 So, welcome to the sixth lecture of this course. Before you know starting that I just want to because at the end of the fifth lecture some students asked questions about this polarization phenomena. So, I will explain it little bit in more detail. See when we actually replaced you know when we you know polarization phenomena was represented by this additional induced positive and negative charges in this circle. Effectively what we have done is we have basically eliminated that dielectric in a sense we we do not have to consider that polarization vector and all that that is all now implicit into this additional positive and negative charges. So, again we do not have to bring in that polarization p vector and all that once we have taken their effect into account by this additional positive and negative charges. Now coming to the sixth lecture. So, now we will get into the electrostatics. So, now if you actually take you know a composite dielectric like this of two materials 1 and 2 and one is oil and other is the press board. Oil's dielectric constant is 2.2 press board is 4.4. So, we know the boundary condition they can be easily derived from whatever theory we have covered till now that tangential component is continuous e 1 t is equal to e 2 t and the normal component is discontinuous by the ratio of the two permittivities in the inverse ratio that means e 1 by e 2 is epsilon 2 by epsilon 1. So, it is the inverse ratio. So, now actually for this configuration e 1 by e 2 is e 1 n by e 2 n because tangential component in this case is 0 is it not tangential component is 0 because we are assuming this as a parallel plate capacitor and field lines are all vertical. So, there is only normal component right. So, now so e 1 n by e 2 n that is equal to epsilon 2 by epsilon 1 which is nothing but epsilon r 2 by r 1 which is 4.4 by 2 and that is 2. So, that means oil gets stressed 2 times under uniform field conditions. So, what is this uniform because individually if you see in oil the field is uniform in this solid insulation the field is uniform. Of course, there is a discontinuity of the field value across the interface, but individually the field is uniform in both the dielectric materials. So, oil gets stressed 2 times under the uniform field condition and that is why we define what is called as you know utilization factor which can be defined for this case which we have seen earlier also a lead high voltage lead to ground and suppose you have oil in between as a dielectric medium then you know you will have uniform field would be there when this is replaced by a parallel plate capacitor that means instead of this lead if you make a plate here then it will be uniform field and with the same distance. So, with respect to that here the maximum stress will be at this point and that will be as compared to the uniform condition with two parallel plates it will be more is it not. So, that is why the utilization factor is e uniform by e max that means here the value of the e max is going to be higher is it not as compared to e uniform. So, you know you will have this utilization factor which should be always less than 1 is it not it will be always less than 1. Whereas, in some textbooks and some research paper they talk of enhancement factor where you know enhancement factor is indicative of how much field is getting enhanced due to non-uniform fields. So, then it will be enhancement factor will be 1 over utilization factor. So, in this case the ideal thing is you should have you know utilization factor close to 1. So, utilization factor being close to 1 means what everywhere the insulating material is getting uniformly stressed. So, material is getting all the regions of the material are getting optimally utilized is it not. Moment you have utilization factor greater than 1 sorry less than 1 or enhancement factor is greater than 1 somewhere the field stress is more somewhere field stress is less where the material is same that means material is not optimally utilized in various regions is it not. So, when we do finite element analysis for insulation design and all that we basically do some kind of parametric studies to reduce the maximum stress and to make this utilization factor as good as possible. So, you know one option is you can increase the distance, but that will be you know not economical is it not the distance the clearance will go up the size will go up. Second is increase the radius of this high voltage late right. So, that also that is a good option and that is generally done if it is possible if the standard lead sizes are available in bigger radius and if the space constraint is not there you can have higher radius, but that also may because you are going to put more copper because this lead is a metal. So, the no no the E max the question here is when we increase the radius whether the E max will increase what is done is this distance is kept same and the radius will be increased like this. So, the clearance will be maintained and the radius will be increased. So, the stress will come down here. Third, but the third option is the is one of the best options put insulation of higher dielectric constant over the high voltage lead. So, what is it what this is going to achieve first thing is because of the theory that we just saw moment you have a higher dielectric constant material the electric stress is going to reduce in that. So, at the you know conductor surface inside the stress is going to go down because now there is a solid insulation of higher dielectric constant and now inside the oil the maximum stress is going to be just at the surface of this solid insulation. So, there also the stress value is going to go down because you are going little bit away from the the high voltage electrode. So, the conditions will be you know little bit better as compared to when it was here. So, governing equation for electrostatic is so starting with divergence E is equal to O V upon epsilon E U replaced by minus del V and then you get this well known Poisson's equation del square V is equal to minus O V upon epsilon. Now, here the assumption that we are making is this it is a uniform homogeneous material. Otherwise, if it is not homogeneous material heterogeneous material then this epsilon will come actually inside this you know del operator after the del operator