Welcome back to Electrodynamics and Its Applications. I'm with my teaching assistants Melodie Glasser and my name is Seungbum Hong. So, today we're going to learn about electrostatics with Gauss Law and symmetry argument. So, we just learned that we have two laws of electrostatics, the first being the flux of the electric field from a volume is proportional to the charge inside, which we call it a Gauss Law. The second one is the circulation of electric field is zero, therefore, the electric field is a gradient. As you know from the second law, Maxwell's equations, the circulation of electric field is a function of the rate of change of magnetic field. However, in electrostatics because all the charge are fixed in their positions, there is no change in magnetic field. So, the circulation of electric field will be zero. So, these were the two laws that we just learned before, and now we're going to advance this concept a little further. So, we are going to deal with Gauss Law only, for conductors, and use arguments of symmetry to understand and calculate the electric field from different geometry. All right. Before going into the details of our scope of the lecture, let's take a look at one important concept that is equilibrium in an electrostatic field. So, here we have three charges minus charges at the corners of a equilateral triangle, and in the center of mass of this triangle we have positioned a plus charge. With this example, let's think about whether it is possible to have this positive charge in stable mechanical equilibrium. Before answering this question I'm going to ask Melodie what mechanical equilibrium is, okay? So, Melodie can you tell our students what mechanical equilibrium is? Yeah. So, mechanical equilibrium is where there's a physical object and if it's moved from a certain location then it will fall back to the location, and therefore its energy is minimized at this particular point and it will try to work back to that point. Exactly. So, in order to have mechanical equilibrium, we need no net force, and secondly we need a restoring force that will put back the particle into that position if there's any perturbation. So, with this example, if I know that the net force will be zero, because they are attracting the positive charge with the same magnitude and in a direction that will result in no direction at all. However, if I move this charge a little bit, back and forth then you will see that this will not have a restoring force to put it back to the original position. So, there is no points of stable equilibrium in any electrostatic field except right on top of neutral charge, and we can prove it using the Gauss law that we learned before. So, we're going to do it in the following slides. So, let's take a look at this picture. This picture is about children dragging the rope to their side and see which group is stronger. When they are in equilibrium, they don't move a bit and that's because the net force is zero. So, the first condition for mechanical equilibrium is that the sigma F which is the summation of all the forces involved is equal to zero. However, as Melodie mentioned before, there can be two type of net force zero states, where you have this concave part of energy, where if you change the position of your ball in the ball will exert a restoring force to bring it back to the original position, whereas in this concave type of bowl which is upside down, if you put the ball on top of it, if you push it, you see it will fall down to the unstable position. So, the second important thing is whether we have restoring force. In thermodynamics, we have learned that could be understood in terms of the curvature of the free energy curve. So, the first derivative should be zero which means it's either minimum or maximum, and the second derivative should be larger than zero meaning we have a concave curvature of energy, right. So in this slide, we are going to explain the dilemma in electrostatics and think about why in electrostatic world where we have the same number of positive and negative charges, we cannot find a mechanical equilibrium point. So, Melodie can you explain this to our students? Sure. So, one way you can think about this is if you imagine that there is a field of equally distributed charges for example like positive and negative charges that all cancel out. If you put a certain charge in there, like a positive charge, it needs to have, it won't feel any forces which goes for the first part of our equilibrium because everything is perfectly balanced. However, in order to have a restoring force, there needs to be something and we know this from Gauss's Law, there needs to be something like this where all the negative charges, or the negative electric field goes towards the positive charge to hold it in place. The only way to do that is to have a negative charge right on top of the positive charge here, and they would have to occupy the same space. Exactly. So, that will break the assumption that we just had. So, if I add a little bit to what Melodie already did a good job, is that for any test charge the positive charge P naught, if you want to have restoring force, you need to design electric field to point toward this charge. Based on the Gauss Law, we know if you want to have this, which is net flux is non-zero and minus in fact, then you need to have additional minus charge on top of plus charge. So, it is impossible to balance a positive charge in an empty space. A positive charge can be in equilibrium if it is in the middle of a distributed charge. So, you can think of this as plus nucleus distributed with electrons around it. So, that's an atomic model we have for the modern physics. Of course, the negative charge distribution would have to be held in place by other than electric co-forces. So, in the next slide, we're going to think about such an example. So, what if there are pivots or other mechanical constraints? We can devise a situation where the charges in stable equilibrium for sideways motion by placing positive charges at each end of the tube and adding non-electrical forces from the