 Suppose we have a plate full of charge and infinitely big plate full of charges. The question is, what's the electric field going to be everywhere? That's what we're gonna figure out in this video. So let me show you the same thing from a side view. So we have an infinitely big plate. You have to imagine that, even though I've not drawn that. And we need to figure out electric field everywhere. So let's start with a specific point. Let's say we want to figure out what the electric field at some point, at some distance are from the plate is going to be. How do we do that? The first question you might have is, why do we care about infinitely big plates? I mean, is that practical? Well, even in practice, we may not have infinitely big plates, we might have finitely big plate. But then, if you were to figure out electric field very close to it, very, very close to it, we can assume the plate is infinitely big. So whatever we get over here, we can use that values for very close distances. So you can assume, in practice, what we are doing is finding the electric field very close to big plates. Okay? If you go far away, we can't use that. But as long as we are close enough, we can definitely use it. So how do we do that? Well, we can start with Coulomb's law, which you might be familiar with. It says electric field due to a point charge is q divided by four pi epsilon not r squared. But by now, you might appreciate that you can't directly do that. But you'll have to break this up into tiny, tiny pieces and then calculate electric field due to each piece and then add them all up. And that's gonna be a nasty integral, which we're not gonna do. So we're gonna go for Coulomb's law, but instead, you know, you might already guess, we're gonna use Gauss's law. And the whole idea behind why we can use Gauss's law over here is because the electric field is gonna be very symmetrical, as we will see. And because the electric field is symmetrical, we can find a closed surface such that the electric field everywhere on that surface will be the same. And so we can pull it out of the integral and then we can evaluate this expression without having to integrate and calculate what the electric field is going to be. That's the whole idea behind it. And if you're wondering, wow, that's amazing. Can we do that for every single problem? No, we can only do it for three special cases. One is this one, infinitely big plane. The other one, you may have already seen infinitely big line of charge. And the other one is when we have a sphere of charge. These are the only three cases where we can use this, okay? So this is one of them. So where do we begin? Well, we start by figuring out what the electric field looks like everywhere. To apply Gauss's law, to choose a closed surface, the first step is that. So let's start over there. How do we figure out what the electric field looks like everywhere? The steps are going to be very similar to what we did with the infinite line of charge. So if you need a refresher of that, you're ready to go back and watch that. But what we do is, because we want to use Gauss's law, the first step is to know what the direction of the electric field is everywhere. Figure out that based on symmetry. And here's how we can do it. Let me first look at it from the side, so I can see it nicely. Same thing, I'm looking at it from the side. Now what I'm gonna do is, I'm gonna draw, I'm gonna divide this plane, this sheet, into two halves along this line. Okay, along this line, this one. And I can say that the top part of this sheet is exactly equal to the bottom part of the sheet because it's uniformly charged, it's infinitely big. They are exactly same, and so they are mirror images of each other, okay? What can we say based on that? Based on that, we can guess what the electric field looks like over here. How? See, here's how I like to do it. First, start with some arbitrary direction. Let's say electric field is over here this way. Now I can say that's wrong because why would the electric field point upwards? Because the top part and the bottom part are exactly similar, so why would the electric field point upwards? There's no reason for that. So for the same reason, electric field can't point downwards, electric field cannot point this way. So the only way electric field can be pointed is it's neither pointing upwards nor pointing down. The only possibility, there are only two now, either it has to be towards the right or towards the left. And since we know this is positive charge, we can guess that it should be away from the plate and so it has to be. The electric field over here needs to be towards the right. What an amazing argument, right? Just from the symmetry argument. But we don't just stop there. Remember, point P was an arbitrary point I chosen. That point could have been over here and I could have made the same argument. I could have divided into two parts and remember, this is infinitely big. So wherever I divide into two parts, I will always get two halves, the top half equal to the bottom half. And so I could make that argument everywhere and therefore electric field everywhere, at least over here, somewhere on this line, everywhere should be towards the right and over here, everywhere towards left. And not just that, since this point is very similar to this point, there's no difference between these two points, right? I mean, you can kind of say that every point is, you know, I'm looking at the center of the sheet. Anywhere you go, because the sheet is infinitely big, I could say there's no difference between these two points. And so the electric field here and here should also have no difference because absolutely no difference from these two perspectives. So I could also say, not just the direction, but I can also say electric field everywhere over here must be exactly the same. Everywhere over here must also be exactly the same. In fact, if you go at a distance R anywhere, you go top or bottom or out of the screen or into the screen, wherever you go, the electric field must be the same at a distance R. Does that make sense? That's our symmetry argument. So if I were to look from here, just to make that more clear, we could say that if I have to take a plane parallel to our given sheet, anywhere on that plane, the distance is the same from the sheet, right? I'm taking parallel. And so everywhere on that plane, the electric field must be the same. Any plane you take parallel to the sheet, electric field must be the same. Does that make sense? That's our symmetry argument. So now