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# PHYS 201: Fundamentals of Physics II

## Lecture 5

## - The Electric Potential and Conservation of Energy

### Overview

The law of conservation of energy is reviewed using examples drawn from Newtonian mechanics. The work-energy theorem is derived from first principles and used to initiate a discussion of the vector calculus underlying the law of conservation of energy.

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html## Fundamentals of Physics II## PHYS 201 - Lecture 5 - The Electric Potential and Conservation of Energy## Chapter 1: Review of Electrostatics [00:00:00]
q, here’s _{2}q. And instead of these discrete charges, you may have a line charge too, maybe λ coulombs per meter. That guy is here. They are fixed. They are not dynamical. The object of interest to us is this charge _{3}q, which can actually move, allowed to move.The question we ask is, do we know what this charge will do? And the answer is, yes, we know what it will do, because if it’s sitting here and we have already computed the electric field everywhere — you know how to compute electric field. Add the field due to this one, that one, that one, add that. That is some electric field sitting here. Then we know, the force on this charge we’re interested in, this
Now this is something you could have done on the first day, because I think on the very first day I told you about Coulomb’s law and the principle of superposition. If you put them together, you know the force on a charge ## Chapter 2: Review of Law of Conservation of Energy [00:03:50]There, if you have the following question: here’s a spring and here’s a mass, the spring likes to have that length and I pulled it to way over there and I released it, and right now it happens to be here, going at some speed. I want to know what velocity it will have when it is here, so this position is v, this position is _{1}x. I want to know _{2}v. That’s a reasonable thing you can ask. Unlike the electric force, the force on this mass is due to the spring. Apart from that, it’s the same situation. So one way to do that is to say, I know _{2}F = ma, so the instant you start the problem here, I know the acceleration because the force of the spring is -k times x_{1}. Once I know the force, I know the acceleration. I can find the new velocity, new position, keep on doing it, or I can solve the equation, d/^{2}xdt is −^{2}kx/m, and I can eventually end up here. Then I can find the velocity here. But we all know that in this problem, there’s a shortcut, right? Yes? Anybody in the last row want to tell me what the shortcut is? Yes?
d/dt derivative is the v derivative, followed by the derivative of v with respect to t. v derivative will give you mv and then there is that. This becomes F dx/dt.This means in every tiny instant of time, the force times the change in the position is equal to the change in this quantity, which we know is called kinetic energy. So if you multiply both sides by mv/2 = integral _{1}^{2}f(x)dx from x to _{1}x. And this is called the work energy theorem. The work energy theorem tells you something you may expect, namely, when a force acts on a body, it’s going to accelerate it. It’s going to accelerate it, it’s going to change the velocity. You’re going to ask, “How much velocity change do I get for a certain action of the force?” and the statement is, the force pushes the body from _{2}x to _{1}x, then this is the change in this quantity called kinetic energy. This is true no matter what the force is. It’s true for friction, it’s true for spring, it’s true for everything, because Newton’s law is always valid. But if you want to extract from this a law of conservation of energy, you have to do a certain manipulation and that’s what I’m going to describe._{2}## Chapter 3: Deriving the Work-Energy Theorem and the Law of Conservation of Energy [00:08:13]So let’s go back here and try to extract from it a law of conservation of energy. So I’m going to write here k is integral _{1}F(x) dx, from x to _{1}x. Now it’s a fundamental result in calculus that integral of any function of _{2}x from the point x to the point _{1}x can be written as _{2}G(x) - _{2}G(x), where _{1}G is any function whose derivative is the F I’m integrating. For example, if F = x, G = x/2, because derivative ^{2}x/2 is ^{2}x. In fact, it’s not just x/2 . You can add ^{2}x/2 + a constant, but the constant is not important, because they’ll cancel between this and this. So you can pick the solution without a constant, okay?^{2}Now what does that mean? Let’s rearrange the expression so I get G(x) = _{2}K − _{1}G(x). That tells you the combination _{1}K - G does not change as the particle moves, but this doesn’t look very nice. We want to get the old law of conservation of energy in the standard form. You define the function U(x), you simply - this G(x). It’s a very trivial change. Then you can say K + _{2}U(x) = _{2}K + _{1}U(x), and that is the energy that does not change with time. So the relation between _{1}U, which is called the potential energy, and the force, is the following. The force is - the derivative of the potential energy.The - came because U(x) with an integral of _{1}F(x) dx from x to _{1}x, there’s a - sign here. Again, the - sign is present because _{2}F is defined to be - the derivative of U. So you can go back and forth from potential to force and force to potential. If you knew the force, you can integrate it to get the potential. If you knew the potential, you can differentiate it to get the force, and more importantly, you have this great conservation law, which is very helpful in the spring problem, because if you know the potential energy, once and for all, it’s kx/2, you balance the kinetic and the potential, you can find the velocity at one other point. Now why doesn’t this work for friction? Friction is the one dimensional force. Yes?^{2}
