 The goal of this video is to explore Gauss law of electricity. We will start with something very simple but slowly and steadily we'll look at all the intricate details of this amazing, amazing law. So let's begin. So let's imagine a situation. Let's say we have a sphere at the center of which we have kept a positive charge. So that charge is going to create this nice little electric field everywhere, right? Now the question we want to try and answer in, you know, the question we want to start with is what would be the flux through this sphere? What is the electric flux through this sphere? How do we calculate that? Well, let me remind you how we calculate flux. We define flux phi as basically the integral of the electric field dot dA, which means it's the integral of e dA times cos theta. And just to, you know, just to remind you, because we've already talked about it, but just to quickly recap the whole idea behind this is if we have some kind of an electric field flowing through some area, let's call it as dA, then the flux is just the product of the electric field and the area, right? That's how we calculate flux. But then why there is a cos theta over here? Well, the idea is, if you see from the fields point of view, it can't see the entire area. From the fields point of view, the effective area that it's flowing through is only this much, the perpendicular component. And so if the angle between these two is theta, then notice the effective area through which the field is flowing is dA cos theta. From this right angle triangle, this is the hypotenuse, this is the adjacent side, the adjacent side becomes the cos component. And that's where the cos theta comes from. And again, why do we write it as e dot dA? Well, we like to treat dA, the area as a vector, by assigning an arrow mark perpendicular to it. And if you did that, notice if this angle is theta, this angle also becomes theta. Again, this is not something new. I'm just recapping what we all learned before, okay? And if you do that, then we can say theta is the angle between electric field and the area vector. And then we can say, ah, this is the dot product of electric field and the area vector. But of course, if you're not familiar with this, it would be a great idea to go back and watch this video on electric, it's called vector form of electric flux or something like that, okay? So using this, our first goal is to figure out what is a flux through this sphere. And so what I'll do is I'll do some of the work, and then you'll do some of the work. So let's start by zooming in so we can nicely see this. So what I'm gonna do is I'm gonna take a very tiny piece of this particular sphere over here, and let me call that tiny piece its area to be dA. And let's imagine that the electric field at that point is e. The question is, what is the flux through the entire surface? So let me first look at what's the flux over here. What is the angle between the electric field and dA? Let's think about that first. Well, because it's a sphere, we might know one of the properties that the surface will always be perpendicular to the radius. And since the electric field is along the radius, this dA is perpendicular to the electric field. So what's the value of theta? It's not 90, because remember, theta is the angle between electric field vector and the dA vector. The dA vector is this way, right? I mean, you can draw dA vector inwards also, but the convention is we always try to, we always draw, you know, dA vector outwards for a given closed surface. So now, what is the angle between the two? They're in the same direction, so theta becomes zero, okay? And that's not just here. That will be true anywhere. So if I were to take areas anywhere, you will find that the theta will always be zero. Always the electric field will be in the same direction as that of the dA vector. So that's nice. That means that everywhere, let me just write that down, everywhere this is zero. That's great because cos zero is one. That's one simplification. Another simplification that happens is notice that the electric field will be the same value everywhere on the surface. Why? Because the point on the surface is that the same distance from the charge are. And therefore the electric field value should be a constant, right? So this is a constant. And that's also great because constant numbers can be pulled out from the integral. So let's see what happens to our integral now. So if I were to calculate flux phi will be the integral. I'm going to pull out the electric field because it's a constant times dA, integral of dA times cos zero, which is one. Okay. So now it's your turn. I want you to try and find out what the final simplified version of this is. Because you already know what the electric field is. You know how to calculate electric field due to a point charge at some distance r. And I'm pretty sure you can try and calculate what the integral of dA is going to be. So please give this a shot and see what you get. Okay, if you've tried, let's see. So the flux is going to be E. What is E? Electric field at any point due to a point charge is going to be from Coulomb's law. Q divided by 4 pi epsilon naught r squared, right? And what's the integral of dA? Well, what is the meaning of integral of dA? You're basically adding up all these tiny pieces over the entire sphere. So that'll be the total area A. But do we know what that area is? What the value of that area is? Yes, the area of the sphere is 4 pi r squared. And so now we can simplify. We can say, look, r squared cancels, 4 pi cancels. And what we now end up with is flux pi equals q divided by epsilon naught. That's our answer. So if the charge is positive, we get a positive flux, meaning it's an outward flow. If the charge was negative, we would have gotten a negative flux, meaning it's an inward flow. Now, at this point, you might say, okay, Mahesh, so what's the big deal? Why, why, this is just some problem that you have solved, right? So what's the big deal? This is where it gets really interesting. Okay, I'm going to start asking a bunch of questions and we'll go deeper into this concept. So to begin with, let's first zoom out. And here's my first question. What would have happened to this value of flux if our sphere had a size of two r, radius of two r, doubled? Can you pause and think about the answer? All right, if you're tried, let's see. Notice that our flux is independent of the value r. There is no r in our answer. That means it does not depend on the size, which means the flux value remains the same. But why is that happening? Well, you can actually see the answer over here. The r-squares are cancelling out. And what does that mean? Because see, flux is the product of electric field and DA, right? So even if the area is increasing, why isn't the flux increasing? Well, the main reason for that is that when the radius doubles, sure, the area increases, it goes up as r-square. So when the radius doubles, your area becomes four times as much. Area becomes four times as much. But what happens to our electric field? It goes down as one over r-square. So our electric field, because now the distances are also doubled, our electric field becomes one by four, one over r-square, right? And so when you multiply, notice the effects cancel out