 Okay, so I already said that the main focus of this chapter was electric fields due to charge distributions. So let me just go ahead and do an example. So here I have, I'll put this at the origin, here is a skinny rod of length L and total charge uniformly distributed of Q. I want to find the, we'll call this X. I want to find the electric field over here some distance X, like that. R I could call it R, it doesn't really matter. I probably should call it R. Okay, so how do we, how do we do this problem? Well, I mean there's no, there's no formula for the electric field due to a line charge. Well there is, but we don't want to start with that. So all we really know is electric field due to a point charge. So I know that if I have a point E equals 1 over 4 pi epsilon knot Q over R squared R hat. Okay, so where Q is the value of that point, the charge of that point charge and R is the distance from that, R hat's a vector pointing away from that. Okay, well I can't, what I can do is to say I don't have point charges but I can break this into point charges. So if I take this and cut it into a little slice of width DX, then I can say, okay, where, I should call this R, I've already seen now that I should call that R, sorry. I can find the electric field right here, I'll call it DE, due to just that one point. And then what I can do is just add up all the different electric fields due to all the different points, they're different distances away so they'll have different values. And then I can find the total electric field. I mean if you wanted to, you could really do a very simple approximation, say this is one point charge Q over 2, centered right here, and another one centered right there, and that wouldn't, it would be a not too terrible approximation. Of course they're broken into four, it'd be better, if they're broken into twenty it'd be even better. If I took the limit as the number of points went to infinity and the length of this went to zero, that would be really great. That's what an integration is. Okay, so I want to find the electric field due to this one piece called DE, integrate both sides, and then I'll have the total electric field. Okay, so let's call this distance R, this piece is thickness DX and at an X value of X. So what's the electric field due to that piece? Well, 1 over 4 pi, epsilon not, I need to know how much charge it is. Right now I'll just call that DEQ, and then how far away is it squared? Okay, well it's going to be, this is going to be R minus X, that's the distance away, and then what's the direction of the electric field? I'll just say X hat, okay, or I could write this as, I could write this as 1, 0, 0. So because I know that if I have a point charge there and this is positive then that's going to make an electric field going that way, so I can write it just like that. Now I can't integrate though. I can't integrate because here X changes, that's a variable that's changing, but my integration variable would be DEQ, that's not good. Okay, so what I can do is I can say if the charge is uniformly distributed along this rod, then the charge to length ratio for this little piece would be the same as the big piece, so I can say DX over DEQ equals L over Q. So now I can just solve this for DX, and I get DX equals L over Q, I'm sorry, I want to solve that for DEQ, so I'd say DEQ equals DX Q over L, and now I can put that in up here and I get, let me just write it as DE, and this is just the X component, the other stuff is 0, I think we can see that, so I'm going to get 1 over 4 pi epsilon knot, DEQ is going to be DX Q over L R minus X squared, and that's it. So now that I have the expression for the electric field due to that one point, I can integrate both sides, and I get the X component of the electric field is 1 over 4 pi epsilon knot, I can bring out this Q over L, and then I have the integral from X equals negative L over 2 to L over 2 of DX over R minus X quantity squared. Okay, so you see here, I'm adding up from here to there, if the whole thing has a length L, then over here this is X equals negative L over 2, and this is X equals L over 2. Okay, getting all, can you still hear me, I hope so, can you still, okay, so now I can just, I just need to do this integral, well I have DX over X over something squared, I'm going to have to do a U substitution, so let me say U equals R minus X, DU equals negative DX, R is a constant in this case, because I'm trying to find, it's kind of a variable in that at the end of the problem I want to be able to say I moved that wherever I want, but it's in this integration, it's a constant, okay, so that means that this integral becomes Q over 4 pi epsilon knot 1 over L, now I have, I can put U in here, and I'm going to leave the limits as X, so I'm not going to put those down there, so I have, instead of DX I have negative DU, and then I have U squared, now I can integrate that, okay, so that's the same as, if you want to use your rule here, U to the negative 2, and when I integrate that I get Q 4 pi epsilon knot 1 over L, I get U to the negative 1, and then I get negative 1 out front, right, no, I get, yeah, I want to take the derivative, right, okay, so I get, I get 1 over U, and I'm going to go ahead and put back in U because that's really what I wanted, so I'm going to get 1 over R minus X, and I need to evaluate that from negative L over 2 to L over 2, okay, so now we get E, X, just the X component, as a function of R, it's going to be Q over 4 pi epsilon knot 1 over L, and then I get 1 over R minus L over 2 minus 1 over R plus L over 2, I can simplify the sum, but I'm going to stop because I don't want this to go too long, but we can still check something, we can check, when we have something like this we can check the units first of all. The electric field due to a point charge is this constant and then Q over distance squared, so I have that constant, I have Q, and then on the bottom I have distance and distance, so it's distance squared, so it is the correct units. The other thing I can check is what happens if L goes to zero, what if it's zero length long, then I get, that's L, Q over L, it should go to the expression for a point charge, okay, well let me, what if I move really, really, really, really far away, the electric field should go to zero, so as R goes to infinity then no matter what this L is, this is going to go to zero, okay, so that's their first example. We could do also an example of the electric field right here, if we want to find the electric field right here we could set up an integral just like that, but it's going to be not a pretty integral, okay.