All right. So, let's cover the flux electric field from this slide, and let's derive a field equation that depends specifically and directly on the fact that the force law is inverse square. So, imagine you have a light bulb like this, right? This is energy efficient light bulb which you can buy from Costco or Home Depot. Imagine you are shooting out photons, and you know photons cannot be created nor annihilated along the path unless you perturb them, right? So if you do that, if you follow the trajectory of each photons, you can put that arrows to infinity, and the errors will not be added or removed along the way, then the number of arrows will really be the flux, right? So, the light intensity, which is the number of photons per unit area, will really depend on the area of interests, which increase as a function of the radius and which is r squared. The radius will increase as a function of r squared and therefore, the light intensity, which is the number of photons per unit area, will be one over r squared, right? So, suppose that we imagine for a moment that electric field represents the flow of something that is conserved, everywhere except at charge. Then, we can prove that electric field is also one over r squared. All right. So, let's take a look at this simple situation where you have only one charge in the world and you have radial function, and imagine as two spheres, let's see this is watermelon. So, Melody, do you like watermelon? Yes. That's my favorite fruit as well. When I was a young kid, there was a conventional market from the place where I attended the school to my home. Every time I went by the path, the owner of the fruit store asked me to take some piece of watermelon, and this was the exact shape that he carved out of the of the watermelon, right? So, imagine this is a watermelon with the shell being very thick and part of that shell, we can imagine we have this. What do you call the shape of this, Melody? I don't actually know. Is that like a cube? No, it's not a cube. Nay. You don't know this? No. Okay. So, let's say this is one type of a shape geometry out of the shell of your watermelon, and think of the flux from the charge going through this entity. So, think about the four faces that are parallel to the electric field, and because they are parallel, the flux will be zero. So, we can disregard all the faces that are surrounding this part. So the only part that are left will be the outermost and inner part of this object. So, in that case, if we do this integration to evaluate the flux, then we end up with only two parts, minus E sub a times Delta a sub a, which is the area of this inner part, plus E sub b times Delta a sub b, which is the outer part. We know that if electric field is one over r squared, then the area is, what? Increasing. Increasing as r squared, right? So, the multiplication of those two will be constant. If you have opposite sign, then the summation of this will result in zero only if electric field fulfills the one over r squared law, right? Imagine now, we are going to tilt the inner surface and outer surface with an angle. Imagine you can do that. In that case, you will increase the area of your surface by one over cosine Theta. However, when you do the dot product, because the normal component will be proportional to cosine Theta, again, the product will be constant no matter which angle you tilt it, right? So, you will see the summation of those component will be zero no matter what type of till you give to that part, and you may wonder why we're doing this kind of manipulation, right? So here's the answer to that. Imagine this is a beer can, and you have a plus charge outside your beer can. You want to know whether the total flux out of the volume enclosed by any surface like this one is zero or not. What we can do is have a laser light from the charge penetrating through this beer can make an arbitrary shaped object. That is, like the one that we see from the watermelons shell with tilted angle, right? If we'd scan this up and down, we can cover the whole volume. You agree, right? Yes. You agree that, from the divergence law, no matter how you cut your object, the flux of the arbitrary object will be the same as the summation of the part that you just cut it, right? Yes. So, each part that you cut will have zero flux and no matter how you scan it, everybody will have zero flux. So, if you sum them up, you have zero flux. So, you can prove that the sphere can has no flux at all. Yes. That's very simple, right? Now, let's develop this idea a little bit further. Let's revisit our watermelon shell and place our charge inside the watermelon. Think of a volume that where you have two pyramids connected through this charge, and think of this as another volume and evaluate what would be the flux inside this strangely shaped object. So, before telling you the answer, let me ask Melody, what do you think, would it be zero or not? I don't think so. Why not? Well, besides that, it says it here. It is kind of like to me if you think of chemistry and if you have a source of generation in chemistry like a chemical reaction, then instead of having a constant amount of solution, there's more product. So, here, since there's a generation shots of charge, then you probably aren't going to have this zero flux. Non-zero flux, yeah. So, that's a very valid point. If you have a source of charge or source of generation for your flux of interest inside the volume of