1

concept

## Electric Flux

5m

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Hey guys. So for this video, I wanna talk about electric flux. It's a concept that is very important in electrostatics. You'll definitely need it to solve problems, especially when we start talking about Gauss's law. So let's go ahead and check it out. So basically, the flux of anything flux is just a measure of how much of something passes through a surface. The way I like to think about this is kind of like if this field here was like a river and you were to stick like a ring inside of it, how much water passes through the ring? That's sort of how I like to think about flux. So when we talk about electric flux, we're gonna be talking about the electric field and specifically how much of this electric field will pass through a surface. OK. So we've got these couple of examples right here, imagine these blue lines represent the electric field and this represents just a surface kind of like a ring I was just talking about. Well, in this situation, basically the flux is how much of these lines will pass through the surface. And in this case, we can say that this ring has sort of like caught all of these electric field lines. So that means the electric flux is going to be all or maximum or something like that. Whereas in this situation, now we're gonna have the ring, but instead of it being upright and all the field lines passing through it, what happens is that these electric field lines will pass directly over it. So imagine you were to turn that ring and instead of it being upright, you were to turn it on its side, no field lines would actually go through that ring that would kind of just go right over it or underneath it. So that means that the electric flux at this point is none or is nothing, there is no electric flux because there's nothing actually passing through the surface. Remember that the electric field that passes through the surface is defined as the electric flux. And then in this situation, we have somewhere in the middle. So some of the field lines are actually passing over it and under it. And then some of them are passing through. But at some angle here. And so in this case, the electric field line isn't an all or nothing, it's actually just some. So it's some partial amount of electric flux. So clearly what we've seen in these three examples is that the electric flux depends on the angle of the surface. Now, the way way we measure this angle is we say the electric field lines make some angle with something called the normal of the surface. And the way I like to think about the normal is if my hand, the back of my hand was a surface, then there is a vector that points directly perpendicular to that surface that's called the normal. And so since the normal is the perpendicular of that surface, then the electric flux is gonna be dependent on the angle that the electric field makes with that surface. And this angle here is measured between the electric fields which is the blue lines and the normal or the perpendicular vector of that surface. And if you have all those three things together, then the electric flux has an equation. It's gonna be E A times cosine of theta. Now there's some units associated with electric flux, you might not need to know them, but you can always get them back from the electric fields in the areas and things like that. And so if you have a bunch of surfaces together, not just one of them, you can calculate something called the total amount of flux, which is what we're gonna use later on in the chapter. So the total amount of flux through a closed surface is just going to be the sum of all the fluxes through the individual surfaces. Now, I want to be very, very careful here about how I explain this but a closed surface, you guys might be wondering what that is a closed surface is just sort of any boundary that encloses some volume. So the easiest one to think of is like a box. So imagine like this, right? So I have a box and yep. So that means that if there were some electric field lines sort of passing through this box, well, this box has six individual surfaces, right? So you have a flux that's going here, flux is going here here on the bottom, on the front. And then also I think I'm missing one uh somewhere over here, right? So you have these individual fluxes from these individual surfaces, but the closed surface represents sort of like that three dimensional object that I've made here. And so the total flux to calculate through this closed surface is just going to be the sum of all the fluxes through the individual surface. And when we're doing that and we're calculating total fluxes, um we know that we can have sometimes you can end up with positives and negative fluxes. Now, usually in physics, positives and negatives have to do with a direction. So let's go ahead and check out the two different cases, you're gonna get a positive flux whenever the electric field and the normal point in the same direction. Now why? Because if you take a look at this equation right here, if we say that these electric field lines are E and this normal or this normal, which we usually represent by a the area vector point in the same direction, then we know that the cosine of the angle is going to be zero. And what's cosine of zero, it's just positive one. So that means it's just gonna correspond to a positive flux. Whereas you're gonna get the opposite and negative flux whenever the electric field and the normal point in opposite directions. Now, you can probably guess why because in this case, the cosine of the angle is 100 and 80 degrees and cosine of 1 80 is negative one. So the way I like to think about this is if the electric field lines is going out of a surface, it's gonna be positive. But if the electric field lines are going inside of a surface, then it's going to be negative, right? That's basically the last thing I want you to know about electric fluxes. Let's go ahead and take a look at a quick example. So you've got the electric flux through each surface of a cube. So kind of like the example that I showed you above is given below. So what's the total flux through the cube? All we have to do is if you have the, if you have a six individual electric fluxes, then the net is just going to be the addition of all of them. So plus one, you know phi one plus P +52, all the way to 56. By the way, these Greek letters right here are the letter P five. So sometimes I'll say that. So basically all you have to do is just add all of these things up. The zeros don't contribute anything. So you just have to do 100 plus 20 minus 40 minus 80 and just go ahead and add all that stuff up. Well, 100 plus 20 is 1 20 and then negative 40 and negative 80 is negative 1 20. So that means that the nets electric flux here. So fine nets is just going to be equal to zero, right? And that's it. So that's basically how you would add up together these electric fluxes. Let's go ahead and take it a bunch more examples in the next coming videos. All right, let me know if you have any questions.

