Hello everybody. Welcome back to Electrodynamics and its applications, lecture eight. I'm with my teaching assistant, Melody Glasser. My name is, Professor Seungbum Hong. So today, we're going to learn about a different approach than those that we took before. So, in the previous slides we were heavily investing our time in thinking about the force in electric field interacting between charges. In this lecture, we're going to take a different approach to think about the potential energy of the whole system, or the whole world in an aesthetic way, and then use that equation to derive some of the forces that we already learned from the previous slides. The reason why we're doing this is because from transferring to moving from classical mechanics to quantum mechanics, this is a more convenient way to tackle some of the most important problems in physics. So, let's re-visit the law of the conservation of energy, one of the most interesting and useful discoveries in the study of mechanics. So, when we think about the total energy, we can categorize different components in different ways, but the usual way we do is to think about the kinetic and potential energies of a mechanical system. That will help us to discover connections between the states of a system at two different times without having to look into details of what was occurring between. That's the field of thermodynamics as well. So, the total electrostatic energy [U] of a system of a number charge is the sum of the terms due to the mutual interaction of each pair of charges. So, I'm going to ask my teaching assistant Melody, in your opinion how can we calculate the total electrostatic energies of all the charges involved in the world that are fixed in their own positions? We can just do a summation. Yes. We can do just summation. Based on what principle? Superposition. Yes. So based on the superposition principle we can do linear summation of the interactions between pair wise electrostatic energy or potential as described here, Qi times Qj, over four Pi Epsilon note Rij. Rij is the distance between the particles of interest, and we are counting all pairs. So, we shall be concerned with two aspects of this energy, number one, the application of the concept of energy to electrostatic problems, and number two, evaluation of the energy in different ways. So we will explain this with more concrete examples so you can understand what we are talking about. So, in this slide, we're going to tackle an example of energy valuation by different means, and we're going to revisit the uniformly charged sphere, and think about how we can calculate the total electrostatic energy of this sphere. So, when we think about this, we have to have a reference point to gather all of this into this sphere form. So Melody, what reference point should we think about before calculating this? So, basically, you have two choices. You can think of it as an infinite, or you can think of the ground. Okay. As your reference point. So, among those two choices we would select the first choice where we have all charges gather around at infinity. In that way, we can think logically that we're gathering them from infinity, to the place where they should be located. Now, this is an abrupt thinking, from infinity and gather all them together to make a sphere. We're going to think about it in more of a step wise approach. So, step-by-step we can think of this as making a snowball or snowman. When we make a snowman where do we start from, Melody? We have a little nucleus ball and then you roll it until it gets bigger and bigger. Exactly. So, we start from a point charge and we cover it with shell. In the course, because it's repetitive, we can take any of the step and think about it and apply that to the whole process. So here's the thing, we will assume that our snowball or the uniformly charged sphere has certain radius, so it's already formed. I'm going to add another shell, the green shell, to make it bigger. So, if we think about that process, we already have electrostatic potential of this sphere which is U, and we are adding a small portion of U which is dU to this system, okay. So, in the process that we just took, we can think of the first one as initial snowball and the other as a shell that we are adding to. Using the Gauss law, we can think of the interaction between the existing snowball and the new shell, and because of the symmetry, we can think of this interaction as spherically symmetric. Meaning, the energy added or a force interaction doesn't depend on the angle of direction, it depends on the distance only. So in that case, what we can do is using the Gauss surface as a sphere, and then the electric field out of this uniformly charged sphere will be the same as the one from the point charge, which has the same amount of charge concentrated in the centre of the sphere. So then, what we can think of is the interaction between QR and the DQ in the shell. So, thus we have QR times DQ in the nominator. Then, we can use the denominator as four Pi Epsilon naught R, and we just integrate this addition from zero to define our