### Video Transcript

What is the capacitance of a parallel plate capacitor having plates with a surface area of 4.50 metres squared and separated by 0.100 millimetres of Teflon? The dielectric constant of Teflon is 2.10.

So we have in this situation, two parallel plates which are both conductors. Theyโre capable of storing charge on them. And each one has an area, we can call capital ๐ด, given as 4.50 metres squared.

Along with that, weโre told that the plates are separated by a distance, we can call lowercase ๐. And thatโs given as 0.100 millimetres. And we also know that the space in between the capacitor plates is not empty. But rather, itโs filled with a dielectric material, in this case Teflon.

We know that, in general, the effect of a dielectric is to allow more charge to build up on either side of the capacitor plates. Our separating material here, which is given as Teflon, has a dielectric constant, weโll refer to as ๐
, of 2.10.

Knowing all this, we want to solve for the capacitance between these parallel plates. To do that, we can recall a relationship that connects capacitance with plate area, plate separation distance and allows for dielectric materials between the plates. This equation tells us that capacitance is equal to ๐
times ๐ nought the permittivity of free space multiplied by the plate area divided by the plate separation distance.

We can recall that ๐ nought the constant is equal to 8.85 times 10 to the negative twelfth farads per metre. And as we look at our equation for capacitance, we recall that if we had no dielectric at all, ๐
would effectively be one. And the equation would read ๐ nought ๐ด over ๐. But of course in our case, ๐
is not one. We do have a dielectric in between our plates with a constant of 2.10.

So now knowing ๐
as well as ๐ nought ๐ด and ๐, we will plug in to solve for the capacitance ๐ถ. Having done this, notice that weโve taken our value for ๐ the plate separation distance and converted it from units of millimetres to units of metres. This is to agree with the units in the rest of our expression.

We can see that now this distance is expressed in metres. That unit along with all the other metre units in the expression cancels out. Weโre left only with the units of farads, the units of capacitance.

And when we calculate this whole expression, to three significant figures, we find a result of 0.836 microfarads. Thatโs the capacitance of this parallel plate capacitor.