tube wall. So, if I put plus and plus here, they will exert a repulsive force. The plus inside this tube, you can imagine this is like carbon nanotube, then the nanotube will exert repulsive atomic force or repulsive quantum mechanical force onto this charge. So, in this central point, you will not have the possibility to move away from this position and then you are in a stable mechanical equilibrium. So, here we are at the point to think about a little bit more complicated problems, equilibrium with conductors. So, the question here is, can a system of charged conductors produce a field that will have a stable equilibrium for a point charge? That is to say at point other than on a conductor, and the short answer is no. This is hard to prove but we can now try to think why that's the case. So, if charge producing a field are stationary then there is some direction for which moving a point away from zero field point P naught will decrease the energy of system. Any readjustment of the charge on the conductors can only lower the potential still more. So, what does that mean? Can you elaborate what I just explained? No. Probably it's a little bit too difficult. So, I'm going to draw this here. So, imagine you have a piece of conductor here, and imagine that you are going to put a charge, a test charge here, right. This test charge will create electric field that is going outward, self-created electric field. The thing is because this is the only charge we have right now in the metals, the existing charge like electrons will try to mitigate that to remove this field. However, as they mitigate this field, there will be depletion of electrons from those sources. So, you have to have some plus charges somewhere which will create additional electric field. Because of the additional electric field, the plus charge want to follow that path. If that happens, then there's no way this will be positioned where it was originally. So, this is a qualitative way to understand this. So, in this slide, we're going to advance our understanding on equilibrium by thinking about equilibrium with conductors. So, the question here is, can a system of charge conductors produce a field that will have a stable equilibrium point for a point charge? In other words, at a point other than on a conductor, meaning the surface of the conductor. The short answer is no. So, I'm going to try to explain this as far as I can to help you understand this. If charge producing a field are stationary, right, then there is some direction for which moving a point away from zero field point P naught, will decrease the energy of system, and any readjustment of the charge on the conductors can only lower the potential still more. So, that means you cannot reach a stationary solution with nanoconductor. Now, I'm going to ask Melodie if she can explain it a little bit more, please. Sure. Yeah. Okay. So, basically, how you can think about this is if you have a material like a metal, and you place a charge on it, then the charges within the material are going to become attracted to that charge, and so you'll have negative charges that appear to try and mitigate this positive charge, but those negative charges have to come from somewhere. So, there'll be creation of more positive charges, and these will try and move around to come up with the least amount of energy. Yes, and there's no conversions on that solution. So, the other thing you can think of is, if you have a metastable state as we mentioned about this ball on a convex surface, the only way you can maintain the position is by moving this mountain left and right to balance this. So, this is dynamic equilibrium. So, is it impossible to balance the charge by electric forces? The answer to this question is no. But it is possible if the charge can be held in one spot by electric fields if they are variable, right? So, if it is static it is impossible, but if you can vary the electric field, yes you can do that. So, before moving further, let's make a mental note on some concept that we just mentioned. So, before moving further, we're going to make a mental note of what we just learned, and we're going to categorize the equilibrium that we just learned. Okay. So, we're going to distinguish two things, static equilibrium, dynamic equilibrium, and then we're going to think about steady-state. Okay. We're going to think about steady-state. So, let's start with static equilibrium. We just learned static equilibrium can be mentally pictured as a ball in a bowl. Right. Ball in a bowl where you have a gravitational field downward and at this point you have net force to be zero, and you have perfectly restoring force to put it back on to that position, and from a macroscopic view, this position is stationary, meaning it is not a function of time, doesn't change. No flow. No flux. Nothing. No motion. That's static equilibrium. Right. Now for dynamic equilibrium and steady-state, I'm going to use a bath tub analogues to help you understand. So, imagine you have a bathroom and you have a bathtub, and you fill the bathtub with water to certain level, and you close the drain and you close the faucet. So, there is no input nor output, and in this saturated case, you will reach dynamic equilibrium when your water evaporates at the same rate as the steam is condensated onto the water. In that case, the water content as a function of time is not a function of time. So, there's no change in the physical parameter here. However, in steady-state, which is a larger category of including the static and dynamic equilibrium, you can think of the same bathtub, but now you are pouring water onto the bathtub and you are draining it, and if you pour the water in the same rate as you drain it, then macroscopically the volume of water doesn't change as a function of time. However, you have a net flow. You have a net flow direction coming in and coming out. In this case, you don't have a net flow, but you still have flow up and down. In this case, you don't even have a flow. So this is kind of a mental picture I want you to have when you think about those terminologies. Is