ah ha ha they just feed all of us really teach you the last time we had almost reached to the definition of the displacement vector and today we will finally reach there let me review what we done we had a material some non conducting material a uniform electric field is applied and because this material is made of atoms the negative charge in the cloud of the electrons moves in the direction of the electric field and so there is a layer of negative charge at the top a layer of positive charge at the bottom and we have already worked out that if you have equal amounts of negative and positive charge a uniform electric field appears that is in the opposite direction this is what I will call an induced electric field now how much was this induced electric field where we work that out we say that supposing the slab of electrons got shifted upwards due to this electric field by how much will it get shifted it gets shifted upwards by an amount D where D is here the nucleus you have the net electron cloud and they are displaced with respect to each other that distance is d so you take this distance D multiplied by the charge of the electron with a minus sign and then you multiply it by the number of electrons per unit volume that is a number density that gives you the amount of charge per unit area that is present why because minus E is the charge and n into D is the volume so this is the charge density so there is a charge density equal to minus n EE D and as in charge density equal to plus n e D at the bottom and it is this charge density that causes an electric field we can apply Gauss's law supposing we had only this charge no applied electric field then Gauss's law tells us there is no electric field outside there is an electric field inside an electric field inside is directly related to the amount of charge enclosed so this is equal to e induced he IND so en is negative in sign that is it is pointing upwards that is not very useful because we do not know D we know the charge of the electron we know the number density of electrons in the material but we do not know D however from a model of what happens to an electron cloud we know that the stronger the applied electric field the more this cloud shifts so we can plot saying applied electric field E external and we can plot the amount of D and we find that is approximately linear it is very linear for small values of applied electric field but for larger values it deviates from being Li so in this region where it is in fact truly linear we can treat this D and therefore this induced electric field is proportional to the applied electric field now there are materials where you can apply an electric field this way and the induced electric field will be in some other direction these are called anisotropic materials we won't touch anisotropic materials right now so we will assume that if you apply an electric field the electron cloud shifts in the direction opposite direction to the electric field and an isotropic material would be one where I have an electric field this way my nucleus is here my electron cloud shifted in this direction even though the electric field is this way the electron cloud deanship directly backwards it shifted sideways this can happen because of a crystal structure of the material and these are more complicated materials but very very important from an engineering point of view for now we will decide to study only systems where the induced electric field is exactly opposite in direction to the applied electric field so we can write e induced is equal to minus e n times D and D is a vector now is equal to some coefficient I called it Alpha last time times the external applied electric field this alpha is clearly negative because the induced electric field tries to cancel the external electric field let us get back to the original problem I have a material and I apply an electric field due to this applied electric field some kind of charge has developed actually charge is developed everywhere but the only place were charged does not cancel out is the top and the bottom now supposing we knew how much this charge was if we knew exactly what this charge was we can actually solve for the electric field we can say that the electric field E is equal to 1 over 4 PI epsilon naught times the total charge integral of Rho total divided by R minus R prime squared times R minus 1/2 times unit vector DV what I mean by this is if I actually knew how much bound charge was developed by the applied electric field I can go back to Coulomb's law and Coulomb's law is always correct if I knew all the charge the only problem is I do not know all the charge all right now there is one other thing if I take this II induced it is proportional to the electric field so I can take the divergence of both now this expression tells me divergence of E is equal to the total charge density divided by epsilon naught no matter whether the material or not if I could somehow calculate the total charge not just the charge I put there but including the charge that is induced then this is true let me take divergence of this so if I take divergence of this equation I will get divergence of e induced is equal to divergence of alpha e which can be broken though it is not very useful to do it alpha times divergence of E + P dot grad alpha the only reason for doing it is we can see that divergence of E is already related to something the road total over epsilon not so you could write it as alpha Rho total over epsilon naught plus e dot gradient alpha now what is more useful is to look at this picture picture we've been drawing for a while now for example to look at this and say that the only place charge