Hello everybody, welcome back to electrodynamics and its applications. My name is Sibonhom, and to my right side I have my assistant. Hi, I'm Melody. Okay. So today, we're going to cover the fourth part of our lecture. So in this lecture, we're going to cover statics which is a very simple world, where everything is fixed in time. Before doing that, we're going to revisit the Maxwell equations which contain all of the electromagnetism that we need to learn from this class. So the first one, is about the divergence of electric field where you can see is equal to the charge density over D permitivity in vacuum. The second one is about circulation of electric field which says the curl of electric field is equal to minus round B or round T. The divergence of the magnetic field will always be zero because there is no monopole. So you don't have any source or sink and the fourth one is about the curl of magnetic field which is created by either the change of electric field as a function of time round the overall t or the current through a wire which is depicted, described by j which is current density over the permitivity of the vacuum. The situations that are described by the equations here even if you think about the second equation is very complicated. Because you can see the change in magnetic field can create circulation on the electric field which is counterintuitive because when we think of electric field is the gradient of the electrostatic potential, but if you think about that the circulation shouldn't result in any gradient but still you can do that by having a changing magnetic field. So, we will firstly study relatively simple situations as I just mentioned, and learn how to handle them before we take up more complicated ones. So Melody, let me ask you one question. When we say statics, what do we mean? It doesn't change as a function of time. Okay. So what doesn't change as a function of time? Is it the position of charge or is it the motion of charge? What specifically are we talking about? Well, there's actually two types of statics. So there's electrostatics and magnetostatics. For electrostatics, it's the charge that doesn't move at all. The magnetic field doesn't change as a function of time, and the electric field doesn't change as a function of time. But actually in order to get a magnetic field we need some kind of electric field for these equations. So, in the second case magnetostatics is the current that doesn't change as a function of time. So the current will generate the magnetic field but the magnetic field doesn't change as a function of time. It's constant. Very good. Very good. So we're going to flip over to the next slide where you can see the static case as pointed out by Melody, nothing depends on the time. But here, we need to be very specific. So we have two cases, electrostatics and magnetostatics. For the first case, electrostatics, all charges are permanently fixed in space. So in this case, you won't have any electric field change as a function of time, nor will you have any magnetic field. Because you don't have any current, you don't have any electric field change as a function of time. Or if they do move, they move as a steady flow in a circuit. So rho and j are constant in time. So I'm going to draw a picture with let's say, imagine you have four plus charges in your world, and if they are not moving at all there is no current. You may argue you cannot have only four plus charges because of charged mentality. Good. So I'm going to add four minus charges into our world, like this. If you imagine, if they are fixed in space you don't have any magnetic field. No magnetic field at all because you don't have any current. So if you want to create a current then we're going to make a steady-state in a loop in circulation like this, and for minus charge we're going to make a circulation like this. If we make circulation like this constantly, then the charge density at any point will be constant roughly. But still you will have a current in one direction in circuit so now, you have non-zero j current which will contribute to magnetic field circulation. So in that way, you can imagine in magnetostatic world everything is rotating hand-in-hand. But in electrostatics everything are frozen. Everything is frozen. So with those pictures in mind, first let's see how the first and second question change. So the first question, the first equation will not change it will be the same. The divergence of electric field will depend on the charge density at the point of interest. So that goes as it is. For the circulation because we don't have any change in magnetic field because we have either constant current or no current at all, so this will be zero. So the curl of electric field will be zero, so that means we don't have any circulation of electric field. For magnetostatics as we mentioned, the curl of B will depend on the change of electric field as a function of time but because we don't have any change in charge density as a function of time, this will always be zero, and will only depend on the steady-state current which is denoted by j in this case. So we will have Delta cross B equals j over c squared C epsilon not and for the divergence to c. So once we rearrange our equations like this, what we find is that for electric field you have two independent separate equations, for a magnetic field you have two independent separate equations. Not like in this case where everything is coupled. You see electric field is coupled with magnetic field, magnetic field is coupled with electric field. But in static case everything is separated. So, mathematically, it is more easier, it's easier to solve the equations. So here as written E and B fields are not interconnected which means that electricity and magnetism are distinct phenomena so long as charge and currents are static. Okay so Melody, in practice, what kind of examples can you think of where we can apply statics, the equations in statics without losing too much of the accuracy