 Okay, so just to stall for time so people have a chance to get here for the quiz, just a concept to review, alright, so what have we done so far in the course? There are two kinds of electric charge, positive and negative, they exert forces on each other, okay, so we have a force in nature that we're exploring. Positive and negative electric charges are the sources, if you're positive you're a source, if you're negative you're a recipient of electric field, alright, so when you sketch a positive charge in its electric field, the lines radiate outward, when you sketch a negative charge in its electric field, the lines radiate inward, okay, so those little pictures you can always use those as the building blocks of your problems. If fields and forces are vectors, if you want to find the total of either the electric field or the electric force, then all you have to do is add as vectors the individual fields or forces due to individual charges. Today we are going to exercise me at the front of the class and then you amongst the class, we are going to exercise the extreme version of this statement, we add as vectors by doing bump bump bump calculus, okay, so how many of you are more than a year out from calculus at this point? Raise your hand high, don't be afraid, it's most of you, I know it's most of you because I know what the pre-health advising is like, okay, it's actually, I like having you guys when you're juniors or seniors in this class, you tend to be more mature, no offense to the first years in sophomores, you tend to be more mature, you tend to have your time management together, but the calculus muscle has atrophy, you don't need calculus in biology, you certainly don't need it in organic chemistry, okay, you need calculus though because calculus, I mean calculus was invented to solve problems with a whole bunch of things in them, okay, so the minute you have more than let's say five to ten things that you need to add up, you desperately need calculus. These days, okay, Leibniz and Newton who were the co-inventors of the calculus method, they didn't have computers, if they had had computers in their day, they may not have invented calculus at first, but it turns out that some calculations are just a whole lot faster even in a computer if you use the rules of calculus to shortcut the brute force adding of things, brute force adding takes time in a computer, you can shortcut that brute force process by using the shortcuts implicit in calculus, so we're going to exercise that today, okay, I brought this just to scare you, right, so because it's big and heavy and it makes an awesome sound when it lands on the table and it's, that's my calc book from college, so you can tell I didn't use it that much, I got it used to begin with and I'm old, all right, so you probably, I mean if this was the second edition in my day, assuming this book even exists anymore, it's probably in its fifth or sixth edition by now, all right, so with that in mind, are there any questions before we get started on the quiz? Anything in the reading people would like to ask about? Anything in the video people would like to ask about? I don't have to do a demo today, so we have a little bit more time for discussion and questions. Okay, if you think of them later, you can always just bring them up and we'll do them later, okay, so I'm going to hand up the quiz. It's the red folder. Okay, so put your notebooks away, your pen out, you know, a pencil, notebook back, all that good stuff. I know a lot of professors don't take advantage of it, but I don't understand why because it makes life a lot easier for you and for me as well. Okay, so a few things, let's start off with some announcements here. So the next assignment, which is due for the next class, so that's Tuesday. I'd like you to read chapter 22.6 and 22.7. This is all about charges and how they respond to external electric fields, focusing on single charges and dipoles. So we're going to exercise dipoles going forward. And then I've got a video that accompanies that and at homework two is assigned today. It should have appeared on Wiley 11 minutes ago. It's due next Thursday by 9.30 a.m. There are two concept questions and then about five problems after that. Some of them are shorter than others. Okay, so I wanted to cover topics involving singular electric fields, dipole electric fields, and I wanted to throw one of these, one of these problems up for you today as well, you know, as a delightful treat. So you have something to look forward to. Oh yeah, by the way, try to have fun over the weekend, all right? Teams are final as of today. I've heard no requests for switches. Okay, so starting tomorrow, meet with your team at least once per week outside of class. If you don't have contact information for your team, let me know and I can share email addresses with everybody on the team. Okay, so since I have your email addresses for university accounts, I can share that information. If you guys have other email addresses you want to use, that's your business. All right. Your homework for now is a team and I hope this is straightforward. A, select a team name and please follow the guidelines. I've said them before, be polite, be respectful. I'll look up words I don't recognize or think I might recognize on the Urban Dictionary. Email that name to me and please CC everybody on the team. So the other thing you should do, and this is in the guidelines as well, is that your team should pick somebody called a lead editor. The lead editor's job is not to do all your work for you. And if I find out that's going on, I'll be very displeased and I'll take it out on your team on the final exam. Okay, what I want the lead editor to do is basically be a coordinator. In science, when we have many collaborators, there's often a lead author or a lead pair of authors. Okay, so you may have five, six, seven collaborators, I have 3000. So for a single paper from my experiment, typically a group of 20 people has actually done most of the work on the content, the meat of the analysis effort that's published. And maybe two or three of those are listed as lead editors for the paper internally within our experiment. So everybody on the experiment knows who are the people that were coordinating that effort, or at least being asked to do the bulk of things like coordinating. So identify a person on your team who feels that they are good at things like this, organization, cracking the whip, that sort of thing to get everybody moving cheerleading. Okay, that's stuff. And also somebody with at least a basic understanding of English grammar, structure and writing, because they're going