develops is where this factor-alpha changes abruptly because we look out here this is air alpha equals 0 there is no induced charge this is my glass alpha equals something this is air again alpha equals zero so if you look wherever alpha is constant nothing is happening but where alpha is zero changing from constant to zero that is where there is charge developing so we could think of this as representing a charge itself wherever e induced changes abruptly that is a place where there must be charge and this picture agrees there because I apply Gauss's law he induced went from zero to positive this charge and closed he induced remained constant there is no charge enclosed he induced went from positive to zero this is negative charging closed so it looks like divergence of e induced represents some sort of charge in fact it represents the induced charge so I will write this these two equations divergence of E is equal to total charge density over epsilon not divergence of e induced is equal to Rho bound power epsilon not the B means this is the induced charge so it is it is bound to the individual atoms of the material it is not something V placed there it is something that appears only in the press is of an applied field the problem in engineering the problem in all of electromagnetics is we do not know it this row bound is simply not known to us so what we are going to do is instead of working with row total which is a sum of the charges we have placed which I am calling a row free and the induced charges row bound or working with E induced which deals with row bound I would like to work with some kind of electric field that only relates to row free that is only row I know about the remaining rows are self consistently to be determined so I am going to take the difference because I can see I can cancel out row bound so I will take divergence of e minus e induced is equal to row free divided by epsilon not what is more we have already written down that we induced is equal to some alpha times e now alpha is not the symbol that is used but I am just show keeping it there for placeholder so I can write this as divergence off and I am going to take this epsilon naught to the other side it is harmless it is a constant times 1 plus alpha times the electric field is equal to root 3 now this probably looks like magic but what it is is actually a kind of a fancy sort of cheating and I will go over it again so that you feel more comfortable with what we have done but we have what we have achieved as a result of all this is we have identified a new kind of field it is related to the electric field but it has some material properties in it this alpha represents the relation between the applied field and the induced field in other words it is talking about the slope of this line now this is a it is a electric field like quantity because it depends on electric field but it is multiplied through by some material properties and it's divergence is equal to Rho F only this quantity is what we call the displacement vector the name is purely historical but its usefulness is extremely high because if you write this equation down you know the right-hand side completely you place the charge there so you know exactly what charge of placed and therefore you know mostly what this vector field is going to look like the problem of course is then how from this displacement vector do I obtain the electric field now this all comes down to finding this alpha and if you look back to see what is this alpha II induced is nothing but minus sorry he induced is nothing but minus e and D okay this is the Q bound and from Gauss's law we found that E is equal to minus E and D and E and n a material he is a constant n is just the number of atoms per unit volume d has to do with the materials tendency to deform in the presence of an electric field so all of this is can you put down as fixed material property of the medium so what does that mean it means that the relation between D and E is something we can put down in a handbook and just use it isn't something to be solved for if I make a plastic then a manufacturer or the plastic four will subject that plastic to tests he will find out if I apply one volt how much is the induced charge if I put 2 volts how much is the induced charge and he will give me a plot of e induced versus E and if the material is a linear material that is he induced is proportional to e he will give me this whole number and this number is called the dielectric constant now let us take go back go back and review if we apply an electric field to a material induced charges appear these induced charges for a field for a slab like this form at the top and the bottom of the material in a middle they tend to cancel out if you ask how much is formed it has to do with the thickness of this layer of charge the thickness has to do with the amount of electron cloud shifted and the amount the electron cloud shifted experimentally has been seen to be linear there is no proof that it always has to be linear that it does not always have to be linear there are non linear portions to this curve but where most engineering applications happen it is a linear curve so I can write the induced electric field as something times the applied electric field start with this this is the first equation now the second thing is if I knew every charge present in the entire universe including the charges I deliberately placed and the charges that are induced on surfaces of materials in the bulk of materials when you all those charges then Coulomb's law is still true because Coulomb's law says when you all the charges just write down Rho over R squared with appropriate direction and