of understanding the phenomenon? I guess we talked about before in the crystal lattice the are ions of different charge like NA has the positive charge and Cl has a negative charge. So they are in their fixed position because they're in the crystal lattice. So in that case we can think of a case where we can apply statics, right? All right. So electrostatics and magnetostatics are ideal for learning about the mathematical properties of vector fields. So that will be our practice learning. Electrostatics, you have zero curl and a given divergence. For magnetostatics, you have zero divergence and a given curl. So, our road-map of learning electromagnetism is schematically depicted in this picture, where we start from electrostatics with more focus on divergence and we will split this category into two; where number one, you knew the positions of all charges that we are interested in. Then, we can use Coulomb's law and do integration. Second case, a little bit more difficult is, we do not know all of the charges positions or only we know part of them with some boundary conditions. That will be two separate sectors that we will learn in electrostatics. Then, we will move on to magnetostatics with emphasis on curl. Once we learn both of them, we'll move to the most complicated case and which is our ultimate goal in electromagnetism. Okay. So let's start with electrostatics. We are going to start with more familiar equation, the Coulomb's law. Coulomb's law is an approximate equation. It is not the accurate equation. It is well-suited for electrostatics, but once everything starts to move, then we need some correction to this law. We will start, as we mentioned, the Coulomb's law and comeback to Maxwell equations for electrostatics. We will see how they're connected, mathematically. So in a nutshell, the two equations that we just mentioned for electrostatics, where you have divergence of the electric field and curl of the electric field, will give you the Coulomb's law. Let's revisit the Coulomb's law. So, Melody, can you explain to our students what Coulomb's law is? Yeah. So, basically, it's the force. Right here, you see the force between two charges, q_1 and q_2, is some constant multiplied by the magnitude of the charges and then, it's inversely proportional on the radius squared and then you multiply it by the unit vector, which gives you some direction. Exactly. So like Melody said, it's the pre-factor here, one over four pi epsilon naught, like in the case of gravitational force, times the quantity of charge involved here, q_1 and q_2, and the distance between those two charges squared. So, the force that is felt by q_1 has the opposite direction of the force felt by q_2, due to the action-reaction law. The pre-factor, one over four pi epsilon naught, is very important because it will give us the information about the speed of light. As in the special relativity theory, the speed of light is constant no matter whether you are in a moving framework or a static framework, unless there is acceleration. So, that has been already predicted from the Maxwell's equation, so that's very impressive. You will see the unit of this pre-factor is either newton meter squared, or coulomb squared, or volt meter per coulomb. You might ask, how accurate is this pre-factor, to which scale? If you're interested in that, you can see many of the lecture or textbooks showing that this could be very accurate down to very small scales, like atomic scales. Now we're going to introduce the concept of electric field in order to understand Coulomb's law. But in this case, we're going to change our perspective from the broader view to very narrow scope, where in this case the electric field is force per unit charge, instead of a physical perimeter as a function of space and time. In that case, you can see E of one describes something about the 0.1 even if q_1 were not there, assuming that all other charges keep their same positions. So, E of one is equal to one over four pi epsilon naught, times q_2 over r_12 squared, times the unit vector e_12. You can see the numerator q_1 disappeared because E(1) is the force per unit charge. If you think about this in vector form and rewrite it, you will have three components E sub x, E sub y, and E sub c. If we think about one component out of that, we can do this algebraic manipulation where this one will be from Pythagorean theorem, and this one is the x component in the position. All right. So we know we can use superposition, which means the resulting electric field is summation of each component, linear summation. So the force or electric field on any charge is the vector sum of the coulomb force from each of the other charges. In mathematics, we can use this summation. We can use this to each component, X and Y and Z, and you will see equations to your right side. Now we're going to move from discrete concept to continuous medium, because we want to use integral and differentiation where you need to have continuous function. The rationale behind this change, from discrete to continuum, comes from the fact that all the charges have a very small volume or dimension. So in our macroscopic world, you can approximate it to a smooth materials or continuous materials. So, often it is convenient to ignore the fact that charges come in packets like electrons and protons, and think of them as being spread out in a distribution, as we just discussed. Of course, it's not valid to a very small scale, like in the atomic scale or sub-atomic scale, because we already know they are discrete in that level. But in millimeter scale or meter scale, you can use it. If I use it, then we have to introduce a new concept called charge density. So, if I say charge density rho, it means we can define the amount of charge at the point of interest per unit volume. So, by definition, the small charge, delta q_2, will be equal to rho two times delta V_2, which is the volume of the small cube of interest. Then, by putting this equation into the equation that we just