Now, we have assumed polarization is uniform in the dielectric. What if the polarization p is not uniform inside dielectric? So, let me draw a very simple picture so you can understand what I mean. So, in this lab we assumed we have uniform charge density. Now, we're going to create a gradient, let's say we have four pluses, three pluses, and two pluses. So, we have higher density and lower density as you go down. Let's make the same case with minus charges, minus four, minus three, and two minuses. Now, I'm going to overlap those two slides and then slightly displaced and see what happens. So, if I displace them a little bit, we have four pluses, top two minus button, we have more minuses and pluses in the bulk. So, the net number of minuses will be two here minus two and here minus two, and here plus four. So, no more is the case that we have neutral in the box. So, if we have non-uniform polarization, we can imagine we will have bulk charge density. Okay, so that's the main message. So, if P is not constant, we can find a charge density and the volume because more charge come into one side of a small volume element then leaves it on the other as I can as depicted in this very simple example. The charge moves across any surface element is proportional to the component of p perpendicular to the surface. So, in more general way sigma sub polarization is equal to p.n. In our case, n and p was parallel, so sigma polarization was P. However, if it is perpendicular then signal polarization will be zero, and in between you have this cosine rule. If we're thinking of an imagined surface element inside the dielectric, the charges moved across the surface doesn't result in a net surface charge. Why? Because they are equal and opposite contribution from the dielectric on the two sides of the surface as you can see in this picture. So, now we're going to solve the problem of volume charge density inside the dielectric. The displacement of the charge can result in a volume charge density, as we have already discussed. Total charge displaced out of any volume V but a polarization, let's say that this minus delta Q sub polarization. This is the integral of the outward normal component of P over a surface S that bounds the volume V. So, take a look at this picture and imagine I am pumping out some charge out of this volume and we are equating that to be minus delta Q polarization then the delta Q polarization is equal to minus surface integral of P.n da. Why? The P.n was the surface charge density. Right? If I use the fact that the whole change in the charge is equal to the volume charge here because I am pumping out of here. Then it is volume integral of rho sub pol dv and if I equate them we will have this equation, the rho sub pol integration of this one over volume is equal to minus surface integral of p.n da. So, if you agree with me then we can try to apply that to our known example. Let's try to apply this into our known example of dielectric slab in between parallel plate capacitor. So, we have kind of Gauss theorem that relates the charge density from polarize materials to the polarization P, which is this one and with Gaussian surface S, again this dotted line we can write down like this, so sigma sub pol dv is equal to, so if you look at here in the air you don't have anything, so it is zero. So, what are left is only surfaces, surface of the dielectric surface of your metal. So, for the surface of our dielectric, we have sigma pol times delta A. So, there is no volume polarization only a surface polarization. Therefore, the volume integral becomes sigma polarization delta A, which is the area of Gaussian surface. Since P, only exists within the dielectric layer and is perpendicular to the interface, the surface integrals becomes P delta A, this surface integral becomes P delta A. So, if I equate this one and this one then sigma polarization becomes P for a parallel plate capacitors. So, this equation works well, well-known example. So, in this slide we're going to expand our equations applications to more general case using Gauss mathematical theorem. I'm going to ask my teaching assistants Melody what Gauss mathematical theorem is? Okay, so I think we have the surface integral here, our electric field times da which is equal to the phi integral this gradient over this factor times dv. Exactly. So, as you can see from what Melody wrote, the flux of a vector field, in this case electric field flux can be written as the surface integral of the normal component of electric field which is equal to divergence, the volume integral of divergence of the electric field of a vector field. So, we can use that also for this polarization flux so P.n da, which is the first term here can be replaced by divergence of polarization volume integral. If you do that then you can see from these two terms that row pol which is the volume density is equal to divergence of polarization and you put a negative sign there. So, if there is a non-uniform polarization, which means divergence is non-zero, it's divergence gives the net density of charge appearing in the material. But if it is uniform- Meaning in and out you have the same. So, divergence is zero then there is nothing. So, we emphasize that this is a perfectly real charge density. We call it "polarization charge" only to remind ourselves how it got there. Okay. So, in this slide, we are going to discuss the electrostatic equations with dielectrics. So, we're revisiting the Maxwell equations. Now, let's combine the above result with our theory of electrostatics. So, fundamental equation is which is the first equation of Maxwell equations. Divergence of electric field is equal to the charge density divided by the permittivity of vacuum. Now, the rho is the density of all electric charges. We call rho sub pol the charge due to unit non-uniform polarizations, and call rho sub free all the rest. Usually, the rho sub free is the charge we put on conductors like electrons or holes or at known places in space. So, let's take a look at this equation again and we're putting all kinds of charge density including free and polarization, and from our previous equation, we're replacing rho sub pol by the divergence of polarization and minus. So, we have this equation here, and if we rearrange the latter part to the left side of the divergence, then we can have this neat equation where electric field plus polarization divided by epsilon naught. That's a new vector field which we will call our displacement