integrate over all space which and from Coulomb's law we can always derive the divergence theorem divergence e equals the same charge density divided by epsilon naught so now we have this equation where this total charge is Rho free plus Rho Brown that is Rho the charge we placed plus the charge that was induced now the induced electric field is clearly related to the induced charge because that is where it came from so divergence of E induced is nothing but the bound charge divided by epsilon not but we do not know what it is but I take these two equations and I say I can eliminate Rho bound you subtract the two equations I have got rid of Rho bound that is what I do I take divergence of e minus e induced and it is only related to Rho free and when I write out induced in terms of the applied electric field I get this equation divergence of some other vector is only equal to Rho free so crucial equation and it it is a bit of a tricky equation the sense that you can just write it down and use it without realizing what it represents let us take an immediate example and see just how this applies it's a trivial example but it is worth understanding I have a charge Q and I place it inside a sphere this is at the center here it is air here if glass and here it is air the inner radius of the glass is a outer radius is B I want to know what is the electric field and the potential as a function of R alright now there are two ways of going about it one way of going about it is to say well up to this point I am completely in air so I ignore the electric field from zero to a nothing but Coulomb's law now this electric field when it enters this meat material is going to cause induced charges and I can drive in write down a relation between e induced and the applied electric field and I can try to work out what is going to happen inside the material however there is an easier way of going about it the easier way is to simply say divergence D is equal to Rho free so if I take if I apply Gauss's law not to the electric field but to D I can take any radius sphere the integral of divergence D is nothing but surface integral over that we surface of that volume d dot d s but this is equal to volume integral of Rho free now I do not know what is happening inside this material but whatever is happening inside this material is induced response that is the charges are all induced the only free charge I know about is this Q so it is equal to Q say spherically symmetric problem so surface integral d dot d s is nothing but 4 PI R square D of R D in the r direction which is a function of R it is equal to Q so I know Dee Dee has only an R coordinate and the d r is Q over 4 PI R square I haven't solved anything else but regardless of what the dielectric constant of glass is regardless of whether I have one glass or many glasses as long as the problem is symmetric in all directions spiracle asymmetric I have already solved the problem I have found out D sub R how does it help me well I will look up my handbook for the glass and I will find out that this handbook tells me epsilon glass is equal to two point two five or it could be two point five or it could be three it's some number it is written by the manufacturer of the glass so in that case I know that D sub R is equal to epsilon glass P sub R so let us just look at the plot and make it clear this is the radial direction this is a this is B if I plot D D is nothing but Q over 4 PI epsilon sorry Q over 4 PI R square so it is going to be 1 over R square this is d what is e well instead of talking about what is e I am going to talk about epsilon naught e epsilon naught E is going to be identical to D in these two regions but in this region where there is glass I know that epsilon naught E R is equal to D R divided by two point two five so this part of the curve alone is going to be a smaller value so this is epsilon naught here what has happened is surface charge is developed here and surface charge is developed here so that the electric field has been reduced the applied electric field has been reduced by some induced electric field so the net electric field is smaller inside the material then what you would expect it to be once I have got this electric field then getting the potential is easy potential Phi is equal to integral from minus infinity or I should say from R to infinity of ER d R which is going to be equal to Q over 4 PI epsilon naught R for R greater than B because this is nothing but Coulomb's law up to here for R less than B is going to be Q over 4 PI epsilon naught B that is the value at this point plus the integral of this curve because this is what the electric field is so there will be Q over 4 PI epsilon naught epsilon naught times 1 over R minus 1 over B this is for a less than R less than B similarly when you go inside or equals a the point of this example was to show you the usefulness of displacement vector it is not just a arbitrary field that we invented we created the displacement vector because when we worked out what divergence e is we find divergence e contains an unknown charge so we constructed a field which depends only on known charge now this field therefore is complicated it is related not to Coulomb's law but to something else luckily for linear materials the relation between D and E is very straightforward D and E are related to local material properties and that is very important you can see here in this picture the material properties are changing I have air glass air however D in here D is equal to epsilon naught e out here D is equal to epsilon naught into 2.25 into E and D is equal to epsilon naught so point by point the relationship between D and E is only through