Alright, guys. So we're a little bit more familiar with capacitance and capacitors. In this video, we're going to check out the most common type of capacitor you'll see called the parallel plate capacitor. Let's go ahead and check it out.

We already have the relationship between the capacitance for any given capacitor. It's just C=QV. But the most common type of capacitor you'll see is called the parallel plate capacitor, where you have one surface of charge that's going to be positive Q and another surface of some negative charge that's going to be minus Q. These are separated by some distance, which we'll call d. Now, these don't necessarily have to be rectangles. They can be circles or some odd shape, but they will have an area. That area A is going to be the same for both of them.

So, if you have this kind of situation where you have a parallel plate capacitor, which you have some area and some distance, then we actually have another formula for finding out what the capacitance is. It's ε_{0}×A/d, that vacuum permittivity constant, times the area divided by the distance between them.

These capacitors, these parallel plates, produce an electric field in between them because the electric field lines want to go from the positive charge over to the negative charge. You should remember the electric field between the plates is uniform. In other words, it's the same always everywhere. No matter where you look inside this parallel plate capacitor, the electric field is going to have the same magnitude. Whereas, if you go outside, if you look anywhere outside of the parallel plates, the electric field very quickly approaches zero. The equation for the magnitude of the electric field is E=V/d, where V just represents the voltage.

The electric field between a capacitor has nothing to do with the capacitance and the voltage. All it depends on is the charge and the area involved. This is actually an interesting consequence of Gauss's Law. The electric field and the equipotential surfaces have to be perpendicular, which means the electric field always points in the direction of decreasing potential.

Let's go ahead and check out this example right here of a parallel plate capacitor, and solve some problems. We're told what the area of these plates is in centimeters squared, the separation distance, and what the voltage is. So our first step is just going to draw a quick little sketch of what's going on. So, we have these plates and separation distance, in other words, d=0.01 meters, we have to change everything into meters, so we have the right amount of units.

Now, let's figure out what the charge on the capacitor is using Q=C×V. The capacitance for a parallel plate capacitor is C=ε_{0}A/d. Plugging in our values, we get Q=4.43×10−11 Coulombs.

For part b, we can figure out what the magnitude of the electric field between the plates is both ways. The electric field is equal to Q/ε_{0}A which works out to be 10000 Newtons per Coulomb. Alternatively, E=V/d which gives the same answer. So we just got the same answer using two different approaches.

Let me know if you guys have any questions with this, and I'll see you guys in the next one.