1

concept

## Gauss' Law

10m

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Hey guys. So in this video, I want to cover a concept called Gospels 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 God'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 goes a long time ago, he did a lot of calculations between charges and flux is, and what he came up with with this was this law. And basically what it says is that the net flux that goes through a closed surface, which is really important. They're 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 admits electric field lines that go to the surface of this box. And instead of using, you know, some equations like e a Cosine theta or whatever to calculate the flux, basically what God's law says is that the total amount of flux going through all of the surfaces. So I'm gonna call this fine net here is directly proportional to how much charge is in the box. And the equation for this is pretty straightforward. It's fine. It equals Q enclosed over epsilon. Not. So this epsilon here is just 8.85 times 10 to the minus 12 is a constant that we've seen before. But that's basically all there is to it. So this charge here that's enclosed within the box, we call this Q enclosed. Alright. So what I want to do now is show you the three most common types of problems that you'll need when using Gauss is lost. 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 this Gospel slaw equation. The only way we can calculate and defy is by using an old equation, which is E. A. Cosine of theta. So what happens is these point charges will admit electric fields like this? They'll have electric field lines like this and we have to calculate at each point, what is the a cosine theta? The problem is is that this E depends on our right. So it depends on the distance. And what happens is this 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 are different than this one. And also what happens is that the 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 ea using ea cosign theta. But now that we have Galaxies law, we can have a much more straightforward relationship between flux and charge. So this is finance and this equals Q. And closed over epsilon knots. 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 a five. Cool um and a negative three Coolum charge. So, if you kind of group these together and combine them, the total amount of charge I have in the surface here is going to be five columns plus negative three columns And you'll just get too cool. Um so there's basically two columns of charge inside of the sphere. So we just pop that into this equation. So what God's law says is that fine? It is equal to two divided by 8.85 times 10 to the -12. So when you work this out, what you're gonna get here is 2.26 times 10 to the 11 Newton meters squared per Cool. Oh, so this is the power of Gaza'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 A. Cosine theta equation anymore. All right, 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, 123 and then four sides, the sort of underside like this. It doesn't matter which one is labeled, which it really doesn't matter because remember that God'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 five to 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 eight. So this is eight plus negative 15. And you're just gonna get 23 this is newton meters squared per cool. Um So what I'm gonna do here is I'm just gonna replace my fine it with 23 then I have to multiply by this, my epsilon, which is 8.85 times 10 to the minus 12. And this is gonna be my cue and closed. And when you work this out, what you're gonna get is 2.04 times 10 to the - columns. So this is kind of the opposite here, you're using the flux to work backwards and figure out. Well, if I know all the flux is and that means that there's there's basically 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 on 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. This one's a little less straightforward. So, let's take out our problem here. We're gonna use gasses law to write an expression for the electric field due to a point charge at some distance R. So for example, I've got my charge like this. What if I wanted to figure out the electric field here at this point, what I'm what I'm gonna call our over here. Well, we actually already know the equation for this. So remember that the equation for the electric field of a point charge is just Q. Is K. Q over R squared. So really what we're doing in this problem is we're gonna use Galaxies law to confirm this equation here. Alright, that's basically what we're gonna do. We should hopefully theoretically get this answer. So, let's start off with God's law, God's Law says that net flux is equal to Q. And closed over epsilon. Not now we want to calculate the electric fields, where does that equation pop? Or what is that variable? Pop up in our equations, remember it pops up in five net. So this is basically saying that E a cosine theta is equal to Q. And closed over epsilon knots. Now we have some kind of an area and a cosine theta, but we don't have a shape, right? We don't have like a pyramid or a circle or something like this. So the most important sort of the trickiest part of these problems is that when you're whenever you're solving for the electric field, when you don't have a surface, you're gonna have to choose one, you're gonna have to choose what's called a Gaussian surface. And this is an imaginary surface. It's not a real thing, it's not like I actually have a sphere that's enclosing this thing, it's an imaginary thing that I'm using so that I can evaluate what this electric field is. Now, here's the most important thing. You're gonna pick a Gaussian surface with symmetry where the electric field is going to be constant everywhere, there's three main types of shapes that you're going to see what you're gonna be boxes, cylinders, or spheres. So if you think about what's happening with this one, I need to choose a Gaussian surface. Do I choose a box? But what happens is if I choose a box. Some parts will be farther from the charge than others. So the electric field is not going to be constant everywhere. So it's not gonna be a box. What about a cylinder? If I use the cylinder, then I run into the same problem. Some parts in the cylinder are gonna be farther away and the electric field is still not gonna be constant. So what happens here is that the best sort of shape to use is actually going to be a sphere? So imagine I have a sphere that's sort of enclosing this charge. Again, it's imaginary, it's not a real sphere. What happens is that we know this charge is going to emit electric field lines like this, they go outwards everywhere. And what happens is my sphere right, is going to be at a constant radius like this. And what happens is my area vector, the normal, the one that's perpendicular is always going to be parallel to the electric field. So this E and this a the angle between these is going to be zero everywhere. So this actually makes my equation a lot easier because basically what happens is that the cosine term will just go away. Remember the cosine of zero is just one. So, if I want to figure out my electric field then all I have to do is just move the area to the bottom here. Now remember the area of the sphere is equal to four pi R squared. So basically what I end up with here is E equals Q enclosed over four pi R squared epsilon knots. Now this may not look exactly like this equation over here, but that's because we have one last thing to do this variable K. That we see in our K. Q over R squared is actually equal to 1/4 pi epsilon. Not. So what this means here is that E is equal to Q enclosed. Actually this is KQ enclosed over R squared. And now basically gotten back to our equation here. Alright, so using Galaxies law, we got the same equation for the electric field of a point charge. Alright, so that's it for this one. Hopefully this makes a lot of sense. Let me know if you have any questions.

