 The first subject that we need to cover now as we explore the concepts of work, electric charge, and electric fields in more depth is the idea of the work required to move a charge or charges in the presence of an external electric field. I'm going to set up a very simple example, but the observations that we make in this simple example will apply universally to all movement of charges through electric fields and the work required to do that. Let me begin with the first concept, which is that now what I'm talking about is work done by a force applied to the charge that isn't just the electric field. So this will be denoted as F applied, and it has a corresponding work due to the applied force, which you can write as W applied, and I'm going to shortcut all of this and write W app. Okay, so this is the work done by an applied force, and this is to be distinguished from the work we've been talking about so far, which I've, when being very thorough, denoted as work with subscript field and have simply shortcutted to referring to as W, the work done by the field. As you'll see, these are related but distinct concepts. They can be related to one another, but it's important to make the distinction between them at the outset of a problem. Let me begin by drawing a very simple electric field. I'm going to draw a nice uniform electric field with vectors E, with magnitude the letter E, which is a positive number, and direction I hat. And then what's nice about this is that I should just be able to clone these electric field vectors over and over and over to build up a very nice uniform electric field. So let me go ahead and do that with a few of these, and now we're good to go. We've got our nice uniform electric field cloned from a single electric field line. And into this, I am going to put a charge. So here is my charge right now, and I will label it Q. Now for now, I'm not going to specify what the sign of Q is. Q could be a positive charge, Q could be a negative charge. We'll leave that just as an algebraic symbol Q for now. And the next thing I'm going to tell you is that an applied force will act on this charge and will move it along a line that's horizontal but points back against the field like this. And then on another line that points vertically down at a right angle to the field. And then finally along another line that points against the field to the left. So we basically have a path that has three steps. Step one, go to the left some distance. Step two, go down some distance. And step three, go to the left again some distance. For step one, I would like the path length to be equal to something like one centimeter, which is 1.0 times 10 to the minus 2 meters. I'm going to leave this as an algebraic symbol D just to make the point. But you can imagine putting a number in here like one centimeter. In step two, I will also move the same distance D, but I will do so in the downward direction. So in step one, we moved in negative i hat against the electric field. In step two, we moved in the negative j hat direction. And then finally in step three, I'm going to go one more distance D and I'm going to again do that in the negative i hat direction. The question we want to answer is what is the work that has been done by the applied force in moving this charge along this path in the presence of this electric field? Well, work is a scalar quantity. And so if you want to know the work total that has been done by the applied force, one need only sum up the work done by the applied force in path number one. The work done be the applied force in path number two. And finally the work done by the applied force in path number three. This is where we can begin to relate things like the work done by an applied force to the work done by the field if the field were the only thing acting on it. If there was no force due to the electric field, that is if the electric field were switched off in this region of space, there would be no external forces acting on the charge. And because of that, it would require no work to move the charge anywhere in this space. But there is an electric field and that thing acts to move the charge, for instance, to the right if this is a positive charge or to the left if it's a negative charge. We know that the work done by the field, which I'm just writing as W, will be equal to the electric force, which I'll denote as F subscript E, dotted into the displacement of the charge at any point. So what we can do is we can think about the work done by the field moving a charge in this space. And because it's the only force in town, besides the applied force, we know that whatever the work done by the field is, it must be the opposite of the work done by the applied force, because the only force that an applied force would have to act against to move anything around would be the electric field. So whatever energy goes into the electric field comes out of the applied force and whatever energy comes out at the expense of the applied force goes into the electric field. Conservation of energy dictates that you can't ever create or destroy energy in a closed system. You can only change it from one form to another. So whatever energy the applied force uses up, moving something around, must correspondingly be gained by the electric field in the form of a potential energy and vice versa. All right, well let's go ahead and think about this. So the work done by the field along path one is just going to be equal to the electric force, which is given by q e vector dotted into the displacement vector along path one. The electric field is uniform everywhere. So e vector will on path one, path two, and path three always be the same number. Q, of course, is not changing here. The only thing that changes in any of these work equations is going to be the displacement vector given the problem that I've set up here with a nice uniform electric field. Well we need to write that displacement vector. So q e i hat is the electric field and that's going to be dot product with whatever d one vector is. Well we know the magnitude of d one. It's just d and we know the direction of d one. It is minus i hat. So we have a number q. We have a number e. We have a number d and we have two vectors, i hat and negative i hat. So we can pull all the numbers out in front, q e and d. And then we just have i hat dotted into negative i hat, which I hope by now looks familiar. This should just be a negative one. i hat dot i hat is one and there's a minus sign in front of that product, so you wind up with minus one. So we wind up with the work being done by the field is negative q e d. Let's think about this for a second. If the charge were positive, if q is greater than zero, then we know that the electric field wants to move the charge from the left to the right. That's the direction that the electric field would want to move it. But that's not the direction that the electric charge moves. No, in this problem, the charge was moved from right to left. In other words, this positive charge of q is greater than zero, is being moved by the applied force in a direction that takes it from lower potential energy to higher potential energy. So the change in potential energy in this particular situation has been