[Music] to speech [Music] [Music] you you good morning today we shall complete up a few more examples on electric fields and then go on to the important concept of electrostatic potential so let us first look at one interesting problem supposing you have to line charges this is coming out of a problem in your textbook problem 2.21 you have to line charges each line charge has let us say a charged rope columns per meter the line charges are at a distance D apart now we would like to know what is the force per meter exerted on line charge to call this to do two line charge what well in a previous class we derived what the electric field was due to a line charge and the answer we got was by taking points above and points below and integrating from Z equal 0 to Z equals infinity and what we got was that the electric field was along the radial direction it is proportional to this row it was divided by 2 pi epsilon not and it scaled as 1 over R so for a distance D this R becomes D so it's equal to along the direction radial direction so Rho over 2 pi epsilon not 1 over D so this is the electric field now this electric field acts on 1 meter of charge because I want the force per meter how much charge is there per meter well it's nothing but Rho itself so the force per meter is equal to Rho times the electric field both are vectors I already have the electric field so the answer becomes the force is along the radial direction and it's equal to Rho square divided by 2 PI epsilon naught D so if you keep two line charges and you each of them has a charge row per meter this is the kind of force per meter that they exert on each other which means that the total force between these two line charges is infinite so line charges are actually very exotic things you cannot actually create them in practical situations however if you talk about wires say your overhead wire power line well those lines can be a few meters apart and distance of 100 kilometers and therefore such very long closely placed wires are actually a very good approximation to this picture so your overhead power lines exert force on each other when they charge up to different voltages and exert different electric fields okay another example I would like to know the force the electric field due to a disc the disc has a radius e and it is at a height the charr the point where I want the electric field is at a height Z the entire disc has a charge let's say Sigma Coulomb per meter square and I want to know what is the electric field above this disc Zed meters above this disc and what is its direction well last time I did half of this problem so let me just repeat that half what I will do is I'll take a ring the ring has a radius R it has a thickness dr now every point on this ring is equally distant from the point where I want the electric field because every point has a distance that is equal to Z squared plus R squared square root this is just Pythagoras theorem because it's a right angle triangle so it's the hypotenuse squared is equal to base squared plus height squared however the electric field due to each of these charges is pointing in a different direction in fact the electric fields describe a cone as I move the charge around the circle the electric field moves in a cone around this point so how do I add it up and find a net electric field what I do is I take diagonally opposite points of charge if I take two points of charge that are diagonally opposite then what happens is that the two directions in which they point R will tend to cancel in the XY plane but will tend to add illicit plane let me try and make that more obvious I'm going to draw the circle looking from about this is my ring of charge I am considering a bit of charge here amount of charges Sigma times the thickness D R times the theta Direction thickness which is Rd theta at a point up here the the field that it will cause will be actually pointing somewhere out I am just projecting it down so it will look like this for example if I took a piece of charge here it would point in this direction but now if I took a point diametrically opposite on that ring same amount of charge Sigma D R or D theta now what's going to happen this charge is going to produce a force it's again up here so the force due to this is going to look outwards force due to this is going to look outwards but if I project it down the projected parts are going to be equal and opposite they are going to cancel so the only electric field that's left will point straight out that is able point in this picture in the upward direction so what do you get as a result for the electric field you get de which is in the Z direction is equal to this amount of charge Sigma D R times all D theta divided by 4 PI epsilon 0 this gives me the the times one over Z squared plus a squared radius is that I can say it's R square this gives me the length of this arrow but I don't want the length of this arrow I want this length the vertical length so there is an additional cos theta I'm afraid I'm using the same symbol twice so let me use call this caution Phi Phi is this angle whereas theta is this angle so now I want to integrate this Dez around the circle so I get integral 0 to 2pi Dez integrated in theta alone the theta integration gives me 2 pi times Sigma D R times R divided by 4 PI epsilon not 1 over Z squared plus R squared times cos Phi now if you look at cos Phi cos Phi is nothing but Z over the hypotenuse equal to ver divided by square root of Z square plus R square so when you put it all together what you get is de R de