2

Problem

The electric flux through each surface of a cube is given below. Which surfaces of the cube does the electric field run parallel to?

Φ_{1} = 100 *Nm*^{2} /*C* Φ_{4} = 0 *Nm*^{2} /*C*

Φ_{2} = 20 *Nm*^{2} /*C* Φ_{5} = −40 *Nm*^{2} /*?*

Φ_{3} = 0 *Nm*^{2} /*C* Φ_{6} = −80 *Nm*^{2} /*?*

A

1 and 2

B

3 and 4

C

5 and 6

D

1, 2, 5, and 6

3

example

## Flux Through Angled Surface

2m

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Alright, guys, let's get some more practice with this electric flux stuff. So we have this electric field and we have the area of the surface and we're told to find with the magnitude of the electric flux is through the surface depicted. So let's go ahead and start with the equation. Three. Electric flux is just going to be e magnitude of the electric field times the area times the cosine of the angle feta. Now, if you take a look at this, we actually told what the electric field is. We have that and we're told the area of the surface is just 1 m squared and it's in the right units. So the only thing we have to find is we have to find it with the cosine of the angle is, And what I want you guys to do is remember that we were never We're dealing with co signs of angles, especially with this electric flux. This angle represents the angle between the area or the normal of the surface and the electric field. So if we sort of extend these electric field lines out like this than what we really need to Dio is we need to find what the normal of the surface is. The normal of the surface is always perpendicular to the surface itself. So it points out in that direction. And the reason we're told to find the magnitude of the electric flux is because we actually have no idea. We have no information of whether the normal is pointing this way or this way. So we're just gonna go ahead and assume that it's positive by calculating the magnitude of the electric flux anyways. So the angle that we really need is actually the angle between this vector and the normal of the surface, which is not the same as this 30 degrees. This 30 degrees is the angle between the electric field and the surface itself, not the perpendicular of that surface. So really, we're not using this 30 degrees. Instead, we're just going to use 30 plus 60 equals 90 so this angle is actually 60 degrees, which is this angle as well by the geometry, right? These things are all opposite angles. So what we really need to dio is that the electric flux is going to be 100 Newtons per Coolum. The area is just 1 m squared. And now we have to do just to co sign of 60 degrees. Now, cost of 60 is just one half. So that means the electric flux is just half of 100 which is 50. And that electric flux is Newton meters squared per cool. Um, just in case you needed the units. Alright, so that's the answer. Let me know if you guys have any questions.

4

example

## Flux Through Cube

3m

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Alright, guys, let's work this one out together. So we have a cube and its side length is two centimeters and it's placed in a uniformed electric fields. We've got two centimeters right here. We're supposed to figure out what the electric flux is through each side of the Cube. So let's go ahead and see what sort of visualize what's going on here. First we're gonna do is gonna draw the normal vectors for each one of those sides. So the top one has a normal vector like this. This one has the one that goes sort of like outwards like this. Sort of like out towards you. Think one has one to the right thing. One's got one to the left and then this one. There's some bottom one that's gonna be all the way down here That's gonna point that way. And there's gonna be a back one sort of on the back side of the Cube. Sort of like we can't see it right now. Um, Anyways, each one of these surfaces now may or may not have an electric field that goes through it, so we know we're gonna have to calculate the electric flux for all of these. So just remember that it's e a co sign of data for the electric flux. Okay, so clearly we can see that for the right side. There is definitely going to be some electric flux here, So the right side is just going to be e times a times the co sine of the angle. But in this case, we have the e A. The cosine of the angle is just going to be zero. And so this is just gonna be e times A. Whereas on the other side, we could basically make the opposite arguments that you have the same exact area because it's a cube. Except in this case, the flux on the left side is just gonna b e a times the co sign of 1 80. And so it's gonna be equal to negative E times a okay, and then everywhere else, let's go ahead and see what's going on for the top side. You have an electric field that points in this direction, and you have a normal vector that points in the upward direction. That means that the co sign of this angle right here that it makes is equal to zero. So that means that over here the flux at the top is just equal to zero. And by the way, if that's the case, then that means the flux in the bottom is also equal to zero because of the same reasoning. And now the only two of the rain on the front and the back side. So now you have an electric field that points in this direction, and now you have an angle, a normal that point straight out, as if I was pointing directly at you. So this angle is also still 90 degrees, even though it's just a different direction because of our three dimensional coordinate system. So that means that the fire of the front So that means that fire on the front which, by the way, is gonna be equal to the five. The flux on the back side because this is symmetrical, kind of, you know, a cube so on the backside is also going to be equal to zero. So in other words, we have to find the front to find the back. The bottom and the top are all equal to zero, and the Onley, too, surfaces that actually have and non zero flux going through them are gonna be defined. The left and the queue on the side, on the right side. So let's see, we've got this is gonna be equal to We've got 100 Newtons per Coolum and now we have a side length of two centimeters. So the areas over here, So each one of these little area vectors that was perfect. Eyes equal to two centimeters times two centimeters. That's gonna be four square centimeters. But you have to be very careful because any time you want to convert this thing 2 m, you have to actually shift the decimal places four times because you have to do this conversion four times. So this is actually gonna be 40.4 on that's gonna be in Meters Square. So the electric field is going to be negative times the area, which is 1000.4 and we get an electric flux that's equal to Let's see, we've got negative 0.4 So that is the electric flux over here on the on the left side and then on the right side, we just got the positive of that number, which is gonna be 0.4 Right? Because if you do the opposite on bits, just gonna be positive. Then that's gonna be your electric flux. Okay, so that's the flux is for each one of those things. Let me know if you guys have any questions.