greater than zero. And that must mean that the work done by the field, which is equal to the negative of delta u, is correspondingly less than zero. That is, work has been done on the field. Work has been done on the field. Now, the good news is that that means that we have increased the electric potential energy, u, of the charge. The applied force did some work. And now it has stored energy in the form of potential energy in the field by raising it to a higher potential energy location in the field. And this is equivalent in the gravitational analogy to taking a tennis ball and pushing it up away from the center of the earth. You are raising it to a higher potential energy. And to do that, you must apply an external force to work against gravity because gravity wants to move it to lower potential energy. The electric field wants to move charges to locations of lower potential energy. And in this case, the applied force is moving in the opposite direction, that the field would like to move it. And that must mean that the field is gaining energy. It's having work done on it. So work has been done by the applied force on the field. Now, instead, if q had been less than zero, and we have that same picture of the electric field and the charge moving to the left, well, if this is a negative charge, it wants to move against the direction of the electric field arrow. So if q is less than zero, in this case, delta u is less than zero. We've gone from a region of high potential energy initially to a region of low potential energy finally for a negative charge. And that's exactly the direction that the field wants to move it. So then we know that the work done by the field in this particular situation, which is, again, equal to negative delta u, is the negative of a negative number, which is a positive number. So for a negative charge, we say that the field has done work on the charge. So one of the key things that you're going to have to really kind of stop and pause and think about is what the work represents and how that representation changes with the sign of the charge. For the picture I've drawn, where the charge is displaced against the direction of the electric field, if the charge is positive, it means the applied force has done positive work and it has stored energy in the field, which means the field has done negative work. And in fact, in this case, as in all cases, the work done by the applied force is the negative of the work done by the field. That's also true down here if this charge is a negative charge. The only difference is now that, in this case, because it's a negative charge and the displacement is to the left, that's the direction the field would like to move a negative charge. And so it's going from a correspondingly higher potential energy to one that's lower. That means that the work done by the field is positive and the field has done work on the charge and the applied force in this case is negative. So this is equivalent to you lowering a tennis ball with your arm in a gravitational field. In principle, what's happened is that the energy that had been stored in the tennis ball at the higher altitude is now put back into your arm by lowering the tennis ball in the gravitational field closer to the center of the earth. The field has done work on the ball in this case. And that energy has been put into you. You had to do negative work in order to go from a high potential energy to a low potential energy situation because that's the way the field would like to push the ball. Similarly, for a negative charge in this picture, negative charge wants to move to the left. It's going from high potential energy to low potential energy. The field has done positive work. The applied force must correspondingly have done negative work. So the result of this calculation is that regardless of what the charge is, that the work done along path number one is just negative QED. We don't know the sign of Q could be positive or negative. E is positive and D is positive. What about work two? The path number two, recall, was a vertical path down in the negative j hat direction by a distance d. So all we have to do is rewrite our force times displacement equation. Well, the only force that could be acting on this is, again, the electric force. And we're dotting that into d2 vector. Well, the electric field still has the same definition as above, QEI hat. That's where it points. And it's dotted into d times negative j hat. So if we group the constants together out in front, we have QED. And that is all multiplied by negative i hat dot j hat. Now, this is intriguing. This dot product is the dot product of a vector that points at a right angle to the other vector. So here is negative j hat. Here is positive i hat. Because these are along coordinate axes, by definition, this is a 90 degree angle. And recall that if you have a generic dot product, a vector dot b vector, that's the magnitude of a and b times the cosine of the angle between them. The cosine of 90 degrees is 0. So if theta equals 90 degrees, then a dot b is 0. And there is no dot product. It has zero length. i dot j, these are two right angle vectors. Their dot product is 0. And so the work required by the applied force or the field to move the charge along this path perpendicular to the electric field is 0. No jewels. It requires no energy to make this movement. And in general, this is going to be true. If you ever move at right angles to the electric field, it costs the applied force and the field no energy, because there's no resistance to motion. This is very similar to how in a gravitational field, if you move the tennis ball from left to right, parallel to the surface of the earth, never changing its displacement above the center of the earth, then no work is done against the gravitational field or by the gravitational field. It costs no energy to do that. So finally, we have w3, the work done on the third path. And we can kind of speed through this. We have qEi hat dotted into negative d i hat. And so we wind up again with negative qEd. And if that isn't a very comfortable calculation for you at this point, pause and make sure you can do this one very quickly using what's been learned in the calculation of work 1 and work 2 using dot products and the numbers that multiply them. So the total work done by the field is equal to work 1 plus work 2 plus work 3. Let's see if I can get my parentheses correct here. Work 3. Work 2 was 0. And both of these were the same for this problem. Negative qEd. So the total work done is negative 2 qEd. And this will be the negative of the work done by the applied force. So this is the subtle thing that I wanted you to learn about by going through this exercise. First of all, the work done by the field is equal to the negative of the work done by the applied force for conservative forces. You can figure out the work done by the field only by considering the displacements that are along the field or have some component along the field. Any displacement perpendicular to the field will cost no energy and therefore will have no corresponding work. So what's interesting about this is that the energy required to make the journey that I drew here, where you go left, down, left like this, is