Z is equal to Sigma R Z D R divided by twice epsilon not Red Square + R square to the power three-halfs this answer still depends on the our and we want the answer for the electric field due to a disc that is we want to add up rings of different values of R all the way from R equals 0 to R equals a so the electric field finally is in the Z direction and it is the integral 0 to e of Sigma R Z divided twice twice epsilon 0 Z squared plus R squared to the three-halves dr this is essentially the same equation it is the same equation that i came up with for solving the electric field due to a plane i ended the last lecture that way that's because if i take the radius of this disc to infinity the disc becomes the plane so if i replace this ie by infinity then I have the answer for a plane well you can see this is a very simple integral to solve because you've got RDR up there and Z squared plus R squared is your dependence so you define U is equal to Z squared plus R squared D U is equal to twice RDR and you get the answer Zed hat I can pull the Z out I can pull the Sigma out over twice epsilon not one more factor of 2 integral Z squared to Z square plus a squared off do you / u ^ three humps how I got that was I removed the comic constant pieces out twice RDR is D u so RDR is D u divided by 2 the denominator has Z square plus R square to the power of 3 halves Z square plus R square is nothing but U so it became u to the power of three humps this is easily solved so you get that this is equal to Z hat Sigma Z over twice epsilon naught u to the power of minus half divided by minus half the half cancels out so you get minus U to the power of minus half going from Z squared to Z squared plus y square so the answer finally becomes that the electric field is in the Z direction it depends on Sigma Z over twice epsilon 0 times 1 over Z minus 1 over square root of H square if you compare with the expression for the plane when the plane is nothing but the disc with a becoming very large when a becomes very large Z squared plus a squared is huge and this huge number is in the denominator which means 1 over Z squared plus a squared square root goes to 0 so this term will be missing and you'll just get Sigma Z over 2 epsilon naught Z and Zed itself will cancel out so this was the result I had given you earlier which was but if you have a plane an infinite plane of charge whose charge was Sigma and you went a distance Z from this plane and tried to find out what the electric field was well it didn't matter how far you went out however far you went out the answer was always Red Hat Sigma over 2 epsilon naught now how can this be there seems to be something very wrong with this how can it not matter how far from the plane you are the electric field is always the same well there is a reason for it and the reason is actually not that difficult if we look at a particular theta from Z it strikes the plane on a circle and if you ask how much charge is there between theta and theta plus D theta so the bigger angle is theta plus D theta well it will be the charge that is there in this ring how much charge is that well it's 2 pi times this radius times the thickness so we if this is Theta and this is Z this height is that sine theta so it will be the DQ would be equal to 2 pi times Z sine theta times D of Z sine theta now what is interesting about this is that it is proportional to Z squared so the amount of charge that is present increases the further out you go amount of charge between theta and theta plus D theta the amount of charge is not constant if I have a one meter away I get so much charge if I'm 10 meters away between 30 and 31 degrees there's 10 times more charge not sorry 100 times more charge however how far away is this charge the distance is Z squared plus Z squared sine squared theta is proportional to R square so square roots she's proportional to Z you can pull out Z squared out of the square root so you get an answer that's proportional to Z so the amount of force exerted by this charge goes like force goes like D Q divided by R squared well DQ is proportional to Z squared R is proportional to Z so R squared is proportional to Z squared equals constant so what it means is the further out you go the more of the charge of the plane you see how much more it quadruples for every doubling of your distance at the same time the distance is increased and therefore the effect of that quadruple charge has become one-fourth so two effects cancel which is why no matter how far away from a plane you go the force is the same so the answer is force remains constant now I hope these kinds of examples give you a taste of how one does simple integrals to calculate the electric field now in a certain sense Poisson equation is all you require if you knew whether all the charges are because sorry not poisons equation Coulomb's law is all you require if you knew where all the charges are because Coulomb's law tells you that the force is equal to 1 over 4 PI epsilon zero integral Rho of R times DV prime divided by so if I knew exactly what charge density was everywhere I just do this integral and I get the answer end of electromagnetic theory the only problem is we usually don't know where charges let me give an example supposing you have a metallic ball and I put one Coulomb of charge on this ball well the charges are free to move around because it's a conductor and we know that inside a conductor charges can move which means the charge could be anywhere on the surface and it could be anywhere inside so we can't use this