vector in the future, the divergence of that vector is equal to the free charge density over epsilon naught. The equation for the curl which is the second equation will not change. We will have the same no circulation at all for electrostatic because there is no change in magnetic field as a function of time, and using the fact that polarization is linearly coupled to electric fields through susceptibility chi and permittivity of vacuum epsilon naught, we're going to replace this equation using this equation and then what we have is divergence of kappa E is equal to rho sub free over epsilon not, and these are the equations of electrostatics when there are dielectrics. So, now in the world of electrostatics where we only had point charges we're introducing materials. So, now material science kicks in in another side. There are in a form which is more convenient for computation in cases where low sub three is known and P is proportional to E. So, let me ask my teaching assistant again. We have two assumptions here and the second assumptions states that p is proportional to electric field. In material science do we have a case wherein p is not proportional to E? Yeah. I think sometimes with certain materials with certain hysteresis loops. Exactly. Right? That includes. Ferroelectric. Ferroelectric yes, and ferroelectric P is not proportional to the E, and in fact I can show you. If you see the P E curve of ferroelectrics, it has hysteresis as Melody mentioned, and you see the P is no more linear function of electric field. So, therefore this cut class of materials is more complicated and we're not going to cover that in this material. Note that we have not taken the dielectric constant kappa out of the divergence. We can only take this out if this is not a function of space and is constant. So, here we're going to introduce the new vector field displacement D, and there is a matter of some historical importance which should be mentioned. In the early days of electricity, the atomic mechanism of polarization wasn't known and the existence of low polarizations was not appreciated. It was explained by analogy to fluid mechanics. The laws of free was considered to be the entire charge density. In order to write Maxwell's equation in a simple form, a new vector D was defined as follows. We can see D is equal to epsilon not E plus P. So, the divergence of D the displacement is simply the free charge density and the curl of electric field was always zero. If we use the fact that polarization is linearly proportional to epsilon electric field by susceptibility, then we can rewrite that like this and then D becomes kappa times epsilon naught E and kappa epsilon is epsilon which is permittivity so D is equal to epsilon E. So, this is an attempt to describe a property of matter which is far more complicated, but we are simplifying sometimes and use this as a constant and then solve the problem with some approximations. So, Permittivity epsilon, epsilon is equal to kappa dielectric constant times epsilon naught and is equal to one plus chi times epsilon naught. So, this is another constant for describing the dielectric properties of materials and we have simpler equations in a vacuum and if we exhibit in every case all the charges whatever their origin equations are always correct. Alright. So, we're going to think about the fields and forces with dielectrics using the Maxwell's equations. So, we will now prove some rather general theorems for electrostatics in situation where dielectrics are present. We have seen that capacitance of a parallel plate capacitor is increased by epsilon r which is relative permittivity, and we can show that this is true for a capacitor of any shape, provided the entire region in the neighborhood of the two conductors is filled with a uniform linear dielectric. So, now we're expanding our understanding from a very simple case of parallel plate capacitor to capacitor we would initiate. Okay? So, without dielectric we know from the Maxwell's equation number one and two divergence of electric field is equal to rho sub free over epsilon not, and curl of electric field is zero, and with the dielectric present we know we have to replace e naught by kappa E and this stays constant as zero so you can add kappa here without difficulty. So, then the Maxwell's number one and the Maxwell equation number two becomes the same, but with replacing E naught by kappa E. So, Melody if I replace a function with some parameters that has linear relationship like this, in this case epsilon naught is replaced by kappa E, and kappa is larger than one, then what happens to the E inside this one? Will it increase or decrease? I think it should decrease, right? It should decrease. Yes. So, for the given same equation, if you put a parameter or coefficient in front of it, which is larger than one, then effectively, you are reducing the other one. Okay. So, the force between two charged conductors in a dielectric, as you can see will be decreased. So, we therefore have the same equations for kE as for EO, so they have the solution kE is equal to E0. In other words, the field and voltage are everywhere smaller, by a factor of one over kappa, than in the case without the dielectric. So, C in the case of an everywhere uniform dielectric, is increased by the factor kappa and that was already understood for parallel plate capacitor type. What would be the force between two charged conductors in a dielectric? In order to solve this problem, we need some assumptions. But before explaining the assumptions, we're going to explain the big picture. So, Melody, do you remember virtual work? Yes. Can you explain to our students the main concept of virtual work? So, the concept of virtual work is you take a certain force that's constant, and then your delta W is equal to this force constant times delta X, and if you know the work and the change in distance, you can find the force. Exactly. So, you are making a work that is virtual, never happen by changing the distance. If you know the relationship, then this will give you the force because the derivative of the work with respect to the displacement is exactly the force. So, that's what we're doing here. So, the force between two electrodes inside dielectric can be calculated using virtual work that is written here. So, minus round U over round X where U is the potential energy change, and you can see it's minus Q square over two times round over round X times one over C. Since the dielectric