comes the question. Now that we know this, what kind of Gaussian surface would you use, would you choose to use Gauss law? Okay. I want you to pause the video and think a little bit about this. Should be a surface such that that integral becomes nice, like nice and easy. Here's Gauss law again. The integral should be nice and easy. So what surface would you choose? Pause and give it a shot. All right. Let's give this a shot. Let's see. My first instinct is that whatever surface I choose needs to be flat in front of it or behind it. Why? Because we already saw such flat surfaces, parallel surfaces will have same electric field or over it. And that we can use to our advantage. The second thing is whatever surface I choose, it needs to go through the sheet. It has to pass through the sheet. Only then I can enclose some charge. So putting these two together, the surface we can choose is a cylinder. Same thing if I show from the side view, the cylinder would look somewhat like this. Should have the same length on both the sides, R on the right and R on the left as well. So now we can use Gauss law. We can equate the left hand side, we can simplify the left hand side, simplify the right hand side and go ahead and calculate it. And so again, before I do this, great idea to pause and see if you can try this yourself. Because it's nothing new. We've all studied about flux and we've done this for infinite line of charts. So it'll be great idea to pause and really, really, really try yourself first. All right, if you've tried, let's see. So let's start with the flux. What's the total flux through the entire cylindrical surface? Well, I can find three distinct surfaces. One is the front surface, which let me draw that over here. The back surface and the curved surface, right? Let's start by drawing the, let's calculate in the flux through the curved surface. How much would that be? Well, notice the electric field lines everywhere over here is parallel to the curved surface, right? Everywhere it's parallel. And when you calculate in flux, you're doing a dot product. And so the DA vector, wherever you go, the DA vector is going to be outwards. Here it's going to be outwards or here it's going to be downwards, right? So what is the angle between the DA vector and the electric field vector? It's 90 everywhere. Wherever you go, even if I take a tiny piece over here, the DA vector is going to come out and that's going to be the angle will be 90 degrees. And so that means wherever you go on the curved surface, this value is going to be zero. Electric flux is going to be zero and that kind of makes sense. Nothing is flowing through the curved surface. No electric field is passing through the surface. So the curved surface gives me zero flux. So the flux only gives me a value on the front surface and the back surface. So what's the value over there? So let's come to the front surface. Let's assume that the electric field over here is, I don't know, some value E. We already know it's going to be this direction. And since the whole area is nice and flat, what will be the direction of the area? The area vector again, normal, outwards. Ooh, notice, area vector and the electric field vector are in the same direction, same direction. So when you do the dot product, cost zero would be one. And so the dot product would be just E into DA. And this entire DA is my A. And so if you were to do this, you just get the flux over here as E into A. So flux here would be just E into A. That's the flux through the front surface. And the same flux through the back surface. The story is the same. Which means the total flux, the left-hand side would be two times E into A. A being the area of that front surface. I'm just going to choose that as A. Let's use blue, A. So that's our left-hand side. So that should equal the total charge enclosed. What's this charge enclosed? We didn't say anything about the charge. So I totally forgot about the charge. Okay, so first thing is total charge is infinity, right? Because this is uniformly distributed and this is infinitely big. And so whenever we have such cases, you know what one thing we can mention is we can talk about how much crowded the charges are. So we like to talk about the charge density. And since the charges are distributed over a surface, here we like to talk about surface charge density. And so let's say the surface charge density provided to us in the question itself, let's call it a sigma. It's given to us. And think in terms of units. Imagine in standard units, it'll be sigma coulombs per meter square. So that means I'm saying that every meter square of this piece has sigma coulombs of charge. Let's say that's given to us, okay? Now, given that, what would be the electric, what would be the total charge enclosed over here? Well, I know that each meter square encloses a charge sigma. But we have a meter squares. This is a meter squares. How much would that enclose? So one meter square encloses sigma. Two meter square encloses two sigma. A meter square would enclose A times sigma. So the charge enclosed would be A times sigma. Does that make sense? Divided by epsilon naught. And so now we can do the algebra. The A cancels out. And so the electric field turns out to be sigma divided by two epsilon naught. Ta-da! We're done. Couple of steps, no integration. Done. That is our answer. Now before we close, do you see something interesting in this formula? Hopefully you see something interesting. The interesting thing is there is no R in the formula. It's independent of R. What does that mean? Independent of R. That means electric field does not depend on the distance. Regardless of how far or how close you are to the sheet, the value is the same. And as we saw, that means electric field everywhere should have the exact same value. Uniform field. That means a sheet of charge, infinitely long sheet of charge produces a uniform field. And what does this mean for our practical case level? Like just we saw before, that means if you have a charged plate, an actual finite plate, then as long as you're close to it, somewhere close to it, you can say that, hey, electric field is pretty much uniform, somewhere close to it. Near the center, you have to be near the center, close to it, you can assume it to be infinitely big and you can use this value. But of course, if you go far away, then of course, electric field dies off. And this will be useful for us in the future. So electric field due to a metallic sheet close to it or infinitely big sheet would be sigma divided by two epsilon naught.