to be the literal editor for the final document that you hand in to give a homogenous read. So it doesn't sound like five people wrote it. Okay, so select that person as well. I'll add that to the slides when I post them online. So select a lead editor. I want that person to email me the team name, but be respectful to your teammates and carbon copy them on the email. So I don't think that the team, the lead editor on a team has decided to change the name of the team at the last minute to team whatever their last name is. Okay, team ego. Alright. So then the question I want you to think about over the next week, in preparation in February for having to meet with me as a team, it'll be later in February, but you should start at least a small conversation now about this. How does past class material inform ideas for solving the grand challenge problem? Okay, does anybody not know how to find the grand challenge problem or not have a copy of it was handed out on day one of the class? Okay, it's available on the course website. So if you don't know where it is to shoot me an email, I'll send you the link. No big deal. Okay, but I want you to read that over and I want you to think about is there any hints, clues, keywords, suggestions, do you know something already about magnetic resonance imaging that may already touch on something we've talked about in the class, just basic things. What stuff that's been mentioned so far in the book, in video or in lecture, could you use to start trying to answer this question? Okay, I remember I have to come up with three independent outcomes for this patient. So that to be sort of three different phenomena that they're based on. So just pick fun seeding find one now as a group. And then in February, when we meet, I'll expect your team to have at least a few ideas. So spitballing that you've done about what you might write about, and we'll try to nurture those ideas in our meeting. And then you'll run with one of those to start off with. Okay, at our second meeting, I'll like in March, I'll expect you to have two ideas, something like that. And then certainly by the last meeting we have before this is due, I'll expect the ideas to be basically written down and maybe you're working through the math details of it before writing it out. Okay, that's the rhythm of this problem. And remember, it's 25% of your grade. So take it seriously. Okay, so for the quiz questions, let's look at the answers to those. So what is an example of an electric dipole in nature, as mentioned in the video lecture, is it an oxygen atom, a water molecule, a hydrogen atom, or an oxygen molecule? Anyone? Water? Yeah, water molecule was the example given. In fact, none of these other things by themselves is a dipole. So water molecules, the correct answer, we're gonna come back to water molecules in a bit. By the way, did anybody wonder who those cute kids were in the video? I forgot to mention them in the lecture video, I was an idiot and that out, even though I did two takes on that part of the lecture, which was painful. Those are my nephews, their twins, they were born at the same time. They're technically the fraternal twins, although they look identical if you don't stare at them long enough. So they were born about six to eight weeks premature. And they had the surfactant injected into their lungs when they were born so they could take their first breath and then their second one afterward. Okay, they were on respirators for a little while, they came home with heart and breathing monitors because that's what they do with hospitals with premature babies when they go home. So they were they were these adorable tiny little two hand sized things with like eight wires coming off them going into a box. They look like little board children from Star Trek. It was really creepy and sort of adorable to a scientist. So what are you going to do? Okay, second question, which of these is a true statement about the electric dipole moment? It points from the positive charge to the negative charge with magnitude QD. It points from the negative charge to the positive charge with magnitude QD. Its total magnitude is given by the magnitude of either charge Q in the dipole, or its total magnitude is given by the separation D between the charges. 123 or four, anybody? Okay, one, anyone? Anyone else? Two? Okay, two? Anyone else? Any other votes? Yeah, okay, wow, there's a lot of not going to commit on this one. It's two. Okay, this is another convention. It boils. It falls out of the convention that we've been talking about where our vectors have to start on the charges that exert the force. It naturally falls out of that business. But if you look at the dipole as a whole entity, that p vector points from the negative charge to the positive charge. And it tells you the orientation of the dipole in space. It's going to get used a bit in the class. So that's something you want to drop down in your notes and emphasize. And its magnitude is always charge times the separation of the two charges. Okay. Which of the following is true about the dipole electric field 123 or four. So one, along the line connecting the charges, it weakens with distance faster than a point charge field. Two, along the line connecting the charges, it weakens with distance slower than a point charge field. Three, it is zero, the fields of the two opposite sign charges cancel out for it's the same as that of a single point charge. So 123 or four, any ideas? One, okay, so it weakens with distance faster. Anyone else? Anyone for 23 or four? No one wants to commit on that? Okay, so one is the correct answer. Those two charges are slightly separated so they don't completely cancel. But as you get further away from them, especially along that, that, that axis that connects them, the effect of one positive charge and one negative charge sort of neutralize each other a little bit, and that electric field strength falls off as the cube of the distance and not the square of the distance as you go to very great distances, okay? Alright, so the reason by the way I'm going and giving you these quiz questions and I'm giving you the answers and so forth is because I will draw from these quiz questions for the exams for the multiple choice bits at the beginning. So you're going to want to feel comfortable answering these questions. They may be phrased slightly differently, you know, I might pick one or two of these from the classes and rephrase them. So I want to know that you actually understand the question and you don't just memorize it's 123 or four to the question asked in class because I could flip the question around and ask the inverse