exactly equal to the energy required to make the following journey. Keeping in mind that this is a displacement d. This is a displacement d vertically. And this is a displacement d, again horizontally. This journey of 2d horizontally has exactly the same work required as the one shown on the left. All that matters are the end points of the path. And in fact, if you wanted to go ahead and make this complicated, you could draw one more picture. And I'll leave this as an exercise to you. Just use some trigonometry to figure out the components of the vectors. And you can see that this is true. If this is 0.1 and this ending point over here is 0.3 in this journey, and I draw 0.1 and 0.3 again, this path diagonally from 0.1 to 0.3 also has the same total work as this path over here, which we've calculated explicitly. OK, well, you know that this distance here is 2d. And you know that this distance here is d. So go ahead and calculate the length of this vector. This is the displacement vector in this problem, d vector. Go ahead and calculate the length of it. You have to write its components. It's got an i hat and a j hat component. Take the dot product with the electric force and see what work you get. And I promise you, if you do your math right, you're going to get exactly the same total work that we got here for this stepwise path. So you'll find that these are the same. And this is a critical observation. The work required to move an electric charge through an electric field from a starting point A to a final point B is the same independent of the journey you take through the field. Let me repeat that again. If you start a journey on point A and you want to go to point B in an electric field someplace else, any path you choose, as loopy or as straight as you like, will cost you the same amount of energy. The same work will be done by the field and correspondingly by the applied force to make you take that journey. This is why energy is so powerful, especially in the presence of conservative force fields like the electric force, where energy is conserved. You can take any path you like through the field, from A to B, whatever A is as your starting point, whatever B is as your ending point, it doesn't matter. Pick the simplest path you can write down mathematically because the work required to achieve that trajectory through the field is the same as the work going from A to B you take on any other path in the field. Very powerful concept. And it will help us to understand the potential of the point charge electric field and what happens there. Now that we've developed these concepts involving work and changes in potential energy and seen how they begin to relate to the electric field, we can begin to consider certain cases that we have already looked at before in a new light. And the very first thing that I would like to consider is the point charge electric field. And specifically, what I'd like to explore with you is changes in electric potential as one moves through the electric field of a point charge. So specifically what we're really interested in are these delta V's, these changes in potential energy per unit charge or electric potential as we move through the field. Let's begin with a picture. We can have a positive charge with its associated, radiating, electric field lines. This is just a single point charge. Because this is a point charge, we know what its electric field looks like. The electric field is equal to k times the charge that's emitting the electric field or receiving it if it's a negative charge, divided by the distance at which you're observing the electric field squared. And this whole thing has a unit vector r hat in front of it. Now let's explore some pieces of this as they relate to the point charge. If we are observing the electric field at a point p, we know that the convention is that we must draw the vector r from the source of the electric field to the point where we're observing the electric field. So that's r vector. Now a point charge has something that we can refer to as circular symmetry. That is, if we move to another point that is someplace else, for instance over here, we can call this p2. And the vector that points to this point p2 has the same magnitude as the original vector r. Then what this tells us is that the electric field at point 2 has the same strength as the electric field that our original observing point p. Why is this interesting? It tells you that there is a circle of radius r that goes around the point charge. And on that circle, the electric field has the same strength everywhere. Now the vector r is not exactly the same at every point. Although its magnitude may remain constant, making the electric field magnitude remain constant, its direction certainly changes. It is not true that r hat is equal to r2 hat. They clearly point in different directions. However, we can define a useful coordinate system for objects that have circular symmetry like this. Rather than using Cartesian coordinates, x and y and z, it's convenient to use something called circular coordinates for these particular problems. Now circular coordinates are exactly what they sound like. If I write down a Cartesian coordinate system, x and y, and then I inscribe a circle around the origin, for instance, and I pick a point p on the circle, of course, in a Cartesian coordinate system, I'm free to write a vector r, and its magnitude will be equal to the square root of x squared plus y squared. So that's the Cartesian relationship between the radius of the circle r and the coordinates in a Cartesian space, x and y in this case, because I'm only considering a two-dimensional space. But on the other hand, there's a perfectly useful alternative coordinate system which you can go ahead and think about. And in fact, it's very important to be able to use this coordinate system in problems with circular symmetry, and that is circular coordinates. So circular coordinates, as the name implies, are only useful when you are dealing with circular things or semicircular things. And circular coordinates, much like Cartesian coordinates, you have two coordinates that you must use to specify a location on a circle. You must have the radius r, and you must have the angle that you are at with respect to the x-axis, theta. So that's this angle here, theta. Now what about unit vectors? So these are distances. You can go one meter out from the origin, and you can go 35 degrees up in a positive counterclockwise direction from the x-axis. So we know that this r is always positive or 0, because r's smallest value that it can take is 0, putting it at the origin. And of course, its largest value it can take is infinity. Theta is positive in a counterclockwise direction. So as we move away from the x-axis like this, this takes us in increasing theta direction, from theta equals 0 to finally theta equals 2 pi. You are perfectly free to go clockwise. But if you do that, now you're going in the negative theta direction. So you can specify a point on a circle using circular coordinates, r and theta. And then there's one more thing that we have to define here, and that's the unit vectors that go with this coordinate system. There are two. r hat points along a radial direction, and