discussed for superposition, then we can change the sigma, the summation, into integral. If you do that, we come up with a very general form of the electric field of one as, one over four pi epsilon naught, which is the prefactor, times the integral of rho sub two, times r_12 dV_2 over r_12 cubic. You can imagine some kind of charge density with an arbitrary shape with a charge density function, rho, where the small portion here, DV_2 can contribute independently to the point of interest in one, okay? So, from the previous slides, we learned how to calculate the electric field from arbitrary distribution of charge, assuming they form a continuous medium. So, here, the question, given the charge, what are the fields? If I asked that, then Melody can bring up her computer and type in this equation, which is general as you can see, and think about the distribution of the charge density that are there and the point of interests here one, as well as doing the integral overall space using the computer. Then she will be able to answer my question. So, that's really computer's job. So, in our roadmap, when we are going to learn about those electrostatics and magnetostatics, we're not going to think about the whole complicated world. We will focus more on simple distribution and see how we can calculate the fields intuitively without relying on the computer. So, in order to do that, let think about two important concepts. Number one; potential energy and number two; electric potential. So, Melody, what is potential energy? So, I guess, it's the simplest form of potential energy you can consider is like the classic example of a rock on a ledge, right? So, the rock on the ledge will release some amount of energy when it falls. So, its potential is really, how much energy it took to get the rock up to the ledge in the first place. Exactly. Exactly. So, potential energy is one important form of the total energy where you have kinetic energy or work done by the system. So, potential energy determines how much work you can do. So, by using the example of the rock on a ledge, if you lift a rock on the ledge, you are doing the work against the gravitational field and store that on to your rock, and then when you release it, then the stored potential energy is converted into kinetic energy. So, if we think of this example and try to apply that to electrostatic world, then it could be defined as work done against the electric forces in carrying a charge along some path from point a to b as depicted in this figure. Mathematically, that is W which is equal to minus integral a from a to b, F.s. So, usually when people say you have a great potential, but sometimes it might be not a compliment if you are not working. So, one of the things we have to always have is the balance between potential and the work we really do. The same as in this electrostatic world. So then, if potential energy is the energy that is stored by doing work against the system, then what is electric potential? Yeah. So, we talked about earlier how we have the electric field, and we want to get an idea of what the field is doing without actually knowing about the charge we are going to put in there. So, we're going to do the same here with electric potential where we divide by a unit of charge and then we can get the potential instead of just the work. Perfect. Yes. So, in order to use the concept that we developed before to see the concept of field, we're going to use the same concept here. So, work done against the electric forces in carrying one unit of charge along some path from point a to b will be the electric potential. So this is really what we used in everyday life like our power outlet has 220 volt of voltage or in the United States, they have 110 volt of power outlet. So, those will be all electric potential. Because we are thinking of one unit charge, so we are dividing force by the charge, it becomes electric field instead of force in the equation. Then let's think about this important question; can we get work out of a field, electrostatic field, in this case, carrying a charge to be along one path and then back to a on the other? In other words, does the integral depends on the path we take? So, Melody, what would be the answer? I would think, no. No. Yes. So there's no way to get energy out of electrostatic fields if the principle of energy conservation holds true. So, if this integral will be path independent. However, in dynamic world like the one that is depicted on the right side on this slide, you may have a case where you can really have the way to get work out of a field by carrying charge from one place to the other. In this case, you're moving magnets up and down changing magnetic field through this loop. So, from the second law of Maxwell's equations, you create circulation of electric field which will light up your bulb if it is connected in the loop. In that case, you're moving energy from here to here. So, the global energy conservation is being hold but the locally, you are transferring energy from one place to the other, okay? Alright. Now, let's think about Coulomb's law of force and a single charge q. So, this will be the simplest case. You only have one charge in the world. In that case, what would be the characteristics of your field? So, it will be the radial field. Exactly. It's outlined in the previous lesson. Exactly. If you only have one charge in the whole world, you will create radial field. Meaning, it doesn't matter which direction you are looking at the charge, it just matters how far away you are from the charge. In that case, we can think of plus charge being put in this point and all of the points of interests that has the same distance from the charge will feel the same field. Let's imagine we're putting a test charge with charge quantity plus one, and then move it from a to a prime, which is the arc, the part of this circle, and then we move from a prime to b. If we do that, what will be the work done on the test charge in carrying it from a to