of it. And you might try to answer the same thing and I don't know, I didn't really get that concept. So I'm looking to see if you master concepts. And if you're not comfortable with these, then I want you to go back and review look in the book. Scan back in the video so you can find something about this business and see if that helps cement the concept in your mind. Okay. All right, so what I'd like to do now is demonstrate how to set up and solve problems involving complex charge distributions. And specifically, using calculus. I know yawn, right? It's just terrible. Even it's okay. Everyone can yawn at calculus. It's fine. I do this too. Alright. Why should you care about this? Well, the world is full of complex charge distributions. Not all of them are simple shapes. The blob I mentioned in the end of the video is a very complex shape. It looks like a big lump of Play-Doh or mashed potatoes or clay or dirt or something like that. You know, there are lumps like that that get a net charge. And if that lump happens to be important to something you're doing, like if you charge up a tumor before you do something to it, like attach a drug using a using the charged up tumor, get some charged drug to stick to it. And so the drug will be delivered directly to the tumor. I don't know. That's a crazy idea, but somebody's probably tried it. You have to care a little bit about what the electric field looks like from the blob from the tumor material because it's going to affect what the drug does in response to that field and what the field does to tissue around the tumor. So all of those things play an important role in accurate drug delivery if you're developing some novel drug delivery technique based on electromagnetism. And they do exist. Okay. So I have here a picture just of a whole bunch of coils of copper wire. Copper will be prominently featured in this course as it is one of the most common conductive materials used by our civilizations. It's relatively plentiful, although because it's in such high demand now, the price of copper per unit mass has skyrocketed over the last few decades. So you may hear at least a few times a year, especially in a city like or a metro complex like Dallas, that and this was very common during the peak of the housing crisis. Lots of empty homes. What kind of piping are they full of? Copper piping. So people would break into the homes, gut the walls, steal the copper and sell it on the market, make a ton of money off of that. Okay. I know I just told you basically how to make a ton of money. Please don't do that. Have some common sense here. All right, but copper is valuable is valuable for many things. And because of that, its price has gone up. And so that makes it a more interesting target for theft. So that's a lot of copper, right? I'd love to get my hands on that and retire. But you can put a charge on copper. Okay, copper by itself just sitting there is electrically neutral. But if you expose it to an electric field, like I exposed that aluminum can to an electric field, you can induce a separation of charge and thus create a net charge somewhere on the conductor. Charges are free to move on conductors. They're stuck on them. But as long as you stay on the conductor, they can move anywhere they like. And almost no resistance. And we'll explore that a little bit later in the course. So imagine a net charge is deposited on, say, a long line of copper wire, like a cord, a lamp cord, or, or there's copper inside of this, this camera cord here, maybe this, maybe if I strip off all the plastic insulation and expose the copper underneath and then dump a charge onto it using tribal electric effect and then put the charge on it, that thing will then gain an electric charge. You may be interested in knowing what the electric field looks like from that. But the problem is, is that we're not talking about one charge or two charges or 10 charges. We're talking about maybe something approaching Avogadro's number worth of charges, 6.02 times 10 to the 23rd. And good luck adding that up by hand. You won't live long enough to add that up by hand. Okay. So it while it is true that a charge distribution may be made at its heart of elementary charges like electrons with negative 1.6 times 10 to the minus 19 coulombs. Nonetheless, there may be a whole lot of them and in any terrestrially sized object like this stapler, there's about Avogadro's number worth of atoms. And each of those atoms, you know, this is a lot of hydrocarbons in here because it's made of plastic and there's some metal here as well, it's probably steel. Okay. So there's iron in here, there's oxygen, there's hydrogen, there's carbon. Each of those have lots of electrons associated with them. So you may have 20 times Avogadro's number worth of electrons floating around in this thing. 20 times Avogadro's number doesn't really change Avogadro's number by all that much. But nonetheless, it's a big freaking number. Okay. You're not going to add that up by hand. So you have to use calculus. And where calculus is most effective at first is with simple shapes. So let's do a simple shape. And this simple shape has a deceptively large amount of exercise, practice, and information built into it. Alright. So I'm going to step through this a bit methodically as best I can. I'm a physicist, I'm a mathematician, but I'll do the best I can. Okay. So let's imagine we have a charge Q, big charge, and it's made of many elementary charges. Right now let's not care about how many. And that charge Q is deposited uniformly on a long, long, long, long, long straight line. So imagine this is the x-axis. And at every point on the x-axis there is a tiny little elementary charge. And all the neighboring elementary charges are uniformly spaced with respect to one another. So uniform distribution means no clumpet. Alright. So what's a non-uniform distribution look like? It might look like this. Okay, that's a non-uniform distribution where there's very little, very little, very little, big clump, very little, very little, very little, big clump, and so forth. Okay. A uniform distribution of chalk, in this case, would look something more like this. I lightly press on the chalk and I roughly uniformly distribute the same number of chalk atoms, if that's a thing. Okay. It's calcium mostly. On the board as I go. Okay. So uniform, non-uniform. Pain in the ass to deal with. We'll deal with this one. Okay. Uniform distribution. Alright. So a concept that's very convenient to define when you have an extended object with some thing deposited on it, in this case chalk on a blackboard. Okay. Is something called a charge density. Alright. So