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Anderson Video - Electric Field Energy Density

Professor Anderson
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Hello class, Professor Anderson here. Let's talk a little bit more about the parallel plate capacitor. Okay, we know exactly what a parallel plate capacitor looks like, you take one metal plate put a bunch of charge on it, take another metal plate put a bunch of the opposite charge on that one. So if I put positive up here and I put negative down here and this is my parallel plate capacitor, we know that electric field must point from top to bottom. All right, it leaves positive charge and heads towards, let's refresh this pen here hold on, points towards negative. Okay, that's what my E field looks like. And we also know that there is a potential difference here right? This is V Plus, this is V minus. The gradient of V goes up which means the electric field goes down. All right, but let's think about this capacitor in terms of something called capacitance. What is the capacitance of this device? Well, if I put positive Q total charge on the top plate and I put negative Q total charge on the bottom plate then there is a very nice relationship between capacitance, charge, and voltage which is just this: Q equals C times Delta V. Whatever this potential difference is that goes here, C is some inherent capacitance of this device, Q is the amount of charge. So capacitance is the ability to store charge at some particular voltage and this is the relationship: Q equals C Delta V. All right, but if I put a whole bunch of charge on this thing it must take work to do that and so I must be able to store up energy in this device. It must be, right? Because if I have this device and I hook a wire up to it, it's going to drive current and so there must be some energy stored in the device. And we know what that is, the energy stored in a capacitor is what? Well, let's just think about it for a second. If I took a little bit of charge and put it on there, I have to work against all the other charge that's on there, and as I add more and more charge it becomes harder and harder to do that. Turns out if you do this carefully and then you integrate it, you'll end up with the energy in a capacitor equal to one-half CV squared. Okay, there's a nice little proof of that. Now if that is the energy stored in the capacitor because I had to do the work to put it in there, can we now say something about the energy in the field that's there? Maybe, let's think about that for a second. What is the field E inside a capacitor? Each plate gets me Sigma over 2 epsilon naught, so since I have 2 plates the field is just Sigma over epsilon naught. Okay, don't worry about the fact that it's pointing down or not, that just means we've got to put a minus z hat on there, but let's just worry about the magnitude of this thing. What's the magnitude? It's Sigma over epsilon naught. All right, but I can relate that to the voltage, right? Sure, because we know that the electric field is negative del V or V equals negative integral of the electric field, but the electric field in this case is just a constant. Okay, and so that can come out of the integral and this whole thing just becomes E times the separation D with a negative sign, where this is the separation between the plates D. What's the voltage? It's just E times D. That looks pretty cool, I wonder if I can relate that back to this stuff over here. Umm, maybe? Let's see. We've got work equals one-half CV squared but we know that C is Q over V, so this becomes Q over V times V squared, which is just one half QV. But we also know that V is equal to E times D and so this becomes one-half Q times E D, don't worry about positive or negative for right now. Okay that's getting somewhere but we still are dealing with this Q right here and in our equation for E we had Sigma. What is Sigma? Sigma is in fact charge density, it is Q divided by the area of the plates, if this plate has area A, this is the charge density how much charge per unit area. All right, so that means I could put a Sigma right here if I just divide by A. Let's rewrite this. So we've got 1/2 Q, Q is equal to Sigma times A, and then we have E times D. Let's rearrange this a little bit, this is 1/2 Sigma times E, times A, times D, and now let's put in for this guy right here, this Sigma from this. If I rewrite this I get Sigma equals epsilon naught times E. So if I put in for that Sigma right there, what do I get? I get epsilon naught E, times E, times A, times D, which is one-half epsilon naught E squared, times A, times D. This is the energy in that capacitor now written in terms of the field. All right, but maybe this has some important consequences, right? If the energy is this, what can we say about the energy density? The energy density is just energy divided by volume. All right, but there's our energy, what volume are we talking about? We're talking about the volume in this region right here. The volume between the plates, but the volume between the plates is just the area of the plate times the separation D and so look what happens, we have one-half epsilon naught E squared, times A D, and we are going to divide by the volume of that region of space which is A D, and we just end up with one-half epsilon naught squared. And this is the energy density for E. And it doesn't just apply to the electric field between parallel plates, it applies to any electric field. When you have an electric field in space, what is the energy density? It is that thing right there. All right, hopefully that's clear. Cheers!