 Okay, so I'm going to do two examples with Gauss's law, there might be two videos or one that let's just see how long it takes. So Gauss's law says this. So this says right here, if you take some closed space, it doesn't have to be real, okay, it could just be fake, and most of the times it is, and then you integrate the electric flux over that surface which is E dot n hat, n hat is a vector pointing away from the surface, and dA is the area element, okay, because those things can change, all those things could change over the area. If you do that and do it for the whole complete, this means a complete surface area, then that flux is going to be equal to the total net charge inside that area divided by epsilon naught. So let's look at a plate from the side with positive charge on it, and we already did the electric field due to one single plate, so this is another way to get it. Now you may be angry afterwards to see that this is easier, but I'll tell you there's a trick, okay. So the trick is to say I have to already know what the shape of the field looks like, okay, so I can't use this if I don't know something about that. So in this case, I know that if the charge is uniform, and it's really big, and I'm close to the plate, then the electric field is going to be constant and pointing away. This is not a line, this is a plate, okay, so, but I've only shown one dimension of it, it's like that. So constant electric field all the way, okay, so it's a really big plate. Now in order to apply this, I have to pick a surface, a volume, I have to pick some shape of volume over which to do this calculation. Then the key for using Gauss's law effectively is to pick a shape that makes things easy. You don't want to pick things that make it hard. For instance, if I picked, let me draw it up here, if I picked a sphere like that as my volume, that would be a poor choice, why? Because even though the electric field's constant on the surface in magnitude, in hat at different areas on that change. So how would I do that integral? It wouldn't be so easy, okay, because you have to integrate over this surface area, even if E is a constant magnitude, E dot in hat changes, so it wouldn't be easy. You could do it, it's not undoable, it's just not very smart, okay. So let's choose something smarter, I thought I had a blue marker, oh well, okay. So let's choose a cylinder, so it's like that, and it's got, I'm just going to make it as simple as possible. It's going to be a cylinder, let me redraw the cylinder, it's like that. And this is an area A and a length L, okay. A lot of books call this a pill box, that makes you happy, okay. So now if that's my surface, I need to do this integral over that whole surface. Well, it's really three discrete sides. I have this whole cone, let me show you, you know, if this was my cylinder like that, then I have, there you go, then I have this end right there, I have the whole side, I can choose one piece and this end, so I'm going to have really three fluxes. So the net flux is going to be equal to, I'll just call this phi one plus phi two plus phi three, and so this is one, two, three, okay. Let me start with phi two, it's going to be equal to the integral, now it's not a closed integral, right, because I'm just going around that one surface area, it's not the whole thing. So it's just going to be E dot in hat dA. Now I don't want to integrate because I want to make it such that this is simple. But if I look at my side piece right here, and anywhere along there in hat is that way, down here in hat is that way, right here in hat's coming out, okay, so it's always perpendicular to that surface area, but E is that way, here E is that way, here E is that way, on the other side E is that way. But in all of those cases E is going to be, if I call this the x direction, E is going to be in just the x direction, either positive or negative. In hat's going to be in the yz plane, so when I take E dot in hat, they're always perpendicular so I always get zero, so this whole integral just becomes zero. Think of this as rain, if it's raining this way how much rain hits the side of that, none. So phi two is simple. Now what about phi one, now we have this in cap over here, now there's E and in hat is the same direction and it's constant, they don't change, right, over that whole surface the direction of the area doesn't change. So if I do phi one it's going to be E which is constant magnitude, so I'm going to bring out that, and then I have x hat dot x hat, right, they're both in the same direction so I get one, and then I have dA. So now I just have to integrate over this area, but I just have the integral of the area, so I just get the area, but that's not a vector, right, because I took the dot product. So I get EA, now I need to do phi three. What's different about phi three is now, if I'm on that side of the plate the electric field is going this way, but so is in hat. So E is in the negative x direction, and hat's in the negative x direction, so I still get the exact same thing. So now if I put all this together the total flux is going to be 2EA, now I just need to find out how much charge is inside there, okay. Let's say that, maybe I shouldn't have used A, okay, well let me say the whole plate, you can do this two ways. Let me say the whole plate has an area of A prime, I shouldn't have used A, and a charge of big Q. Then if I assume the charge is uniformly distributed then I should say that the charge in here inside that little piece right there is going to be, I'll just call it lowercase Q, lowercase Q over A would be the same ratio as the total charge over the total area, Q over A prime. So the Q inside is going to be equal to AQ over A prime. So if I put that in up here I get 2EA equals AQ over epsilon not A prime, these A's cancel and I get the magnitude electric field is going to be Q over A prime over 2 epsilon not. That's the magnitude of the electric field near the surface of one plate, and remember in a capacitor we don't have that 2 because we have two plates next to each other, okay, but that's the electric field. And you see that we didn't have to do any complicated integration, but we did lose something. We had to sacrifice something. In this case we had to sacrifice ideas about the direction electric field. This is the magnitude electric field, not the electric field vector. When we did it before we broke this into essentially little pieces and treated each one as a point charge and we got the direction, okay. So when you use Gauss's law you have to already know something about the direction. So you can't really, it's not necessarily easier, it's easier if you already have an idea about what you expect. Okay, that took longer than I wanted so I'll stop and then I'll do a thin line of charge. I'll use the same idea.