this is a concept. It's in the book, but I'm going to motivate it here. Okay. I've distributed a fairly uniform line of chalk across this board. And so what that means is that the number of chalk particles per unit length of line is a constant. The number of chalk particles deposited per unit length of this line is a constant. So if I take that slice right there and I go in and I count the number of chalk particles in here and I get out a ruler and I measure the length of this line segment. Okay. So I'm going to get some number of chalk particles in one when I count. I'm going to go into the microscope and count and then you won't see me for 30 years. Okay. Which may be good for you. Okay. And then I'm going to get out a ruler and hopefully in less than 10 seconds I'll be able to measure this in inches or centimeters or whatever. Okay. And then I calculate n1 divided by delta x and I'm going to call this delta x1. Okay. Delta x1. I will get some number which I'm going to write as the Greek lowercase letter lambda. And I'm doing that because this is how we write a linear charge density, notationally in physics. Okay. Now let's say I go and pick another segment with some different number of chalk particles and some different length, delta x2. And I calculate n2 over delta x2. If I have deposited uniformly chalk particles along the line on this board, I will get lambda because this thing is a constant. This ratio is a constant. That's what it means to be uniformly distributed. No matter what piece of the hole you look at. When you add up the number of things in that piece and divide by its length or if it's a two-dimensional surface, its area or if it's a three-dimensional volume, its little volume element, you will get a number. And if you grow the volume you're looking at and recount and take the ratio you'll still get the same number. In physics that's what we mean by constant uniform distribution. Okay. This is great because if this is true you have a very powerful piece of information that no matter what piece you look at, as long as you dutifully add up everything in the piece and take the ratio you'll always get the same number. And that's going to come in handy. Okay. Any questions on that concept? Great. So we have a charge q that has been deposited on a line. I declare in this problem the line goes from negative infinity to positive infinity. It is infinite in length. That is or isn't simplifying depending on how you want to look at this. Okay. But that's how I chose to set this problem up. This will affect us later. But basically what that means is there is an equal amount of charge deposited to the left of where this point p intersects the line as there is to the right because that goes off the positive infinity, that goes off the negative infinity. The links on either side of point p are identical and so this because this is a uniform distribution of charge the same number of charges are to the left of point p as are to the right of point p. What is the electric field that point p which is a distance d from the line? It's a fixed distance that won't change. Okay. So what's the electric field due to that line of charge? Now already there is some intuition we can have about what this electric field might or might not look like. There is in this problem something called symmetry. Symmetry is basically some kind of equal behavior around a point in space and the equal behavior here is that there's just as much charge to the left as there is charge to the right. So this point p, the line is symmetric around it. In other words if you were to chop off half the problem put a mirror here and look in the mirror you recover the other half of the problem. You only need half the problem to do the other half. That's what's great about symmetry. Okay. Now because we have a symmetry let's think about electric fields. I haven't said whether this is a positive charge or a negative charge down here but for a moment let's assume that this is a positive charge on this line. What is the electric field going to look like if I select one single point charge right here? What's the electric field from that point charge going to look like at point p? Where is it going to point? Any ideas? So picture just one charge ignore the rest of the line just one charge, one point charge. What's the electric field for a point charge do as you go away? Oh come on, not really. It gets smaller, it weakens and it radiates outward. Okay. So basically all you have to do to figure out which way the electric field from a positive point charge points up there at that dot is just draw a line from the point charge through the point p and it will point up into the right. So we'll have a y component and we'll have an x component and those will add together to give you a resulting vector that points up into the right. What if I take the symmetric twin of that charge over here? Now what direction does its electric field point? It's equidistant from the middle but it's just on the other side of point p. So what direction does its electric field point? Again it's positive up into the left with the same magnitude because they're the same distance from the point p but the direction has changed a little bit. So you'll have one vector pointing this way, one vector with equal magnitude but a slightly different orientation pointing away. What happens to the x components of these two vectors like x-men now? This is great. They cancel out. What about the y components? They add up. So the x components cancel out. This is great. And the y components add up. Okay. And now you've got the physics dance. Okay. Nothing. Nothing. I just you know I just came up with that right now. No appreciation for how hard it is for a physicist. If you watch the Big Bang Theory. Okay. I should just assign that for homework. It has so many problems within so many directions we can talk for hours about it but it's a great show. So all right. So we kind of expect if you do some sketching. Okay. Pick two charges that make up this line at equal distances from where p intersects the line. You'll find that the x components you expect will cancel out. But the y components will add up. And so if you were to do that over and over and over again, pick a point here and a point here. It's symmetric twin. Pick a point here and a point here. It's symmetric twin. And keep doing that all the way out to infinity. You'll find that every one of those pairs, the x components cancel, the y components add. And so at the end of the day, calculus had better tell us that the x components all cancel out and the y components add up to something non-zero. All right. So that's just