 So today what we're going to do is we are going to, you'll notice there's no demo equipment today because today is We're going to just dig our hands into the guts of the cadaver that is calculus and just do it We're going to own it today. All right, so we are So we're going to solve problems involving complex distributions of charge That's going to be our task today, and I will attempt to demonstrate in itty-bitty steps how to do this I hope I will instill in you two things one not be afraid to look stuff up That is okay. Just tell me what you did. I used Wolfram Alpha to do this interval I looked in wikipedia and got this number just write your source down Okay, just say what you did in science. You don't have to reinvent the wheel every time you want to do something complicated You just have to see if anyone else has done it before and acknowledge those okay, and then just go ahead and use the Results so I want to drill that into you of course I expect you to be able to set up and solve problems. That's a life skill But you know the details of how you get that final number grinding through some hypergeometric function that you have to integrate I promise I'm not going to make you do that. I promise. Okay, those are all we make grad students do those and they hate us for it That's been solved. Okay, probably like 13 Russians 40 years ago went into Siberia and figured out for you So you don't have to do it anymore. Okay, somebody has done it All right, there are lots of smart people in the history of the world and we have books full of their information All right, we should use it and you should learn to use it and not be afraid to use it as long as you say what you did So let's go over the concepts again that we've kind of touched on so far These are the core ideas. We've walked away with it all fits on a slide. Okay? There are two kinds of electric charge positive and negative they exert forces on each other Positive negative electric charges are the sources or recipients of electric fields So by convention electric fields emanate from positive charges and they go into negative charges Okay, that's our picture and that's how we describe nature involving charges present in nature. It works very well Okay, electric fields and forces are vectors. That's because they have both a magnitude and a direction We saw that kind of dramatically last time with the Vendagraph generator I had that little piece of Mylar in my hand and as I moved it around the tip of the Mylar just kept pointing at that sphere Okay, you can really see the vector nature of the force there There's a direction and a magnitude to this thing So we describe it using an mathematical object called a vector to find the total electric field or electric force We add as vectors the electric fields or forces due to the individual charges So imagine you have ten charges, okay that you can do that by hand That's awful, but you can do it by hand to get the total force that the other nine exert on the tenth You would sum up the forces from the other nine on that one So you calculate the force of one on ten two on ten three on ten four on ten and so forth So you wind up getting a sum from I equals one to nine of the forces the Coulomb forces of each charge on that 10th one and similarly if you just take the 10th away and ask what's the electric field at the point where the 10th charge was Due to all the other nine charges again You would just write down the little point Coulomb electric field equation and you'd sum those up as vectors And you get a big vector at the end. Okay, we have magnitude. We have direction All right any questions on those concepts. We're going to exercise this more today. All right All right, so so let's do some announcements. So you're next assignment for class You're going to keep reading chapter 22. I think we're going to finish it out. We're going to start talking about The actual Effects that forces have on charges electric forces have on charges We're going to start throwing f equals ma into the next we're going to start looking at velocity and acceleration We're going to do all that 1307 stuff, but now we're going to use this force to do it Okay, you had a little sneak preview of that with the uranium nuclear Nucleus decay in the first homework problem set. Okay, we're going to build on that And one of the things we'll do is I introduced in the lecture video this thing called the electric dipole That's just fancy talk for I have a positive charge. I have a negative charge They have the same magnitude in the separated by some distance. Okay, so that's what a dipole is okay two poles dipole So one pole another pole equal magnitude charges opposite sign separated by distance. I Introduced to you a typical dipole that exists in nature dipoles are everywhere We owe our existence and are near death to them. Okay, so I gave the example in the video of The alveoli and the little layer of water that fills those sacks in your lungs And if you don't have a surfactant in your lungs to lower surface tension the dipole force in the molecules of water on the Surface of that thin layer are enough to prevent an infant from taking its first breath So as long as we collapse it would not be able to reinflate the dipole force is that strong and we have to secrete a chemical called a surfactant to overcome that Prematurely born infants don't have that so if you're born six seven eight weeks premature or more You do not naturally produce the surfactant It's one of the last things that's made by the body before you're born So artificial surfactant has been around for decades and it's simply sprayed into the lungs when the infant is born So that they can reinflate their lungs after they take after they exhale their first breath Okay, so it's not the inhale That's the problem. It's that first cry and then they can't reinflate their lungs and that's really bad Okay, so premature infant several kinds of issues physics plays a role and probably all of them in some way or another and specifically in this case the dipole force plays a significant role in your Chance of survival three seconds after you're born okay You have to overcome that dipole force the body has a way of doing that But if you're born too early, it's not ready yet and you can't so thankfully medicine has a way to overcome that We've learned how to do that. All right Okay, so this actually this statement here Well, we'll get to this one for his homework to is assigned today should already be up on Wiley You should have an electronic copy of it again. Look at your numbers in Wiley plus It's off the website and there is a problem again that I wrote myself tacked on to the end and I want you to Hand in solutions to that one too. Okay That's again doing a week Now here's the start point teams are final star as of today So starting tomorrow what I want you to start doing is meeting with your team once per week outside of class But I had to do a little last-minute reshuffle. We had a late drop in the class and There's been some consternation expressed about their late last reshuffle So I'll probably move people back around