That does not mean you cannot find the work due to friction. If you know it’s moving to the right, then the force is to the left. You can integrate that force, as long as it is moving to the right. But if it’s going right, left, right, left and oscillating, then you cannot integrate it once and for all. That way doesn’t work for friction. Now here’s an even more powerful example of the law of conservation of energy. You have this famous roller coaster and you are here. I say “you are here,” because you’ll never get me on top of one of these guys. So you are here and you got some speed, and I want to know your speed when you are here. Again, it follows from Newton’s laws. You can calculate from Newton’s laws, because at every point, wherever you may be, there’s the force of gravity acting this way and there is the perpendicular force from the track acting this way, and the perpendicular force is such that it cancels this part of gravity, and this part of gravity will accelerate it and you can find the velocity here. Keep doing this, you can find the velocity at the end. But that’s not how we do this. We use the fact that the gravity force allows us to define an energy, and I’ll come to that in a moment. Then we say kinetic + potential is kinetic + potential in the beginning and at the end, and we can simply find the velocity here if you knew the potential energy here. All right, so the law of conservation of energy is also a very profound result, which has survived all the revolutions of quantum mechanics. After quantum mechanics, we don’t like to think in terms of force, and we don’t like to think in terms of trajectories. You don’t even know for sure where the particle is. But amazingly, the total energy is always conserved in the problems we study. So the law of conservation of energy is very profound and it’s also very useful for computations. Now the next question I have is, if I go to two dimensions — and it’s the same as in three dimensions. I’m just going to go to two dimensions — what does it take to get the law of conservation of energy? So I’m going to mimic this derivation in 2D and see if I can pull out a law of conservation of energy. So here’s what I will do. So in two dimensions, let me start this time with the kinetic energy. Kinetic energy is ½ v has got 2 components, x and y, therefore it’s ½mvx + ^{2}vy, but I’m going to write this ½^{2}m v⋅v. Then let me ask, what’s the rate of change of this kinetic energy?You’ve got to take derivatives. Now you may be a little queasy about taking derivatives from this dot product, but you can check by expanding the component, just like ordinary product, and you can take derivative of this guy times that guy, + derivative of that guy times this guy, and they will both be the same, and you will find it’s equal to K is _{1}m times dr — let me see — I’m sorry. I want to keep the dv/dt here, then I want to write this v as dr/dt. That’s right. So let me do the following, excuse me.Let me do it again. K = integral of _{1}F⋅dr from the starting point r to the ending point _{1}r. In other words, the change in kinetic energy in one dimension is just force times distance for tiny distances. In higher dimensions, like in 2 dimensions, if the particle is moving in the 2 dimensional plane along some path and it moves a distance _{2}dr, and at that point, the force is in that direction, you should not simply take the product of the length of the force times the length of the distance, but that times the cosine of the angle.Why should it take that? Because that’s what gives you the rate of change of kinetic energy. Just go to the kinetic energy and it suggests to you that you take this product and define it to be the work, because if you define it that way, just like in 1D, the work done by the force will be the change in kinetic energy. You guys with me now? In spite of this little mess up here, this ma is my force. The K and this is the integral. All right, so now it looks pretty good, except for the complication. You have a dot product here. You have this nice result that _{1}K - _{2}Kis the integral _{1} F⋅dr from rto _{1} r. So let’s imagine where _{2}r and _{1}r are. This is say where you started, and maybe you ended up here at _{2}r. You’ve just got to integrate this _{2}F from r to _{1}r. Let’s imagine there is no friction. Let’s imagine _{2}F depends only on the position. What’s my problem now? What problem do we have? Yes?