2

example

## Electric Field Within Spherical Conductor

2m

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What's up, guys? Let's gets more practice with Gasses Law by checking out this example together. So we're supposed to use gas Is law to figure out what the electric field is inside of a spherical conductor with some charge? That's negative. Q. Okay, so what I want to point out is that this conductor here remember that the property of a conductor means that any net charges will always move and distribute themselves along the surface of that conductor. So if you have some conductor right here and it has some net charge, negative Q. And basically all of these negative charges will always distribute themselves evenly among that surface because they want to maximize their distance away from each other. So now we also know what we're working with. The third type of problem of gasses law figuring out what the electric field is. So we have to choose a Gaussian surface with symmetry. Now, if you're working with a spherical sort of like conductor or spherical geometry thing, the best thing that you're gonna use is also a sphere. But we have to use this field. We have to figure what the field is inside of this conductor. So the Gaussian surface that I'm gonna choose right here is gonna be within the conductor. Okay, So, basically, how do we relate this gas is law to the electric field. We have that the total flux, which is e a times the cosine of theta is equal to the Cuban closed, so I can either use big cure Little Q Q closed over Epsilon. Not now. What happens here? I've had this surface. I've chosen this Gaussian surface within this conductor, and it's it's, um, arbitrary distance. So I'm choosing it could be here or here, over here somewhere. And I have to figure out what the enclosed amount of charge is now. We said that for a conductor, all of the charges, all the net charges will always distribute themselves evenly among the surface. So that means that the amount of charge that's enclosed within the surface here is zero. There is no charges here. There's no in charges within this Gaussian surface that I've done. There is a charge if I chosen this Gaussian surface outside, but that's not what we're asked, were asked to figure out what the inside electric field is, so that means that this whole entire term is just equal to it's just equal to zero because Q enclosed is equal to zero. Now what happens is that what does that mean for the electric field? What We know that the area is not equal to zero because the area of our gas and surface is just gonna be whatever we choose that surface to be. It could be. It's, um, Radius are something like that. So what that means is that the electric field has to be equal to zero, and that actually makes sense. Think about when we're talking about the electric fields inside of conductors. We said the electric field inside a conductor always has to be zero. So the gas is law actually helps reinforce this idea that if there are no charges within this Gaussian surface, then that means the electric field has to be zero. And that is the power Gasol's law. Let me know if you guys have any questions with this

3

Problem

Rank the flux through surfaces A, B and C in the figure below from greatest to smallest.