again. So I think it's teams Delta and Echo You'll be getting more emails from me. Okay, so we'll sort that all out so people are happy So let me let me say this if there's anybody on team Echo that wouldn't bind mind being moved to a different team Email me that will make my job a whole lot easier. Okay, so team Echo was the big one But we had a drop from another team I'd like to kind of get the teams all equally sized at the beginning of the of the course Of course, I can't predict future changes, of course But if anyone on team Echo would like to be moved to team Delta or wouldn't mind being moved to team To team Delta I gotta move somebody so just let me know that you know, oh, I don't care. It's fine. I'm cool Whatever I'll do whatever you tell me. I'm you know sheeple. It's fine. So all right So your job remember your homework for next week next Tuesday is just to select a team name. Oh, yeah, Sophie Yeah, I what I'm gonna do today is I'm gonna email all the teams so that everyone has everyone's email address. Yeah Yes, so I will comment next week on team dynamics and Leadership and then we'll address some issues about people who are contributing and people who are not contributing and we'll talk a little bit about that Okay, all right, because you know, we're all human beings. We all commit different levels of things to activities It's fine to get it, but I gotta give the teams a way of dealing with this, right? So So you're gonna select a team name remember the guidelines. I'm gonna urban dictionary things I don't recognize email it to me and everybody on the team has to agree So whoever emails it to me is see seeing all the other members of the team I'll get you your email lists for your teams today. I Want when you meet okay, so some time between now and next Friday you're gonna meet as a team in person face-to-face And what I want you to just chat about is how does anything you've learned in this class or even the previous class inform? One idea one avenue of a solution That you could apply to the grand challenge problem. Okay, that's it That's all I want you to do in your first meeting and if you come up with nothing No problem. You got other meetings and we're gonna talk as a team to mentor Kind of thing at the end of September. Okay Any questions? No, okay. Well, then let's do quizzing so that we can get to Nothing Okay So We're gonna check a claim At the beginning here But let's go through the quiz questions first and discuss the the answers to those so let's look at the first one as Mentioned in the video lecture. What is the approximate typical number of charges in a macroscopic a human scale something in this room? Blob matter. Okay, is it one a million? Okay, two a billion Three whatever the bad number is But really I have no idea what that is. I just wrote it down four of a God knows number Okay. Yeah, all right. So many of you went for a God knows number and yeah, that's the number And I think I think I actually I put it wrong in the video I think it's 6.5 times in the 23 in the video So I'm gonna have to go in and do some careful video editing on that one and re-upload that YouTube video But that ain't happening anytime soon. So I promise you that's the correct number if I did say 10 to the 23 I apologize and you know what? Yep, this is the jar. This is the famous error jar. I owe $4 into this now. I already put one in and took one off the I owe you last week Oh my lord, okay, so that's the mistake either way you're getting a buck out of this So I'll fix that in the thing. Okay So in that case then we'll just have to correct by factor of 10 something I write on a later slide No problem. I should really not second-guess myself at 730 in the morning. That's a bad tactic So I'm gonna recommend everybody just sleep past 730 in the morning and don't do what I do get up early Okay, good. Yeah, I've got at least one take around that. All right. Good. All right So that will get fixed and I will not have to do any careful editing in the YouTube video And if this is all recorded so all my colleagues can see why I showed up my PhD taken away from it All right, so question two, which of these is a true statement about the electric dipole moment? One it's total magnitude is given by the magnitude of either charge q in the dipole takers for one To its total magnitude is given by the separation d between the two charges takers for two Three it points from the positive charge to the negative charge with magnitude qd q times d and takers for three Okay, and four it points from the negative charge to the positive charge with magnitude q times d Lots of takers for that. Yeah indeed group think worked here as well All right, so it's number four. It's um it falls out of the derivation But the convention if you just want to memorize it is that this is one of the weird things where something starts on a negative Goes to a positive dipole moments you start in a negative charge and go to the positive charge And that's that p vector which is equal to q times d vector and d just points from the negative charge to the positive charge in the problem, okay All right So how do we determine the electric field of an arbitrarily shaped large number of charges one? We always do the sum by hand putting in one term for each charge even if this takes forever No one's going to jump on that one. Okay, all right to we utilize integral calculus to perform a large sum more quickly Few people all right fine three. We can always just use Coulomb's law for the electric field and stick in the total charge q No one fell for that. Okay for we can't there's no known way to handle large numbers of charges. Oh, thank God Okay, good. All right. Yeah, it's two. We're going to use integral calculus. In fact, we're going to exercise that today Okay All right now I made a claim which I've misstated and then restated correctly Thanks to help from all of you and that is that this Apogadro's number has something to do with sort of terrestrially scaled things All right, so so let's kind of test that claim a little bit right at the beginning All right, so we're going to solve problems first of all involving Complex charge distributions. That's our goal for today and I'll demonstrate how to set up and solve a Basic problem along this line and I'll give you a variation on that problem to try to do yourself. Okay So let's talk about copper. All right, let's you know, we're thinking about matter We're thinking about charges, you know copper is made of copper atoms We know that now thanks to hundreds of years of study of matter You know, this is a actually a very quite lovely picture of lots of copper wire laying around in little spools Okay, a little copper tubing something like that So each of these is you know a few meters long and and you know a few maybe 10 millimeters in diameter something like that It's pretty much solid copper for the most part. So it's it's fairly pure copper Coppers quite valuable as a commodity these days Well during the housing bust When houses were abandoned one of the most common things that was stolen from abandoned homes There's copper wiring and copper