[Music] so I want to introduce the idea of flux to you here and uh this is frequently deal with fluids but I'm I'm going to just kind of go at it just with electric Fields here imagine I've got a an area that somehow defined maybe it's a sheet of tissue paper or something like that and I'm going to consider the electric field that passes through that area so these purple arrows here represent the electric field lines and there's a certain number of those field lines that are going to pass through that area and it's also going to pass through at a certain angle flux is simply defined as the product of the the field the area and the cosine of the angle between them or more precisely it's an integral of the the field dotted with the the area where the dot product brings in that cosine Theta now that second definition there is something that if you've taken a multivariable calculus you might recognize but we're not going to quite get into it that far instead we're going to deal with things that are a lot more kind of straightforward than that in any case what this means is I have an electric field that points through this surface and if it's at some angle Theta compared to the perpendicular to the surface What's called the normal to the surface then I will have the flux being EA cosine Theta now what this means is the stronger the field is the greater the flux is but also the more the field points straight through the surface in in other words parallel to Da perpendicular to the surface if you will the greater the flux will be and way that a lot of people will think about this like I said is with fluids that if you say for instance held out a bubble wand with with a um uh bubble solution in it and you held it into the wind that if you held it facing into the wind so that the wind blew straight through it you would blow a bigger bubble there' be a larger amount of flux of the wind flow through that I don't like that a heck of a lot in this case though because it implies something's actually moving through the the area and in this case the electric field is not moving it's just there and it's cutting through this area and that's that's a sort of more subtle definition here in any case the flux is EA cosine Theta as far as we're concerned and I want to work out an example here to show you why it's important to us let me imagine first off I've got an infinitely long line running straight across the page and I'm going to ask what the field is a distance a away from it and this is a problem that we've kind of Beat to Death by now but let me go grab a couple old friends here the field as an X component when I'm a certain distance away from a infinite a line of charge on the x- axis and the Y component as a function of a as a line of charge on the x- axis as well the first of these two the ES subx if I look at it if x goes to Infinity in both directions in other words X1 is negative infinity and X2 is positive Infinity then what's going to happen is the denominators on those guys are going to blow up on me and it'll go 0 minus 0 which means there will be no field in the X Direction and this makes sense there's no reason to go left or right if all the charge is equally spaced to both sides of me e sub y on the other hand is going to be a little more subtle to deal with so let's take a look at e sub y I've got X2 2 over the < TK of a^2 + X2 2 - X1 over the < TK of A2 + X1 2 and I'm going to rewrite those in terms of the angles sin Theta 2 and sin Theta 1 I've drawn that angle out for you here and the reason I do this is because then when I take this line out to Infinity basically what I'm saying is that Theta is going to approach 90° in each Direction in other words Theta 1 will be90 or negative pi or negative pi over two and Theta 2 is going to be positive 90 or positive pi over two and therefore the S of pi over 2 is one the S of piun /2 is -1 and that thing in parentheses winds up being two so the electric field is just 2 K Lambda over a now what that means is that's the electric field strength at some distance a away notice I wasn't specific about where that was left or right so that means every spot a distance a away has the same field strength all the way out to Infinity in both directions of course you never quite reach infinity but you get the idea it also means it's the same thing on the opposite side a distance a away there I'm going have an electric field strength that's 2 K Lambda over a pointing the other direction and in fact if I swing this around and get an end view on it don't ask me how you get the end view of something that goes to Infinity but what you can see is the electric field points radially outward everywhere and every time I'm a distance a away from The Wire I'm going to have an electric field strength of 2 K Lambda over a so this points out sort of in a starburst if you will at any point on that dotted line Circle I'm going to have the field strength to be 2K Lambda over a now let's go back to looking at Edge on here and what I'm going to do is kind of slip a toilet paper tube over it if you will this toilet paper tube has a length of l and a radius of a so therefore everywhere on that tube I'm going to have an electric field that points straight outward with a magnitude of K 2K Lambda over a now this gives an easy way of calculating the flux to this thing because what that means is the electric field is everywhere perpendicular to the surface everywhere parallel to Da if you will and therefore the EA cosine Theta is just e * a in this case I'm just going to multiply the field strength times the area and better yet the field strength is the same thing everywhere so when I look at the area of this cardboard tube all I'm going to have is 2 pi a the circumference of the circle times the length of the tube I'm not counting the the the pi r squ the pi a squ I should say that is the end because there's no there's no electric field coming out of that it's all coming out of the the the round part of the tube so if that's the area and I multiply it by the electric field strength what I'm going to have is 4 Pi K Lambda * L and I'll look at 4 Pi k k is after all 1 over 4 Pi Epsilon KN and Lambda time L is just going to give me the the Q that's inside the tube so what I'll have is EA is Q over Epsilon not now this is actually a pretty fundamental thing what it is it's our first statement of gauss's law the flux that comes through any closed surface is proportional to the charge that is enclosed inside it and the proportionality is just this 1 over Epsilon KN in other words if I calculate five the flux over some closed surface in this last case the toilet paper tube then I'm going to just be measuring how much charge I've trapped inside that thing now this can rather complicated if the electric field is varying or if the area in the electric field don't line up well but if I deal with situations with really great symmetry such as my toilet paper tube in another example I'll do in a second here then what I can do is I can use this to actually work to my advantage to calculate the mag the magnitude of the electric field in other words if I know the electric field is all the same on some surface then I can use the charge inside to figure out what that electric field strength is let me give you a really really simple example something we know the answer to and that is let's think about just a point charge let me put a point charge there and I'm going to surround this with a a sphere a balloon if you will that has a radius of R now I'm not really surrounding with a sphere what I'm doing is I'm creating what's called a gaussian surface it's an imaginary surface that goes around this thing but on that surface that allows me to define a DA and the electric field is everywhere going to point outward and there's no reason this electric field is going to be stronger on the right or the left or the top or the bottom so what I've got here is a situation of great symmetry e is the same all the way around and it points the same way with respect to the surface all the way around namely perpendicular outward so therefore I'm just going to have e * a is equal to Q over Epsilon KN where Q is that charge sitting inside here the a in this case is the surface area of a sphere 4 pi r s and the E is going to be e subr the the r component the radial outward component of the electric field if there were a Theta or F component if this thing actually had electric field that turned One Direction or the other this would be a much more complicated problem but it also would be hard to explain because the charges at the center it should have spherical symmetry in any case let's take this 4 pi r 2 and divide it through I get e subr is Q over 4 Pi Epsilon r^ 2 and then finally 4 Pi Epsilon leads us back to k k q over r^ 2 e subr that's simply just just a restatement of the electric field as calculated for a point charge now one thing that's interesting about this that you probably wouldn't have guessed is all that really matters here is that I've got spherical Symmetry and that the charge is inside that dotted sphere that I've drawn there that Gan surface in other words if I don't have a point charge but instead I say for instance have a ball of charge that is all contained inside that dotted sphere but maintains spherical symmetry the answer is still the same so as long long as that sphere is all inside my sphere that I've drawn where I am standing out there at a distance R then the electric field is the same whether it's a point charge or whether it's a balloon that's blown up to meet me and this is not something most people would would guess right off the bat but it's one of the fundamental aspects of Gus's law that makes it really important in in electricity and magnetism and more important even in advanced electromagnetic Theory something you'd get to later on if you're if you stay in a physics career