it's positive in the outward bound direction. So to sketch this out, if I have my circle here, and I increase the size of the circle by some amount such that the radius moves outward in an outbound sense away from the origin, if this was r1 and this is r2, then I can say that here the circle has expanded in the positive r hat direction. That is, it has expanded in a way that makes the radius move outward from the original radial position. If the circle shrinks, on the other hand, so if instead I had started with this and then declined to something smaller, so now if I track a point on the radius of the circle from 1 to 2, this is moving inward, and inward equals negative r hat. Now similarly, if one is talking about angles, so if we look at an angular position of a point p, we can define a theta hat direction. And theta hat is positive when a point moves in the counterclockwise direction, and theta hat is negative when a point moves in the clockwise direction. So if I shift this point over here to p2, and I originally am at theta 1, and then I go to theta 2, I can say that the point p has moved here in a positive theta hat direction. It's moved counterclockwise around the circle. If, on the other hand, I had moved my point p2 down in a clockwise sense, then I would say that it's moving in the negative theta hat direction. So you can define directions and unit vectors, just like in Cartesian coordinates, but you have to bend your brain a little bit. Now instead of moving out along a horizontal axis or moving out along a vertical axis, now you're moving out along a radial direction, or you're moving in a circle around the origin. And if you're moving in the radial direction, you are said to be traveling in our hat or negative our hat, depending if you're going out or coming in. And if you are traveling in a circle, you are said to be going in a positive theta hat direction if you're moving counterclockwise and in a negative theta hat direction if you're moving clockwise. With all of this in mind, we can come back to the point charge. And again, we have its associated electric field. Why I've just bothered to go through all of this is because now it's crystal clear. When I talk about the electric field and I write our hat here, you know immediately that what I'm saying is that we're talking about an electric field that at any point around here on the same radius circle. So if I take the collection of all points at equal radii, r, r, r, r, r, r, r, r, and r, the electric field either points out away from the charge in the plus r hat direction. And it does that if q is greater than 0. Or the electric field points in the negative r hat direction if q is less than 0. So if we have a negative charge, the electric field lines point radially inward. They point in toward the origin. That means the negative r hat direction. And if we have a positive charge, they point outward away from the origin. And that's a positive r hat direction. And you can see already why circular coordinates are really useful for this problem. Because you do have circular symmetry in a point charge electric field. And so it's extremely convenient to simply be able to define things like a radial direction or an angular direction around the point charge. Now with that in mind, let's look at a simple question. What happens if I move some other charge q from a point A that's closer to the positive charge and then outward on some crazy path to point B that's further away from the positive charge in this picture? The question I would like to answer are energy related. What is the change in potential energy for this motion from A to B? What is the change in electric potential going from A to B? And these are all questions I can answer using the electric field of the point charge and some basic concepts that we've built up already with work and energy. Let's begin with the change in potential energy. We know from our discussion of work and energy that the work done by the field is equal to the negative of the change in potential energy moving from point A to point B in the field. So for instance, if q is a positive charge, little q, and I move it from A, which is closer to the big charge, to B, which is further away from the big charge, independent of the path that I take, all that matters is the difference in potential energies between these two points. This is a point of higher potential energy for a positive charge close to this central charge, and B is a point of lower potential energy further away from the central charge when q is positive. So in this case, I'm going from high potential energy to low potential energy. And since delta u is the u at the end of the trip minus the u at the beginning of the trip, the difference in those potential energies, I know that for q greater than 0, this will be a quantity that is less than 0. So the work done by the field will be positive in this case, because I'm taking a negative number, multiplying it by a minus sign to make a positive number, and that gives me the work done by the field. So in this case, the work done by the field will be greater than 0. And again, that's for this specific case. If I change the sign of the charge, you can go ahead and verify that these relationships flip. The difference in potential energies becomes positive, making the work done by the field negative. That is, if you have a negative charge here, you're actually moving it in a direction that the field doesn't want it to go in. It would like the negative charge to move closer to the positive charge, not further away. Now the nice thing about this is that we can relate the change in potential energy from A to B by the work done by the field. So let's go ahead and do that. Work, in general, is equal to force times displacement. But we have a super crazy path here. So what's the first lesson we have to take away from the discussion of energy earlier? Work, in a conservative force field, is independent of the path you take. So why take a crazy path? Why not take a simple path? Why don't we just pick a path that takes us from A to B in a nice, simple way? So for instance, it might be that we, A and B, lie along the same radial line. And so we can just move outward from A to B in a positive r hat sense, nice and simple. But we have another problem here. Unlike in the previous example, where the electric field strength was the same everywhere in space, here the electric field does not have the same strength everywhere in space. As you move radially further outward, the electric field strength declines. So we can't just take force and multiply by displacement. If this is the displacement d in the direction r hat, we don't get to just multiply these two things together. We have a bit of an issue here. So how are we going to handle this? What we have to do is break the path into tiny pieces. And we have to sum up from one to the total number of pieces, the force on piece i dotted into the displacement vector for piece i. Now, I could make these displacement vectors very tiny. So I could make these little delta d's. And you see where this is going. Bump, bump, um, calculus. So to be a little bit careful here, what I'm going to do is I'm going to note that the displacements are all little pieces