that clear? Yes. Okay good. So we'll then move to the next part. Okay. So now let's think about the stability of atoms. So, we have just mentioned that when we tried to explain to you in electrostatic world, you cannot find a stable position for a positive charge. Right. Only in dynamic world, you can do that. So, let's think about the stability of atoms and also the model of atoms that was developed throughout the modern physics. Right. So, the first picture, the physical picture of atoms was to have minus charge in the center and plus charge distributed over the minus charge. Of course, we know this is not correct. So, they went through the Rutherford-Bohr model where we have electrons revolving in orbits like planetary motion around a nucleus, but now we also know what the problem is here. So, unlike the planets, which is electroneutral. So, the acceleration that we have through this orbital motion doesn't affect the energy. The kinetic energy of the planets. However, for charged particles we also know from Maxwell equations that acceleration of charged particles will lead to emission of rays. So, emission of rays will reduce the kinetic energy of those charged particles. So they will spiral down and collapse with the nucleus. So, now with the quantum mechanics now we know that these electrostatic force is balanced by uncertainty principle or in other words quantum mechanical effects, and we can only think about as a probability function. Density function of waves. Right. Still, we think that pluses in the center and minus are distributed over that part, and that's the only way how we can have a equilibrium state in our atoms. All right. Now, we're going to cover some of the examples that we just promised to think about using the argument of symmetry in Gauss Law, and based on those logical thinking, we can easily calculate the electric field from for example a line charge. So, let's first think about the field of a line charge. Imagine like in the picture you have a line with infinite length. If we do have this line, and if we deposit charges in a uniform way, then Melodie. What would be the symmetry of this field out of this line charge? That would be cylindrical. It will be cylindrical. Right. Why? Because the actual field will be canceled out by the same amount of charge sitting on the opposite sides of the line charge. Right. So, we know from the argument of symmetry that the electric field will have a cylindrical symmetry and they will only have radial component not eggshell one. Okay. In that case, what we can do is to draw a fictitious cylinder. Right. Fictitious cylinder here, and then think about the Gauss law. So, what is Gauss law? Melodie. It is a EA is equal to q over epsilon note. Exactly. So in other words, it is the electric flux out of a closed volume is equal to the net charge inside that volume. The electric flux out of that volume is equal to the net charge inside that volume divided by the permittivity of the vacuum. Right. So let's think about this in this example. So, you have a cylinder. Fictitious one, and the line is with infinite length, but we just cut it arbitrarily, and let's say this is length L, and let's say this is r, the radius of this circle of the cylinder. Then the electric field flux, we can only think about the peripheral area because the top and bottom part will have no flux. Right. So then the electric flux as Melodie mentioned is electric field times the area because it's uniform, then this electric field times two pi r, which is the length of this circumference times L, which is the length of this side. Right. So this would be the flux. Just to say. One cannot see this. So, the flux E times two pi r L will be the flux, and then the charge inside this cylinder. Lets say that's q and divide it by epsilon naught. Okay. Now, because this is infinite line, we cannot just give a finite number for the total amount of charge, but we can define the charge density charge per unit length. Right. So, let's say if I move this L to this right term, then it becomes q over L epsilon naught and if I define this as a line charge density, which is lambda, then it becomes lambda over epsilon naught. Right. Then with this knowledge, if I move 2pi r in front of our electric field to this part, then we will have the field equation here, which states electric field from a line charge with infinite length and uniform charge density has a dependence of lambda over 2pi epsilon naught r. Lambda is constant, 2pi epsilon naught is constant, r is the only variable, which means the distance from the line charge. So, you can imagine the strengths of electric field from this line charge follows 1/r dependence. That's different from 1/r2 dependence from the point charge. So from our everyday life examples- So, with the knowledge of electric field from a sheet of charge with infinite area has no r dependence. We're going to extend this idea to two sheets with equal and opposite charge densities. As you can see from this picture, it resembles the capacitor structure, where if you charge your capacitor the top surface becomes positively charged and the bottom becomes negatively charged with the same charge density, right? So, let's use the Gauss law that we just learnt. So, I'm going to draw it again here, you have plus charges here in the infinite plane you have minus charges here in infinite plane. From the argument of symmetry, if I make a Gauss surface like this to include both planes then I have no net charge inside, right? No net charge. Therefore, there will be no flux out of this, there will be no flux. One way to fulfill that is to have no electric fields on both sides. Or, you need to have electric field that comes in and comes out, comes out and comes in in the same way, right? That's the only way you can think of it. But, let's assume this is the case, no electric field. Then what we can extend this is, then if I make a Gauss volume like this, because I already know the electric field outside this one is zero, right? And because I know net charge is non-zero, we need to have electric field to the right which is