A

Φ

_{C}, Φ_{B}, Φ_{A}B

Φ

_{A}, Φ_{B}, Φ_{C}C

Φ

_{A}, Φ_{C}, Φ_{B}D

Φ

_{B}, Φ_{C}, Φ_{A}4

example

## Electric Field due to Hollow Shell

7m

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Hey, guys, let's go ahead and check out this problem. This is actually a very common type of problem that you'll see in Gaza's law. And it's very important that you know that. So watch this video a couple times, just in case you didn't catch everything. But basically what we're gonna dio is there gonna be evaluating with the electric field is in these particular three regions surrounding this conducting spherical shell. One of them is gonna be inside of the shell, but outside the point charge, one of them part B, is gonna be within this conducting spherical shell, and then part C is going to be outside outside of this radius, this outer radius be here. So if we're trying to figure out what the electric field is, then one thing we can use we have, We have charges and we can use gas is law. God's law helps us figure what the electric field is. But first, what we have to do is we have to draw a Gaussian surface and that Gaussian surface for a point charge. Hopefully, you guys know by now is going to be a perfect sphere. We covered in the last in the previous video. But basically the reason for that is that the electric field lines for this point charge point radiantly outwards. And that's also where the normals of the surfaces also points. So basically, these electric field lines and the normals always point in the same direction, no matter where you are on the Gaussian surface. So that means that if we're trying to evaluate the flux, which is going to tell us what the electric field is due to the charge that's enclosed, that's gonna be the electric field times the area of this Gaussian surface times the cosine of theta. But what happens is because thes things always point in the same direction. That means the cosine of the angle is just one, because data at anywhere along the Gaussian surface between the electric field and the area vector, since they always point in the same direction, that data is just equal to zero. So coastline zero just one. Now, gasses Law tells us that the total amount of flux through this Gaussian surface that I've constructed is equal to the charge and closed, divided by epsilon not, and what we're supposed to find is what this electric field is. So the only thing we have to dio is figure out what the area of our Gaussian surfaces. So I'm just gonna go ahead and say that the Gaussian surface has a radius off our little art. It's just some arbitrary thing I picked. If I picked it out here and drawn it a little bit wider, then that would be our right. So it's whatever basically you pick and the area of that surface is just four pi r squared because that's the surface area of a sphere. Now all we have to do is just move that area over to the other side and we get that the electric field is equal to Well, the charge enclosed is just gonna be this little Q right here. So it's little Q divided by Epsilon Not, but we have to divide it by this one over four pi r squared. So one of four pi r squared, and we can actually rewrite this as one over four pi epsilon, not times Q. Over R squared or another way that you might see that is just cake you over our square. All of these are valid representations of the electric field, which is exactly what you would expect for just a point charge, because that's the in charge. That's the enclosed charge of this Gaussian surface. It's just a point charge. Okay, so this is the area does the answer to part A in part B. We're now going to be looking at inside of the spherical conducting shell. So we have to construct another Gaussian surface, and the Gaussian surface is gonna be a sphere just like it was before. So now we can actually use a shortcut, because if we're trying to figure out the electric field and we know that we're inside of a conductor, remember how we talked about the electric field inside of a conductor is always equal to zero. This is an important fact that you must remember if you ever trying to value the electric field and you're inside of a conductor, that's going to be equal to zero. So some of you might be now confused. So if this is inside a conductor so this is actually answer, by the way. Now, some of you guys might be really confused because you say the galaxy in surface that we have just constructed right? If you imagine this Gaussian