piping because if you sell that for scrap on the black market or really any market Just to scrap you can make a lot of money very quickly Copper is very valuable per unit, you know mass basically it's because we need it so much for Electrical devices these days it's so commonly used everywhere in computers and wall wiring and plumbing because of its thermal conductivity Which is very much related to its electrical conductivity. It's a very valuable metal. All right so if If you have some net charge big cue that's deposited on the copper wire or copper is a very very good conductor So if you put free charges onto neutral copper They will be free to move anywhere they want in the copper material And they'll kind of move until they come to an equilibrium place where all the charges are equally pushing on one another And they can't move anymore, and they stop so you have to wait a little bit Maybe a billionth of a second or a millionth of a second or something like that But eventually the charges will settle out an equilibrium and they'll be distributed over the copper Primarily on the surface, and then they'll just stop moving they can't leave the copper But they can move freely on it or through it, and they'll do that until they come to equilibrium Now that charge cue that we've deposited on here It could be made from some huge number of elementary point charges, and you know remember the elementary charge is about 1.6 times 10 to the minus 19 coulombs, which is a tiny number Okay Now are there really typically like Avogadro's number or thereabouts of like atoms and charges just in stuff That we find around us all the time so this morning using the wrong number for Avogadro's number It was we'll correct that in a second. I went and check that out. Okay. I mean I've checked this before I've done it a variety of ways this will come up again in the course later with an exercise we do later on But you can just check this you can you know back to the envelope very quickly on a small piece of paper Kind of look up some numbers and work out is this true All right, so as the famous physicist Richard Feynman once said shut up and calculate if you want to answer this question Just get some numbers and do some math. All right, so let's get some numbers and do some math All right, so let's consider a lamp cord right typical cord, you know much like this camera Cable here. It's got some braided copper wire inside of it That's used to carry information from the camera back to the device and actually some information from the Device back to the camera because there's a little light on there that comes on when it's recording and so forth. Okay Now, you know imagine that the typical lamp cord is something like I don't know a couple of meters long So I said two for the purposes of this problem I just pulled these numbers out of nowhere this morning. Okay, and it's entirely made from copper metal keep this nice and simple It's pure copper So the typical thickness of like an electrical cord and you can look this up if you want on the internet But in the United States we use something called the American wire gauge Which is a bunch of standard numbers that represent diameter thicknesses of wire and it's a very standard thickness for house wiring is 14 AWG American wire gauge and that corresponds to a roughly 1.6 to 8 millimeter diameter All right So house wiring if you were to you know got the walls here and look at the electrical cables behind the wall You'll find that the majority of them in this building are 14 AWG We'll figure out why that is a little bit down the line with some homework But that's a very standard number in American construction and you'll see why later. Okay, it actually has a physics reason for that Now the density of copper is 8.96 grams per centimeter cubed at standard temperature and pressure Do you think I just know that number off the top of my head? No good Weston, right? Yeah So how do you think I got that number? I googled it That's exactly right, and I found Wikipedia and Wikipedia told me the number Okay, you know Wikipedia can be wrong for some some political things, but for like science things. It's basically pretty solid So don't be afraid to use Wikipedia to look up basic information about the cosmos like the density of copper There's not a lot of debate about that anymore in our society. Okay, because you can you know measure it and stuff All right, so I just looked that number up Of course that probably needs to be converted to kilograms per meter cubed And I think if you do that it comes out to 89 60 kilograms per meter cubed. Okay, so it's pretty heavy It's a lot of kilograms for a meter cube the copper copper is not white Okay, if you've ever tried to pick up any significant amount of copper, you'll know that okay, so We've got a density, but we want to get a mass and once we have a mass if we know the mass of a copper atom We can figure out how many copper atoms are in that cord So we got to get the volume of the cord. We got to use that with the density to get the mass of the cord Okay, so the volume of the cord if we just Approximated as a long two-meter cylinder with a cross-sectional area of radius one half the diameter You can plug in numbers and solve so volume will be area times length and That's for a perfect circle going to be pi r squared times l and then if you plug in the numbers You should get if I did this right four point one six times ten of the minus six meters cubed so a teeny tiny volume Not a big surprise. It's a very thin cord. It's long, but it's very small. Okay, so it's small in meters cubed I mean a cubic meter is you know big, right? The neck cord doesn't fill a cubic meter. Okay All right So we have that and we have that the total mass then of the copper metal is just the density which is Kilograms per meter cubed times meters cubed will give us the mass of that cable And we find out that the mass of that cable is about you know 37 grams so 0.0373 kilograms So not that heavy, but you'd be able to feel that in your hands. Okay All right, so carrying that number with us on to the next salt slide So that's just repeating when I wrote on the last slide right there We can keep going if we want to find the total number of copper atoms in this cable We've got to divide by the mass of an individual copper atom and that will give us the total number All right, so the mass of a single copper atom is sixty three point five four six atomic mass units or am you? Now did you know you can do numerical conversions on Google? How many people knew that? Okay a lot actually yeah, so you just you can even talk it into Google, but I just typed it I said sixty three point five four six am you in kg and it converts immediately to one point oh five five times ten of the Minus 25 kilograms so what Google has done is behind the scenes they're running a program called units UNITS It's a standard program on the Linux or macOS Unix operating system So if you have a Mac computer and you pop up a terminal and type units you'll get a little prompt And it'll say what do you want? What do you have and what do you want? And