because what is finally del of something is because there are partial derivatives with respect to x, y, z there. So, if dielectric constant is a function of x, y, z then you know that dielectric constant should be here and not there. So, in the FEM analysis we will worry about such things more closely. So, in in this case if volume charge density rho V is 0 then we go get well known Laplace equation and let us you know understand little bit on this. So, if Laplace equation when expanded this you know del square operator x, y you get this. Now, again consider a simple parallel plate capacitor problem although you know it looks very simple lot of concepts are there even in this to understand. Now, the equipotential lines are all horizontal and E lines E field lines are all vertical and they are orthogonal to each other. Now, here you know the first question is when you actually solve this Laplace equation for this and these of course trivial here you know you do not have to really solve anything this is a trivial solution that you know lines would be like this. But for the sake of argument suppose we have to you know model this and solve in FEM then we will have to actually solve this Laplace equation. Then you know we have been always you know talking about voltage difference and we are imposing the boundary condition the voltages are being defined is it not. But here now you have charges like this positive and negative charges and the corresponding voltage is there. So, when you actually solve through FEM you do not actually worry about this charges that the source finally what is source? Source in all electromagnetic there are only sort of two sources one is charge another is the current. Actually really if you see the source is only charge because current is only manifestation of moving charge is it not. So, but you know for the sake of you know discussion we consider two charge two sources one is charge and other is current. So, here basically this charge distribution is what is giving this potential V ok. So, now, but when we do FEM analysis since in finite element method we solve it as a boundary value problem is a boundary value problem we define only after drawing the geometry and doing meshing and all that we only define the potentials. We do not worry about the charges because charges are implicit in the voltage definition the boundary condition is it clear. So, the question asked by one student is what is this implicitness. So, basically this charge distribution is what giving the voltage V now these charges have come from where some you know battery has charged this capacitor and those charges got deposited. Now, the charge distribution would be such that it will give the corresponding voltage V right. Little bit more you know when I show you something else it will be more clear when we consider the end effects. So, when since we are not considering the end effects here right. So, you have by getting uniform field and know that at edges the field is uniform right. So, now before you know going to those end effects we are coming to something called as you know boundary conditions. So, top plate you are defining voltage V and it is called as non-homogeneous diraculate boundary condition. Whenever you specify voltages right then it is called as diraculate boundary condition and if the voltage is non-zero it is called as non-homogeneous. If the voltage is zero it is called as homogeneous diraculate condition. So, these terminology you will quite often see in the textbooks on FEIM and electromagnetics. So, bottom plate is zero. Now two vertical boundaries two vertical boundaries what effectively is happening we have defined homogeneous Neumann condition. Neumann condition means derivative of field with respect to the distance. Now, what is this N? N is the unit normal that means this is the vertical boundary the normal to this will be in x direction is it not. So, that is why it is daba v by daba n equal to 0 effectively it is daba v by daba x is equal to 0 is it clear. Now because actually we have defined it as homogeneous Neumann condition and this thing effectively the field has become uniform right. If suppose you want to analyze the end effects what you have to do we will see in the next you know slide. So, this daba v by daba n equal to 0 another point to note is those who have used FEIM and you know used solve some electrostatic problem they will you know realize that they never define this daba v by daba n equal to 0 they just define this top plate potential and bottom plate potential. These vertical you know homogeneous Neumann conditions that derivatives being equal to 0 are not explicitly defined right. So, that actually get automatically defined during the FEIM analysis how we will see when we do the FEIM understanding of FEIM theory. So, there I will explain how they get automatically imposed even if you do not actually define them in even commercial softwares right. Now the point to be noted here is that we have defined homogeneous Neumann condition here that means you know the field what does it mean daba v by daba x is equal to 0 on this vertical boundary that means what here suppose you take this vertical boundary and this is that daba v by daba x is equal to 0 that means voltage at this point and voltage at this point is same then only daba v by daba x will be equal to 0 is it not that means you are making this whole line as equipotential right. If suppose and this was you know this top plate was very close to this then in fact you know this is a sharp point. So, field the equipotential lines are going to turn like this is it not sharply. So, in fact daba v by daba x is not really going to be 0 at if you want to consider the end effects is this point clear. So, by imposing this you know not considering the outer boundary and imposing this either implicitly or explicitly this condition homogeneous Neumann condition we have ensured that the field is uniform at the ends. The question as by one of the students here is when we do not consider the end effects what will the potential distribution even above and beyond the plates. So, we will understand that through you know so this slide here shows a charge density distribution on top and bottom plates of capacitor of a capacitor and this is a three dimensional distribution and with all