formula this formula requires us to know exactly where the charge is whereas in practice what we know is the total charge and we know the shape of the conductors but we don't know exactly where the charges so even though we've done quite a bit in writing down Coulomb's law we have not actually got to the really useful parts of electrical engineering in electrical engineering the most useful components are inductors and capacitors and a capacitor is nothing more than a metallic pair of metallic plates which are kept close to each other so in order to tackle this problem we need to introduce some new concepts the first concept I want to introduce is work done in order to move a charge and when I was studying in college I think work was probably the most difficult concept I ever encountered angular momentum was about equally hard but those two concepts drove me quite crazy so let me spend a little time trying to understand this word you should have learnt it thoroughly in mechanic's but I did mechanics too and I didn't understand see the real confusion that comes is as human beings when we carry a heavy ball or we carry a weight and we just stand there if I'm just holding this chalk and just standing there we feel we are doing work we feel our body is burning fuel with oxygen producing carbon dioxide so that our muscles can hold up this piece of chalk therefore we are doing work and are quite correct our bodies are doing work however in physics the word work means something else the word work does not mean holding up a stationary object if you have a stationary object that is not moving no matter how much force there is on that object no matter how heavy the object is we are not doing work a weightlifter when he lifts the weight and reaches the top he is not doing any work I mean he he'll probably hit you if you told him that is he's not doing any work but he according to physics is not doing any work he is just standing there all the work he did wasn't lifting that weight but keeping it stationary above his head may have caused a lot of problems to his muscles but he was not doing any work the way it was not going any higher so work according to physics and according to engineering means motion of objects against applied fields and the words are all important you must be moving you're not moving there's no work it must be an applied field for example supposing I have an object that he's under some strain maybe I've got a compressed spring and I move the whole object I'm not doing any work if I'm if I if I move the object for the objects internal stresses are the only forces present I am not doing any work the forces have to be applied from outside and the motion has to be against this field for example supposing force due to gravity is downwards and my weightlifter have got his weight and he starts walking the poor fellow is doing a lot of exerting a lot of effort but he is not doing any work because the weight is neither going up nor down and up and down are the only directions that are against the applied force going sideways doesn't get you anything may get you an Olympic medal but it won't get you any work so now that we know something about work how do we calculate work you have to use that definition so there is motion and the motion is against a field so if I move if I have a gravitational field mg and I move in some direction the part of that movement that is sideways doesn't count that part of the motion which is like the weightlifter walking around does not do any work it's only the part of the motion that is parallel to the direction of the field that does any work so the amount of work that is done DW is given as the force dot the vector distance traveled so if this is d s and this is force mg you take the dot product of the two if you take the dot product of the two then only the part that's parallel to the force counts the part that is 90 degrees to the force doesn't count and that contributes to DW this is the work done by the force on the object and when work is done on an object usually it implies kinetic energy so the object starts moving faster and faster and as you know from this concept we can say since total energy is conserved the object has lost potential energy and the potential energy was nothing but MGH and gained kinetic energy so this is exactly what we have to do in electric electricity and magnetism as well because we have forces now the force is called the electric field instead of gravitation and we have to find out how much work charges do when they move in an electric field is the same idea but let's see how the idea works out so now I have a charge capital Q I know that if there's any other charge small Q the force on that charge is along the line joining capital Q to small Q and pointing away from capital Q now I decide that I want to move this charge from the point A to a point B it can be on any path whatever I that I like I want to know is there any concept of work done is there any concept of kinetic energy of this charge and does this charge gain or lose any potential energy we know that in gravitation the concept is there if this was the earth this was a satellite the force would point the other way and if the satellite decided to move like this it would have picked up kinetic energy and it would have lost potential energy is there a similar concept in electricity and magnetism well at every point on this orbit we can work out what the electric field is you just have to draw straight lines the lines are getting larger as you get closer to the charge now you