you can convert just about anything to anything else and if you try to convert like meters cubed to meters squared It will tell you what the correct conversion factor is to handle that okay. It's pretty cool And it's got hundreds if not thousands of units that you can transform So if you need to get from one unit to another and you don't trust yourself going from centimeters cubed to meters cubed in Your head or even on a piece of paper check it with units and see what you get okay? Check it with Google and see what you get All right, so so that's the number I got then if I just divide the mass of the wire with a mass of a single copper atom I find out that I have three point five times ten to the twenty three copper atoms Which is within a factor of two Or thereabouts of Avogadro's number right correct me if I'm wrong six point oh two times Yeah, you're not in your head Western because you know such on such an idiot that I screwed that up This is good. All right. Yeah, so that's within about a factor of two six point oh two times ten to the twenty third Which is Avogadro's number I learned today Okay So I said 20 here because I'm a moron and I'll fix that in the slides before I upload them all right All right, so that ain't bad for approximating factor or two You know for that many things in something we just kind of winged it on in terms of I just went for two meters long It's got 14 AWG. Here's the mass of a copper atom go to town. Okay So if you then ask well fine great So how many charges are in there how many electrons or how many protons you can look that up in wikipedia to a stable copper has 29 electrons and 29 protons so the Yeah, I think that's right the total number of electrons. I really shouldn't do things before eight in the morning a Total number of electrons or protons then was just 29 times the number of copper atoms and that comes out to be one ten to the 25 Okay, and that's a factor of that's again within a factor of about 20 or so of Avogadro's number, so that's pretty good for this kind of quick quick work So when somebody asks you how many atoms are in that table you can just say what about Avogadro's number? How many atoms are in that bottle about Avogadro's number? It's basically right to within some factor of 10 or 20 or two or three or something like that I mean you basically be right and that's just because we're Macroscopically sized in this number is meant to represent the rough number of atoms in a Macroscopically sized objects that no accident. Okay by the way that Avogadro's number is defined All right, so that claim kind of holds some water All right, and we can use that especially going forward if we're going to estimate things if somebody Asked you to estimate something about a piece of material and you don't know how much stuff is in it You can just wing out Avogadro's number and go from there and just say look I assumed Avogadro's number We can correct that later if I'm wrong All right, so what happens if you have Avogadro's number worth of charges that you have to add up and get an electric field for that's the problem We're going to tackle today, and we're going to do it using integral calculus integral calculus is just a shortcut trick using the basic ideas of calculus to Divide up a problem into many many equally sized small pieces Add them all up send the piece size to zero and see if you get a finite result a non infinite or non zero result when you're done Okay, so what we're going to do is we are going to look at sort of a lamp cord problem again All right, so now imagine I have a bear lamp cord the copper is just exposed and I it's infinite in length All right, and so why do I get to do nonsense like that? Why do I get to say infinite? Well, you can approximate very long things especially if the scale of everything else in the problem is teeny tiny compared to it You can approximate the size of the big thing as infinity and see what kind of answer you get and then later on if somebody says well There's no such thing as infinity in nature at least not that we've really encountered so far, so Really, it's finite in size. What happens if you assume it's actually finite in length It's only 10 meters long instead of infinity long well as long as the the problem You were originally kind of considering how to small scale in a big scale and that big scale is way bigger than the small one You'll find out that the corrections to your assumption are fairly tiny less maybe less than a percent level So who cares at that point and why the plus won't even care at that point Okay, I won't care that you're off by tens of a percent it cares that you're off by a percent all right So calculus lets us Along with just sort of the basic tricks of mathematics the little toolkit of mathematics It lets us take a seemingly complicated problem bust it into many equally sized small pieces Add up all the pieces and get a result And if you tried to do this by hand, which you're welcome to try doing adding Avogadro's number worth of charges together If you try to do it by hand, you'd never solve the problem You can't count to Avogadro's number in your lifetime even if you count it as fast as you humanly possibly could okay, so Let's see if we can tackle a problem that involves a large number of charges and throw calculus at the problem and try to make progress That way so this is the picture We have deposited a charge q Uniformly on an infinitely long copper wire all right, so we throw this charge onto the copper wire We wait a moment. We let it all settle out into equilibrium So it's nice and uniformly distributed over this this shape this line this copper wire Okay, and now what we want to do is we want to just pick a point in space Call it P point P is going to come up all the time all right We're gonna pick point point P and we're gonna ask what is the electric field due to all that uniformly distributed charge at that point P All right Well if this was just a point charge with magnitude q located here, you'd know exactly what to do What would you do anyone anyone want to take a crack at that? Winston If you just had a point charge sitting below the point P and you wanted to get the electric field What would you bust out to do that? Coulomb's law yeah Coulomb's law now ask for points You always are going to use some some version of Coulomb's law and for the electric field I'll just jot this down here the electric field for a point charge is equal to k Times the charge of the point charge q over your distance from the point charge squared Times a unit vector that points from the charge q to where you are observing the electric field So that's Coulomb's law for the electric field It's very similar to the original Coulomb's law for force But we've divided out the second charge the one that would be sitting at point P the test charge Let's keep that in mind. Let's hold on to that equation. I'll leave it up on the board Okay, we're gonna we're gonna need that again in a moment We don't have a point charge We have a bunch of point charges of the Avogadro's number worth of them distributed Uniformly over this very thin copper