surface, what is the charge that's enclosed? It's this plus. Q. So how could the electric field possibly be? Zero? Well, what? Turns out what happens is that the enclosed charge is equal to zero. Because even though you have this positive point charge that exists in the center of this galaxy in surface, what ends up happening is that the electric field gets set up inside of this conductor so that there is a basically a distribution of charges that equals negative que that gets set up inside of the inside walls of the conductor. So what happens is you have a plus Q from the point charge, right, so this plus Q is the point charge that's at the center. But what happens is that point charge basically sets up an electric field inside of the conductor that cancels it out. So this negative que here is on the inside wall of this conductor. So that means that if you take a look at this Gaussian surface right now, the total amount of charge that's enclosed within this Gaussian surface is the negative. Q. That's distributed evenly on the walls of the conductor and the positive. Q. That's a dissenter of this Gaussian surface. So what's the total amount of charge that's enclosed? It's just zero. So that's why the electric field is zero. This is a very, very important thing. Watch that explanation over again. So the positive Que sets up a field inside of the conductor that cancels out the electric fields to the electric field to zero. And because of that, there's a distribution of negative que that gets set up on the inside walls. Now what happens is that due to charge conservation, the fact that you can't just generate a negative que out of nowhere, that means that a positive Q has to get set up on the outside walls. So this is what happens. It basically just polarizes these charges so that they cancel out the electric field and it's equal to zero. Alright, so go ahead and watch those videos on the electric field inside of conductor. Just so you make sense of that. Okay, so basically, the answer is that inside of the electric inside of the conductor, electric field is equal to zero, as it always should be. All right, in the last part. Now, as we have to figure out what the E at part C is so er greater than be right. So we have to set up our galaxy and surface. The galaxy and surface that we're gonna use is gonna be all the way out here, right? And it doesn't matter how far we make it. But what we do know is that this distance out here is just gonna be our And so this is the Gaussian surface that we've that we've constructed. So that means that the fi, the total amount of electric flux is e a co sign of data. Right? We know that data is always just gonna be equal to one. And all we have to do is just that this e a here is equal to the Cuban closed over Absalon. Not so. Let's take a look. What is the amount of charge that's enclosed within our Gaussian surface? We have a plus Q that gets set up on the outer wall of the conductor theme. Inside of the wall gets a negative Q. And then we have this positive Q. That's at the center of that sphere. So that means that this Q eyes actually let's see this Q and closed right here is just equal to Q minus. Q plus Q. So let me go ahead and remove myself here. So I've got Q minus Q plus Q. So that means that these charges just go away and you end up with just a positive Q over here. So that means. But the electric flux is just gonna equal to, Let's see, the area of the spherical Gaussian surface that we've constructed is four pi r squared. Just it always was. And that means that the electric field, actually let's see, means you wanna move this over. That means theological field is equal to plus Q over Epsilon not and then divided by one one of the four pi r squared. So that means that that's just equal to cake You over r squared, just as it is for a point charge. So it turns out that this spherical conductor doesn't actually really do anything other than set up a charge or set up a distribution of charges so that it cancels out the electric field inside of this. But then anywhere else after that, it basically just starts to look like a point. Charge again. Right. So make sure you go un understand each one of these parts here. By the way, this is actually supposed to be part C over here. And let me know if any one of these processes or steps confused you just drop me a link in the comments and I'll answer your questions. All right, let me know if you have any questions. See you later.