b? So, from a to a prime, what happens? So, actually from a to a prime since it's the same distance from the charge, there's no change in potential. So, there should be no change in the work. Exactly. So, there will be no work done. No work done. Only from a prime to b we'll have work done. So, the integration here, you can change from a to b, to a prime to b, and use the same equation. If you do the integral of one over r squared function, it results in minus one over r. So, by doing the math here, you will be able to get the answer, which is minus q over four Pi Epsilon naught times parenthesis one over r_a minus one over r_b. Now, let's expand this discussion a little bit. So, work done along the path comprised of arcs and radial parts. If we think of a more complicated case as depicted on the right side, instead of just one arc and one linear part, what if we have multiple arcs and multiple linear parts? Then again, there is no work done along an arc, and only work done along the radial parts. If you do the math, because the starting point become the ending point in between, they will be cancelled out. So, only the real starting point and the real ending point will matter. So, no matter how you design the path, that is comprised of multiple arcs and linear parts, if you have the same starting point and ending point, the path along those are different ways will be the same, right? So, what about the smooth path? So, like in the case of spherical coordinates, we can think of doing more chopping, more smoothing out, and still the principle holds through. So, therefore, the work done in carrying a unit charge from a to b is independent of path, right? So, in this picture, we see we have another point called p naught, instead of a and b, and we just learned the work done in carrying a unit charge from a to b is independent of path. So, it means it's like state function, where electric potential difference between a and b will be constant no matter what path you take. Now, why do we need another point p naught that we call reference point? So, before we were talking about work, and work is actually only from a to b. However, if we have another reference point, then we can give some quantitative amount. So, if you think of the analogy before with a rock, it's the difference between using sea level as your height for judging the potential energy of the rock. Exactly. So, we can give absolute number of electric potential to each point, and that is very convenient in everyday life. Okay. So, W depends, W unit means the unit charge work, depends only on the end points, and it can be represented as the difference between the two numbers, right? So, in fact, it is a state function of let us start with potential Phi B and Phi A, the difference, and let's choose a reference point P naught and agree to evaluate the integral by using a path that always go by way of point P naught. Then the W unit will remain the same. However, we can determine Phi for any point in space. So, Phi becomes a scalar field, which we call electrostatic field. So, for convenience, we will that P naught be the point at infinity y. If you make r infinite, then this becomes zero. So, naturally, the electrostatic potential at infinity will be zero and zero is a good reference point, right? So, for a single charge at the origin, the potential Phi is given for any point x, y, z as follows. Phi of x, y, z is equal to minus q over four Pi Epsilon naught times one over r. Then we can assign numbers to any point of interest. Now, I just mentioned that we can put zero volt at infinity. But in practice it's hard to connect infinity, right? So, where can we find zero volt in our everyday life. The ground. Yes. As Melody points out we can find zero volt from the ground. Then you may ask, why does ground has the same potential as the point at infinity? Melody, can you explain to us? Yeah. Can I have this? Sure. So, if you think of the Earth capacitance of the surrounding environment, and at infinite, there's zero volts, right? So, the change in voltage for a capacitor is equal to the change in charge over the capacitance. However, since this is a very large system, the capacitance is very large essentially infinite. So, if your Delta V is equal to zero, then on the other side of this capacitor, which is the earth will also be zero. Very good. In addition, if you connect anything to this large source of charges, then everything will balance out. So, if you connect something here, has a positive, there's a lot of negative charges that can make this zero as well. Perfect. That's a very good explanation. Okay. So, now we're going to think about some mathematics again, with a focus on why we introduced this electrostatic potential or electric potential, and why we add this concept in addition to electric field. Okay? In order to do that, let's first check if the superposition principle can be applied to electrostatic potential as well. So, we will start from the definition. The electrostatic potential of arbitrary point P is equal to minus p naught is the reference point, P naught to P, E_subtotal dot ds all right? E_subtotal dot ds. Now, we know the true electric field acting on that position of interest can be decomposed into components because we can apply superposition to electric fields. So, then we can replace this E_subtotal by Sigma of each component j, okay? Then, we also know we can interchange the Sigma and integral here. So, if we pull this out of the equation, and put Sigma to the right side, then what happens is, this will become the potential contributed to by each component of charge in the world, right? So, we will call that Phi j. Because now we only have Sigma, it means all of the charge around the world that is acting on this point, they have their own share of potential, can be a linear sum of those potential. So, we just proved that superposition principle can be applied to electrostatic potential as well. So, complete formulas for the potential Phi at a point one risen from a group of