5

Problem

Where does the normal vector point for a spherical shell?

A

Radially inward

B

Radially outward

C

Perpendicular to the radius

D

A normal vector does not exist

6

example

## Flux Through Spherical Shell due to Point Charge

3m

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Hey, guys. So for this example, we're gonna build off of something that we talked about in the last example. So were asked to find out what the electric flux is through a spherical shell of radius r do some point charge there that's in the center. So we're asked for the electric flux. Let's go ahead and start with our electric flux formula. We've got E a times the co sign of theta in which this data is between the normal vector and the electric field at that specific point. So the first thing is, first, where's the electric field? Points due to a point charge? Well, at any surface here, it always points away from that point charge. Remember, the electric field lines always points outwards, but at the same time, the normal vector of a spherical shell also always points outward directly at the surface. So this perpendicular lines here, the normals are always gonna point away from that spherical shell, which means that the cosine of the angle right here this data, wherever you look along the surface, thes these field lines and the normal always point in the same exact direction. Since and since the data is always equal to zero. That means this CO sign of data is always just going to be equal toe one, no matter where you was that you're looking at. So that means that the total amount of electric flux is gonna be the total amount of electric field times the total amount of area. So the electric field, let's see theoretic field due to a point charge is one or remember, it's k times what we're gonna use little Q right now divided by r squared. So at some are distance or actually, it's gonna be That's gonna be big r squared. So that's the electric field. Due to this point charge in the area of the spherical shell, the surface area of a sphere is just four pi times r squared. So if you go ahead and put those two things together, that means that the electric flux is gonna be K times Q divided by whoops. I got big r squared times four pi and I forgot to forget to make this little are a big are. So if you caught that, if you caught that I was supposed to write big are instead of little our That was good. So that's four pi big r squared. So anyways, I've got cake over r squared times 44 pi r squared. Now what happens is the r squared will cancel. And so we end up just getting that. The flux is equal to four pi times K Times Q. And this is the answer for pie. Que times Q. There's actually another way that we could write this because we know that this K has a relationship with that epsilon, not that permitted ity constant. Or remember that this K is equal to 1/4 pi times. Epsilon. Not the reason we wanna make this substitution is because now the four pies will cancel, so this actually will turn into this will just turn into five equals. Let's see, the K will turn into this. The four pies will cancel, and we'll just get Q divided by epsilon. Not this is actually really important is a really important result. We're gonna talk about it much later when we get to Galaxies law. That's just another way you could express this, by the way. So both of these things would actually be perfectly valid on this if you were given this on a test or anything like that. All right, So this is the answer, or this is the answer. Let me know if you guys have any questions.

7

Problem

What is the total flux through the two surfaces depicted in the following figure? Note that surface 1 has an area of 50 cm^{2} and surface 2 has an area of 100 cm^{2} , and E = 500 N/C.

A

513 N•m

^{2}/CB

3.74 N•m

^{2}/CC

3.16 N•m

^{2}/CD

5.13 N•m

^{2}/CE

5.44 N•m

^{2}/CAdditional resources for Electric Flux

PRACTICE PROBLEMS AND ACTIVITIES (3)

- A hemispherical surface with radius r in a region of uniform electric field E→ has its axis aligned parallel t...
- A flat sheet of paper of area 0.250 m2 is oriented so that the normal to the sheet is at an angle of 60° to a ...
- A flat sheet of paper of area 0.250 m2 is oriented so that the normal to the sheet is at an angle of 60° to a ...