end effects considered. So, you can see at the ends of the you know at all the four corners of top and bottom plates you have sudden and sharp rise in charge density distribution positive charge density on the top plate and negative corresponding negative charge density on the bottom plate. So, at the ends of both these plates four ends you have sudden increase in electric field intensity because of sharp corners and therefore those increase values of electric field intensity have to be supported by corresponding charge densities higher charge densities. So, it is very clear from this three dimensional distribution of charge densities the corresponding end effects and high rise of electric field density and charges in case of a capacitor when end effects are considered. Now the same thing of course these I have shown in 2D right that in the PPT that I showed it was a three dimensional distribution, but here on the paper I can show only nicely only 2D. So, now what I have shown the charges in the middle they are sparsely distributed and charges at the end they are highly you know concentrated because electric field intensity is high that is what we saw in the previous that power point presentation. So, the FEM so in order to reproduce exactly those you know high fields and high charges at the ends you need to have very fine discretization is fine FEM mesh at the ends and you know how to achieve that and all that we will see when we start looking at FEM right. So, and also remember that when we are actually here this whole since we want to take the end effects into account we have enclosed this capacitor configuration to into a bigger box. So, these are like you know bigger you know rectangle and in 3D it will be a cuboid right and these distances again from these plates to these box should be sufficiently high so that those boundary conditions do not affect the field distribution right so that is one thing. So, going further and also remember the field distribution will not just be like this, but there will be some field lines like this some field lines like this also if you actually do the FEM analysis you will see that right. So, the field is everywhere in this so although the field will be much less in this part, but field will still be there because we are not constraining. So, we will take one more this last point and then we will go to the next topic. So, now we will see the conductor dielectric interface which is very important and there are some very interesting applications. So, first is you know if suppose you keep a isolated conductor in electric field ok. So, now this electric field is in this direction vertical direction upward and you are keeping the isolated conductor. Now, moment you keep it in the E field is everywhere here the positive charges will get pulled up and negative charges will go down because again this is positive this is negative right. So, now the electric field lines will basically be like this and electric field lines here will terminate on negative charges. So, now remember there is no volume charge here because all the charges got displaced towards the surfaces. So, there is no volume charge there is only surface charge. So, that means you know in our first Maxwell's current divergence d is equal to rho v there is only volume charge is there, but now since there is no volume charge the corresponding that equation gets sort of replaced by dN d normal or epsilon 0 E normal is equal to the corresponding surface charge density. So, this is the same this is the manifestation of that first Maxwell's equation only. It turns in a different form because now instead of volume charge you have surface charge right and and it it it makes sense because rho s has the unit of coulomb per meter square. So, that matches with the unit of d which is also coulomb per meter square. So, that is why d is equal to dN is equal to rho s right. So, the the internal E field now the field also has to be 0 inside this conductor because there are no charges within the volume of that conductor. If there are no charges in electrostatics typically when currents are not flowing if there are no charges you take any Gaussian surface inside this conductor and calculate integral E dot T s is equal to you know epsilon integral E dot T s you will get identically 0 because if the charge encloses 0 and then if you calculate this E also will be coming to 0 right and that has to be because there are no charges inside but charges are the electric field is only outside the plate and why electric field is 0 that can be also understood by noting that internal E field gets cancelled because the internal E field is from positive to negative is it not it is vertically down. So, that will exactly cancel this external field. So, E internal from positive to negative inside the conductor will get will cancel the E outside. So, that actually ensures this condition that E is 0 inside the conductor because there are no volume charges. So, both things are in sync with each other. Now there is one very interesting example of electrostatic shielding and you know Faraday cage and all that. So, what is that let us understand. Now suppose you know you have a conducting shell. So, maybe you can assume a spherical conducting shell right which is you know from inside there is a space inside and there is some charge inside that spherical shell that is plus q. Now that plus q is going to induce negative charges when I say induce that means it is going to you know pull the negative charges by attraction right and this will get negatively charged this will get positively charged right. So, now if you take any again this conductor this is a conductor conducting shell. Now here E is equal to 0 by the same logic that we just now saw E is 0 here and that is also can be reconformed by taking a Gaussian surface here and now then the net charge enclosed is 0 is it not because this positive q and negative q will cancel each other. So, the net charge enclosed will be 0 in the Gaussian surface taken inside that within that conducting shell is it clear. So, that means you know what you have ensured is the E is 0. But then these you know when you take a practical application you cannot have these positive charges remaining there you have to drain them to ground because otherwise the if you put it like this