r _______._{1}
r. Anybody know what’s the problem? Yes?_{2}
r. Because if it had this form, we are done. You agree? If it had this form, then again you write_{2} K+ _{2} U = _{2}K + _{1}U. But no one’s telling you that is true. If it was true, you are in business. But it’s a very unreasonable thing to ask of this force, that its integral from some point to another point, does not depend on what path you integrate it on. In other words, you can take that path, break it up into segments _{1}dr, and on each segment, the force is some variable. You take F⋅dr of that little segment, add it up over all the segments and make the segments vanishingly small. That’s the meaning of this integral. And you’re supposed to get the same answer, no matter what you do, including going like this, or anything else. How can they all give the same answer? In fact, I’ll give you a simple example where it’s simply not true, so the answer depends on the path. In fact, just about any force you pick randomly will not have the property, the answer does not depend on the path. Answers typically will depend on what path you take. So let’s take a simple force, F = iy. I can think of more complicated ones, but I don’t have time to do them.So here’s a simple one: force depends on position. It points in the
Now let’s take the other path, go like this and then like this. On this section, we have the same problem. Force is to the right. You’re moving straight up. So this is a force for which it’s not true, and I bet you that if you just randomly selected forces in the 2 dimensional plane, they will never have the property that the integral is path independent. So it looks like we are onto a hopeless task, that the answer cannot depend on the path. So maybe there is no problem in which you can derive a law of conservation of energy. Of course, you guys know, you can derive a law of conservation of energy for a family of forces. In fact, the answer is, there’s an infinite of numbers of forces you can devise for which this is true, namely, the answer does not depend on the path. Before writing those forces, let me explain to you one other way to write the condition. The condition, namely of a conservative force, is that the integral from here to here is the same as the integral from there to there, from 1 to 2 along path So what I’m saying is, The theorem — if you want, you can call it a theorem, you can call it whatever you like. It’s a fact — number 1, here is a recipe for finding conservative forces. Take any function y) times cosh(x), doesn’t matter. Take a function. Then the force that I want is the property that F = -_{x}dU/dx and F = -_{y}dU/dy.In other words, I’m going to manufacture a force whose x derivative partial means keep y as a constant, take the derivative with respect to x. That’s 3x and ^{2}yF will be -_{y}2x. ^{3}3x and -^{2}y^{2}2x. Let me see if I got that right, because you guys have been catching me too many times. Okay, I think that’s all right. This force is guaranteed to be conservative. I will show you why this is true.^{3}yWhy is this magic working? But let me write it in another way that’s more compact. So the force I want to get, written as a vector, is i times −du/dx + j times −du/dy. That’s the force I claim is conservative. We’re going to write that as - grad U. This is called the grad and it means gradient, and this symbol is shorthand for this. There’s nothing more to it. Gradient is a machine. Derivative is a machine. You give it a function, sin x, it gives you an output, cosine x. you give it x squared, it gives you 2x. Gradient is a different kind of machine. You feed into it a function U of x and y, and it gives you a vector in all of the xy plane obtained by taking the two derivatives, one along x and one along y, and assembling them into a single vector.So let’s see for this particular j times 2x. Amazingly, it is claimed that this force, when you integrate it from one point to another, answer will not depend on the path; it will depend only on the end points. So we have to understand what makes this work. Before I show that, I will make another statement. I will show you why this works, but the second statement, I will not prove, which is that all conservative forces are of the form - grad ^{3}yU. In other words, not only is this an example of conservative force, that’s it. There are no more examples. Every conservative force you cook up will always be — yes?