interest, then you probably will have net flux because that's intuitive. But beyond being intuitive, let's see if we can prove it mathematically. So, think about this. The surface here has the outward normal, and surface there has also outward normal. So, the the faces parallel to electric field will not contribute to the flux. So, all of them will be zero. We only have to think about the external surface of one pyramid and external surface of the other pyramid. Right? And because both of them has the same polarity for the surface normal, as well as they have the same direction of the electric field, they will have positive number both sides. So if you sum them up, you will have non-zero value. So, that's why we have non-zero positive flux out of this charge in this case. Now, so with those two information, if the charges outside the arbitrary volume you don't have any flux. If the charges inside a volume you have non-zero flux. With those two information, we're going to evaluate the case of an arbitrarily shaped volume, and that's fascinating. Because, think of a potato from Idaho shaped like this, it's an arbitrary shape. Imagine now you're going to put your point of charge inside this potato, and now I'm asking Melody, can you evaluate the flux out of this charge through the shape that I just mentioned if it is an arbitrary shape? And to some students, this might be daunting task. But to our surprise, using those two information that we just learned, the answer can be very simple. So, let's try to start figuring out how we can answer that question. So, we're going to carve out part of the potato in a shape that we can understand, in a very easy term. What would be the shape? Like a circle. Circle. So, we're going to carve this out in a sphere type. So, no matter how we cut the volume out, the flux from that volume is the same as the flux from each part. You agree, right? Yeah. So, the arbitrarily shaped potato, and I'm going to call it where we have the charge in a spherical shape. So, I'm going to pull it out here. So, I have a sphere plus potato without a sphere. So, if I want to know the flux out of this arbitrary shape, I just add the flux from here, flux from there. Right? Now, for this part, because my charge is located inside a cavity, it is considered to be outside my volume. Do you agree? Yes. Because it's inside a cavity, it is not inside the carved out volume. If it is not inside the carved out volume, what is the flux? Zero. It's zero. So, this is zero. Now, the part that I'm now interested in is this spherical shell. And because it's a sphere, is very easy to calculate the flux. All this is a radial field, right? So, the electric field on the surface of this sphere will have the same magnitude with radial vector. So, the En will be 1/4pi epsilon naught times q r2 that we learned times the area of the sphere is 4pi r2. So, if I multiply them, you end up with qr epsilon naught. So, flux from the sphere is q/epsilon naught, flux from the potato without the sphere is zero. So, if you add them up, no matter what shape you have, the flux through the surface as prime is q/epsilon naught. Isn't that fascinating? Yeah. That's really cool. That's really cool. So, again if you have charge outside an arbitrary shape is always zero, but if you have charge inside an arbitrarily shaped volume, there's q/epsilon naught. Yes. Okay. So, we didn't use any complicated equation here, it's all easy. Right? Yes. Good. So, we call that Gauss Law, the divergence of E, it is different from Gauss theorem, but it is related. What if we have more than one charge? For example, two charges. Then we just add them up. Right? Because, we can use superposition law. So, here are three examples that I'm going to ask Melody. Now, think of a volume with an egg shape, and imagine you have two charges, q1 and q2 outside that volume, what would be the flux out of this sphere? No, Because there's no charges inside, it will be zero. Exactly. Now, the second example. Let's put one of the charges, which is q1 inside this egg-shaped volume, what would be the flux out of this? q1/epsilon naught. Good. And the second case, where we put both of them inside the egg, Q1 and Q2 both, what would be the flux? It would be q1+q2/epsilon naught. Very, very good. So, with this knowledge, what we can do right now. Imagine you have a sphere of charge, it's smeared out, and it has the same density, charge density row. What would be the field inside the sphere, electric field? And you may think, ''Oh, this is a very difficult problem.'' And you may think, ''what is the relationship between this egg plate and this one?'' Right? Well, let me show you how this is related. So, you know from the spherical symmetry of this charge, this should be radial field. Yes. Right? This should be radial field. If it is a radial field, the electric field that is felt by this point will be shared with all of the sets that makes the sphere inside the sphere. Am I right? Yes. And the electric field that is experienced with the sets of field is due to the charge inside the sphere, not outside the sphere. Yes. Right? So, if I know the distance from the origin of the center of the sphere to the point that I'm interested in, and denote it as small r, and if I denote the distance from the origin to the outer surface as large R, then I can start to figure out what the electric field here is. How? Let's start. What is the charge inside this yellow sphere? You know the density, Rho. Volume. Volume, it