[Music] so I'm going to walk through a complicated little uh scenario here where I have two lines of charge they're both of length L and I've got a position P out there it's at L and then halfway up in the y direction so it's at half l in the the y direction I'm going to ask where that that point p is what what is the field strength there if each of these red lines has a charge distribution Lambda spread over top of it so let's take this one thing at a time first thing I'm going to do is look at that vertical segment and the vertical segment runs from yal Nega L over2 to positive L over two if I'm considering my field point to be right at the center so I'll take that and split it up into little chunks of charge DQ and each one of those little chunks of charge DQ is going to create its own electric field de which is K * d Q over R 2 in the r hat Direction and it's this R hat that makes life a little tough for us because what we've got to do here is integrate that DQ the DQ extends up and down the Y AIS so I'm going to replace DQ with Lambda time Dy that should cover all the charges that spread on that red line but the r hat I still have to worry about so what I'm going to do is take that R hat and break the E into components so look at e subx and e suby what is d x and d y and to do that I needed to use a little trig here de X is going to be de * the cosine of that angle Theta and de Y is netive De * the sign of it the negative sign there is to denote that when Theta is positive on the left side de Y is negative on the right side the cosine is L over R the adjacent over the hypotenuse and the S is y over R the opposite over the hypotenuse so I can solve for de subx and D suby in terms of k k Lambda L and R cubed and Dy so let's take those guys and what we're going to do now is we're going to scan up and down or basically scan from bottom to top of that red line there from negative L over2 to positive L over2 which means we're going to integrate it we're going to take our de subx and our de suby and integrate them to figure out what the total electric field in the X Direction and the electric field in the y direction is let's start off with the E subx first so I'll take e subx and I'll point out first that K Lambda and L are all constants in this case they don't need to be inside the integral I'll also point out that R is really just L2 + y^2 where Y is varying in this integration so I've got an integral there and this is actually an integral I did for you before in terms of substituting some trig functions but for now I'm going to say let's just go grab a a table of integrals there's actually one in your textbook and I can look that one up and plug in an answer so I have K Lambda L and then 1 over l^2 y over the square < TK of l^2 + y^2 and I evaluate that between y1 and Y2 pull that l^ s out of the denominator inside there and plug in my y1 and Y2 and I get this form which I'll go ahead and pull aside here and kind of highlight for you because that is a general form for what the electric field would be if I'm standing a distance L away from a charge distribution lined up along the Y AIS from y1 to Y2 I haven't plugged in anything yet that is specific to our situation aside from the fact that it's a vertical charge distribution and I'm distance L away from it in our case y 1 is negative L over2 and Y2 is positive l over2 so I can plug those in and do a little bit of juggling by algebra here to figure out what e subx should be putting it in a standard form of the root of 5 over 5 * 2K Lambda over L okay now let's look at the E suby similar sort of situation here I'm going to take this integration I've got the K and Lambda still in there the one thing I can't pull out is that y though that y that's in the numerator has to stay behind so it's a slightly different different integral but still one I can just look up when I look it up I get that solution there and I plug in my values of y1 and Y2 and I get this as an answer and I'm going to bring that up and actually the first thing I'm going to do is multiply through by L over L and my reason for doing this is so that the thing out front the K Lambda over L is the same for both of those so now what I have in these two boxes are standard solutions for e subx and e suby for a charge density of Lambda stretched on the Y AIS from Y2 to y1 to Y 2 when I'm standing a distance L away from it these are totally General solutions for this kind of geometry that I've got here now I want to point out that i' I actually kind of did this before in a horizontal pattern and wound up with these cosine and sign functions I just want to show you how that relates again that e subx is K Lambda over L * the S of theta 2us sin Theta 1 if you look at that Y2 over the square root that's just the sin Theta 2 and y1 over theare root is just the S of theta 1 likewise on the bottom it's just going to be cosiness so in other words I can express this in terms of the Y's x's and L's all I want to or I can express it in terms of thetas the reason the thetas come in handy is if I were to take the thetas and go to very very large angles in other words if I made this line segment go to Infinity an infinitely long line then Theta 1 would go to 90° and Theta 2 would go to positive 90° and what I would have is e subx would just be 2 * K Lambda over L and E sub y would go zero that's kind of a standard geometry to do in an intro physics class well now that I've got what the field is and my vertical segment e subx isun 5 over 5 * 2K Lambda over L and E Sub yal 0 I'm going to go back and look at what the horizontal segment should do for me now so looking at that horizontal segment the geometry is slightly different I mean first off it's extending on the X Direction but also I'm going from negative L to zero my de is going to be K * lamb DX over R 2 R hat and when I go to Breaking this thing up into SS and cosiness the sign is going to go with the X component now and the cosine with the Y component I still have that negative sign on De X because again as Theta is positive down below de X is negative above and vice versa so I get Dex and Dey just like I did before the one thing I've done here is i instead of saying the distance to the line is L over2 I've replaced that with a just so that it makes the the algebra a little bit cleaner to do and also allows me to point out something when I'm done here so let's grab our de X and Dey and plug them into some integrals here and the first one I'm going to work on is ex so the ex when I integrate that now what I have is K negative K Lambda factors out front and I have this integral of xx2 + a^2 to 3 power yeah you'll notice that is actually the same integral that we did when we were messing with ey earlier so I wind up with an answer that looks similar except instead of having L's and y's in it I have A's and x's and I can pull this thing to the side multiply top and bottom by a and what I get is a standard form for the charge at some distance a away from a line along the x- axis now plugging in the numbers for our particular case that a is equal to l over2 and X1 is equal to negative L and X2 is equal to zero I get that answer for what e subx is now I'm GNA go ahead and solve for E sub y for this case as well e suby is the same integral I was doing for e subx in the vertical case and what I get is a similar looking answer again only swapping the X's for y's and the A's for L's this again is a general form for E suby when I'm standing near a line of charge on the x axis a distance a away so grabbing this and plugging in our particular values X2 = 0 X1 = l and a is equal to l over2 what I get is an answer for E sub y these exact answers for e and ex aren't really that interesting to me instead it's the equations that I'm most curious about here let's go ahead and stick these things together if I put together the vertical components and the horizontal components now what I have is the vertical was all in the X Direction in other words the field due to that vertical segment of charge over there only points in the X Direction where the horizontal uh um charge led to a field that pointed sort of up and to the to the right I put these two guys together and I get as a total field that's 2K Lambda over L * xat plus 2un 5 over 5 y hat there's a lot of algebra that went into cleaning these things up and I I guess you could do that on your own later on but I'm not really concerned about it the thing I want you to most take away from this is this right here that when I looked at the vertical segment of charge the one that ran up and down the Y AIS the the answers that I got for the electric field strength in the X Direction and Y Direction were in those blue boxes there so if I ran from y1 to Y2 with my charge and I was standing a distance L away each of those gave me what the the electric field was in The X and Y directions when it was horizontal I had the x-axis where my charge distribution now and I was standing a distance a away and what you'll notice is if you look at e suby in the green box and compare it to e subx in the blue box those are very similar looking equations where all I've swapped is x's and y's and A's and L's this makes sense standing a distance away from a charge should be something that I can just rotate my axes and it should work in any case these equations would work for a vertical or horizontal or in some sense any charge linear charge distributions when I'm standing some distance away from it and you don't necessarily need to rederive the integrals every time so much as you could start at this point and pick it up