When you go to physics seminar, you never assume the speaker has a problem. I mean, you assume that you don’t have a problem. If you don’t follow it, it’s a problem with the speaker. That’s what makes seminars exciting. It’s the closest thing we have to gladiator fights in the old days. You bring a speaker from somewhere, maybe Harvard, then you put them on the stage and you just roast them for the whole hour. Now you can do that to me just fine, because this is not even my discovery. Somebody else did it hundreds of years ago. You should relish that challenge. If this was a small seminar, you will have time to do it. I don’t have that much time, but I also don’t have that little time that if you don’t follow me, you cannot intervene. Okay, you have a question? No. Okay. Now we are going to understand why a force manufactured in this fashion is conservative, okay? So let’s calculate integral r. What is _{2}F ⋅dr? It is F + _{x}dxF. That’s the meaning of the dot product. I hope you know what I mean by this. I’m going from here to here, that vector _{y}dydr is some amount of dx and some amount of dy. It’s got jdy and idx. And F is some other thing, pointing in that direction, so F ⋅dr is this. Just look at this part. This is what you’re integrating. That becomes, except for this - sign, dU/dx times dx + dU/dy times dy. Now what is that? What is that? What do you think it stands for? In 1 dimension, if I took df/dx times dx, what am I calculating? Come on guys, you know what the derivative means. When you multiply the derivative by some dx, what do you get? Pardon me? Yes, go ahead. Take a shot. Yeah.
dx, but for small ^{2}dx, that’s all it is. That’s the meaning of the rate of change. If you multiply the rate of change by the change you get a change in the function. If you’re in 2 dimensions, if you move from one point to another point, you’ve got a function U that depends on x and on y. Therefore it changes, because you change x and it changes because you change y. It’s changing because of two reasons. This is the change due to change in x; that’s the change due to change in y. For very small, infinitesimal dx and dy, that is the change.That is just the change in the function U. Therefore integral _{2}F ⋅dr from 1 to 2 is U - _{1}U. So the magic is the following: you pick a function _{2}U, you can think of the function U as measured perpendicular to the blackboard. It’s like a height or something coming out of the blackboard. You cooked up a force which is related to the rate of change of the function U, so that if you add all the F ⋅dr’s, you’re getting the change in the function between two points. And that change in the function is independent of how you got there.Think of As long as you’re keeping track of how many feet have I climbed, you’re going to get only one answer. So not every K, therefore we get _{1}K + _{1}U = _{1}K+ _{2} U. In other words, the _{2}U that you began with is the potential energy for that problem. So what you’re manufacturing is pairs of potentials and forces. For every U that you dream up, there’s a force, which is obtained by gradient, - gradient of U. And in that problem, where that force is acting on particles, the energy that is conserved will be the kinetic, + the function U at the starting point, will be equal to kinetic plus potential at the end point. And the interesting thing is, there are no other conservative forces, except forces you can get by taking a U and taking its gradient.So all conservative forces fall under this category, and they all have a conserved quantity where the potential energy is the U(x) is - integral of _{1}F. The only difference is the integrand has become this F ⋅dr, rather than the F. So you can get conservation of energy in higher dimensions for those forces that come from a potential. _{x} dxU is called a potential. U is called a potential energy.So now we can ask the following question: suppose I give you a force. I know if it’s conservative or not. How are you going to find out? Now there are two options. One is, you are so smart, you can look at the function and say, “Hey, this is the gradient of some other function x^{2}y^{2}, F is -_{y}2x^{3}y, maybe you can guess that U is x^{3}y^{2}. But for more complicated functions, you won’t be able to do it. So there’s a process for testing any force to see if it’s conservative. And the process is the following.It argues as follows: dU/dx. F is -_{y}dU/dy. Then consider dF/_{x}dy, which is -d/^{2}Udydx. Then consider dF/_{y}dx, which is -d/^{2}Udxdy. And it’s the property of partial derivatives that the cross derivative, second cross derivative, is independent of whether you take dydx or dxdy. So let’s go to our test case, U = x. ^{3}y^{2}dU/dx = 3x and ^{2}y^{2}dU/dy = 2x. Now I’m saying, take the ^{3}yx derivative of this y derivative. That’s written as d/^{2}Udydx. y derivative of the x derivative gives me 6x. Then I say take the ^{2}yx derivative of the y derivative. This gives me 3x. That means 6^{2}x. So cross derivatives are always equal. Yes?^{2}y