can see that you're not you're hardly moving in the direction of the charge of the field so if you move on this line the amount of work that you will do is only the dot product of the field and the distance so we can work out I will call this one two three four all the way up to point M so total work done on the charge that is this is the work that would have made the charge move faster and faster it would have given kinetic energy to the charge is equal to R 1 2 dot QE r 1 let's say plus r 2 3 . QE of r 2 plus etcetera i keep adding them all up till I get to r n minus 1 n dot QE r in - I have said R 1 R 2 etcetera I could have easily said R 2 R 3 some point the electric field on some point along this arrow if I add this all up this should be the total work done and there is a shorthand way of writing this some just as we changed Coulomb's law and made it look like an integral we can make this also look like an integral we can say that this this whole route that we went on we are going to call it a path and typically symbol C is used contour C for contour so I will say that I am going to do an integral it is a vector integral along this route C and on this route I am going to do Q times the electric field that's the force dot d R why do I say this if I take R 1 to R 2 3 r 3 4 up to r n minus-1 n I have actually built up this entire curve so in a certain sense I if if I want to know how much work I've done by moving on this curve it is like summing up all these little pieces and you know there is some consisting of very small steps is nothing but an integral but it says it's a different kind of integral it is what is called a line integral it is not your standard kind of integral let's write one just to see what we mean by the difference for example if I have x and y and I wanted to know going from 0 to 1 what is the value of f of X which is equal to integral 0 to 1 sine X DX that is the kind of integral we have learnt in mathematics very straightforward you learn what it is it's the integral of sine X is minus cos x and you put it between limits and you get that this is equal to 1 minus cos of 1 so this is an integral along the real line but I could equally well ask I want to know the integral or going around a circle I want to know that the integral as I go along the circle of integral let's say sine of x squared plus y Square D theta or if you like let's leave it like D theta now what am i doing I am saying that I have a circle and I am dividing the circle into lots of little pieces each of these lengths is nothing but all D theta which is equal to D theta because R is 1 it's a circle with radius 1 and I want to integrate a quantity called sine of a square root of x squared plus y squared well square root of x square plus y square itself is nothing but R squared which is equal to 1 so it's a constant so I have taken a trivial example the point is this also is an integral but it's not an integral on a real line it's an integral on some curve in x and y and more generally it could be a curve in X Y Z T could be a any general curve if you talk about integrals on general curves those are what we call line integrals and the particular example I gave which is QE dot the variable which I chose dlc is an example of a line integral and what it means is if I have a curve I break the curve into many small pieces on each per piece I find the vector corresponding to DL at each piece there is also an electric field so I can form a dot DL that's now on number and I'll just add it up some I equals one to n qei dot d Li and if I make these deals very very small this sum becomes an integral that's what it means and this is so important I think it's important that you think about it and you properly get comfortable with it we'll keep coming back to this it will it will come back to haunt us unless we are quite comfortable with the idea but what is the result now the total work done on the charge okay which is in a certain sense equal to minus of change sorry change potential energy we assume total energy is conserved the energy gained kinetic energy gained by the charge must be the loss of potential energy of the charge it's equal to integral from the starting point is the ending point of Q e dot d now that's interesting it's just a definition but there's one more interesting thing about it if I have a charge Q and I have a curve here to be at any point the electric field points in the radial direction it points away from the charge the distance I move DL does not necessarily point in the radial direction but you can always break DL into two parts there's a part which I'll call dr and a part that is the rest of it and it's obvious that no matter what we do it's only the part dr that can do any contribution to this integral because dot product only counts the portion of dl that is parallel to e so you can rewrite this equation as going from integral a to b but q ER dr that is at each point instead of keeping the full direction of L I am taking advantage of the knowledge that E is always pointing in the r direction so the only the are part of L matters so I only keep dr what does this do for me well if I've got rid of all the other directions I don't have to keep a and B as points I can say going from R to R B and furthermore I know an expression for ER so let me write that out it's our a to our be Q times capital Q over 4 PI epsilon not er is nothing but square dr integral of 1 over R squared is minus 1 over R so it becomes Q capital Q over 4 PI epsilon naught times 1 over R a minus 1 it's a very strange result