a different constant so the advantage of this is I have got a new field which is essentially the electric field if I knew that field I know the electric field but this new field satisfies a very simple equation it is satisfies an equation whose right-hand side I know completely and that is an enormous advantage and you can see it in this example because I know Q I do not know the induced charge but even so I found the answer all right now I want to go on to one more problem and that problem is supposing I had my slab and I had the electric field coming at it we worked out if there is a induced charge and we know that there is a reduced electric field now if we apply Gauss's law we have an incoming electric field and an electric field inside the material so I will call this e to have called this e1 so e to minus e1 it is normal so I do not have to keep the vector notation is equal to Rho bound divided by epsilon not is that clear that is the amount of induced charge is the amount by which the electric field reduced but I know that this e 1 is e 2 is related to the displacement vector because if I look at the displacement vector nothing changed displacement vector didn't even notice this material because there is no free charge as only bound charge so we know that if we scaled up this second electric field by the dielectric constant epsilon of the material and we scaled up T 1 by the dielectric constant of air okay so these are there is nothing but D 2 is nothing but D 1 these two cannot really be any different because the same Gauss's law I can apply to divergence D I can apply divergence D is equal to Rho 3 and I know that Rho free is zero I haven't placed any charge there whatever charge has developed has developed because it was induced so I know that these two must be equal because if they are not equal I would not get 0 when I worked out the total amount of enclosed charge let me repeat because I think is a subtle point the amount of electric field entering an amount of electric field leaving a Gaussian cylinder an imaginary cylinder is not equal which means there is charge enclosed how much charge is that that is nothing but the induced charge but if I apply the same thinking to D then I have that D 2 minus D 1 is equal to Rho free divided by epsilon not but Rho free is zero so it is equal to zero so we get this special condition if you want to be more accurate and we do what we should say is that supposing the electric field is coming at some angle we do not exactly know what angle the electric field will be inside we know there will be less we can draw a cylinders and when we draw the cylinder if this direction is led eetu Z minus e1 Z would be Rho bound over epsilon not it's only the normal component that matters if you look at the sideways component there is flux entering but the same flux will leave so only the top and bottom contribute similarly when we do the argument here we get D to Z minus D 1 Z is equal to 0 for D 2 Z is equal to D 1 z the way this is put is normal component of D is continuous across interfaces I say it's continuous across interfaces but it's really continuous everywhere okay it is continuous here also it is continuous there also but that is assumed to be true but in a place where things are changing abruptly you can see that the electric field is not continuous e to is not equal to e one why because there is this row bound if e two are equal to e 1 e 2 Z minus e 1 0 would be 0 but instead eetu Z minus e 1 z is equal to the induced charge divided by epsilon not however even in such places the displacement vector is continuous for the displacement vector to be not continuous you should have placed charge on this surface so let us just emphasize that and go on to next point supposing I had a slab with an epsilon say glass epsilon equals to 0.25 and supposing I deliberately put charge how much are let's say you know ten to the minus six columns per meter square some charge I have put so I know this charge and therefore this charge is row free because it is it is charge I can account for it is present even if there is no applied electric field now I apply an electric field this applied electric field will create an induced charge so there will be a small amount of negative charge and small amount of positive charge so on the top after these this negative charge and this positive charge combined these two will cancel so what you will be left with a little bit of negative charge when you try to work out the pro this problem you will have divergence D is equal to row three which means D to Z minus D 1 Z is equal to row free but row free is no longer 0 row free is 10 to the minus 6 Coulomb per meter square so you will find that now your displacement vector is not continuous it is not continuous because we deliberately put charge there if we hadn't put charge there then we would have had a continuous displacement vector there are quite a few problems we're working with displacement vector will actually simplify your analysis and that is why it is an extremely important formula now let me remind you of the other condition that we always have been using which is that integral e dot dl from any point any point is independent of path and i mentioned in a class earlier and if I have a point two and 0.1 and I take two different paths going from one to two is like minus of going from two to one so this is nothing but integral 1 to 2 T dot DL minus integral 2 to 1 e dot DL is equal to 0 ok sorry plus integral 2 to 1 is equal to e dot is equal to 0 or if I take any loop whatsoever and integrate II around that loop I get zero this statement which we made in the very beginning to obtain potential and this statement