wire Okay, so we're going to treat this copper wire as Infinitesimally thin that is it's only one dimensional it has no extent in the vertical direction or out of the page It only has extent in the horizontal direction. So we'll treat this as a perfect one dimensional thing Okay, an infinite straight line Now when you uniformly distribute charge over something you can ask well What's the density of charge per unit dimension of the problem? And the reason it's useful to do that is Because this is often a constant in problems that you'll get for this course Okay, so you can ask in this case because you have a line about the linear Charge density This is denoted by the Greek letter lambda lowercase lambda Okay, it kind of looks like a upside-down y and This is just equal to charge per unit length Okay, so it has units of Coulomb's per meter If you have a two-dimensional surface and you have a uniform charge density on that you have a surface Charge density then it's Coulomb's per meter squared if you have a volume It's Coulomb's per meter cubed and that's denoted by sigma the Greek little lowercase Greek letters. Sorry. That's a denote by row Sigma is the surface charge density All right, well we can ask since this is a uniformly distributed total charge q Over the length of this thing whatever it is. Okay, which we're going to approximate is infinite Let's for a moment just say that the length is some big number L whatever that is So I would write this as q over that big length L which later on I'm going to approximate is infinite. Okay, no matter what scale I Look at that line if I chop off a Little piece of the line right here Okay Let's call that Delta X So if instead I look at all the charge that's distributed on a little piece of the line Delta X I Then add up all the little charges in that little length Delta X and I'll get some piece of this thing We'll call it Delta Delta q Okay, if this is a uniform distribution I should find that this the total charge total linear charge density Because this is a uniform distribution and this thing is a constant no matter what scale I look at it It will be the same number So if I pick a small piece of the line and look at all the charge in that small piece I should still find that this equals Lambda No matter what scale I look at a uniform charge distribution. I always get lambda for a line This is a helpful trick. This is how we're going to attack this problem later on This is how we're going to go from one little charge to understanding something about the whole distribution of charge Okay, so let's keep that in mind linear charge density is a constant no matter what scale we look at this problem Okay well, let's Let's go deeper Let me take a little section of the line here So this thing shoots off the positive infinity over the right. It shoots off the negative infinity over to the left okay, so I'm going to take a look at an infinitesimal tiny little piece of this So let's just kind of denote that as This thing here. All right, this little sort of shaded thing here All right, so this little thing here is an infinitesimally small piece of the line This is known as a differential Okay in calculus speak And it has a notation that goes with it If we are denoting this for instance Here as the x-axis, okay Well, let's put our line of copper with its charge on it on the x-axis The little piece of length here will be DX Now that's not D multiplied times X. It's a singular symbol and it means differential of X Okay So that's a single symbol now Sometimes you can manipulate these differentials like their for instance parts of a fraction People often get themselves into trouble with derivatives by thinking that like numerators and denominators will just cancel for no reason whatsoever There are rules With derivatives and there are rules with integrals that can often make the problem simplify as if it were Involving little fractions, but you have to be careful to use the rules before you just go blindly canceling things out, right? So if you see something like this DX over X do not do that. Don't say oh hey the X cancels out. I'm left with D Don't we're about to get into very bad territory at that point Okay, and then then I'll have to then we'll have to meet outside a class and figure out what your solution went Like that right afterward. Okay So treat that as a singular symbol a single unit. All right, just as you would say delta X Okay, delta X is a symbol that means a small change in X You don't cancel the X you don't put delta X over X and cancel the X's and we shouldn't do that here either Now this little piece of the line this little differential of the line DX carries with it a little piece of the total charge differential Q little DQ okay, and Let's imagine that we've looked at such a tiny piece of the line that this DQ Is a point charge For all intents and purposes. We've zoomed in so far. We're looking at like a single charge Inside the matter inside the material That DQ represents a teeny tiny itsy-bitsy little point charge Right, maybe an elementary charge with size 1.6 times ten of the minus 19 or something like that. Okay, really small So what we're gonna do is we're not gonna bust this line into a huge number of teeny tiny little pieces We're gonna look at one representative piece We're gonna figure out all the stuff about it just like you do with Coulombs law you have to figure out Q and R squared and our hat You're gonna figure out what those pieces look like for the representative part of the line And then you're gonna add them all up at the end using integral calculus. All right, you're just gonna integrate You're gonna sum them all up So let's do that That little DQ let me redraw the line over here, okay So here is my little DQ it's DX and has a little charge DQ and Let's say Here's my point E Now it's a little charge a little charges make a little electric field Okay, so I know that if I ignore all the other pieces for a second that little piece of charge is Going to make a tiny little electric field up here at P And if I could just figure out all the little electric fields Contributed by all the little pieces at point P. I could add them all up and get the total I could do the superposition principle add up all the electric field as vectors get the total field Okay, that is our strategy That defines our strategy So let me begin by dropping some coordinates on the problem. Here's the y-axis We've already said that the line is the x-axis Okay, and I've decided to put the origin of my coordinate system straight below point P. So point P lies on the y-axis Let's keep going. We have a few more things we can draw here. All right So we're going to have an R vector That goes from the thing emitting the electric field to the place where we're measuring it That's the convention our vector our hat they point from the thing emitting the electric field To the thing to the place you're measuring that field All right, so let's just go ahead and draw an arrow There we go. That is the R vector for that little piece dq Just that one Does every piece on this line have the exact same R vector? Why not