 Okay, I gotta head start to save time So this is going to be the using Gauss's law to find the electric field due to a thin line of charge Which you which we already did we're gonna do it again. Okay, so here. I drew the line of charge twice Here it is looking at it from the side and if you had the if you had it coming straight out It would look like this and we need to sketch The pattern of the electric field so that we can then You know pick a fake surface to calculate the electric flux through So in this case we're trying to pick something that would give some type of symmetry That would make our integrations easier. I mean if I picked a box here It's not going to work because it will work. I take that back. It's not going to work easily Because it's not going to give us a simple integration and that's what we want, but if I pick I'm going to pick a cylinder just like the last case like that You can see that again. I'm going to get the electric fields perpendicular to the area Okay, and but in this case it was perpendicular to the I'm sorry parallel to the area vector And in this case the ends are going to have a zero flux Okay, but I'm going to pick the same shape so if I draw it like this and This is going to have a radius of r and then if I draw that same one up here So we're close to the We have to have a long rod and we have to be close to it so that we can assume and the electric field is constant like that If you have a if you have a finite rod these electric fields start pointing out and it it's not this true anymore Okay, so this is going to be a length L and A radius r. I'm not saying the area. I'm saying the radius now So again, we're going to have to do three fluxes the end caps plus the sides, but in this case Let's look at the ends Here I have the electric field like that So here in hat is that way in hat is that way on the end caps The flux e dot in hat is they're perpendicular. So the flux is zero So they did the flux on the end caps is zero. So all I need to worry about the sides so The flux on the side. It's going to be e Dot in hat da Okay, so again like I said if you look right here in hat is Perpendicular to the area pointing away and in all those locations. It's in the exact same direction as the electric field so when I take E dot in hat I just get e The scalar value the magnitude of e. So this is going to be e Da and so see we picked something Such that that would work. This also assumes that the electric field is constant around the magnitude constant Or I couldn't pull it out But if I pick a cylinder then these are all the same distance away, so they would have the same electric field Now I just have to find the area of that side and and that side if you unroll it is A rectangle so it has a length L and this is the circumference of the circle So this is just going to be e times l times 2 pi r Where r is just the radius of that cylinder So that's the that's the flux now how much charge is inside? Q in It's going to be If I look at the same thing if I say the ratio of the charge inside of here to the total length Which again, I'll call L prime which I messed up. I shouldn't have done that. Okay Then if I say Q in over L that'd be the total charge Q over the total length L prime You sometimes problems will give this as the linear charge density The charge per unit length, which is this whole thing Okay, so that's Q in so if I put this all together I get e 2 pi L r equals Q in is going to be L Q over L prime That's going to end so we get L Q over L prime and then we have an epsilon naught Okay, so now I want to solve for the magnitude of e first thing you can see is this length cancels with that I get e equals Q I'll put this Q over L prime up here over 2 pi r Epsilon naught and that's the electric field due to a line of charge R is still there. I didn't cancel so whatever the radius as I get the Radius of the cylinder larger the magnitude electric field goes down. That's exactly what we had before Okay, but again, that's the magnitude. It doesn't it's not the direction It's not the electric field vector So if I want to find the vector then I have to I have to know something about the configuration