increases the capacity by a factor of kappa, all forces will be reduced by the same factor as you can see from this equation as well. In the statement above is true only if the dielectric is a liquid. Why? Because kappa will change if in the solid, the structure will change. So, we have to assume the structure doesn't change as we have displacement. So, that's the complexity. All right. So, electromechanical energy in solid dielectrics. Any motion of conductors that are embedded in solid dielectric changed the mechanical stress conditions of the dielectric and alters electrical properties. So, let me repeat it. In real solid, if you have any motion, if their lattice changes parameters or if you make a strain, electric properties like dielectric constant will change as well, so it's not that simple. So, using this rigorous approach, we can calculate it, but based on our knowledge acquired from this lecture, this is beyond our scope. So, we will think of a simpler case like moving the conductors in a liquid. In a liquid, it doesn't change the structure of the liquid as you know. Liquid moves to a new place, but it's electric characteristics are not changed. So, it's much simpler. In that case, we can use the equation that we just showed in the previous slide with safety. Okay. So, we will discuss the limitations of Coulomb's Law for dielectric materials. Again, as mentioned, in solid materials, the kappa will change as a function of distance, so this will not be true in general. It will only be true for a world filled with either gas or liquid where the change of the distance between constituents or components will not dramatically change it's property, okay? It depends on the fact that kappa is a constant, which is only approximately true for most real materials. It's done on approximation. It is much better to start with Coulomb's Law for charges in a vacuum, which is always right for stationary charges. Now, what does happen in solid? In fact, this is a very difficult problem, hasn't been solved because it's indeterminate, but nowadays, people are using a lot of softwares and other types of methodology to tackle this problem. All right. Let's tackle one other question. Why does a charged object pick up pieces of dielectric? For example, why does your computer attracts a lot of dust in the air? The answer has to do with the polarization of dielectric when it's placed in an electric field. This is a schematic. Let's say, this is a dust, and you have electric field around it distributed like this, and assume you have stronger electric field in this front rather than the backward. Then, you can see you will have attraction. So, there are polarization charges of both signs and there is a net attraction because the field nearer the charged object is stronger than the field farther away. Okay. If they are the same, there is no movement, okay? So, you can imagine. In parallel plate capacitor, if you put a dust inside, do you think there will be attractive force? No. There will be no force, no net force. So, what happens? It will have only gravitational force. So, your dust will be collected in the bottom electrode. However, if you make your parallel plate capacitor replaced by one AFM tip, which has larger concentration near the tip, what happens? Then, there'll be attraction towards the tip. Exactly. So, a neutral piece of paper will not be attracted to either plate inside the parallel plate of a capacitor as I just mentioned. However, if you have a local concentration due to the geometry of one of the electrodes, then you will have attraction. So, the variation of the field is an essential part of the attraction mechanism. So, again, why a lot of junks are attached to Atomic Force Microscopic tip can be shown here. As you can see from the paper published by Jungwon Woo, you can see the electrodes static field. Neuro tip is very high as you can see here and becomes very small as you proceed further away from it. So, you can imagine any dust will be attracted to the tip. Now, what is the force proportional to? So, one can prove that for small objects, the force is proportional to the gradient of the square of the electric field. Again, gradient of the square of the electric field. Let's take a look at this step-by-step. The induced polarization charges are proportional to the field, right? Right. Check. For given charges, the force are proportional to the field, right? Right. Check. So, here, you see the force, which is Q times E, is depending on E square because Q is linearly proportional to the E and E is proportional to E, so E square. Now, there will be a net force only if the square of the field is changing from point to point, right? Right. So, we need to have gradient of E square. So, that's why the gradient of the square of electric field. So, the constant of proportionality involves the dielectric constant of the object, the size, and shape of the object. But basically, the gradient of the square of dielectric field would determine the strengths of the force. Okay. All right. So, as a last example, we're going to think of a case when Faraday was inserting the glass slab between the conducting plates. What would be the force acting on the slab, right? So, we will use the virtual work as well as you can see and as a homework, we will ask you to solve this, but we will give you the explanation and derivation of the equation, so you can do it yourself. So, we will assume Q is constant on the conductors. So, we're injecting constant charge and then we isolated them. Then, we're now going to insert it from right to the left and you can imagine the force will be attractive, but you can prove it. Then, you will have two places, one place where you have fully air filled and the other place filled with dielectric, and this is in parallel, in parallel circuit. So, if we merely used the principle of conservation of energy, we can easily calculate the force using the virtual work as we just showed before. We can calculate the charge on the left side, which is kappa epsilon V over D times W, that's on the right side, and then epsilon V over D times L minus X times W, which is this air part, right? Using the fact that C is equal to epsilon W over D parenthesis kappa X plus L minus X and putting that inside our virtual work equation, we'll be able to get the right answer of the force that is attracting this path inside the parallel plate capacitors, okay? So, with that, we're going to wrap up our lecture here. Bye-bye. See you again.