equal to because we only have one phase, E times A equals Q over epsilon naught. If I divide Q by A, then it becomes sigma. So, E becomes sigma divided by epsilon naught, which is two times the field created by single sheet of charge, right? The same argument applies to the left or right side of the sheet, right? Here, we already know to the right side, the electric field is zero, right? We have non-zero minus charge inside. So, the only way we can imagine is to have electric field going into the side and it has the same magnitude, right? So, this is one way to understand the electric field between the capacitors outside the capacitors. So, now I'm going to ask Melodie to explain what I just explained from different perspectives. Okay. So, if we redraw the same picture where we have positive charges over here and we have negative charges over here, we know that the electric field from positive charges should go in this direction and we also know that the electric field from negative should go inward. So, you have an additional electric field like this and so before, our plate electric field was equal to sigma over two epsilon naught, and now we have two of them that are kind of additive, and so we can multiply that by two and these twos will cancel out and you get the same equation as over there. But, what about the outside? So, if you think of this as well, you know that on this side, the electric field is going to be playing in because it's negative. However, you still have the electric field from over here going out which is positive and then you would have the summation which is a positive and then you would subtract the negative electric field over here and that would equal zero. Exactly. Yeah. Very good. Thank you Melodie. Okay. So, we just learnt from what Melodie told us or my argument that the electric field between the sheets with equal and opposite charge densities, will be sigma which is the charge density over epsilon naught. Outside them, we have zero. So, you can kind of understand that the charge outside the capacitors is shielded. Meaning, you don't have any electric field emanating from capacitor. All right. So, now we're going to revisit the problem of sphere of charge using the Gauss law and using what we just learnt, the argument of symmetry to understand what the electric field inside a sphere of charge is as well as outside the sphere of charge is. So, this computation can give a good approximation to the field inside an atomic nucleus. Because, atomic nucleus if, think of a set of protons and neutrons being squeezed together with nuclear force and they are uniformly distributed. Despite the fact that the protons in our nucleus repel each other, they're spent nearly uniformly throughout the body of the nucleus due to the strong nuclear force, okay? So, let's take a look what it's looked like and if we understand this problem, we can also understand the gravitational force inside our earth, okay? So, we are you going to catch two birds with one stone, okay? So, let's try to solve this problem by thinking of our sphere of charge with charge density rho which is defined by the total charge Q over the volume of the sphere, which is four over three pi, let's say the radius is capital R. That is R power of three, that's the denominator and then Q will be numerator, okay? With this in mind, let's think of any point inside the sphere. Let's say, this is a point with the distance from the center of the sphere to be small r, right? Then from the argument of symmetry, we can understand the electric field will be radial, meaning it doesn't depend on the direction, right? To fit this radial, then I can find a set of points that will have the same electric field. Those set of points will make a small sphere with radius of small r, right? We know electric field E will be the same for every point of them, pointing outward in radial direction. Therefore, we can use the Gauss law, right, where the charge inside this fictitious sphere will contribute to the flux out of the sphere, right? So, that will be electric field times four pi small r square, will be equal to the charge inside this which is, four pi over three r to the Q times rho over epsilon naught, right? If I do this math correctly, then four pi is out, r square will remove this one, right? So, I'm coming up with rho r over three epsilon naught which is just written here. So, that's for a case when this small r is smaller than or equal to the large R, capital R. So, meaning, the electric field is a linear function of r. So, it increases as you move away from the center of your sphere of charge, and will hit the maximum when you are at the surface. Once you're at the surface what happens is, outside the surface, right? Let's say this is outside the surface, then the Gauss surface will always include the same amount of charge. It's fixed. So, Q doesn't vary no more. It varies no more with the radius. Therefore, it will be constant. So, you will understand the E times four pi r square in this case, we'll have Q over epsilon naught where Q is just a constant. Therefore, you will start to have one over r square dependence which you have there. It is decreasing, right? So, this is the way you can use the argument of symmetry and Gauss law to calculate the electric field inside a sphere of charge which is a very difficult problem to tackle if you start from scratch, okay? All right. Now, there is another question here, what about a thin spherical shell of charge? Meaning, what would be the electric field inside a thin spherical shell of charge? Melodie can you tell us the answer? I want to guess zero. It is zero, why? If you make any Gaussian surface inside, it will contain no net charge, all right? So, that's how you can understand. So, if we carve out all of the materials inside our Earth and we only have crust, then inside you will have no gravitation, right? Okay, good. So, in fact, imagine this is the distributed mass and imagine this is gravitational force, then it will follow the same line. So, meaning your gravitational force will be zero at the center of your path, okay?