Hey, guys. So now that we're a little bit more familiar with capacitance and the relationship between capacitance, charge, and voltage, we're going to see in this video how that all relates to the potential energy stored. Because remember, the capacitors separate charges between 2 plates in a parallel plate capacitor, and that separation leads to some potential energy that's stored. Well, how much energy? Well, I'm not going to show you the derivation, but basically, the energy stored by any capacitor, not just a parallel plate capacitor, is given as this equation:

U = 1 2 C V 2But what we can do is we can use the relationship between charge, capacitance, and voltage to change this equation into 3 common forms that you'll see. So using 12CV2, we can sort of use Q=CV to write this also as 12QV. And we can also write this as 12Q2/C, the capacitance. Now, basically, all these three equations are perfectly valid for any capacitor, not just a parallel plate capacitor. And the choice as to which one you're going to use really just comes down to what variables you're given in the problem, and that's basically it. So this is the amount of energy that is stored between capacitors with charge, and that potential energy we can use in some energy calculations.

Now another important variable to know is the energy density, which is given by lowercase u. And the energy density is basically just the amount of energy per unit volume. So in other words, we have u=U/volume. Now, I'm not going to use capital *V* because we use voltage for that, so I'm just going to write here *volume* instead. Whereas the volume is what? Well, the volume is basically the area of these parallel plates. So for instance, the area of these rectangular plates times the distance between them, which is that capital *D*. It's basically the volume of the gap between those parallel plates.

So in other words, the volume is equal to the area, which if you're given a rectangle is just going to be length times width, times the distance between them. And so what we can do is we can take this lowercase u and write it in a different way. So we're just going to write this as 12, and we're just going to use the first form, which is CV2. And now we just have to divide it by the area times the distance.

Now one of the things we can use is we can basically use the relationships between Q=CV, and we can use the relationships between capacitance, which is in a parallel plate capacitor, epsilon not *A* over *D*. And if you go ahead and manipulate these equations and stick them inside for this equation, we can actually find that the energy density of a parallel plate capacitor is something very important and very simple, which is 12epsilonnaughtE2. So in other words, if this is a parallel plate capacitor, then that means that the energy density is only related to the electric field that exists between the plates. Alright.

So these are basically the 5 equations that you need to know. So there's all 3 or all 5 of them. And, basically, just to get familiar with these, we're going to go ahead and work out a bunch of example problems. Okay. So let's get started. We've got 2 parallel plates, and we've got the area between them, 50 centimeters squared, with the separation distance. And both we're supposed to find *a*, which is the energy stored, and also what's the energy density. So in part *a*, we're going to be finding *U.* And we know that the three forms for *U,* big *U,* are 12CV2 equals 12QV equals 12Q2/C., so you can choose which equation to use based on the given variables. Alright, guys. Let's go ahead and work out the second one here, which is what is the strength of an electric field in a capacitor storing this energy per cubic centimeter? So now what we're being asked in this question is we're asked for the strength of the electric field in a capacitor. We're going to assume that it's a parallel plate capacitor as well. So this right here, this is a joule, so this is millijoules per cubic centimeter. So in other words, the energy, which is big *U,* is equal to 2.5, and we have millijoules. So that means that it's_times 10 to the minus 3,_that's joules. And now we have per cubic centimeters, so we have centimeters cube right here. That's the volume. So we can figure out what the energy density is, but how do we relate that back to the electric field? Remember that in the If we're trying to figure out what the electric field is equal to, then we have to use the little *u* is equal to 12epsilonnaughtE2. So now this is our target variable. Let's go ahead and manipulate this. We're just gonna move the one half over to the other side, the epsilon naught over, and we're gonna get 2×u/epsilonnaught=E2. So that means that E=2×u/epsilonnaught. Okay? So let me know if you guys had any questions with this, and let's go ahead and get some more practice.