why not? Because they move right so some of those pieces are going to be really close to point P like the one dead dead underneath it That's going to be the closest any piece gets some of them are going to be really far away Okay, so that hypotenuse of that right triangle you can kind of see there the The height of it is the y The length of it is the x and there's the hypotenuse of the right triangle that hypotenuse is going to stretch as You change position Well, let's dig just an inch deeper into the ice on this one. Let's see what lies beneath What is it that changes when you pick a different charge on the line? Does the y coordinate change does the x coordinate change which is it Weston? Which do you think you think the x coordinate changes? Yeah, and you can you can kind of see that So if I pick another piece, let's say way over here Okay, and just for a moment I draw it's our vector I'm going to erase this in a second so we don't get confused We see that it still has a component whose height is still Why the distance of the point above the horizontal axis? But this x is different than this x so all that changes as you pick another Thing on the line is the x coordinate shifts This is helpful because it helps us think ahead that eventually what we're going to have to do is set up an integral Where we sum over x coordinates over the pieces in x y is going to be a constant y never changes Why should not be part of your integral? Okay, because y is a constant it never moves unless you change the point P Manually and move it here, but then all the y's are still the same for all the pieces All right So you're free later on once you get the answer to move y around you can make y 1 meter 2 meters half a centimeter Whatever, but for the purposes of the integral y doesn't change because when we're adding things up. We're adding things up only along x This is helpful. Okay, this will help simplify our solution later And so let me let piece get some space back for myself here Okay So we have a tiny little electric field contributed by a tiny little charge I'm going to write that tiny little electric field as differential e vector Equals K. It's due to the dq the dq is our squared distance away from the point P And there's an our hat vector for that dq that points from the charge to the point P The next step with a Coulomb's law problem once you've got your coordinate system set up and you've written down Coulomb's law is to do what what are we how we're going to attack this this formula? I just wrote down any ideas Plug in what we know. Okay. Do we know K? Yeah, it's a constant never changes, right? Now dq. Well, we just said dq is a piece of the line, okay, but We've also got ours and our hats and we don't know how to relate Like charge and distance so that eventually we have something consistent. We can sum over here I mean right now we're kind of summing over charges and each of them is at a different geometric location But I don't know how to get dq and R to talk to each other so I can add this up in a consistent Covenant way. I don't know how to add up dq is when I've got ours floating around So we've got to find a way to relate R and q all right. That's that's a problem. We don't know how to solve yet So let's see if we can attack each of these pieces and By doing so come up with a relationship a geometric relationship between what charge you're looking at and where it is Okay, so let's start with R squared That's usually the easiest piece to deal with it's just going to be the length of a hypotenuse squared Well, I've written my coordinate system. I already told you before you've got a right triangle with a height y a base length x Okay, and a hypotenuse there all right, so if I draw the components of our vector For the piece. I've shown you here. It has a piece that points to the right and a piece that points up and The length of that is just going to be The length of it squared is going to be the x-coordinate squared Plus the y-coordinate squared That's really all I can do at this point. I know that my point p is some height y above the line That's never going to change for this problem I know that this particular Charge is located some distance x away from the origin So I write that down and that's about it. That's about as far as I can go right now Let's write down our vector We need that in order to get our hat. So we might as well do that next So our vector I've already kind of sketched what it's going to look like here It's got a piece that points in the positive I hat direction It's got a piece that points in the positive J hat direction So it's a distance x in I hat plus a distance y in J hat And again, that's about as far as I can go with this right now Okay, finally We have to do our hat and to get our hat. I divide our by its magnitude Well, this is going to look pretty I got x I hat plus y J hat over the square root of x squared plus y squared I feel like we've gotten in a worse position now. We don't really know what's going on here. All right, so let's let's write down what we've got Clean some of the chemistry off the board here last class All right, so let's see. Let's see what we've done. I think we've made the problem worse We've got de vector equals k. Okay, we know what that is dq. No idea what's going on there We got our squared in the denominator So we got x squared plus y squared down here, and then we've got this our hat vector, which is x I hat plus y J hat all over the square root of x squared plus y squared Now we can go one step further and kind of group some things here. We've got we've got an x squared I totally agree Lucy. I've got an x squared plus y squared. It's fine. Trust me That's what's happening inside of me right now as I look at this. Okay, I got an x squared plus y squared here I got a square root of x squared plus y squared here I hope this looks familiar to people, but you can rewrite the square root in a suggestive way You can take x squared plus y squared raise that whole thing to the one half power and that's equivalent to the square root And when you do that you see that now you're multiplying something with a power of one times the same thing with a power of one half And so you can rewrite this as I'm gonna pull this over here De vector equals k dq all over x squared plus y squared to the three halves And if you're not sure what happened there review your fractional review your addition of powers when you multiply things together It's a fun weekend activity for Labor Day Okay, and then you have your your remaining numerator in the in the unit vector x i half plus y j half Now I feel like we've made this problem worse, but we are on the cusp of a breakthrough Okay, because we have one piece. We haven't dealt with dq Is there a way we can relate the little piece of charge occupying a little piece of length Can we relate the charge to the length somehow through a number any ideas? Javon got any thoughts on this Just trying to fly into the radar Yeah, they go linear charge density linear charge density rides in and saves us Because this is a charge per unit length and we know we've got a little