charge and distribution charge will be this one. The physical meaning is that the potential energy which a unit charge would have if brought to the specified point, in this case, P in space from reference point, will be equal to this one, okay? A group of charge. Again, if we smear this out, we can use the smooth function and integration where now we assume it's a distribution of charge. You may wonder, why do we do this? Because when we didn't have computer, it is easier to deal with scalar functions, scalar field instead of vector fields. Because if you're going to do superposition of vector fields, you have to have three equations all the times. But in case of electrostatic potential, you only have one set of equations to do the addition. Then, what you do is just differentiation, and then you can have your resulting vector field. So, along the way, you do a lot of calculation, you deal with scalar function, and at the end of the road, you come up with a vector that's much more easier than dealing with vector fields from the starting point to the endpoint, okay? So, we're going to revisit the equation that we covered in the previous slide. Electric field is equal to the minus gradient of electrostatic potential. Melody. Do you remember what type of characteristics do vector field have when it can be described as a function of gradient of a scalar field? The curl is zero. Yes. You're right. For the gradient of a scalar field, it is always zero. So, we may revisit the fact that the curl of the gradient of a scalar field is always zero, and you may remember the staircase argument. When you have a staircase type of function, when you complete the loop, you cannot complete it because it is incompatible. Now, with that in mind, so in electrostatics, this is like staircase function, and we're going to take a look at this function from two different perspectives. One from the point derivative, the other from integration. We are going to limit ourselves to very simple situation where you only have one dimension. Very simple. So, take a look at here. You have one dimension, x-axis only, and we're going to think about two points, x, and x plus Delta X where Delta X is an infinitesimal distance between those two points. Now, I told you why we are using Phi, the electrostatic potential. It's because E can be obtained easily from Phi, and there's only one integral for Phi, while there are three integral for electric field, because it's a vector. Let's take a look at this simple case. I'm now moving from x to x plus Delta X. By doing so, I'm doing work, Delta W, on the test charge. That Delta W is just the change in the potential energy where you can see y and z is fixed, only we think of x and x plus Delta X. If we do that, then we know this is round five or round x times Delta X, that's what we learned. If we bring up the definition of Delta W, which is the potential energy, again, which is the work done against the field for a unit charge in case of this electric potential, then it is E.ds, but because it's just one dimension, it is minus E sub x times Delta X. So, E sub x is Delta W or Delta X, am I right? So, in that case, if I equate it with this one, then E sub x becomes round Phi or round x and minus. In one dimension, this is the gradient. So, this is the end of proof to prove that this equation holds true. So, we just learned that from the equation, electric field can be derived from the derivative of electrostatic potential by using the gradient. We can also think of the other way around where the electrostatic potential is defined by the fact the electric field, if you integrate them from path p zero to p, then you can get the electrostatic potential as well, so this is vice versa. Now, the idea here is, which way is more convenient to calculate the electric field on the point of interest. Whether you want to go through the route to gather all the information of the electric field from the charge in the world that act upon your point of interest, and then do the summation, or if you want to change that into electrostatic potential world, and then do all the summation there and do the derivative. So, our argument is that it is easier to use electrostatic potential because it is scalar and one over r is usually a little easier to integrate than x over r cubed. There's only one integral for Phi, while three for E. If you look at those equations, you can go around and come again. So, from the yellow mark, this is the definition of the electrostatic potential that we just mentioned, and the red arrow really is from the linear integral theorem that we just mentioned. If we integrate the gradient of a scalar function, then it will be a state function. By comparing these two, you will get the insight into the equation between E and minus Delta Phi. So, from vector calculus, if electric field is minus Delta Phi, which is the case for the E field from Coulomb's law, as Melody mentioned, as Valerie mentioned, the curl is zero. So, there is no circulation whatsoever. So, that will be Maxwell's second equation for electrostatics. The work done around a closed path is zero in electrostatics field, the work done, because there's no circulation, and that's the Stokes' theorem. For any radial force, the work done is independent of the path, and there exists a potential. In another words, for any radial force, the curl is zero. So, Melody, what type of force is radial force, but not one over r squared force? Maybe inter-atomic forces. Van der Waals force, yes, or nuclear force. There are many types of force that have the radial characteristics, but doesn't follow the inverse square law, and those will have curl free. Why? Because the work done is independent of the path. So, the existence of potential, and the fact that the curl of E is zero comes really only from the spherical symmetry and radial directional electrostatic forces. So, with this one only, we cannot derive the Coulomb's law because we need another condition which is, you need to have one over r squared.
