hello class professor anderson here let's say we want to calculate the electric field of a line charge using gauss's law so first off what does a line charge look like well it's quite simply a line of charge okay we usually denote the charge on it by a positive lambda where this thing is charge per unit length okay so some sort of charge density but it's in terms of linear length okay so just imagine you have a whole bunch of positive charges on that thing and those positive charges are now going to create an electric field well what does the electric field around it look like by symmetry we could say that the electric field must be pointing out in this direction okay but if this is a positive charge of course it's going away from all the positive charges everywhere and now you might ask yourself why is it going perfectly radially outward from this line charge how come it's not pointing to the right or the left and the answer is a little bit subtle it has to be an infinitely long line of charge okay if it is an infinitely long long line of charge then any region of that plus lambda will have a mirror image of another plus lambda over on the other side and therefore if i'm calculating the e field right here at this point what would i get i would get a little component going that way i would get a little component going that way and those two would add up to get me something perfectly vertical okay so this is the idea with this infinitely long line charge by symmetry everything has to be radially outwards it can't be pointing to the right or to the left it has to be radially outwards okay so that's good because it's going to simplify our problem quite a bit and let's see how that works remember when you have your picture and you write down gauss's law integral e dot d a equals q enclosed over epsilon not the next thing you want to do is draw a picture of the surface that you're looking at so let's try blue what would our surface look like remember you want to draw a surface that takes advantage of the symmetry this thing doesn't have spherical symmetry anymore because it's not a point charge it has to have cylindrical symmetry if i draw a cylinder around this thing the e field lines are poking everywhere through the sleeve of that cylinder here's the top of it right here okay let's draw it sort of the way we drew the last one where we had dashed lines for the e fields so if i dash this one out that means it's internal inside this can and it's poking out and so we'll put a little hole right there where it comes out of the cylinder the e field lines are coming out radially what about d a for this particular surface well d a is going to look like this it's a little chunk of this can and the d a is also radially outward from the center so e and da are in the exact same direction almost there right we're almost there except we have to worry about the closed surface right this is an integral over the entire closed surface and so we have to worry about the end caps the end caps will have a d a that is pointing in that direction and over here it will be pointing in that direction and those are orthogonal to the electric field and so we're going to see that they don't in fact contribute to our integral so what does the integral become we have two now integrals we have to integrate over the sleeve of e dot d a and sleeve just means this portion of the can but we have to also add the end caps and that's what the left side of this thing becomes everything on the same right side keeping closed over epsilon not all right this is looking pretty good right integral of e dot d a over the sleeve how do we do with that how do we deal with that well it is e d a cosine of the angle between e and d a at the sleeve which is zero degrees right they are parallel and so those two little vectors are in the same direction therefore phi is zero degrees and so that's going to cancel out real nicely what about this part right here e is radially outward but d is in this direction along the line of the charge and so we're going to get cosine of 90 degrees that one's going to go away entirely and then all of this is still equal to q enclosed over epsilon not all right this one becomes integral of e d a and let's just be clear that we're talking about integral over the sleeve now this one becomes 0 because cosine of 90 degrees is 0. it's all equal to q enclosed over epsilon not so let's deal with this and then we'll worry about q and close e is a constant if we are a fixed distance everywhere from that surface and we call that distance s then the magnitude of e has to be the same everywhere around that can right by symmetry it has to have the same value so e comes right out of the integral e times the integral of d a over that sleeve all right but we know that the area of the sleeve is very defined it's 2 pi s times how long the sleeve is so let's just give the sleeve a length l what is the area of that sleeve it is 2 pi s times l all right once around on that circle and then move that circle this distance l and that's the area all of that has to be equal to q enclosed over epsilon nine but q enclosed is related to this lambda of course because this was charge per unit length so how much total charge is in there it is just lambda times how long the wire is l and then we're still dividing by epsilon naught and now this is really cool right because the l's cancel out and look what we get we get e is equal to 1 over two pi epsilon naught times lambda over s again gauss's law gets you a scalar quantity it gets you a magnitude of e it doesn't tell you the direction for the direction you have to go back to the symmetry of the problem that we started with and so now we can write down the final result what is e in terms of a vector field it is 1 over 2 pi epsilon naught lambda over s and it's in the radial direction s hat okay this is the electric field of a line charge if its charge density is lambda all right hopefully that's clear cheers