a little bit of the sketch. A little bit of the physics intuition. Let's dive into this. And I'll begin again with a linear density and a uniform distribution. Let's look at that line. I'm going to take a little piece of the line, okay, with some length. And I'm going to consider just a segment of it. So a little slice through this cylinder I've just drawn that represents this thin line of charge. Contained within this slice is some fragment of the total charge which I'll write as delta Q. So big Q is the total charge deposited on the line. Delta Q is the little fragment of charge we find in this little piece. And it has a length delta x. Because this is a uniform distribution, I know instantaneously that if I define a charge density along a straight line, charge per unit length, and take delta Q and divide it by delta x, I will get that number lambda that is a constant. Because this is a uniform distribution of charge. And it doesn't matter what piece I pick and how much charge is on it, that ratio will always be the same number. Okay, so here comes the first calculus tidbit. Imagine that I take an infinitesimal slice, just a tiny wee little slice that contains maybe one elementary charge. That little slice will contain just an infinitesimal fragment of charge, DQ. This is a differential. It is not two symbols, it is one symbol. And it basically means an infinitesimal bit of charge. That's the job of that symbol, is to tell you just wee little bit. Okay, so tiny it almost doesn't matter, except that there are as like Avogadro's number of those things when you add them up, they do matter. And that tiny, tiny little bit of charge occupies a correspondingly tiny, tiny little bit of the x-axis differential of x or dx. Okay, so this is looking terribly familiar. This is looking like, oh my god, he's about to say derivative. He is about to say derivative. Okay, derivative. Alright. If I take the ratio of DQ to dx, since this is a uniform charge distribution, what will this be equal to? Lambda. Yes, thank you. Because it doesn't matter what segment of the x-axis I look at, as long as I dutifully add up all the charge in that segment and take the ratio, I will always get lambda. Okay, and again, this is a thing. This constant here has units of coulombs per meter. Charge per unit of coulombs per meter. Okay, so this is great. This is a finite piece, and this is an infinitesimal piece of the line. And in calculus, especially if one wants to add stuff up, one has to do integrals, one has to do anti-derivatives. We are going to have to add up all the little electric fields due to all the little infinitesimal charges on this line. And to do that, we need to define things like a coordinate system and think about the geometry of the problem. Because ultimately, when we add this thing up in an integral, we want to integrate over something sensible, like a coordinate in a coordinate system that has a definite minimum and a definite maximum. Okay, so that's what we're going to want to do. That's the sneak preview of what's about to happen. So let's make that happen. Okay, this is in my notes, but I'm going to wing this. Then I'm going to remember the definition of the integral, like the core, raw, basic calculus. You know, it leaves you bruised on the floor kind of definition of the integral. That's the derivative. My greatest student, my greatest failure. Now, so the tangent, the slope of a tangent to a point on a line is the derivative of the function that gives you the line at that point. The integral effectively undoes the derivative and returns the original function to you. So that's its fundamental role in mathematics, but it has the following definition. The integral from a to b of f of x dx is equal to the limit as something called the partition size goes to zero of the sum from i equals one to n of f of x i delta x i. Okay, what the hell is going on here? All right, if I have a function like that and I partition it three, four, five into five pieces, okay, and here's a and here's b. All right, a, b. Those are my starting and ending points for where I'd like to add this thing up. In calculus what you learn to do is you learn to partition this infinitely so that you're slicing this so finely you're essentially sampling individual heights of the function as you go. f of x is just, so this is f of x here, okay, f of x at any point x is just a height of the line. So here it's that point and here it's that point and here it's that point. It keeps changing, the height changes as the function moves. This means send the partition size to zero. So what I want to do is keep infinitely slicing up the partitions but send the partition sizes each down to zero. If that limit exists for a function it is integrable and in visits for you guys we're only dealing with integrable functions. I'm not interested in giving you things you can't actually integrate what's the point, okay? Mathematicians adore that stuff. I don't like it so much, okay? They're trying to teach you math, I'm trying to teach you about the universe. Universe likes integrals that actually are integrable, okay? Otherwise we wouldn't exist. I think, I think that's a true statement. All right, so we're going to have n partitions so you have to sum the value of the function times the width of the partition over all those partitions. So, so that's, let's say here, that's delta x i, that's the width of partition i, and you have to calculate the function f of x i in there. So that's height times width is area. So this is just the area of that box and you sum all the areas and you get the total area which is the area under the curve. All right, so that's my four-minute Calc review of fourth chapter of that book, okay? Why do I write this down? Because we have already partitioned our line into a whole bunch of little infinitesimal dqs taking up little infinitesimal dx's. And so it ought to be true that in the limit that the partition size goes to 0 for that sum i equals 1 to n. My function, okay, my function is lambda, right? So I have the same charge density lambda in every single partition. This is the beauty of the constant. And then I multiply by the size of the partition delta x. And this ought to be equal to the sum from 0 to x of lambda dx. Okay, no magic here. I've just used the definition of the integral to write this down. Well, this is a constant. So what can I do with a constant that doesn't participate in the integral itself? Pull it out. I can take it out. Okay, it's just a little algebra. You put that lambda outside the integral. Don't bother with it. All right, so actually let me let me make this go from 0 to l the length of the