of a radial path. So I'm going to change my notation here a little bit. We're going to have the force on each piece dotted into the little piece of a radial line that each chunk of the path represents. And we're going to integrate these from point A to point B. And that is going to give us the total work done by the field. So we have to bust out calculus to answer this problem. All right, well, what's the force? The force that big q exerts on little q is just given by Coulomb's law. Nothing magical there. It depends only on our distance from the big charge q and the direction in which the direction that points from big q to little q. So this is just Coulomb's law for a point charge. Nothing magical here. What about dr? Well, every chunk of the path, which I can write up here, takes us in the positive r hat direction along some little piece of the path dr. So we can separate direction from magnitude for each of the pieces and just multiply them. We're always moving in the plus r hat direction and always by a little distance dr. So now I have direction and magnitude separated out in my vector. Let's return to the integral. The work done by the field is equal to the integral from the origin of the path point A to the end of the path point B. Well, the beginning of the path is just at some radial distance r a from the big q. And the end of the path is just at some radial distance r b from big q. We know the form of the force. It's kqq1 over r squared r hat. And this whole thing is dotted into r hat dr. So we're nearly there. These three things are constants of the integration. These are unit vectors that are dotted into one another, and they point in the same direction. So their product is 1. And that leaves us with a very simple integral to do. Pull the constants of integration out in front. We're going from the r a to the r b. We have 1 over r squared and dr. And you can go look this up or you can remember how to do this integral. But the bottom line is that the answer is negative kqq1 over r evaluated from r a to r b. And then finally, writing the work done by the field and all of its glory, moving around a point charge, we have kqq, don't forget that minus sign in front, times 1 over r b minus 1 over r a. And finally, we know that this must be equal to negative delta u between a and b. So let's go ahead and write delta uab as kq, little q, 1 over r b minus 1 over r a. That's the change in potential energy from any point a to any point b in a point charge electric field. And in fact, you can pick points that don't lie along the radial direction. a and b don't have to lie along the same radius. They can lie along different radii. Anytime you make a movement along your path along a radius, then you have some work. Anytime you make a movement along the path perpendicular to the radius in a circular direction, theta hat, no work is done because theta hat and r hat are perpendicular. And so f vector dot displacement vector is 0. So it doesn't matter whether you start out at points a here, which is r a in this direction, and then end over here at point b, which is r b in this direction. In order to get from here to here, you have to make a bunch of radial movements outward to b. And you have to make a bunch of movements perpendicular to the radial direction out to b. The perpendicular movements along theta hat count for nothing in terms of work. They're all free. It's only the movements along the radial direction. And so all you have to do is know the initial radius and the final radius and plug them in here. And it doesn't matter whether you started to the left of the positive charge and end somewhere over on the right of the positive charge, the work is the same as if you had started on one radial line and moved along that radial line by the same amount from r a to r b. All right, well, this is great. So we have the change in potential energy. And then finally, we can do some definitions here. So let's do some definitions. It is convention to define the potential energy at a radius of infinity, that is u of infinity at r equals infinity as 0. So u of infinity equals 0. This is similar to the convention in gravitational potential energy. When you are an infinite distance away from a mass, you have no gravitational potential energy. So just as in gravitational potential energy in electric potential energy, being an infinite distance away from the charge causing the electric field means you have no potential energy with respect to the field of that charge. And this is convenient because then we can define all potential energy differences relative to infinity. So for instance, if I had moved from r a equals infinity to r b equals r, some finite radius away from the charge, then delta u a b equals u b minus u a. Well, u a, by definition, is u infinity in this case. And that's 0. So this is just equal to u b. And that is all equal to k q q. We have 1 over r b, which is just 1 over r minus 1 over infinity. And 1 over infinity is 0. Very convenient. That's why you choose infinity to be the 0 point of energy. And so we wind up with k q q 1 over r. So the potential difference between infinity and a finite point r away from a point charge has a very simple formula, k q q over r. And you see it's a scalar. It has no direction. It shouldn't have direction. It's just energy. And this is very helpful because then we can think about changes relative to infinity rather than changes relative to some other location in space. And this will come in handy in a moment. Now, let's return to our friend electric potential difference, which is the change in potential energy divided by the charge that you're moving around in the electric field. So we can bust out our formula. We have k q little q over q 1 over r. And again, this is delta u with respect to r equals infinity, or r a equals infinity, our starting point. And we see the q's cancel. And so we get this very nice formula that the change in potential energy between, let's say, infinity and some finite point r is just k q 1 over r. Now, of course, if you have delta v between some finite points a and b, then this is going to be, of course, k q 1 over rb minus 1 over r a. Now, let's look at what we've learned. And we'll begin by making a table of numbers involving the charge of the magnitude of the charge, the sign of the charge that's causing the electric field, the little charge q that's responding to the electric field, the relationship between r a and r b, which I'll represent here with a blank in between, the work done by the field, the change in potential energy, and the change in electric potential. So let's imagine we have a positive charge causing the electric field, and we have a positive charge moving in the electric field. If r a, if our starting point is greater than r b, so that is if r a is a much bigger number than r b, and we start out far away and then move in this charge, little q, closer to big q, then if you look back at the work done by the field, which we can find right here, if q times q is a positive number, this is all positive out in front, and then the minus sign makes it negative. Now, if r b is a number that is smaller than r a, 1 over a smaller number is bigger than 1 over a bigger number. So if r b is smaller