are identical as saying the same thing what I am saying is I will go this way and then I will come back this way since both gave me the same answer adding up the two answers this is the negative so this is nothing but integral 1 to 2 e dot DL minus integral 1 to 2 e dot DL equals 0 but this is a very interesting equation because supposing I have my slab and let us say my electric fields have got a sloping electric field and I want to know what this electric field is going to do is going to do something what I will do is I will take a loop like this and I will try to integrate e dot DL on this loop this way the vertical pieces I will keep very very thin so I have only large horizontal pieces and if I take this large horizontal piece what I get and call this side one this side too I get e one this direction is Z this direction is X so e 1 x times DL minus e two x times DL is equal to zero that is the vertical parts do not contribute much because every thin II one the X component multiplied by the length of this - E - X component multiplied by the length is equal to zero that is what this equation says or it says D 1 X is equal to e 2 X so I have two equations and let me put them prominently I have that D normal is continuous and E tangential is continuous these are the equations we have which describe everything we know from the start of this course from the start we obtained that if you have charges you can derive from the electric field potential and the relation is electric field is minus the gradient of potential from whatever we have done now we know that there is a new field very useful field called displacement vector which is the electric field times the local dielectric constant epsilon is dielectric constant furthermore looking at the displacement vector and applying Gauss's law we know that the normal component at an interface of D is continuous and looking at the fact that the electric field can be derived from potential through e equals integral minus integral 1 to 2 of 5 DL sorry Phi is equal to e dot DL using this I have also obtained electric field tangential component is continuous let me finish the class by trying and using these equations supposing I have a slab with epsilon and I have an electric field that is arriving at a slope this angle is Theta so the normal electric field e normal is equal to the magnitude of e say the magnitude is e naught cos theta and E tangential is equal to e naught sine theta so this is e normal this is e tangentially now I know that if this is e naught and this is e 1 I know that V naught cos theta times epsilon naught that is the normal component of D is equal to epsilon times e1 times cos theta whereas I also know that II not times sine theta is equal to e1 times sine theta sine theta 1 so I have two unknowns V naught e 1 sorry II not is known a 1 and cos theta 1 I have two equations if I divide these two equations I will get epsilon naught and theta is equal to epsilon naught inverse epsilon inverse and theta 1 all I have done is I have divided this equation by this equation so sine theta or cos theta is tan theta sine theta 1 over cos theta 1 is tan theta 1 and epsilon naught an epsilon going to the denominator so I put epsilon naught to the power of minus 1 and epsilon to the power of minus 1 this is not quite the same thing but it is like Snell's law essentially saying the same thing now we can also work out what happens to e 1 itself for that all we need to do is we need to eliminate the theta 1 so supposing we divide the first equation by epsilon square it square the second equation and add them up what do you get the right hand side will be e 1 squared cos square theta 1 plus e 1 squared sine square theta 1 cos square theta 1 plus sine square theta 1 is 1 so it is equal to e 1 square but what happens on the left hand side I divided this equation by epsilon so I get epsilon naught over epsilon times e naught whole square cos square theta plus he not square sine square theta so it is a mess here but the mess does not matter I know what theta is I know what Epps not is so left hand side is a known thing is equal to e1 square so essentially if you look at what this is saying it is saying that the portion that is tangential portion that is along X does not change it contributes fully to e1 with a portion that is pointing into the slab that part gets reduced in the proportion epsilon naught over epsilon so if you finally solve this you get e 1 is equal to e naught square root of epsilon naught over epsilon 1 squared cos square theta plus sine squared so if I had normal incidence which is theta equals zero sine theta goes off cos theta is one that will say P 1 equals epsilon naught over epsilon 1 e naught so the electric field inside the material reduces in proportion to the dielectric constant which is exactly what we had come up with earlier because we found that induced was proportional to the applied electric field and D and the quantity we came out with the displacement vector was a minus e induced so obviously if D is continuous then you must have a reduction in e wherever the dielectric constant is large whereas if theta is large then what will start happening is this term will dominate this term and you will have that II 1 is equal to e naught this example should tell you the sort of an introduction to the idea of what we call matching conditions and the equations here these two are matching conditions for electrostatics namely the normal component of D is continuous tangential component of E is continuous this more or less establishes all the building blocks we need to solve problems in electrostatics you you you you
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