5

example

## Surface Charge Density

4m

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Alright, guys, welcome back. We're gonna work out this example together. We're supposed to figure out what the surface charge density is on the inner and outer surfaces. Now, before you watch this, just make sure that you've seen the last video on how to figure out what the electric fields are at different points throughout this conducting surface. Because you need to know what happens with the charges that are set up on the inside walls, things like that. Just make sure you watch the last video and you're good to go. So we do know from last video that due to gas is law. There's an electric field inside of here, the electric field zero within the conductor, and that exists again on the outside. And this is because whatever charge exists over here gets set up and basically distributed around the inner walls of the inside of this conductor. Here. We know that if this is three cool homes thing, this has to be a negative three cool arms that gets set up on the inner walls. And then on top of that, there has to be a positive three cool homes that gets set upon the outer walls because of charge conservation, right? So I'm just gonna use plus signs everywhere. So basically we're supposed to dio is what was supposed to figure out what the surface charge density is on these inner surfaces. Now that surface charge density is given by the letter Sigma and that Sigma has an equation. It's the total amount of charge divided by the total amount of area. So the I want to be very careful here because this Q is not the enclosed charge of like what we using Gaussian surfaces. So Q is not the Q enclosed. Where we're basically doing is we're just figuring out Okay, If all of these charges here get distributed on the surface of this tiny little thin shell right here on the inside walls of this conductor, then basically, what is the total amount of charge divided by the total amount of area here? So just don't make the mistake and saying that this is the total amount of enclosed charge that's different. So Thesiger MMA, we're gonna have to figure out what the charges are and divided by the areas. All right, so that means that sigma on the inside wall is gonna be the total amount of of charge that's on the inside wall. In other words, the negative three columns divided by the area of the inside wall. Now the area of this inside wall here is just equal to four pi. Times are inside squared and we actually have with the surfaces are we know that each of these we know this is three centimeters and this is five centimeters in the outer wall. In other words, a inside is just four pi, and now we have to do three centimeters. So we have to convert that we have to square that. So that means the area of the inside wall is going to be. I've got 1.13 times 10 to the minus two. That's in square meters. So basically, I can go ahead and just plug that in here and figure out what the surface charge density is. So Sigma on the inside is gonna be Q, which is negative three cool OEMs divided by 1.13 times 10 to the minus two and my sigma inside equals. Let's see, I've got negative 265 and that's gonna be cool OEMs per meter squared, right? So that's basically how you figure out what the surface charge density of the inside wall is now. To do the same thing on the outside wall is just gonna be the exact same process. So to figure out what Sigma out is that's just gonna be cube divided by a the outside wall. So this a of the outside is just gonna be four pi. And then now we just use instead of using three centimeters, we're just gonna use five centimeters. I'm gonna plug that in. So I've got 0.5 and then we're gonna square that. So that means that this outside area is just equal to, uh, let's see, I've got 3.14 times, 10 to the minus two, and that's just gonna go straight in this equation right here. Okay, So that means that the outside surface charge density is just the let's see, we've got three cool owns distribute on the outside divided by the area, which is 3.14 times 10 to the minus two, and that is equal to 95 cool OEMs per meter squared. Notice how this surface charge density ends up being positive, and this one ends up being negative because that actually has to do with the amount of charge that's divided by this area here. So those charges actually could be negative or positive. All right, let me know if you guys have any questions with this, I'll see you the next one.

6

Problem

A spherical, conducting shell has a charge of –6C. If a 4C charge were placed at the center of the shell, what is the electric field at 4 cm? At 12 cm?

A

E

E

_{4cm}= 2.25 × 10^{9}outwards;E

_{12cm}= 1.25 × 10^{8}inwardsB

E

E

_{4cm}= 2.25 × 10^{13}outwards;E

_{12cm}= 1.25 × 10^{13}outwardsC

E

E

_{4cm}= 2.25 × 10^{13}outwards;E

_{12cm}= 6.25 × 10^{12}outwardsD

E

E

_{4cm}= 2.25 × 10^{13}outwards;E

_{12cm}= 1.25 × 10^{12}inwardsAdditional resources for Gauss' Law

PRACTICE PROBLEMS AND ACTIVITIES (21)

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