than r a, then you have a big number minus a small number, and you get a positive number inside of these brackets. And so overall, the work done by the field will be a negative number, which is what we expected. The change in potential energy will be positive, and the change in the electric potential will also correspondingly be positive. Now, what if you have two positive charges, but r a is a smaller number than r b? That is, you start out closer to big q, and you end further away. Well, you'll find the signs flip. Of course, the work done by the field is positive now, because this is a direction that the positive charges electric field would like to move the other positive charge. And then the change in potential energy is negative, and the change in electric potential is also correspondingly negative. What if q is positive, big q is positive, and little q is negative? And again, we start out at an r a that's greater than r b. Well, in this case, you wind up, if we go back and look at the work. So the work done by the field, you now have a positive number times a negative number. So k q q is negative, and there's a minus sign in front of it, which makes it positive. You have r b, which is a smaller number than r a. So this is a big number minus a small number, and you get a positive number, and overall everything is positive. So the work done by the field is positive. And correspondingly, the change in electric potential energy will be negative. Now let's look at electric potential difference, because this is where things get interesting and may feel a little bit counterintuitive at first. The change in electric potential between points a and b is given by this formula here, k q times the difference between 1 over r b and 1 over r a. Now for this example, q is a positive number. We have that the distance b is smaller than the distance a. So 1 over r b is a bigger number than 1 over r a. And so we have this difference being positive. q is positive, k is positive. And the delta v between a and b is positive. What's going on here? Well, you can understand this from the formula delta v equals delta u divided by q. If delta u is a negative number, and q is a negative number, well, then delta v must be a positive number. So what we see here is that the sign overall in front of delta v is controlled by two things. The charge of the electric charge that's causing the electric field in the first place. So if that's a positive charge, then q is a positive number. And if it's a negative charge, then q is a negative number. And the other thing that controls it is the direction of the journey. If we go from a point that's further away to a point that's closer by, then we wind up with a positive number overall on this difference. And those two signs, the sign of q and the sign of this difference, will multiply together and give you the overall sign of delta v. So you do have to be a little bit careful. Delta v doesn't just track the sign of delta u because delta v involves the division of a charge out of the change in potential energy. And if that charge is negative, it can change the sign of delta u to give you the sign of delta v. So however you choose to attack this, keep in mind that you do need to think carefully about signs in these problems and what it means to have a change in electric potential energy versus what it means to have a change in electric potential itself. We can complete this picture by looking at the positive charge being the source of the field, a negative charge being in the field, and the negative charge taking a journey that goes from a distance that's closer by the positive charge to a distance that's further away. Well, this is opposite the direction that the field would like to move the charge. So we know that the work done by the field will be negative because the work done by the field is the negative of the change in potential energy. We know that this will be positive. And finally, we can look at the change in electric potential itself. Again, q is still positive, but now we're going from A which is close by. So 1 over rA is a bigger number than 1 over rB. And this overall gives you a negative number when you subtract these two ratios. And so you wind up with A negative deltaV. We can look at the case of a negative source charge and a negative electric charge that's moving around in the electric field of the source. We can consider the case where we start further away and then move in close. The negative charges don't want to be close together. So that's the opposite direction that the field would like to move them. So we wind up with negative work done by the field. And in this case, we wind up with a correspondingly positive increase in potential energy. And again, if we look at the equation for the change in electric potential, we have a negative source charge. So that's a negative number. B is smaller than A. So we wind up with a big number minus a small number. This is positive. And so we wind up with an overall negative change in electric potential. And then finally, we have two negative charges. And this time, we go from a close by point to a point that's further away. This is the direction the field would like to move the negative charges away from one another. That's a corresponding drop in potential energy. And again, if we look at the equation for the electric potential difference, we have a negative charge on the front. So this product is a negative number. And inside, we have the difference of a small number and a large number, which gives us an overall negative quantity, and so delta v is positive. So the message here is that you do need to pay careful attention to the signs of different quantities involving energy. If the charges move in a way that the electric field would like to push them, work will be positive. If it's the other way, work will be negative. That's the work done by the field. Correspondingly, if the work done by the field is positive, the change in potential energy will be negative. And if the work done by the field is negative, the change in potential energy will be positive. And then you need to look very carefully at the relationship between delta u and delta v, especially because charge enters one more time in that relationship. And you need to be very careful about moving your signs through these equations as you calculate changes in potential energy and work done by the field or work done by an applied force. For the final subject of this lecture, I'd like to discuss something which can be generically called the energy of configuration. Basically, the question we would like to answer is the following. If it takes work from an applied external force to make an arrangement of charges, what energy has been stored in that arrangement of charges? So if I have to do work with some applied force, which we know from conservation of energy for a conservative force field is equal to the negative of the work done by the field, I want to know what energy is stored in the charge configuration. That's the question we would like to answer. To begin with, it's important to remember how energy adds. Let's say that I want to move a charge. And we'll call it charge one. And let's say that moving charge one carries a penalty