bit of charge Occupying a little bit of length So lambda is going to be equal to q over l whatever that is some constant It will also be able to dq over dx that little bitty piece of charge over the teeny tiny piece of length that it occupies That's a constant and So now we see we can relate Charge and geometry Charge is related to geometry position in the problem and that's good Because now we see Something very suggestive happened at the end of this De vector is k and we're going to kind of rewrite these things up a little bit so you kind of see the Calculus of it x i hat plus y j hat dx So I put the dx even though I could have put lambda dx up here like I did here I just moved it and these are just numbers that are multiplied I can move them anywhere in the product and it doesn't matter okay So I just moved the dx off to the right suggestively because you know you may be familiar with with Seeing something like you know dy equals a dx and if I want to get y Y is the integral of dy which is the integral of a dx Okay, so if I want to sum up all the little functions of dx multiplied by a and Get y I just have to put that curly s next to it right and add up So that's what we're about to do Okay, we've got about as far as we can get Setting up this little piece of charge in terms of coordinates Geometry and constants coordinates Geometry and constants once you've gotten it as far as you can go in those directions. It's time to integrate Okay So that's what we're going to do That's as far as you can take that one piece and now it's time to sum them all up all right We've used this Don't worry you're going to get to exercise this. Yeah Oh, I'm just giving that as an example So in you know in a calculus class or in a calculus video or something like this You might see you know something like you know y is some function of x so y equals a times x whatever a is okay, and The differential of y Is a dx and if you want to integrate those Differentials so does anybody know what the inner I mean the this is like the simplest integral to do You're just taking the sum of all the little pieces of y and then it better return the length y when you're done Okay, that's the nicest integral you ever have to do in life That is equal to the integral of a dx whatever that is okay, and that might be tough or it might be easy if A is a constant. It's easy if a depends on x It's not and you probably have to go look up the integral or put it into Wolfram alpha Which is an a web front-end to Mathematica something like that okay? I Don't care how you get the integral once you get to this point Your bike three seconds away from solving the problem because you just have to integrate Okay, and how you do that I leave up to you you can do it yourself. You can look it up in a book Books are a thing anymore. You can use the internet use some program. You purchase to say what you did okay? So to get the total electric field E. I Need to integrate all the little pieces de and that means I need to integrate k lambda over x squared plus y squared to the three halves X i hat plus y j hat dx So all I did to get to this step was put the integral sign in front of it and say whatever that is It's equal to the total electric field That's just the definition of a sum the sum of the parts equals the total Okay, so there's the total. I want that I have that so I need to do something with that Now let me ask a question if I gave you the following a Let's see a x DX, let's do it this way a x plus b x squared so there's some function of x DX and I told you to integrate that What might be the first thing you would do to make your life simple? Yeah, you can take out the axe That will make your life simple because now you'll have x multiplying another function of x awful Because then you'll have to integrate by parts and yeah, don't do that All right, so you've just made your life more difficult You're one to watch. Yeah, okay, no problem What are you gonna you can't but I want you to integrate it distribute the DX distribute the DX Okay, yeah, the DX is just a little you know It's a little infinitesimal number and like anything an algebra it can be distributed So we can rewrite this as the integral of a x DX plus b x squared DX and that just breaks into two integrals a x DX plus the integral of DX squared DX you can distribute So there's really two integrals added together up in that funky looking thing on the right-hand side of the electric field Okay, you've got an integral of x over x squared plus y squared to the three halves And you've got an integral of y over x squared plus y squared to the three halves both DX Okay, so I'm gonna go ahead and write that Just gonna do the same thing I wrote there in simple form like this nasty thing up here Okay, so E Equals all right. Well K is a constant Land is a constant. I'm tired of having my constants inside the integration They're not going to participate in the integral. So I'm just gonna yank them out in front and I've got the integral of X over x squared plus y squared to the three halves I had DX plus the integral of y Over x squared plus y squared to the three halves J hat DX now This is just a unit vector. It's just a constant thing that points. That's all it does just points points in the x-direction It's a constant too So actually let's just pull that out in front just get it out of the integral So we don't get all confused about what the hell that thing's doing in there same with J hat J hats just gonna go in front of that integral And we already see now how you know what's about to unfold in all of this What's about to unfold in all of this is we're gonna do this integral And we're gonna get the x-component We're gonna do this integral and we're gonna get the y-component and I'm gonna put a bracket on the end So they don't look like a slob and then we're gonna be done Okay, so you're like this close to being done with this The the hard part about problems like this and I don't mean to psych you out, right? I mean, I think you guys can all do this Okay, but if you want to sort of set back and go where where do I have to put the most brain power into a problem? It's setting it up. It's getting that that thing to integrate in the first place after that It's just executing the rules of algebra and calculus Okay, which gets wrote after a while the physics of it is going from the picture with charges To something mathematical that you can sum up later So you have to use physics to get the thing you can add up and then you just do math All right, so because I would like you guys to play around with your own problem We're gonna do this like a baking show and then a miracle occurs and the cake comes out of the oven, okay? Before I show you what's gonna happen in one of these integrals. Let's step back and look at this picture again Let's imagine that this is a positive charge that's distributed on this line Okay, so all the little itty-bitty pieces are also positive and let's say I pick a piece over here and I pick a piece in equal distance from the point P over here and they have the same magnitude. There's both dq They're both positive The