line. Alright, dx. Okay, so the limits are 0 to l the length of the line, 0 to l the length of the line. What's this integral equal to? This is like the easiest integral of the universe. x, y. How do you know that? Stephen, did you say that? Okay, because you're right, your mouth is hiding. I see the smile in your eyes. But why? Why is that x? Because you remember from calculus class, which to be fair, 99% of the time is a perfectly acceptable answer. By the way, also 99% of the time which adds up to like more than 100% of the time, it's perfectly acceptable to go and use a resource to look up an integral. I am not. This is not math class. This is physics class. You use math to solve problems. Okay, you want to use Wolfram Alfred to do an integral in the homework? You do that. Okay, you want to use a book to look up an integral? You do that. Just tell me what you did. I just don't want to assume a miracle occurred and somehow you magically knew how to do the integral. This one you can intuit. The integral is the anti-derivative. So the derivative of what function of x gives you 1. df dx equals 1 when f equals, anyone? I heard it. Powder? Yes. df dx equals 1 when f of x equals x. Okay, so if you take the derivative of x with respect to x, you get 1. So if you have a 1 effectively left over inside of this integral, and if the integral is the anti-derivative, you must get back the thing that the derivative originally acted on. So that's the reverse. That's how you run the problem backward in calculus. Okay, but don't worry about it. If you memorize it, that's fantastic. If you didn't go look it up, all right, go refresh your brain a little bit on this stuff. So this is going to be lambda x evaluated from 0 to L, and that means you do L minus 0. So you evaluate at the maximum and subtract the function evaluated at the minimum, 0, and you get lambda L. Well, if the total length of the line is L, and the total charge distributed on the line is q, what is lambda times L going to be equal to? Yeah, it's just q. So if I have q distributed over a length L, that's the whole line, but the constant uniform distribution still applies. Okay, so if I rewrite this as lambda L, that's just equal to q. So basically what have I learned? By summing up all the little charge densities times the little widths that they occupy, I get back the total charge. This is sort of a trivial revelation. If I have all these charges and I add them up, I get back q. Okay, but that's how calculus shows you that it works. Calculus is about taking a problem, busting it up into tiny little pieces, adding up all the pieces, and getting a total result. That's fundamentally what it's about, for these integrals. Okay, so we're going to exercise this concept right now. Let me make this concrete. That's pretty abstract. Let me make this concrete. And I might regret it a little bit. Nope, good. That's your tens of thousands of dollars hard at work right there. Okay, our goal is to find a total electric field. That's what we want. That's what the problem asks us to do. Find the total electric field. In a calculus sense, what this means is we want to integrate, let's just do this as an indefinite integral right now, all the tiny little fields that are due to all the tiny little charges. So that's what DE is. So if I have a charge, let's make it a positive charge, and that charges dq, okay, and it occupies a little space on the line dx, then that little dq is the source of some piece of the total field at point P. So here's point P up here, let's say, and the electric field due to that little charge dq at point P is DE vector. It's a differential of electric field due to a differential of charge. So what we're doing now is we're going to bust that line into pieces. We're going to look at one representative piece, sort out all the things we don't know in terms of coordinate system and geometry, and then at the end, once we've got all homogenous and written in terms of our coordinate system and our geometry, we're going to add it all up. That's it. Bust it into pieces, pick a piece, write down what you know, write down what you don't know, get all that stuff sorted out in terms of coordinates and geometry, then add it all up. That's it. He said with two decades of experience. This is the beginning for you. But that's the mantra you want. Bust it up, pick a piece, make it nice and uniform, add them up. It doesn't really work terribly well, it's like a chance, does it? I know, it's terrible. Somebody can make this better for me. Somebody could think about this from sort of a cheer leading mentality, and I don't care whether you're male or female, just come up with a good slogan for me and extra credit. I don't care. I'm shameless about that stuff. Okay? Alright, so let's look at a representative piece. I know you're, I hear that the E word, extra credit. Now you're like, oh, that's it. You're not even going to listen to me for the rest of this lecture. We're going to be like, let's see, how can I turn this into an acronym? Alright, so here's just a part of my infinite line. I can only draw part of it. Okay? And here's my point P. And I'm going to pick just a tiny little infinitesimal slice with, with DX over here on the left. I'm just going to pick any old slice. It doesn't matter which one. I'm going to draw my coordinate system. I'm choosing to put the point P right on the y-axis. And I'm choosing to put the line right on the x-axis. Not a terrible set of choices. Okay? So step one with your picture. You're going to bust it into pieces. Pick a coordinate system so you can use that coordinate system to say where things are in space. Cartesian works great for this because it's all about straight lines. Step two, write down the electric field equation for a tiny infinitesimal little point charge in terms of this differential notation. DE vector will be due to Kdq all over r squared r hat. So this is just the point charge electric field for some tiny little charge dq. Remember dq is a single symbol. That's not distance times charge. That's a differential of charge. So be careful with that. K constant. We've got that already. dq, what the heck is that? We're going to want some coordinates in this thing later. So we've got to think ahead here. How are we going to get dq written in terms of something we can integrate? Like a coordinate. So we have a mystery there that we have to sort out. Okay, distance squared. Well we're going to use our coordinate systems to get r vector and r and r hat. So quite naturally we're going to get we're going to get this from our coordinate. Get this and get this from the coordinate system. Alright, so when you look