associated with it. That is, I have to apply a force, w applied one, that's greater than zero. And then I want to move a second charge. And that requires an applied force, w app two, which is also greater than zero. It may not be the same number as w applied one. What is the total work done in moving charges one and two? Well, this is the great thing about energy. It's just the sum of the work applied to move charge one and the work applied to move charge two. And of course, if I move more charges three, four, five, all the way up to Avogadro's number, then I can keep going. And in general, generally, the total work required to arrange, for instance, a configuration of charges is equal to the sum from i equals one to the total number of charges, for instance, that you'd like to arrange of the applied forces for each charge to be moved. This is the great thing about energy. Energy is a scalar. Energy is a number. No vectors that you have to worry about. If you want to know the total energy required to do something, you just add up a bunch of numbers. The trick, of course, is figuring out the numbers. So let's do an example. Let's say that I have a region of space where I can organize charges. And let's say that anywhere outside of this box I've just drawn is infinity. So any place out here I'm going to call infinity, just as a conceptual exercise. So if I put a blob out here and I ask you, what is its radius? So what is it? This is blob number one. What is radius one? You would say infinity. If I have blob number two over here and I ask you, what's its radius? You would say infinity. Anywhere outside this box is infinity. I've obviously condensed this not to scale. Now what I would like to do is I would like to take four charges. And I would like to move them inside the box. So let me go ahead and erase these because they're not very helpful. And let me put these labels a little bit closer here. OK, great. So here's my game. I'm going to move each of these charges in from infinity to a place in the box. And just for the sake of convenience, let's go ahead and draw some Cartesian coordinates in here. And I will conveniently use the graph paper to do this. So I have an x-axis and I have a y-axis. Awesome. Now there are no charges anywhere in the box. Zero electric fields inside the box. And because all these charges are an infinite distance away from each other, their potential energies in their respective electric fields is zero. Zero is the correct answer. If you are an infinite distance away from a charge and you are another charge, you have no potential energy with respect to the other charges electric field. OK, so let's do some work. Let's move charge one inside the box. So I'm going to take it on some crazy path. And I'm going to land it right at the origin. Now, here's the million dollar question. How much work was done by me, the applied force, in moving charge one from infinity to the origin? So if this is equal to the negative of the work done by the field in moving from infinity to the origin, charge number one, and that must therefore be equal to the change in potential energy, which is u at the origin minus u at infinity. We already know that this is zero. Now all I have to do is figure out what's the electric potential energy at the origin? Well, let's go back and look at our picture. All the other charges in our little universe here are out at infinity. And the charge we moved is at the origin, an infinite distance away from all the others. There are no electric fields in the box. Therefore, there was no force to overcome. And so I hope by now I've convinced you that this has to be zero as well. And so the answer to the first question, how much work is done by an applied force moving charge one from infinity to the origin in a region that's originally free of electric fields? The answer is zero, the same as the work done by the field, because there is no field to begin with, and the same as the change in potential energy, which is zero at zero at infinity, and when there are no electric fields anywhere along the path that you have to move a charge nor at the location where the charge ends, then there is no potential energy change that occurs. So this is a big fat goose egg, zero. OK, that one wasn't so bad. Well, let's grab charge number two. And let's move charge number two into the box, a distance A away from charge number one. So here we go, in the box, taking some crazy path, and we're going to go ahead and land right there. Now, charge two is a distance A away from charge one. What was the work done by an applied force to move charge two to its location? Well, again, this will be the negative of the work done by the field to move charge two, and this is going to be the change in potential energy for charge two. Well, let's figure these things out. First of all, we ended at this point, we'll call it u2, and we started at a point at infinity. And again, the initial potential energy at infinity is a big fat goose egg at zero. So all we have to do is figure out what's the potential energy now of charge two at its final resting location inside the box? This time, was there an electric field inside the box? If you said yes, you're correct. Of course, we've just put an electric charge inside of the box, charge number one. Now we've moved charge number two inside the box. So when charge number two came from infinity to within some finite range of charge one, it started to feel its electric field. And of course, we've dropped at a distance A away from the electric field, away from charge one. So it's definitely experiencing an electric field of some magnitude, because it's a finite distance away from charge one. And if we dig back now to the formula for the change in potential energy, a particle that starts at infinity and ends a finite distance away from another charge, it's this formula right here. So the change in potential energy is k, q, big q. That's the source of the electric field. Little q, that's the charge that's feeling the electric field, and one over the distance between them. We have that this will be equal to k, q1, q2, 1 over A. And that's it. Well, this is a fun game. So we have now stored energy in the system, because it took work to move charge two from infinity into the game board and place it a distance A away from charge one. That cost energy from the applied force, and that energy has been stored in the field, in the form of potential energy. That's the beauty of a conservative force field. You can store energy as potential energy by doing work against the field. OK, we are two terms on our way to solving this problem. Let's move on to a third term. So now we're going to move charge three. There you are. And we're going to take it on some crazy path. Remember, the path doesn't matter. All that matters is the final destination and the original starting point and their mutual distances away from the charge. In this case, the charge is, because we have just dropped charge three, a convenient distance A away from charge two, and it is this distance, which is conveniently the hypotenuse of a right triangle, away from charge one. Well, I would hope by now you can figure out what this is. This distance here is root two A. To figure this out, you