### Video Transcript

Two parallel conducting plates are
separated by 10.0 centimetres, and one of them is taking to be at zero volts. What is the electric field strength
between them if the potential 8.00 centimetres from the zero-volt plate and 2.00
centimetres from the other is 450 volts? What is the voltage between the
plates?

We have two questions about these
two parallel conducting plates and we’ll start with the first one: what is the
electric field strength at this particular location between the plates? Drawing these two plates in, we’re
told that they are separated by a distance of 10.0 centimetres. And we’re told to consider a
specific point in between these two plates; it’s eight centimetres from one plate,
which we’ll say has zero volts, and two centimetres from the other plate.

We’ll take the plate on the left to
be the one at zero volts and we’ll say the difference between the potential at that
plate and the potential at the right plate is 𝑉 volts. We’ll solve for that voltage in
part two. But for now, our focus is on the
electric field strength in between these two plates.

Knowing that our plate on the right
is at higher potential than our plate on the left, we know that when an electric
field is set up between the plates, it will move parallel to them and it will go
from right to left, higher potential to lower potential. And moreover, this field — even
though it doesn’t quite look that way from our sketch — will be uniform in between
the plates; it will have the same value everywhere.

It’s that field strength, which we
can call 𝐸, that we want to solve for in this part. And to help us do it, we’re given
some information about this point we’ve marked out with an 𝑥. First of all, we’re told that this
point is 8.00 centimetres from the left plate, the one which is taken to be at zero
volts. And second, we’re told that at this
𝑥 point, this location, the potential is 450 volts.

We can say then that we’re given
the potential at this point; we’ll call it 𝑉 sub 𝑝. We’re also given the distance
between this point and our zero potential point, the left plate. And we then want to solve for the
electric field that exists between these parallel plates. Working with these three variables,
there’s a mathematical relationship that ties them altogether.

When we have a uniform electric
field in between two parallel conducting plates, then that field multiplied by the
distance between the plates is equal to the potential across them. Working with our particular
variables, we can say that 𝑉 sub 𝑝 is equal to 𝐸 times 𝑑. Well, remember it’s the electric
field 𝐸 we want to solve for.

Dividing both sides by the distance
𝑑, we now have an equation that will let us do that. All that remains is to plug in 450
volts for 𝑉 sub 𝑝 and 8.00 centimetres for 𝑑 and then we can calculate 𝐸. When we’ve made these
substitutions, before we calculate this fraction, there is one change we want to
make and it has to do with the units of 𝑑.

Right now, 𝑑 is in units of
centimetres, but the SI standard unit for length is metres. Recalling that 100 centimetres is
equal to one metre, that lets us perform the conversion between these two length
units. 8.00 centimetres is 0.08
metres. And when we calculate this
fraction, we see that it’s 5.63 kilovolts per metre. This is a measure of just how much
the potential changes for every metre of distance between our parallel plates.

Now, let’s move on to part two,
where, as we mentioned before, we want to solve for the voltage between the plates,
in other words, the voltage of the right most plate relative to the left plate which
is at zero volts. There are a couple of ways we could
go about doing this.

One way is to take our 𝑉 sub 𝑝
value, well 450 volts, which remember applied at a point 8.00 centimetres from our
zero volt standard, and multiply that by a ratio of distances. On top, we have the total distance
between the plates and on bottom we have the distance from the zero-voltage plate to
this point, where 𝑉 sub 𝑝 is measured. This method is perfectly good and
will give us the result we’re looking for.

A second way we can solve for this
voltage is to reapply our equation 𝑉 is equal to 𝐸 times 𝑑. Recall that in part one, we solved
for that electric field strength in between the plates. So now, knowing that electric
field, we can multiply it by the distance between the plates to solve for 𝑉.

Either one of these two methods
will work and both will lead us to the voltage between the plates. If we do choose the second method,
one thing we want to keep in mind is that we want to rewrite our distance between
the plates in units of metres from its given units of centimetres.

Both of these methods lead to the
same result and that is 563 volts. This is the voltage across the
total distance from one plate to the other.