and the flux does not change. So the reason why the flux is independent of the size of our sphere is mainly because of the inverse square law. If the electric field follows some other law, let's say inverse cube law, or inverse any other power, this wouldn't be true. So truly because of the inverse square law, we're seeing the flux is independent of the size. Okay, now next question is, what if I move the sphere a little bit to the right? Or let's say I move the charge a little bit to the left, off-center, that's important. Now what will happen if you calculate the flux through the entire surface? Will the value remain the same now? Or increase or decrease? This is a little bit more interesting. So again, pause the video and give this a try. Okay, now we need to re-derive it because it's no longer off-center, so this derivation will not work anymore. Okay, that's the first thing. Now if you had to re-derive, see the problems. First of all, can we say the electric field is constant everywhere? No, because here we are closer to the charge, so stronger electric field, and as you go from here to here, weaker electric field. So electric field is no longer going to be a constant. So this will not be true anymore. What about this value of theta? Well, if you were to draw again, these tiny, tiny d-vectors, notice that the angle is also no longer zero. It'll keep changing, which means this is also not true. So can you see how complicated this integral is gonna become now? We can't pull out that E, that theta will not be zero. Oh man, it's gonna become a little complicated now. So we might think that the answer could be different, right, because we'll get a completely different derivation. But what if I told you the answer remains the same? And this is the mind-blowing part. This is the most important part, that even if I move the charge off-center anywhere, the flux remains exactly the same. But I'm pretty sure you'll be wondering how? How is that possible? Can you prove that to me? Yes, I can prove it to you. There's a beautiful proof, but I wanna do that in a separate video because I don't do it slowly. It's a very logical proof. I wanna do that separately, but the flux remains the same. Now, although I won't prove it to you, but I want to give you some imagination, okay? And so the way to imagine this is, instead of thinking of this as an electric field, imagine that this charge is shooting out bullets, pew, like this. And let's say when the sphere was kept this way, when the charge was at the center, let's say the number of bullets that are coming out, one, two, three, four, five, six, seven, eight, nine, 10, 11, or 12, whatever that is. Let's say there are about 12 or 11 bullets are coming out per minute through the sphere, okay? When the charge is at the center. Now here's my question. What if I move that sphere a little bit to the right? Now, how many bullets will come out per second, or sorry, per minute? Well, it has to be the same, isn't it? It's still shooting 12 bullets every minute, and those 12 bullets have to come out through this sphere. I mean, imagine this is an hypothetical sphere, so the bullets will not get stuck anywhere, okay? But the answer should remain the same, isn't it? The flow has to remain the same. Of course, you might say that over here, the bullets will reach quicker, and it'll take longer time for the bullets to reach here. Sure, but over a minute, if I ask you how many bullets are moving out of this surface, it has to remain the 12, isn't it? So the bullet flux remains the same in a similar manner, although not exactly the same, but in a similar manner, even the electric flux remains the same. And you'll find this again due to inverse square law. And now we can go even further. We can ask, why sphere? What if I got rid of my sphere and took some other random surface? What would be the flux of these bullets? It should still remain the same. So you see, we derived this for a sphere, but now we're saying that doesn't matter what kind of surface we have, it doesn't matter where the charge we keep it inside that surface, our flux will remain exactly the same. Same will be the case for the electric flux as well. But of course, there's a small detail over here. What if this particular thing had a hole? Now notice some of the bullets can leak through that hole. Now the number of bullets passing through the surface will reduce, which means there should be no holes in our surface. In other words, our surface must be closed. So as long as we have closed surface, the flux through it, it doesn't matter what surface, it's the most general case now, has to equal the total charge, sorry, the charge inside divided by epsilon naught. All right, we can go even further and ask, what if there are more than one charges? What if there's another charge kept over here? Well, think in terms of bullets. You just add its bullets as well. So if this is charge Q1 and this is Q2, what would be the value? It'll be Q1 plus Q2 divided by epsilon naught. So if there are three or four charges, just add all the charges inside and divide them by epsilon naught. But our last question we might have is what if there's a charge outside? Will it also contribute to the flux? So let's imagine a charge which is outside. What do you think? What will be the flux due to just this charge? Well, let's see. It's so exciting, okay. So if you again look at the bullet principle, now you will see not all the bullets will enter, but whatever bullets enter this particular surface, they will give you a negative flux and the same bullets will also exit that surface, giving you a positive flux. Notice that means the negative and the positive will cancel out. That means no net flux due to this charge. Can you see that? And now it doesn't matter where you keep this charge, how strong that charge is. Even if it's shooting thousand bullets inside, all those thousand bullets will exit. Does that make sense? Meaning this only when you're calculating the flux, you only calculate the flux due to the charges inside. The external charges will not affect the flux. Bullets are there, but when you're calculating the total, they just, the inward and the outward part, contribution of it, cancels out. So now we can write this equation all its grand glory. So let me clear this. Let's write this in all its glory. So this is Gauss law now. So Gauss law states that if you calculate the flux through the closed surface, and the way we write it in the integral is we put, we put a circle over here to represent that. We are calculating flux through a closed surface. Doesn't matter what kind of surface that is. Doesn't matter at all. That value should always, always, always equal the total charge inside that surface. Total charge. If you want, you can also write as sigma if you want. Sigma, okay, whatever. Total charge inside the surface divided by epsilon naught. I don't care about the charges outside, even though they do contribute to the electric field, they do give us bullets. They do not contribute to the flux, okay? And so in all its glory, this is the most general Gauss law for us. So again, the mind-numbing part for me is that we derive it for such a simple case, a charge at the center of a sphere, but look at this. It is the most general case of electrostatics.