this outer you know conductor potential will be corresponding to this V these charges and that can be quite high depending upon what is this plus q here and the corresponding voltage. So, that is why you need to ground this outer one and then it becomes a perfect shielding right and that is the principle behind this electrostatic shielding that you know we have a Faraday cage metallic enclosure and there is a high voltage source inside right and then we ground that Faraday cage and then what happens is the same thing only thing now it is a cube here this conducting shell is cube here with some finite you know metal thickness and now if we ground this this all this positive charges will go to ground and inside this thin you know metallic box with a metal thickness the E is going to be 0 there because of the same phenomena plus q this is say plus q there will be minus q on the inside of this metallic enclosure and if you now take any Gaussian surface within the thickness of that Faraday cage the total charge enclosed will be 0 and that is why E everywhere will be 0 inside that. So, effectively what you have done you have actually since E is 0 inside that metallic plate like here you are effectively isolated the inside and outside environments because there is you know no E field. So, there is some region which is separating inside and outside wherein E is 0 that means there will not be any you know effects of outside voltage as well as inside voltage on vice versa is this point clear. So, that is the principle behind this electro static shielding. So, I will just repeat what we just now said about this electro static shielding you have you know some high voltage there could be some high voltage source which represent this positive charge q and around it you have this conducting shell and then you have this minus charges induced as shown on the inner surface of the shell and plus charges on the outer surface of the shell. And since this charges positive and charges got separated and there is no you know charge inside this conducting shell E is 0 and then that is how these two surfaces they got sort of electrically isolated you can say because E is 0 now. Now, this again interesting thing to you know understand what I just said that these become electrically isolated we just extend this case suppose this is that plus q charge and you have this minus charges induced in this conducting shell and you have this corresponding plus charges on the outside. Now, suppose actually we what we do is we this plus q charge we shift now it is shown concentric here or symmetric if I shift this q charge on this side right hand side what will happen this negative charges on this side will be more is it not and negative charge you know density on this side will be less. But what will happen on this positive charges they will remain as they are right because now there is no sort of connection between this positive charges and this negative charges because they are sort of because E is 0 in between right. But if I suppose you know bring one ground plate in the vicinity of this now set of electrodes now what is going to happen these positive charges will get concentrated on this side right. So when we actually have had initially drawn this figure we assume that there are no you know other objects charge objects in the vicinity and this whole system is electrically isolated right and then that is why then this positive charges become you know they are sort of independent of moment of this high voltage source or plus q source within this inner most you know part right. If we have understood this then as I mentioned in the previous slides after you know putting this conducting shell we should ground it so that this positive charges will get drain to drown the ground and then you know this will be safe electrically even to touch right. So I will show you one interesting application and practical you know demonstration of this. So in this slide what you see is a high voltage source is a high voltage transformer which you know for which 230 volts is the input and 100 kV is the output at this point. Now you can see all these you know electrodes which are having high voltage they are you know smooth electrodes rounded electrodes because otherwise they will lead to high voltage stresses and would they themselves become source of corona or partial discharge right. So all these high voltage equipment this is 100 kV source is enclosed in this aluminum cage right and if you are able to notice this can you see this there is a copper strip running all over you know copper strip here which runs at the bottom level of this pagodic cage from inside and this basically copper strip positively connects all these individual panels of this aluminum cage and this cage is grounded to an earth pit right. High voltage source is in the inside this cage and this is you know mesh here to see what is happening inside. So this cage is grounded to an earthing pit so this becomes an electrostatic case of electrostatic shielding and here you know the thickness of this aluminum cage is really not so much important for the point of your electrostatic shielding right whatever thickness mechanically you know sufficient and commercially available you can use that. But later on we will see when we are interested in electromagnetic shielding so what we just saw is electrostatic shielding from the point of your high voltage. But suppose you are conducting some test inside this setup which generates high frequency EM waves electromagnetic waves like for example corona or partial discharge studies those frequencies you want to shield from going out and affecting some electronic products as well as if there are some discharges happening outside those discharges should not affect the test that are being conducted inside. So either ways you want electromagnetic shielding now this is you know high frequency shielding. So for that later on we will when we study the theory of eddy currents we will see that you need to have finite thickness of the shield depending upon the frequency of interest so that we will study later. So what we just saw was only the electrostatic shielding the electromagnetic shielding we will discuss little bit later. Before ending discussion on electrostatics let us discuss one practical example of condenser