x derivative of F? You want to see if _{y}dF/_{x}dy = dF. If that is true, it is conservative. And it comes from the fact that if _{y}/dxF is the derivative of some function U, then this mix derivative being equal is a diagnostic for that. In 3 dimensions, there are 2 more equations. We’re not going to use them, but I’ll just mention what they are. You get that by moving every index 1 notch. So this x goes to y, this y goes to z, this y goes to z, this x goes to y.There’s one more condition. And another condition, dz. I’m not worried about whether you write this down or not, but I do want you to know in 2 dimensions, you have to remember this condition. So this mathematic extends to all dimensions, but I’m doing it for the case of 2D.So let’s take stock of what has been done. What has been done is to think from scratch on how to get the law of conservation of energy in high dimensions. You start with the kinetic energy and you ask, why does it change? And you find it changes due to the force, and you find the change in kinetic energy is the line integral of But that doesn’t give you a work energy — it gives you a work energy theorem, but not a law of conservation of energy, because this integral can depend on the path. Then you say, maybe there are some functions for which the answer depends only on the end points. Then we get a law of conservation of energy, and we found out that we can manufacture such functions at will, by taking any function of There’s no other conservative forces which are not gradients of something. And the relation between the potential and the force, you should know now. It’s what you learned in 1D, suitably generalized to higher dimensions. Change in potential is this and the forces obtained from the potential by taking the gradient, the change in potential is obtained by integrating the force. If you ever get confused about + sign and - sign and so on, go back to the harmonic oscillator, for which the thing is 1 over — is So let me give you one simple example that we all know, which is the force of gravity near the earth. So here is the earth, so there is some U, is it equal to integral of _{1}F ⋅dr from 1 to 2? You can easily check that.If you go from here to here, straight up, then the integral of - Okay, but now, why did we spend quite a bit of time on conservative forces, because that’s not what the primary focus is? It was electricity and electrostatics. So we’ve got to ask ourselves, does the electrostatic force meet the test of a conservative force? Well, we know it will, because otherwise I wouldn’t have spent this much time building up the stuff. But let’s just verify that. It looks like a very formidable task. Let’s see why it’s a formidable task to verify it’s conservative. I’m claiming that you’ve got all kinds of these charges producing a field here and I want to prove that the line integral of this force on any closed loop is 0. Or I want to prove the integral on one path is the integral on any other path, for any 2 paths between any 2 points. I don’t have time to show all that, right? If the theorem is wrong, do you agree it may not be so hard to show that? What will it take to show it is not conservative? Yes?
E. The line integral of _{2}E + _{1}Eis the line integral of _{2 }E+ the line integral of _{1}E. _{2}E gives 0 on any closed loop, other will give 0 on any closed loop. The trick is going to be just do it for one charge. That’s it. If you can see it for one charge, you get by superposition that it’s true for any charge, because the field is additive and the integral is additive, okay. That’s the two parts. If you’ve got charges, the net _{1}e is E + _{1}E. The integral of _{2}E + _{1}Eis the integral of _{2 }E + the integral of _{1}E_{2}.That’s the way integrals work. If each one vanishes on a closed loop, you’re done. So I’m just going to take one charge. Here’s one charge and let me draw for convenience all the field lines coming out of it. Then I want to see, if I do a line integral, say from here to here, will I get the same answer on some other path? Let’s test that. So the first integral is from 1 to 2 on this path. Then I’m going to take another one, which is different from it, where you can see the answer is the same, then go on making more and more changes, and the answer will not change. So I’m going to take another path where I go tangent to the lines, along a circle. Then I go here, till I get to that radius, then I turn around and do that. I claim the integral will be the same on this path, this one, that one and that one. So I want you to think about why it’s going to be true, okay? Think of it, why it’s okay to do that. You’ve got a new path but it doesn’t make a difference. Ready? Tell you the answer now? How many people don’t know why it’s true? So other people know why it’s true. I’m going to ask the other people. Okay, why is it true?
