Hey guys. So in this video, I want to cover a concept called Gauss's Law, which is a super important topic in electricity, that you need to know now in some textbooks. It's even an entire chapter and it can be kind of confusing because it ties together a lot of concepts and ideas about charge and flux. What I want to show you by the end of this video is that Gauss's Law is really just a very straightforward relationship between flux and charge.

And I'm gonna show you the three main types of problems that you'll need to know when solving problems. So let's go ahead and get started here. I'm just gonna jump straight in. So this guy, a long time ago, did a lot of calculations between charges and fluxes, and what he came up with was this law. And basically what it says is that the net flux that goes through a closed surface, which is really important—the closed part—depends only on the charge that is enclosed within that surface.

So basically, here's what I mean, right? Imagine I have this little box here and in the center of this box, I have a charge. And I'm just gonna pretend that it's a positive charge. Now, this positive charge emits electric field lines that go to the surface of this box. And instead of using, you know, some equations like \( e \cdot a \cdot \cos(\theta) \) or whatever to calculate the flux, basically what Gauss's law says is that the total amount of flux going through all of the surfaces, so I'm gonna call this \(\phi_{\text{net}}\) here, is directly proportional to how much charge is in the box. And the equation for this is pretty straightforward. It's \(\phi_{\text{net}} = \frac{Q_{\text{enclosed}}}{\epsilon_0}\). So this \(\epsilon\) here is just 8.85 times \(10^{-12}\), a constant that we've seen before.

What I want to do now is show you the three most common types of problems that you'll need when using Gauss's Law. We're just gonna jump straight into the first one here. The first one is where you're given some kind of charges and you're asked to calculate the flux. Let's take a look at this first example here. What is the net flux that goes through this surface \(A\) over here. Alright, so we've got this little surface \(A\). We've got these two charges, one is positive and one is negative. So, I want to kind of suffer through this a little bit with you. Show you why this is super useful. Now, pretend for a second we didn't actually have Gauss's Law equation. The only way we can calculate and define is by using an old equation, which is \(E \cdot A \cdot \cos(\theta)\).

So what happens is these point charges will emit electric fields like this; they'll have electric field lines like this and we have to calculate at each point, what is the \(a \cdot \cos(\theta)\). The problem is this \(E\) depends on \(r\), so it depends on the distance. And what happens is this surface over here is going to be at a higher distance than this piece right here, and it's gonna be higher than this one, or different than this one. And also what happens is that the normal vector may sometimes point in a different direction than your electric field lines. Notice how this \(E\) and this \(A\) don't point in the same direction. So the angle is gonna be constantly changing. What this means here is that this is basically impossible. We can't calculate \(E \cdot A\) using \(E \cdot A \cdot \cos(\theta)\).

But now that we have Gauss's law, we can have a much more straightforward relationship between flux and charge. So this is \(\phi_{\text{net}}\) and this equals \(\frac{Q_{\text{enclosed}}}{\epsilon_0}\). So, what this means here is that the net flux through the surface only is directly proportional to how much charge is inside this surface here. So, how much charge is in the surface? Well, I've got 5 Coulombs and a negative 3 Coulombs charge. So, if you group these together and combine them, the total amount of charge I have in the surface here is going to be 5 Coulombs plus negative 3 Coulombs, and you'll just get 2 Coulombs. So there's basically 2 Coulombs of charge inside of the surface. So we just pop that into this equation. So what Gauss's law says is that \(\phi_{\text{net}}\) is equal to \( \frac{2}{8.85 \times 10^{-12}} \). So when you work this out, what you're going to get here is \(2.26 \times 10^{11}\) Newton meters squared per Coulomb.

So this is the power of Gauss's Law. All you have to do in a surface here is just know how much charge there is. And then you can figure out how much flux the total amount of flux that goes to the surface. And you don't have to use this \(E \cdot A \cdot \cos(\theta)\) equation anymore. Alright, so, let's move on to the second problem now, which is kind of reversed. In some cases, you'll be given the flux either through one or multiple surfaces and you'll be asked for the charge. So, for example, we've got the flux through four sides of a closed pyramid and we've got these numbers over here.

So, I'm just gonna draw this out really quickly. Imagine I have this little pyramid like this. And basically, right, so, I've got, you know, 1, 2, 3, and then 4 sides, the underside like this. It doesn't matter which one is labeled, which it really doesn't matter because remember that Gauss's Law is only concerned with the net flux, not the ones through individual surfaces. So here, what we want to do in this problem if we want to calculate, well, what's the charge that's enclosed? Now, now we actually have the net flux and we want to figure out \(Q\). So all we have to do here, I'm actually just gonna go ahead and move this down over here is figure out, Well, what is the net flux? Well, if you were told that the flux through each one of the surfaces, then the net flux is just gonna be adding all of them up.

So I'm just gonna add \(51\) plus \(52\) plus \(53\) plus \(54\). That's the net flux. So this just means that your net flux here is equal to we've got \(10\) plus \(20\) plus \(8\). So this is \(8\) plus negative \(15\). And you're just gonna get \(23\) this is Newton meters squared per Coulomb. So what I'm gonna do here is I'm just gonna replace my \(\phi_{\text{net}}\) with \(23\) then I have to multiply by this, my \(\epsilon\), which is \(8.85 \times 10^{-12}\). And this is gonna be my \(Q_{\text{enclosed}}\). And when you work this out, what you're gonna get is \(2.04 \times 10^{-} \) Coulombs.

So this is kind of the opposite here, you're using the flux to work backward and figure out. Well, if I know all the fluxes and that means that there's this amount of charge that's inside of the pyramid. I don't know how it's all arranged, it could be on the surface, it could be in the center. It doesn't matter. All I know here is that this must be the total amount of charge that's enclosed within that surface. Alright, now let's move on to the last problem here, the last kind of problem, which is a little bit more tricky. So in some problems, you may be given some kind of charges or a charge and you might be asked to calculate the electric field.