is four over three Pi. Is it small r or large R? Small r. Small r to the cube, to the power of three, that's the charge inside the sphere. What is the flux then? This is charge, like an egg plate. You have to put Epsilon naught, that's the flux. Right. Now, how do we define the flux? It's the E field times the area of the sphere. So, it is E field times 4 Pi. Is it small r or large R? I will say small r. Small r. So, it's E times 4 Pi r squared, and if you rearrange this, E becomes how? Four pi is out. r square becomes r, so you came with three Epsilon naught, Rho r. So, what you have is electric field here is linearly dependent on r. All right? Outside this sphere is one over r squared but inside is r dependent. So, if I plot as a function of the distance, let's say this is r, electric field goes linearly up and then goes one over r squared like this. How interesting. You just got it through this egg plate argument, right? You can also imagine the gravitational force, gravity, follows the same curve because it's distributed mass and you have the same form of equation. So, what it means, if you drill a hole from Seoul to, say, Argentina, you will have linear harmonic oscillator because it is like spring. All right? So now, with the egg plate and Gauss' theorem, you can apply that to many other fields as well. Okay. Okay, so Gauss' law; the total flux out of a closed surface is equal to the total charge inside divided by Epsilon naught. So, in a mathematical equation, it is the surface integral or the normal component of electric field around a surface is equal to sum of the charge inside divided by Epsilon naught, and you can come up with this law. So, Gauss' law follows from the fact that the exponent in Coulomb's law is exactly two. Otherwise, you don't have the Gauss' law. It is an expression in different form of the Coulomb's law of forces between two charges. So, that would be the summary. Now, let's think about Gauss' law in terms of derivatives. So, we can have point function. Let's think about the infinitesimal cube again. You can see cube to the right side of Melody. So, she's thinking of cube. There's a cube test, another one, but not here. So, the Gauss' law says the charge inside a cube, which is Rho times dV, over Epsilon naught should be equal to the flux, which is divergence of the electric field times the volume of the cube, dV. Yeah. If I remove those two, then it becomes divergence of an electric field is Rho over Epsilon naught, that's the first law of Maxwell equation. Then if I have the curl-free condition, if I combine them, then it becomes the Coulomb's law of force. Very simple. Now, I just mentioned this before but let's revisit this. Field of a sphere of charge. What is the electric field, E, at a point, P, anywhere outside the surface of a sphere filled with a uniform distribution of charge? We figured out the electric field inside but now we're going to think about outside. So, outside the sphere, it doesn't matter what type of distribution you have, if it is radial distribution or spherical distribution. So, we can condense all of the charge into one single point. If you do that, then we know that the flux out of the surface is equal to the charge inside, contained inside the volume, divided by the permittivity, and because we know the distribution is Rho, then we can come up with the number for the Q is equal to Rho times 4 Pi over 3 times a cubed. So, we can replace Q by this one inside the equation but still, our dependence will be inverse R square. So, it is the same as that for a point charge, Q, and that's why when we think of a gravitational force outside the Earth, we can condense the mass into one point and just think of the distance between the center of mass instead of doing all the integration over the entire volume of the Earth. Now, let's think about how we can describe or picture the field lines. So, we can use equipotential surfaces and that's the geometrical description of the electrostatic field where you have a line of potential, where you have the same potential, then the direction of electric field is always tangent to the lines, that's the gradient direction. The strength of electric field is represented by the density of lines per unit area through a surface perpendicular to the lines. So, the more E field lines you have in a smaller area, then you have higher field. Gauss' law states the lines should start only at plus charges and stop at minus charges, and the number which leave a charge, Q, must equal to Q over Epsilon naught and you can neither create nor annihilate the line in between. Equipotential surface are at the right angles to the electric field lines and are spheres centered at the charge for a point of charge like this. For the field lines for a dipole, minus and plus, will look like this, and we will use a lot of this picture in the next slides or lectures. So, take a look at this and see one or two interesting thing. One is in the center plane, all of the fields are crossing in perpendicular way and the potential here is zero. But potential here, so there's another plane that has zero potential like deferents in addition to the ground and in addition to the infinity plane, it is between the dipole mode. However, that doesn't mean you don't have any field because field is the gradient, not the absolute value of your potential. With that, we're going to wrap up here and see you again next time. Bye. Have a nice day. Bye-bye.