You will agree that if the electric field is conservative, multiplying by a number Any work you’re doing here, you’re doing here, therefore the answer is the same. Then you can see that you can go on adding all kinds of changes, radial and angle, radial or angular, and when you’re done, all the radial integrals will give you that part, and the angulars won’t count. So as you move around, it’s only when you’re moving radially you’re even aware that you’re doing work. Going in the angle doesn’t matter. And the net radial change in going from here to here is that distance, no matter how you move. Now you might say, prove to me that that integral is the same as that integral, because this was made up of radial and angular sections. This doesn’t seem that way, but we can always make a very fine mesh in which every displacement is approximated by radial angle or radial angular. So you can approximate any contour by sequence of radial and angular moves. Angular moves are free. Radial moves keep track of the change in the distance times the force and you can pair them with this reference. Every So we’re going to do the calculation. So we use the formula that U at r is - the integral _{1}e⋅dr from 1 to 2. You see, we know it is conservative. I’ve shown you with this argument. So if you want to find the potential energy between 1 and 2, you can take any path you like between 1 and 2. So let’s pick the points. Here is point 1 and here is point 2. You can pick any path you like, so here is what I’m going to do. I’m going to go radially out and then I’m going to go along the arc. So this is 1, this is 2, but I’m going to start at some intermediate point 3, then go right.So work is done here; no work is done there. So let’s just find the work in going from 1 to 3, which is - Πε. That is _{0}r^{2}F . The displacement is e times the little length of _{r}e. Do you understand that? A tiny section here, _{r}dr, has a magnitude equal to the change in the radius, and a direction equal to the radius unit rate — the radial direction. And you want to do this from r to _{1}r. Okay, _{2}e⋅_{r}e is 1 so I get — I forgot one thing. You guys know what I forgot here? One thing’s missing. Yes?_{r}
So I forgot the ^{2} is just 1/r, so that’s 1/r − 1/_{2}r. Again, sorry, I made a mistake now. The change in potential energy is really _{1}q times e, but q is the test charge, so I’m going to call this V. So it’s hard for you guys to do it on the notebook, because I want to distinguish between potential energy and potential. In other words, the force — now some other test charge q is the electric field times q. And the force is going to be - the gradient of the potential energy. The electric field is going to be - the gradient of the potential. Yes?
r.
## Chapter 4: Electric Potential [00:59:34]So let me correct what I said. I’m trying to find the difference in what’s called the potential. It’s not the potential energy. Potential difference between some point and some other point is the work done on a unit charge. If you’ve got some other charge, you should multiply it by the charge. So the energy of some other charge in this field, the potential energy, is So we’re comparing this expression to this expression, we can conclude that r) + any constant. Any constant, because you cannot immediately say this guy is equal to this guy and that guy is equal to that guy, because even if you add a constant to both, it will still be true. And we can pick the constant any way we like. It is common to pick the constant so that V goes to 0 as r goes to infinity. That’s what we will do. That’s the convention. Then we can write here, what is the V(r).This is the q again, because q is the guy producing the potential. q is the guy experiencing the force. So _{0}q times _{0}V is the potential energy of charge q, of charge q sitting at that point. Now let’s also verify, if you like, that this _{0}V(r) does produce the electric field. In other words, I got the V by integrating electric field. I want to go back, just to get you familiar with this. Getting the electric field from the potential. Let’s try to do that. So E — let me just do _{x}E — it’s −_{x}dV/dx, right? -dV /dx = -q/4Πε. Derivative of 1/_{0}r is -1/r times ^{2}dr/dx. What is dr/dx? r is (x + ^{2}y + ^{2}z)^{2}^{1/2}. dr/dx is just x divided by this whole thing, is x divided by r. So this becomes q/4Πε(1/_{0}r)(^{2}x/r). This is E._{x}So what will be the vector r), times ^{2}r over r. Put