what it says is supposing instead of going this way I had done this I had looked around this queue many times had gone way out come back I had gone this way and come back no matter what I did if I finally landed up in B and if I initially started from a the answer doesn't change the answer only depends on where I started and where I ended he doesn't depend on how I got there and we have meant such kinds of integrals before or if you haven't you should have if you have done any course in thermodynamics you know of things called state variables entropy and variables like that that you encounter when you do this take second law of thermodynamics they don't depend on the path either when you're actually changing the state of a thermodynamic system you go through some complicated path but the initial and final states are defined based on where you landed up what we are seeing here is something similar there is some quantity here the different amount of work done on the charge which didn't care how you got from point A to point B only where you started in where you ended now as it turns out since we are doing electricity and magnetism we are not so much interested on the total work done on the charge but we are interested in potential energy gained by the charge which is equal to minus of this quantity because it is the total work done by the charge this will give you kinetic energy so if you want potential energy you have to take the minus of that so it gives me Q Q over 4 PI epsilon naught 1 over R B minus 1 and typically when we are talking about point charges if you have a point charge Q and we want to use this formula we will take the other point the starting point very far away if you take it very far away then one over our a goes to zero so you can say this is equal to Q cube over 4 PI epsilon not 1 over R B so you have now got an expression for a state that depends only on the final location of the charge and it represents potential energy gained by the charge when you come from very far away to that point as before this is potential energy we would like to know potential energy per Coulomb so we define a special variable a special field called electrostatic potential the symbol we use is a Greek letter Phi and the Greek letter Phi is nothing but potential energy gains divided by Q itself that is the potential energy gained per Coulomb and it's defined for a point charge source as Q over 4 PI epsilon naught 1 over let me recapitulate the important thing about this potential function is precisely that it doesn't depend how you got to the point or however you got to it this potential function is the same so we have we originally had electric field which was Q over 4 PI epsilon naught 1 over R cubed R but now you come to a different function Phi which is equal to minus integral through are from far away he dot DL which is equal to Q over 4 PI epsilon naught 1 over this is columns law and this is what we've just worked out now there's a huge advantage to this function we may have invented it but this is a vector this is a scalar which means that this quantity in verse three separate numbers for every point this quantity involves only one number so Phi is much simpler and he the other thing is if I represent something spy represents potential energy of charge potential energy per Coulomb of charge now we worked all this out for a single charge Q but we know that any interesting problem is going to have many charges and in fact if we have a metal we don't even know where the charges are so what uses this first things first we know that electric field is equal to 1 over 4 PI epsilon zero integral Rho R prime DT prime divided by R minus R prime cubed and x r - so there's superposition working for us for every little piece of Rho DV prime I can calculate potential which means that the potential due to all these different pieces of charge Phi of R must be equal to 1 over 4 PI epsilon naught integral Rho R prime VB prime divided by once again potential is a much simpler function electric field involves a vector and it involves one over R cubed times R so it's going as 1 over R squared potential just involves a simple one over R it's a very very simple function and in a certain sense this potential contains everything because if you knew this potential again if you went to a nearby point and measured potential and took the difference then it must be true that this is equal to e dot d R with a minus sign it's just coming from the definition because this is integral from far away to R Plus D R and this is integral from far away to R and you know if you take two integrals you can take the difference by saying it's R to R plus dr of e dot d dl the minus sign and for very very small lengths of integral it is nothing but the integrand multiplied by the difference of the limits so that's this so it means that if I know this potential function and I take the potential function at very nearby points I have effectively got a feel for electric field because the difference between nearby points of potential is nothing but the electric field itself of course there's a dr in there but is nothing but the electric field so you have got a case where you have simplified the problem we started with the electric field which was a complicated vector field and we are now divided divide scalar field much simpler field it's got a physical meaning to various potential energy of the charge and it does it gives you the electric field after all I will continue next time [Music] you
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