing between the plates is
0.50 millimeters and the capacitor is connected to a 200-volt battery. Find the free charge on the
capacitor plates. Find the electric field in the
dielectric. Find the induced charge on the
dielectric’s surfaces.

Let’s start out by focusing on this
first question: find the free charge on the capacitor plates. We can call this free charge
capital 𝑄. And we know that this charge will
accumulate on the parallel plates of our capacitor. We know that in general when a
capacitor is charged, positive charge accumulates on one side and corresponding
negative charge accumulates on the other. This charge sets up an electric
field between the plates.

In terms of our variable capital
𝑄, we could say that positive 𝑄 is on the left plate and negative 𝑄 is on the
right plate, but the magnitude of charge on each plate is 𝑄. So far so good! Now we’re told some other
information about this capacitor. We’re told that the area of each
one of the capacitor plates is 50.0 square centimeters. And we’re also told that the
distance separating the plates one from another, which we can call lower case 𝑑, is
equal to 0.50 millimeters.

And moreover, these two capacitor
plates are not separated by an air gap. Rather, they have a dielectric
material in between them. And in our case, it’s Teflon. A dielectric material, like the one
we have here, is typically characterized by its dielectric constant, often referred
to using the Greek letter 𝜅. This looks like a capital K but
actually it’s a Greek letter 𝜅. And for Teflon, we’re told that
this value is 2.1. And finally, recall that we’re told
that a potential difference of 200 volts is set up across these parallel plates.
we’ll call that potential difference of 200 volts 𝑣.

And knowing all this, we want to
solve for 𝑄, which is the magnitude of charge that appears on either side of the
capacitor plates. To do this, we can recall a
mathematical relationship that connects some of these quantities we’ve been
given. To start off, we can say that the
capacitance of the capacitor is equal to the charge magnitude on either plate 𝑄
divided by the potential difference across the plates. But there’s another relationship
for capacitance in terms of the geometry of the capacitor plates in the dielectric
between them.

This second relationship says that
capacitance is also equal to 𝜅, the dielectric constant, multiplied by 𝜀 naught,
the permittivity of free space, multiplied by the area of each one of the capacitor
plates divided by the distance separating them. Now this is a beautiful thing
because you see we have a relationship for the value we want to solve for, 𝑄, in
terms of other variables we’ve been given. In fact, the only value in this
expression we don’t yet have is a value for 𝜀 naught, the permittivity of free
space.

That value is a constant which we
can approximate as 8.85 times 10 to the negative 12th farads per meter. Knowing that, let’s now multiply
both sides of this equation by the potential 𝑣 so that 𝑄 is isolated on one
side. We now have a symbolic form for the
charge we wanted to solve for; it’s equal to 𝜅 times 𝜀 naught times the plate area
times the potential difference across them divided by the distance separating
them. And we know all of these values, so
we’re ready to plug in and solve for 𝑄.

One last thing before we do though,
look at the units of 𝐴, the area, and 𝑑, the distance. They’re not in meters, and we’d
like them to be. That’s because meters is the
standard unit for length. And converting these values to be
expressed in meters or meters squared will make them consistent with the other
values in our equation. So let’s convert these values
starting with the area 𝐴. We know that 100 centimeters is
equal to one meter. So therefore, 50.0 centimeters
squared is equal to 0.005 meters squared.

And then, when it comes to distance
in millimeters, we know that 1000 millimeters makes up one meter. So we’ll just write 𝑑 as 0.50
times 10 to the negative third meters. Wonderful! Now that 𝐴 and 𝑑 are in the units
we want, we’re ready to plug them in and solve for 𝑄. In order to do that, let’s clear a
little space by erasing our diagram sketch. With all these values plugged in,
where 𝜅 recall is 2.1 and the potential difference 𝑣 is 200 volts, when we
calculate 𝑄, we find a result of 37 nanocoulombs or 37 billionths of a coulomb. That’s the charge magnitude then
residing on each capacitor plate.

Now that we know the charge 𝑄,
let’s recall the second question in our statement. Question two states find the
electric field in the dielectric. To get started on this question,
let’s redraw our sketch of our capacitor with the dielectric between the plates. If we drew just the capacitor
plates separated by an air gap with no dielectric, we know that the charges on
either side would do their part to create an electric field that’s in between these
two plates. It goes from the positive plate to
the negative plate and is uniform between these parallel plates.