radius. In that way, we can calculate the full electrostatic potential of a uniformly charged sphere. So, the energy is proportional to the square of the total charge and inversely proportional to the radius, and we can interpret the equation as saying that the average of one over Rij for all pairs of points in the sphere is 60 percent, three over 5A, which we will derive shortly. So, take a look at this, the Qr, which is the total amount of charge of existing sphere is equal to the volume density of the charge times the volume of the sphere which is four over three times Pi r to the power three, and if I do infinitesimal change of this dQ, then we know it becomes four Pi r squared dr times the charge density. So, if we put this into our nominator then dU becomes four Pi Rho square r to the power of four dr over three Epsilon naught and if you do integration from zero to the radius of the final sphere A, then we come up with this equation where you have three over five times Q squared over four Pi Epsilon naught A. I'm going to ask my teaching assistant, Melody, what meaning it has, three over five times Q squared over four Pi Epsilon naught A, what does that mean? Melody, how can we visualize this number in other form? So, I think before we learned that the energy is proportional to the amount of charge incorporated within a system. However, and that would be your original q squared over four Pi Epsilon naught. However, in this case, it's around 60 percent of what the energy would be of both of those two charges. Exactly. That's very good. So, if I add a little bit to what Melody already explained, the electrostatic energy between two point charges would charge quantity Q and a distance separated by A, you can understand the energy will be Q squared over four Pi Epsilon naught A. So, this could be a reference to compare with. So, this sphere, which has a total amount of charge equal to one of the point charge here will have the same or almost the same order of magnitude of energy by putting that in a uniformly charged sphere. It is about 60 percent of the energy accumulated between two point charge with the same charge quantity. Thank you. So, the next example that we're going to tackle is the energy of a capacitor and forces on charged conductors. So, essentially, the shape and size are the same as in the case of uniformly charged sphere, but in this case, we are bringing in a conductor instead of nothing, and see how the charge will behave, and how much energy will be stored. So, how much work is done in charging the capacitor is another way to ask this. How much work is done in charging the capacitor with Q, for example. We imagine that the capacitor has been charged by transferring charge from one plate to other in small increments, dQ. So, now, we're going to think of uncharged conductor and bringing charge from infinity and injecting charge small, bit by bit to make it fully charged. So, we are thinking of a small increment dQ and we know that the internal energy or the electrostatic energy of capacitor is equal to the potential times the charge, potential times the charge. So, what we can do is doing infinitesimal change of the energy dU and because the voltage in the conductor is constant. So, Melody, voltage should be constant inside a conductor. Why is that the case? Because there is no movement of charges and there is no gradient in potential because the charges will equilibriate to a stationary place, and therefore the voltage is constant when there's no gradient. Very good. So, if I complement or add a little bit to Melody's comment, in a conductor, charge are free to move. They're free to move. So, if there is any electric field inside, they will not stop moving around so that violates the electrostatic world. So, in order for them to be static, we need to make sure there's no electric field inside a conductor, and electric field is derivative of electrostatic potential. So, if there's no electric field anywhere inside the conductor, there is no gradient of electrostatic potential leading to constant voltage throughout the space and only place they can reside is on the surface. So, the voltage will be kept constant and we will only change the amount of charge that we inject into the system, and that's why it's VdQ. Now, we already proved that capacitance times voltage is equal to the charge. Capacitance times voltage is equal to Q. CV is equal to Q. Therefore, we can replace V by Q over C, that's equivalent, and then if we rearrange the equation, it will be QdQ over C. So, from here, we're going to do integration from zero to Q because we are injecting charge bit by bit from zero to the final amount, Q. If we do this integration, it will result in one over two Q squared over C, and again CV equals Q. So, if I replace Q by CV, then it will be CV square on the numerator and divided by C, so it becomes one over two CV square. We also know that capacitance of a conducting sphere is about is four Pi Epsilon naught A. If I replace it with this one here, then the conducting sphere will have electrostatic energy of one over two Q squared over