at this equation, this is the building block. This is the elementary piece that you're going to sum up with calculus. When you look at this, your goal is to go, okay, I need all of this to be something I can actually integrate at the end. Alright, jump ahead to what you're going to have to do. And to integrate, you're going to have to do an integral over some coordinate like dx or dy or dz or something like that. Okay, so it's going to be necessary to write all of the stuff in here that could change with position in terms of position. Certainly if we picked another slice over here, would the coordinate change for that little slice of charge dq? Have the coordinates changed? Has the relative position to p changed? Yeah, I'm closer now to p. If I pick one right here, this is the closest I'll ever get. If I could pick one over here, now I'm further away. So while dq itself may not vary in size if we look at uniform pieces, certainly the relationship of dq to that point is going to change with location on the line. So there is something that happens as you slide around and look at different differential charges on the line. So you have to be aware of the fact that that dq isn't a constant. It's something that does vary with the integral as you sum up the pieces. Okay, so these are all little red flags that are popping up at this point. All right, but let's break this up and let's attack it piece by piece. And I'm going to start with the things that I think are somewhat easier once you have a coordinate system written down. Let's do r vector. R vector by convention points from where to where. Yeah, source to the recipient. So what's the source in this case of the field? Yeah, dq specifically. All right, so here's my little dq. So there that's the start. And where's the recipient located? p, point b. So great. All I have to do is draw an arrow from dq to p. And that pictorially represents r vector. So that's r vector. Awesome. Okay, so now we actually have to write it in terms of our coordinate system. Well, when I picked this piece over here, it's some distance x along the x-axis away from the origin where point p is located. What's the distance in y that this point is located above the line? What's that? D. Yeah, big D. In fact, if I pick any other point, let's say I pick a point dq here, has x changed? But the y is not. Yeah, so this x is smaller, but the distance vertically that you have to go to get to point p is the same. Same thing over here. This x is different from the one that I picked originally. But that height, that distance you have to go to walk this way and then walk up to p. That's always the same. This is nice. Okay, this is helpful. All right, so let's use this board. So I have to write r vector in terms of my coordinate system. So to get from dq to the point p, I have to walk x in the positive i hat direction and d in the positive j hat direction. x i hat plus d j hat. And I'm done. In terms of my coordinates and things I was given, I've now written r vector. What's r squared going to be? Yeah. I have a question. Oh yeah. So you were not given the distance on the x axis, we just name it x. Just name it x. Yeah, this is a variable. And you know it's a variable because if you were to pick any other random dq, it's the numerical value of x would change. So it's varying along the x axis. So let's, yeah, we'll just use this placeholder x for this. I mean, you can pick any symbol you want. It's convenient to use x because it reminds you that this is a coordinate on the x axis. Okay. Okay, so r squared. x squared. What was that? x squared? Yeah. And then, okay, r itself is like a square root of this. Oh yeah. Okay, so yeah, we can do r. r is the square root of x squared. Plus d. Plus d squared. Yep. Okay, so that's just a little Pythagorean theorem where you can take the dot product of r with itself and take the square root. You get the same answer. Okay. Now what about r hat? How do I get r hat in general? Okay, so right, r vector over its length. Exactly. So you can always get r hat by doing this. And this won't look so pretty. I never promised pretty. And let me go one step further and just remind you that you can write because I'm going to use this like crazy in a second. You can write the square root as the contents under the square root raised to the one half power. Okay. That's a totally interchangeable thing. In fact, I prefer the fraction notation but that's just because it often times what you'll have to do is add up fractions and it makes it easier than carrying around those square root symbols. Okay, so we have r. We have our vector. We have r. We have r hat at least written in terms of things we were given like d and things that we are going to integrate over like a coordinate like x. Okay, that's about as far as we can take that. Let's put these pieces together so far in DE. So coming back to our little Coulomb's law single point charge electric field thing. We've got k, a constant, dq. We're still sorting that out. We'll come to that in just a moment. r squared in the denominator. So that's x squared plus d squared. And then r hat. So we have x i hat plus d j hat all over x squared plus d squared to the one half. Not pretty, okay, but functional. Alright, let's deal with dq because this is great. We've got x here and we can imagine integrating over x from negative infinity to positive infinity to add up all the pieces of charge and all of their corresponding electric fields. But we have this dq thing and we've got to get this written in terms of coordinates. And this is where charge density comes galloping to the rescue. What was charge density? Lambda. It was dq over dx. Now, while this is normally not a fraction, there are conditions under which you can treat the derivative like a fraction and just move its bottom half to the other side of the equation. And that's when it's equal to a constant for sure. Okay, and this is a constant. This doesn't change no matter what part of the charge we look at as long as we look at the corresponding part of the line it takes up. So we can change this and solve for dq in terms of dx little pieces of the x-axis. Well, we're almost done. De vector, I'm going to combine the denominators here. I've got x squared plus d squared and I've got x squared plus d squared to the one half and these are multiplied times one another. Let me be more careful here with my parentheses. So I can combine those and then all I have to do is add the powers above the parentheses. Okay, so let me do that. So we're going to have k lambda dx all over x squared plus d squared to the three halves and then we have this x i hat plus d j hat. We got x's. We