Something interesting happens
though when we insert our dielectric material, in our case Teflon, in the gap. In that case, the electric field in
this space weakens. And it weakens by a factor of the
dielectric constant of the dielectric 𝜅. We can recall an equation that
shows us how this works. The electric field strength between
parallel-plate capacitors is equal to the charge density on those plates 𝜎 divided
by 𝜅 times the permittivity of free space. And we recall that 𝜎 this charge
density is equal to the magnitude of charge on either plate divided by the plate
area 𝐴.

So we could also write this
equation for electric field this way. We could say that the electric
field between the parallel plates of a capacitor when there is a dielectric between
them is equal to the charge magnitude on either plate divided by 𝜅 the dielectric
constant multiplied by 𝜀 naught times 𝐴. This is the equation we can use to
answer the question of what is the electric field in the dielectric. And looking at the values on the
right-hand side of this expression, we see we know all of them already.

We’ve solved for 𝑄 in the previous
part of the problem, 𝜅 is known as 2.1, 𝜀 naught is a constant with value we know,
and 𝐴 is a given value as well. All that remains then is to plug in
and solve for 𝐸. When we do, writing our value for
the charge 𝑄 as 37 times 10 to the negative ninth coulombs, we find a result of
0.40 million volts per meter or 0.40 megavolts per meter. That’s the electric field strength
in the dielectric itself. Now that we know the electric field
in the dielectric, let’s move on to the last part of this question.

In this last part, we’re asked to
solve for the induced charge on the dielectric’s surfaces. If we look back at our sketch of
the dielectric in the gap between the capacitor plates, what we come to realize is
that the dielectric itself has charge which sets up on either one of its sides to
oppose the charge on the capacitor plate side that the dielectric faces. That’s a complicated way of saying
that negative charge will build up within the dielectric on the side facing the
positive charge on the capacitor and vice versa on the opposite side.

This charge build-up incidentally
is what reduces the magnitude of the electric field that’s in between the capacitor
plates. It weakens that field as we saw by
a factor of 𝜅, the dielectric constant. In this final step then, we want to
solve for just what is the induced charge on the dielectric’s surfaces. To figure this out, let’s introduce
a little bit of new notation. We saw that the magnitude of charge
on either side of the capacitor plate itself is capital 𝑄 and that that charge is
spread over an area. In other words, there’s a surface
charge density 𝜎.

We’ll call the capacitor’s surface
charge density magnitude 𝜎 sub c. And at the same time that we
realize there is a surface charge density on the capacitor, we know there is also
one on the dielectric. We can call that surface charge
density 𝜎 sub d, for dielectric. If we focus on these two variables,
𝜎 sub c and 𝜎 sub d, it turns out there is a mathematical relationship that
connects them. That equation says that 𝜎 sub d,
the surface charge density on the dielectric, is equal to 𝜎 sub c, that on the
capacitor, multiplied by this factor of one minus one over 𝜅, the dielectric
constant.

This is good progress, but you may
notice, “Wait a second! We didn’t want to solve for a
surface charge density. We wanted to solve for an induced
charge!” That’s true, but here’s the
thing. Remember that surface charge
density is equal to charge divided by area. This means we could represent 𝜎
sub d, the surface charge density on the dielectric, as the induced charge on the
dielectric 𝑄 sub d divided by area 𝐴. And likewise, we could represent
the surface charge density on the capacitor as 𝑄 — we won’t call it 𝑄 sub c, but
just 𝑄 since that we called it earlier — divided by 𝐴.

Now we’re really doing well, so
let’s start working with this expression. Perhaps the first thing we notice
about it is that this factor of 𝐴 appears in the denominator on both sides. That’s handy. The area of the dielectric and the
area of the plate is the same. That means it cancels out. So now to solve for 𝑄 sub d, which
remember is the induced charge we want to find, we only need to know 𝑄, which we
solved for earlier, and the dielectric constant 𝜅, which is given. When we plug in for these values,
we find that 𝑄 sub d, the induced charge, is 37 times 10 to the negative ninth
coulombs times one minus one over 2.1.

To two significant figures, that’s
19 nanocoulombs. This tells us then that the charge
that builds up on the dielectric’s surfaces is just about half as much as the charge
that builds up on the capacitor plates.