bushings in which capacitive reading is done to improve e-field distribution. We know for a coaxial cable configuration as shown here with two cylindrical electrodes the e-field is non-uniform as given by this curve the electric field intensity is high conductive surface here inner conductor and as we go away from there in this insulation insulation here near the outer electro electrode is less stressed the e-field intensity is less here and in that sense this insulation is not effectively utilized stress is higher here stress is lower here and that is not a good insulation design. So e-max is given by this formula so e-max will be at this point and it is given by this formula. Now how do we make this electric field intensity more or less uniform we cannot make it absolutely uniform but how do we make it close to uniform distribution there are different ways in which one can achieve this one of the common ways we will discuss in the next slide. So let us understand in this slide how capacitive grading in condenser bushings achieves more or less uniform e-field distribution. Now what are condenser bushings they basically connect transformers to transformer high voltage terminals to transition lines a condenser bushing consist of central HV conductor right which on one side it connects to transformer winding on the other side it connects to transition line. So what is done here is you have this HV conductor of condenser bushing that conductor is taken out of transformer tank to this opening and these are the metal flanges here which are at ground potential. So this HV conductor has to be properly insulated from this ground if we put simply you know paper insulation here throughout then it will be non-inherently stressed as explained in the previous slide. So what is done is we insert here electrically floating cylindrical conductors and they basically get potentials on them based on their position here. And in between this poise of course there is paper insulation. Now this figure this actual configuration can be equivalently represented by this schematic. Now this is not to the scale right and you know central conductor is assumed to be at some positive potential at the instant being considered. So there will be some positive charge on that conductor central conductor. So now these floating metallic coils will basically they will get charged to negative and positive values as shown and that too alternately. So here you will have negative charge on this side you will on this surface you will have positive charge again negative and positive charge on the next you know cylindrical coil conductor and so on. So thing to note here is this total charge on this for example negative charge here on this surface will be numerically equal to this positive charge. Of course this positive charge will be again equal to this negative charge you know in terms of magnitude. So now this if we have understood this charge distribution. So let us then go further and understand how uniform E field distribution is achieved. So in 2D cross section these floating metallic cylindrical coil they basically will appear as line charges and for line charges E max is given by this standard formula q upon 2 pi epsilon Rn ln where Rn is mean radius and ln is length of the nth metal coil right. So let us derive this formula from fundamentals. If you consider insulation between any 2 conductors for example let us take this central conductor and this first metal conductor this floating this is at floating for some floating potential rho s will be given by D max and this D max will be normal to the conductor. So in the horizontal direction and what is rho s? rho s is the surface charge as we saw in the previous slide. So rho s is D max and D max is epsilon E max and what is E max? E max is V upon A ln V by A as seen on the fifth slide right. So going further rho s can also be written as q upon surface area. So what is surface area? 2 pi R into L because 2 pi R is circumferential length into L which is length or height of the conductor. So rho s is given by this which is nothing but D max and D max is equal to epsilon E max. So from this we get E max as this which is same as this formula. Now if you replace q by ln q upon ln by rho which is charge per unit length in coulomb per meter then we get this formula which was given in fact in lecture 4 slide 4. So now coming back to how do we get input E field distribution. In this formula if we maintain E max constant in each of these capacitors, now these are capacitors, these are all capacitors. See in each of these capacitors if we maintain E max constant and how do we do that? We just give this R L product constant. q is anyway same for all these conductors as was explained on the previous slide. So as mean radius is going up as you go away from the central conductor what you do is you reduce the height or length of this floating metallic conductors and by virtue of that you can maintain E max as constant and because of that you know the uniform more or less uniform E field distribution is obtained. Now let us understand this more intuitively when we are reducing this height of these layers what actually we are doing is we are reducing the effective capacitance between these conductors. Why? Because surface area is reducing because epsilon A by D is a standard capacitance formula for the parallel plate capacitor of course. So if A reduces D being constant the capacitance is reduced but charge is same so we will increase. So we will increase. Now this is fixed potential. Potential of this conductor is increasing thereby it reduces the electric field intensity between these two conductors right. So this is how E max comes down and then you know it can be maintained E max can be maintained constant by keeping this product as constant. It should be remembered that here we have done some approximations. For example we have it neglected fringing at the ends. If we have to consider fringing then of course we need to do finite element analysis and then how do we do that for a typical high voltage insulation system we will do that kind of understanding in one of the lectures later. So with this we conclude this lecture 6 and we have finished electrostatics part of our course and we will you know in the next lecture we will start with magnetic fields. Thank you.