just have to take A squared, add it to A squared. That gives you two A squared. Take the square root, you get root two times A. So simple Pythagorean theorem. OK, let's figure out what the applied force had to do in terms of work to get charge three onto the game board. Well, again, this is going to be the negative of the work done by the field in moving charge three, which is going to be the change in potential energy for charge three. Now, this is where things get interesting, because charge three is going to land at that point we drew on the game board. It started at infinity, so we know that the starting potential energy was zero, because it was an infinite distance away from all other charges. But now as we move it into the game board, it's experiencing the electric fields of two charges, charge one and charge two. Charge two only had to move against the field of charge one, but charge three has to contend with the fields of both charges one and two. And you see what's happening. It's getting a little harder every time to move charges into the board. Something, some bigger energy cost is happening, whether it's coming from the field or whether it's coming from the applied force. Somebody's paying the piper on this. All right, so the first one's free. But after that, every charge costs you a little bit more. So what is u3? Well, we're working against the fields of both charges one and two. So it can't be that there's only one charge that's going to be involved here. We have to break this into pieces. This is going to be the potential energy of three relative to one added into the potential energy of three relative to two. Remember, energies add. They add like numbers. So all you have to do to get the total potential energy of the system is add the individual potential energies of charge three with respect to charge one and charge three with respect to charge two. And so that's just going to be k, q1, q3, 1 over, this is going to be r. So this is root 2a plus k, q2, q3, 1 over a. So going back to our picture, the potential energy of three with respect to one is going to be k times the charge of one times the charge of three divided by the distance between them, which is root 2a. And the charge of three with respect to two is going to be k times the charge of two times the charge of three divided by the distance between them, which is just a. If all these charges are positive, so if q1 is positive, q3 is positive, q2 is positive, we know the distances are positive, a is a positive number, k is a positive number. You see, every time this is costing the applied force more and more work, more and more energy to do this. All right, well, we're almost there. We just have to get charge number 4 moved on to the game board. So we'll do this one quickly. You get the idea. You see how this works. We move charge 4 in, because I've been constructing a square here out of these charges. That distance is a. That distance is a. And the distance from charge number 2 is root 2a, because that's a different hypotenuse of a similar right triangle. And so all we have to do now to get the work required to move charge number 4 into the game board, which is a negative of the work done by the field, which is delta u4, is just going to be the sum of u4 1, u4 2, and u4 3. And we can just write that down. So that's q1, q4, 1 over a plus k, q2, q4, 1 over root 2a plus k, q3, q4, 1 over a. And then the total work that an applied external force has to do is just the sum of the applied forces to move charge 1, 2, 3, and 4. And each time we moved a charge, we had more electric field to contend with. So this was just 0. That first charge was free, but the next three were not. And if these were all positive charges, it kept costing more and more energy to move a charge onto the game board every time. So you can imagine if you're trying to store charge on a device and you're trying to put like Avogadro's number of charges on the device, the first one's free. If the device has no electric fields to begin with, the first one's free. The second one costs you a little bit of energy. The third one costs you more, and so forth. And by the time you get to that last one, you really have to push. You need to have some kind of device that's capable of pushing very hard to move that last charge. And we'll explore devices like that very shortly. All right, so you can just sum up all of the works. So you have k, q1, q2, 1 over a. That was the work required to get charge 2 on the board. Plus k, q1, q3, 1 over root 2a, plus k, q2, q3, 1 over a. That was the work required to get charge 2 on the board. And then finally you have k, q1, q4, 1 over a, plus k, q2, q4, 1 over root 2a, plus k, q3, q4, 1 over a. And that was the work required to get, so this was 2, 3, and charge 4 onto the board. And you see it grows every time. Every time we move another charge on, we have to contend with, if we move the nth charge onto the board, we have to contend with n minus 1 electric fields, and all the potential energies associated with those. Now of course if some of these charges are opposite sign, then that will help a little bit, because they'll want to attract charges. Like if you have negative charges on the board, they'll want to attract positive charges. And that will actually gain energy to the applied force rather than lose energy from the applied force. But you get the idea. You can see here that there's a lot of stuff that you have to do. And so it becomes convenient, of course, at some point, to turn to calculus to try to help you with a problem like this. And in fact, as we begin to explore devices called capacitors, which are capable of holding a certain capacity of charge on their volume, you will see how calculus will come in handy when trying to deal with moving little bits of charge around a system, especially infinitesimal little point charges and lots of them. So this sort of concludes the discussion, the basic introduction to the concepts of work and energy and changes in potential energy and electric potential difference that we will be dealing with going forward. And the key ideas to walk away with from all of this is that if you want to know the total work, then all you have to do is sum the individual works. If you want to know the total change in potential energy, all you have to do is sum the individual changes in potential energy. And because electric potential difference is related to electric potential energy, the same rule applies. If you would like to know the total amount of electric potential change that you experience, all you have to do is sum the individual changes in electric potential. So this is very helpful because all of these are related to energy in one way or another. All the rules of energy conservation applies. So conservation of energy is a concept we're going to exercise a lot going forward. And again, because work and potential energy and electric potential difference are all related to energy fundamentally at its heart, conservation of energy in a closed system with conservative forces, which is what we're dealing with, is in full effect. And so we're really going to exercise this going forward. That's a core concept that we're going to really beat to death in the rest of the course.