four Pi Epsilon naught A. I'm going to ask again, Melody, what this means as I asked her about what the energy of uniformly charged sphere means. So, Melody, what does this mean? I think it's similar before again. You can think of the energy as being proportional to the number of charges. So, if you had two charges of equal energy like before, then it's about half of what the energy between these two charges would be. Exactly. So, with the same reference now we are in the position to compare the electrostatic energy with the same amount of charge injected into conductor versus forming a uniformly charged sphere. So, in a nutshell, spherical shell versus uniformly charged sphere. We can compare apple to apple. The energy of a charged conducting sphere where a thin spherical shell of total charge Q is injected is just five over six times of the energy of a uniformly charged sphere. So, it means, it takes less energy to charge a conducting sphere than to form a uniformly charged sphere. So, let's take a moment to think about it. If I have the same amount of charge, similar amount of charge and I want to put it within the boundary of a sphere with a radius of a, which way is preferable, Melody? The first one. Yes. It takes less energy than this one, so this will be the preference over this one. Thank you. Now, we're going to change the gear, the topic, and play with another example, which is a parallel plate capacitor, which is used in your smartphones or even for a grid, electrical grid. You have parallel plate capacitors to instantly store a lot of charge. Also in the railway, you can find it. So, it is pretty universal and we're going to think about the force between two plates in the parallel plate capacitors using a different way than the way we used before. So, the way we used before is thinking about the charge amount and the force between them, but here, we're going to think about the energy stored in the system and perturb the system in a way to displace it a little bit, and see how much energy is caused, and using that energy caused, we're going to calculate the force that is acting between those two plates, and we call that virtual work. So, here, as described here, if we imagine that the spacing of the plates is increased by the small amount, Delta z, that's imaginary, we didn't displace it, but if we think we did, then the mechanical work done from the outside in moving the place would be Delta W and that is equal to force times the displacement, Delta z. All right. So, let's take a look at the case here where you have a parallel plate capacitor. The energy, total energy, U is equal to one over two Q squared over C. We just proved that. Now, if I take Delta infinitesimal change of U, then because Q is constant, the charge we put on the plate are staying there, all the charge are staying there. So, Q is constant. So, the change only is coming from the capacitance change, and if you think about a capacitance, and do the Delta operation here, what results is you see F Delta z is equal to Q squared over two times Delta one over C, and one over C is minus Delta C over C squared. So, if I put that here, we have to know the relationship between Delta z and Delta C. So, here's the thing. C is equal to Epsilon naught A over d. Now, if I inverse it, one over C becomes d over Epsilon naught A, and d is the distance between the plate. So, if I take Delta one over C, then this becomes Delta d over Epsilon naught A because the area of plate doesn't change, and the change in the distance between two plates, Delta d, is equal to Delta z. That's the distance that the plate is moving. Therefore, using this fact, you will be able to know that you can equate this here, this one, with this equation and if you do that so as you can see, the distance change of the plate, Delta d, is equal to the distance traveled by the plate by infinitesimal naught Delta z. So, if you replace Delta one over C by this one in the equation on the right side here, you instantly see you can omit Delta z from both sides and if you do that, the force becomes Q squared over two Epsilon naught A. So, without knowing anything about the electric field or the amount of charge interactions, we were able to derive the force between the two plates. So, Melody, what is the meaning of this equation? It looks like Coulomb's law where the force is dependent on two charges multiplied by each other and here you have charges on both sides of the parallel plates, so the force is dependent on the charges accumulated by the parallel plates. Exactly. So the more charge we have on top of each plate, the more force we will experience. So in this slide we will deal with another example that is a little bit more advanced than just a simple parallel plate capacitors. We will think about a torque on a variable capacitors. You can see there is a pivot about which these blades can rotate with an angle theta collectively and depending on the angle, the capacitance of this device will change linearly. And we're going to think about the torque acting on this device by using the same concept of virtual work. So, it is