got d's which is just a constant. We got lambdas which is just a constant. We got k's which is just a constant. We got dx. We are ready to integrate this thing. Okay, so let's do that. We don't need this picture anymore. I'll leave that though because that'll come handy. Okay, so the original goal, get the total electric field at point B. To do that, we have to add up using calculus, integrate all the little pieces of electric field due to all the little point charges. So we finally have something we can put in here for de. We have k lambda dx all over x squared plus d squared to the three halves x i hat plus d j hat. The only thing we're missing are limits of integration. Where does x go from and where does x go to? Well, I said it early in the problem and I said it a few minutes ago. X starts out way at the left at negative infinity and it ends way over at the right at positive infinity. Now, a side note, this looks like a terrifying integral. But please remember that if you have the integral of f dx plus g dx, well let me put it this way. If you have f plus g dx, okay, so two functions of x, f and g, it's totally allowed. You can go back and prove it again using basic mathematics if you like. But this can be written equivalently as f dx plus g dx. You can do the integral of the sum is the sum of the integrals. And that's what we're going to do here. We're going to distribute, well, okay, so we have, oops, good golly, no one said anything. That's dx, okay. We have a constant, a constant. That thing definitely is going to change with x. That thing is definitely going to change with x, okay. That's a constant. And the i-hats and j-hats, they play no role in the integral. They're just pointing in some direction. That's all they do. They just point. So they're constant too, from the perspective of the integral. So we can rewrite this thing up here as the integral from negative infinity to infinity of k lambda x dx all over x squared plus d squared to the three halves. i-hat plus the integral from negative infinity to infinity of k lambda d dx all over x squared plus d squared to the three halves, j-hat. x component, y component. And now you just have to do some integrals, okay. So this is where you look things up, okay. Go to Wolfram Alpha, do the integral function on that, punch this thing. You can punch it in symbolically. It's what I like about Wolfram Alpha. So I went on Wolfram Alpha and I plugged in, we'll do the x component first, okay. So I plugged in that integral from negative infinity to positive infinity and I did the integral of x over that mess, dx. And what did I get? What originally were we supposed to find about the x components of the electric field? I got zero. And we expected it to be zero. So that's a nice check. So it turns out, if you do this integral, this definite integral, you will get zero. Now in my notes I work through the indefinite integral and then I write the definite limits in and you can see how I work through that, okay. So, but you could punch that definite integral into Wolfram Alpha. You could look it up in a textbook. Like the Russians always have the best calculus textbooks like this thick, you know, grad scheme and write check and they have these awesome, like thousands of integrals that they've solved, okay. They're all in there. Every single one of them that you'll ever need is in there. Somebody's always solved something before you have to deal with it most of the time. It's the interesting problems that haven't been solved yet. Those equal Nobel Prizes, all right. So that's what I'm prepping you for, Nobel Prizes, all right. So then you have to do this one and you do this definite integral, okay. And what you find is that this here comes out to be 2k lambda over d j hat. That ain't bad. For all that, that's a lot of sound in theory to get to that. What do we have? We have an electric field, a distance d away from the line and its strength at that point is simply given by 2k lambda. So you just have to know the charge density which you might be given in the problem divided by the distance you are from the line. The strength of this electric field from this line falls off not as one over the distance squared but one over the distance linearly. So if I go 2d away from the line, the electric field strength is now one half what it was at the point p. If I go 3d, it's one-third. So there are electrons running through this wire right now and if I were to put a charge, a net charge on this and let it sit there, I would get a field that falls off in strength linearly with distance from the wire and anywhere around the wire it looks the same. So if I'm six inches over here from the wire, the field points that way. If I'm six inches above the wire, it points that way away from the wire with the same magnitude. If I'm over here behind the wire, it points this way. So I have a sketch of this in the notes which I'm not going to reproduce here because I'd like to move you guys to the student problem. But basically it's just a cylinder that represents the line with a whole bunch of electric field lines radiating out uniformly from it. This is how you get an electric field that diminishes linearly, not quadratically which is what a point charge field does. So what's cool about this is that you take a bunch of point charges, each one of those has a field that diminishes quadratically with distance. You go twice as far away the field is four times weaker. You go four times further away, the field is 16 times weaker. You put them in a line and all their little fields help each other out to create a field which is actually stronger and diminishes more weakly than any one of those. So imagine what you could do if you put them on a plane, put them on a circle. There's a charge disk problem or chapter in the book. I haven't worked through that but I leave that as reading so that you can see what the power is of this method. So what I'd like you to do is I'd like you guys to pair up and I'd like you to start trying to pick at this problem. It's a related problem to the one I just did here. So it's got all some of the same elements so you can recycle some of the stuff you see up here. You now have a finite line of length L and the point P sits just above the right edge of that line. Charge Q is uniformly distributed over this line. What's the electric field at point P? So start talking to your neighbors, start working together, okay? And I want you to chatter and if you're just totally confused, there's your hand but let's get you unconfused, okay? This is the entree into doing this. I'm not expecting you to be masters just because I did this at the board, okay?