[Music] so we're going to talk about uh how you go about calculating the field due to a whole bunch of charges if they're lined up in a particular distribution I've got an example here where I've got just a I don't know 25 or so charges lined up across the bottom of the screen there and I want to figure out what the field is at p and I could go about this the normal way of doing business which is to take each one of them and say okay it's a distance R away and I've got an R hat unit vector and I can determine k q over R 2 for that and then I can go to the next one and the next one and the next one and find out what the field vectors do to every single one of those guys is and add them all up the problem is this is going to be a vector addition it's going to be kind of sloppy and messy to do so if I would take each point charge one by one by one I've got myself a heck of a job by the time it's all said and done so the way to do this is actually to to pull out some calculus instead there's going to be some integrals and things like that flying about today so grab your grab your hats and hold on tight if you don't remember a lot about integration you might want to go back and refresh some of that to get it back into into your head right now but any case what we're going to do is we're going to take this line of charges and instead of considering a whole bunch of different charges we're going to consider it just to be one long band of charge where it's all equally distributed out we've painted it across the whole length of the thing and I've still got little boxes drawn there because I'm going to take it and break it up into chunks but in general this is just one long continuous charge distribution spread across maybe a piece of metal or something like that who knows in any case what I've got now is a whole bunch of charge spread from one end to the other of this bar of length L and I'll say that runs from X1 to X2 just so I can put some coordinates on it and I'm going to cut it up into little chunks that are DX wide and my reason for doing that is because then I can look at that little chunk right there and say what's the field do to that guy well to do that I got to figure out how much charge is in that guy right there and I'm talking about how much charge DQ is in one of those little blocks and if it's uniformly distributed and I have a total charge big Q over the whole thing then Q over L is how much it's spread apart and If I multiply that by DX that tells me how much I have in that one little chunk this Q over L is what we refer to as the linear charge density and we'll actually use the letter Lambda to denote this this is the density charge per length KS per meter of charge spread out on this red line and the DQ the amount of charge in that idual little chunk there is just Lambda time DX so what I can do is calculate the field due to that one bit of charge there K * DQ over R 2 in the r hat direction that really means I'm taking K * Lambda * DX over R 2 and if I want to add up all of them what I do is integrate summing all these red blocks up is the same as integrating over X so I change this into an integral from X1 to X2 of K Lambda over R 2 R hat d DX now this e is a vector remember so this integral that I've got written here is actually two equations in one I have to do it for the X component and the Y component and to make things even sloppier R depends on X and R hat depends on X so I've got things embedded in that integral that aren't obvious right off the bat so let's see if we can come up with some tricks to figure this out first thing I want to do is take that R hat and instead of working it in terms of the unit Vector let's just go ahead and pop it apart and get the X component in the y component the X component if I look that red Vector up top there is going to be related to the S of theta it's negative because it points backwards along the x- axis the Y component is going to be related to the cosine of theta so I have K Lambda * DX over r^ 2 * sin Theta for e subx and time cosine Theta for E sub y now the other thing I can do is look at the triangle I've drawn here and realize that the S of theta and the cosine of theta depend on X Y and R sin Theta is X over R cosine Theta is y over R sooa and all that so I can plug those back into these equations and I have e x is K Lambda time the integral of x r Cub DX and e y is K Lambda * the integral of y r Cub DX and the K Lambda I can pull out because those are constants and don't vary in the integral the r depends on X as x^2 + y^2 square rooted so R cubed in the denominator there became x^2 + y^2 to the three Hales power so now I've got these things written out explicitly in terms of just X the Y is a constant as the integral happens here so I could solve these I could go grab a table of integrals or something like that and work it out and that's where the book leaves you on this and and goes ahead and does that but I want to take it a step further and show you how this really works and to do that I'm going to backpedal I'm going to go back to where I had it in terms of the sign and the cosine so I have e subx is K Lambda the integral of sin Theta R2 DX and e y is K Lambda the integral of cosine Theta R2 DX now first let's take that that one in the bottom there the E the problem is that Theta depends on X or X depends on Theta whichever way you want to look at it and R depends on both as well so let's take R and solve that to say it's equal to Y cosine Theta now I can plug that into that bottom equation and into the top equation and I've gotten rid of the r for my integrals so now I have integrals that have thetas and x's in them and I have to figure out how the Theta and the X relate to each other well I can grab that sin Theta equal x over R up up top there but that's just going to lead me back into putting R back into the equations and I don't want to do that instead what I'm going to do is realize that the triangle also can show me that the tangent of theta is X over Y and now I can say that X is equal to Y tan Theta notice the only place X is showing up in these integrals is at the limits and also as the differential the DX so what I need to do is take DX which means DX is equal to y * D tan Theta and the derivative of tan Theta is secant squar so DX is y SEC s Theta D Theta secant squar is just the same thing as 1 over cosine squ so now I've got DX is y over cosine 2 Theta D Theta and I can substitute my DXs so let's grab that DX and put it into both these equations and they clean up kind of nicely I have sin Theta over y d Theta in in the X equation and I have cosine Cub Theta over y^2 DX in the in the uh in the Y equation which that cosine squ denominator cleans that up to be cosine Theta over y d Theta for e suby now remember the Y is a constant in this so I can pull that on out front and put it with the K and the Lambda so really all I have here is integrals of sin Theta D Theta and integrals of cosine Theta D Theta where Theta 1 and Theta 2 are the angles that I would go to to start and end this line of charge well the integral of sin Theta is just simply cosine Theta so the minus sign that was out front on E subx goes away and the integral of cosine Theta is just sin Theta so here's my if you will final answers on this e subx is equal to K Lambda over y * cosine Theta 2us cosine Theta 1 and e y is K Lambda y * sin Theta 2 - sin Theta 1 where Theta 1 and Theta 2 are the angles I I have to go back to to see the start and ends of this line of charge if you will so let's do a a kind of quick example of this let me let me consider if I'm standing right over the center of this line of charge one thing that's nice about this is I have an expectation if I'm at the center that the X component should go away I'm being push just as much by everything on my right side as I am on my left side and if I look at the angles Theta 1 and Theta 2 there notice Theta 1 is going to be a negative number because it's swept back there behind the the the axis so Theta 1 is NE of theta 2 and therefore cosine Theta 1 is the same as cosine Theta 2 and E subx just goes to zero now e sub y on the other hand the S of theta 1 is just the negative of the S of theta 2 so what happens is the they add together let me imagine for a second that I've got something where this this line of charge is really really really long in that case Theta 1 and Theta 2 are going toach 90° and the S of theta 1 is going to go to 1 and the S of theta 2 is going to go to one so therefore what's going to happen here is I'm going to have K * Lambda over y * 2 and E sub y is equal to 2 K Lambda over y the electric field points straight away from the line of charge and has a magnitude that drops off as one over y as I get farther and farther away this isn't 1 over R 2 anymore but then again it's because I have an infinitely long line of charge that I'm standing next to this this exact problem is done in the book in a slightly different way and if you don't follow what I'm doing here you may want to look at it there