easy to see how the idea is extended to conductors of any shape and for other components of the force and we replace force by the component we're looking for and we replace Delta z by a small displacement in the corresponding direction. In this geometry, in this problem, instead of thinking of the lateral or horizontal translation, we're thinking of rotation. Because we're thinking of rotation, we will replace force by torque and replace displacement by angular displacement Delta Theta. So, if we have an electrode with the pivot and we want to know the torque Tau, we write the virtual work as Delta W equals tau times Delta Theta instead of F times Delta z, where Delta Theta is a small angular displacement. As you can see here, there are multi-layer capacitor and if you think a little bit further, these capacitors are connected in a parallel fashion. Although it seems to be in series because they have alternating electrons from positive to negative, in circuit this becomes parallel capacitor and because of that, the total capacitance of this device will be the summation of each individual layer. So, that's why we have n minus one terms in front of this. If n is equal two, if we have two plates, we have one capacitor. If we have three plates, we have two capacitors. If we have four plates, we have three capacitors. So that's why it's n minus one and then only the overlap area will be the capacitor area. Only the overlap area. So that's why we have one minus Theta over Pi in front of the area of this semicircular, half circular plate. You can see one minus Theta over Pi is in Radian. If Theta equals zero where you have full overlap, you have whole A. If Theta equals Pi which is 180 degrees out of phase and there's nothing overlap, that is zero, and boundary conditions are written here. Theta can only be between zero and Pi and n should be equal to or larger than two. So knowing this, we are going to put this C into this equation where Tau Delta Theta equals minus Q squared over two C squared times Delta C and if we replace C by the equation above, we end up with this equation where Tau is equal to Q squared d over two Epsilon naught n minus one times A times one minus Theta over Pi squared. So, Melody, what is the meaning of this equation? If you look right here you can see the angular dependence of the torque. So if the angle is equal to Pi, which is when the plates are no longer overlapping each other then the torque goes to infinity and so the most comfortable position of the plates is when they're completely overlapping. Okay, from this discussion, can we understand which way the torque is acting? Is it acting to decrease Theta or increase theta? So, you want to decrease Theta. Exactly. You want to decrease Theta and in that way it will make a torque to make it round about and make a full overlap. In addition, we can think of this as a non-contact break if you apply high enough voltage, if it deviates from this plates it will react in a way to resist the motion. So this is another way to think in terms of engineering and think how the knowledge we got from science can be used in engineering field. Okay, so let's take a closer look to what is happening between those two plates in parallel plate capacitor when we apply voltage or if we inject charges on the plate. So, what is the origin of forces between the plates? If for a charge on one plate we write Q equals Sigma A, where Sigma is the surface charge density, then the electric field between the plates is Epsilon Naught equals Sigma over Epsilon naught which Melody has explained a few lectures before. So, in this slide we're revisiting the parallel plate capacitors that we solved using virtual work and see what happens at a microscopic level. So if for the charge on one plate we can write Q equals Sigma A, where Sigma is the surface charge density, A is the area of the plate and we learned a few lectures before and Melody explained it very well that the electric field between those two plates is equal to Sigma over Epsilon naught. So if we think about electric field as the force per unit charge, we can just multiply the electric field from one plate and the amount of charge on the other plate. So that results in Q times E naught which is two times larger than the force we just derive from the previous slide. You may wonder what's wrong with our calculation. The reason that we have one over two in front of this can be visualized using this picture. So if you think about the charge on the plate they are spread out in space like this and you have layer of surface charge Sigma with a finite thickness and the electric field reaching on the surface of the electrode will be shielded by the top-most surface layer charge and be decreased down to zero once the electric field has traveled across the layer of the surface charge. If you think about the situation, in fact, the only charge that feel easier to its full amount is the top most and the bottom most will feel nothing. So, on average, they will have half of the field that is reaching the plate and that's why we have one over two in front of our equation.