electric field from this one points where Does it point toward the point P or away from the point P? It's a positive charge Towards point P because positive charges are the source of electric field So this electric field emanating out in all directions, but if we're just at point P There's a line that connects the electric field to that point P All right now what about from the symmetric mirror twin of that charge over here equidistant from the origin But over on the right hand side, where is its electric field point? Away from point P so like that Towards yeah, because it's also a source of electric field Okay, so the line of the electric field is converging on point P. They are Equal in length Okay, which are their components points in opposite directions horizontal or vertical? Horizontal yeah, the horizontal components are doing this the vertical components are adding up or canceling out We'll do this again adding up Okay, now you're doing the electric field things Nothing guys it's Thursday. We're in like the second week of classes more energy, please I've got 13 weeks left in me. What have you got okay? Every semester all right think about that for a second. It's like it's like it's like hell right? You just do the same thing for 15 weeks every semester Thank you. Thank you for giving me some small symbols of joy in my life Okay, all right. So yeah, so the horizontal components cancel out the vertical components add up So what do we expect that integral over that thing in the x direction to equal? Zero yeah, well you're not going to commit you're not going to let everyone in the back row here you Sophie said zero So just in case we're wondering okay Yeah, so for every piece that's over here on the left There's a twin piece on the right whose x component points in the opposite direction, but it's equal magnitude and so Without doing any math. All right. You just look at this problem goes there is a symmetry There's an infinite line to the left. There's an infinite line to the right So there's always going to be a twin charge over on the left for one on the right that cancels out in x component I expect that integral to be zero and in fact if you plug it into Wolfram Alpha or whatever you want to do You will find out that that integral is in fact if you do it. Let's put some limits in here symmetrically From negative infinity to positive infinity You'll find out that that is a big fat nothing zero No x component in the resulting total field. Yay Now you just have to do this one This one non-zero because all the white components all point in the same direction and add up Okay, now they have different strengths as you go further and further and further away but they all add up and The result of doing this one If you work it all the way through you will find that the total electric field is 2k Lambda over y in the j hat direction Okay, so Y is not something that varies in the integral But of course you're now free to say well, I'm going to move the point P twice as far away Okay, so now we find me let's say I started out at y equals one meter and I move the point to y equals two meters What happens to the strength of the electric field from this line? Does it double does it have does it go down by a four? double the distance What happens to the strength? Fourth of it is there a squared somewhere that I didn't notice to double see doubling it would if you square it would go down by four It doesn't linear so you double it and goes down by two Okay, so it's kind of cool about this is while each individual charge each little point in this thing The force declines as a function of one over r squared when you add them all up They kind of reinforce each other and so in fact the electric field of the total only declines as Linear falls that you go twice the distance that goes down by half you go four times the distance that goes down by four Okay, not 16 All right, so that's just one of these effects when you get a bunch of charges together You have to be a little careful watch your intuition read the equation make sure it says what you think it says, okay? If we had a positive charge density Then the electric field would point in the positive y direction We kind of saw that already when I was doing the dance right if we have a negative charge density some negative q per unit length Then the whole thing flips okay, and you're in the negative j hat direction So lambda can carry an additional sign if it's a negative charge density that would flip it around, okay? So what you wind up having here And I don't think I have a sketch of this But I have it in the notes as if you kind of imagine here's the line and you go anywhere around it a distance Why the electric field for a positive charge density will always radiate outward and it will fall off linearly with distance So this cord if it's carrying a net charge Carries an electric field with it that radiates out uniformly in all directions around the long axis Okay, and it declines linearly And so something to keep in mind is that all devices around you right that carry a net electric charge through them a current They have electric fields that they're emanating and so, you know You have to keep that in mind when you're designing equipment if you're gonna put a patient In an instrument that you know measures something about their electric currents or whatnot You have to keep be mindful of the fact that you can disturb their own currents by using electric fields from your own equipment So things to keep in mind even in medicine You have to be a little bit careful if you're using a new instrument hasn't been properly shielded against its own fields for instance, okay? All right, so here's the problem. I want you guys to do for the rest of the class It's a small variation on what we just did here. All right, so much of this will apply But I want you to try to step through it on your own Now the line is not infinite. It's finite and it's all to the left of the point P So imagine the copper wire is now a length L, but it's only to the left of the point P Okay, it carries again a charge q distributed uniformly over the length L Get the electric field for this at the point P Alright, so your goal your goal is to set up Something starting from this point charge Coulomb's law And you're going to try to get all these pieces written in terms of coordinates and constants and geometry Coordinates constants and geometry can't repeat that enough. That's what I'm looking for right? So you're going to set up this differential Which can then later be integrated like we did here. Okay, so we're together in groups I want you to partner up with two or three, you know people together Work together. Okay, and if you don't like the person you're sitting next to then quietly You say I gotta go to the bathroom and then to shuffle to another row. All right All right All right, and I expect you you're talking. I want you helping each other. Okay, if you have questions Come talk to me and if you need to build your chops in calculus Don't be afraid to meet with me outside of class. We can we can work through that together. Okay?