r over r, position vector over r, is the unit vector in the radial direction. I’m doing this a little fast, because this is something you can go home and check. I don’t want to spend too much time doing it. I claim that if you took this potential and took its x derivative or y derivative or z derivative, you get the x, y or z components of the electric field.Okay, if you didn’t follow this, you should go home and check that this is true. All right, so this is not that important for me, this checking, because we know it’s going to work. What is important for me is for you to know that we’ve now found that the potential due to a single charge leads to a conservative force. The field due to a single charge leads to a conservative one, therefore by superposition, if you add any number of charges, if you find the potential due to all of them, then you take its gradient that will give you a conservative electric field. So what have you got so far? Our gain is the following: we have a law of conservation of energy, which is the following: ½ mv_{2}^{2} + q times v at the point r. And what is _{2}V at any point r? It’s the sum 1/4Πε. Let’s imagine there are many, many charges, _{0}q, _{1}q, each one _{2}, q_{3}at r,_{1} r, _{2}r, etc. This equals _{3}q which is charge number I divided by the length of the vector from _{i}rto the _{i} r where I’m finding the potential. In other words, this is where I want the potential r. This is rand _{1}q is sitting there. This is _{1}r and _{2}qis sitting there and so on. And the potential at the point that I’m interested in is obtained by taking each _{2}q divided by the distance from that point, and this is the usual 1/4Πε._{0}That is the complete story, and the charges are not discrete but continuous you can write an integral. So every problem, you first find the potential by adding the potential from all the charges. Then its gradient will give it the electric field. But we have now a law of conservation of energy with this as the conserved energy. The second advantage is that it is easier to work with
So I’m going to illustrate that with one simple problem and that’s the last thing for today. That’s the dipole. So here is our dipole. Let’s take a charge What we will do instead is to find the potential everywhere, and take the E. Well, let’s do that. So let’s find the total potential due to these two guys. The first one is q/4_{y}Πε divided by 1/_{0}r - 1/_{+}r That’s it. Just wanted to simplify the expression a little bit. So this is true for any separation, but I want the limit in which _{-}.r and _{+}r are much bigger than the separation between them._{-}So let’s define an r. You get that, you’re done. So vector _{-}ryou can see = _{+}r - ai/2 and vector r = vector _{-}r + ai/2. Because the vector ai/2 looks like this, so ai/2 + r should give you _{+}r. You can check that. And you can check that ai/2 + r gives you r. so these are the two expressions. So what is _{-}r? _{+}r is the length of the vector ^{+}r. That’s equal to the square root of the length squared. The length squared of _{+}r squared = root of (_{+}r - ai/2) ⋅ (r - ai/2).
a minus twice ^{2}r⋅a. The other one would be r+ ^{2} a + twice ^{2}r⋅a. And that we can approximate as r times 1 - twice r⋅a/rto the 1 half. I’ll tell you what I did here. I’m going to neglect ^{2}a squared compared to r squared, because r is much bigger than it.But I’m going to keep this r⋅a/r^{2}. Remember 1 + xis roughly 1 + ^{n} nx + dot dot dot if x is small, and x is indeed small, because it’s r⋅a over r squared. So now we can see V = q/4Πε{1/[r(1-_{0}r⋅a/r^{2})] - 1/[r(1+r⋅a/r^{2})]} .This is going to be found in every textbook in the planet, so don’t worry about it. If you simplify this again by taking it upstairs, you’ll get twice a times q is the dipole moment, so it’s p⋅r/4Πεr_{0}^{3}. That’s the final piece, the dipole moment. Dipole moment is a charge times the separation between the charges, which is 2a times q. Okay, so what one should do now is, having taken this V, one can take its x and y derivatives very easily to calculate the electric field at each point. It’s very easy to take the derivative of this, just pick the x or y. It’s one of the homework problems I’ve assigned to you, where I show you the dipole field. So the moral of the story is, add the potential due to all the charges, then take the derivative, because derivatives are easier to take. Yes?
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