Maxwell's Equations - Basic derivation

hello today we're going to do a basic derivation of Maxwell's four equations which are the fundamentals of electricity and magnetism we know from my video on electromagnetism that if we have a point charge let's call it Q and let say it's a positive charge then the electric field will radiate from that charge radially in all directions equally and we know that if we bring another charge let's call it a test charge little Q into the vicinity of capital Q there will be a force between the two charges which is known as the Coulomb force and that is given as we showed in the previous video by Q Q over 4 PI epsilon R squared where epsilon is the permittivity of free space and I also said that the field strength at any point is given by Q over 4 PI epsilon R squared now what is field what is the electric field well it's the capability at any point to do something in other words it's the capability to exert a force on a charge if there were a charge there so if you were to put a charge say plus Q little Q at this point here then you could work out what the force would be because the force would simply be the electric field times Q so in other words the electric field is essentially force per unit charge and forces always measured in Newtons and charge is always measured in coulombs so the value of the e field the electric field is measured in Newton's per Coulomb I now want to consider the concept of electric flux flux is simply the amount of a field that goes through a particular area so if we take our charge again which radiates outwards he field going in all directions and we want to know how much of that field goes through an area or a window if you like of area da so it's a small window da then we say the flux which is represented by the Greek letter Phi the amount of flux that goes through the window will only be a small amount of flux OD Phi is equal to the amount of the field which is the value of the field at that point e times the area da so it's as simple as that flux is the field that goes through the area but mathematically it is written in this form e dot da and the dot is a vector product and it's also equal to e times da times the cosine of theta and let me explain why we need that cosine term you see it all depends on the angle that the flux or the e field meets the window let's suppose that this is the e field in this direction if the window is square on the other words at 90 degrees then the e field will pass through the window if on the other hand the window is flat on then no flux will go through the window it just bypasses it so the critical thing is what angle is the window to the e field and that's where this cos theta term comes in and what theta is is it is the normal to the window the angle between the normal to the window and the e field so this is the e field this is the normal to the window in other words 90 degrees to the plane of the window and theta is the angle between the two and when theta is equal to 90 degrees cos theta is equal to zero so this whole term becomes zero and so the flux going through the window is zero when the angle between the know which of course will now point in that direction and the e field is zero cosine theta will be one and so you now get maximum flux EDA passing through the window so now let's consider a charge once again it's a charge Q and we're going to construct round it a sphere now this is not a metal sphere if you like you can regard this as a an imaginary sphere or if you like a sphere made out of plastic so it's just a sphere around the charge and the sphere has a radius R well if you ask yourself what is the amount of flux passing through an area da on the surface of the sphere then the answer of course we've just arrived is that the flux going through that little window will equal e times da we don't have to do dot da in this case because of course we know that the field will always go out radially and we know that the radius of a of a sphere always meets the surface of the sphere at 90 degrees so you're always going to get maximum flux who a window which is on the surface of the sphere and so if you integrate this to find out the total flux going through the entire sphere you get that Phi total flux going through the entire sphere is the integral of e dot Da and that is e times 4 PI R squared since as I've said you don't really need the e dot Da it's EDA because the cosine term will automatically automatically be 1 but right at the beginning of this we showed that e is equal to Q the charge divided by 4 PI epsilon R squared and so if we substitute that for e in this equation here we get that the total flux is equal to e which is Q over 4 PI epsilon R squared times this term here 4 PI R squared and you'll see that the four PI R Squared's cancel out leaving you with Q over epsilon so the total flux passing through the entire surface of this sphere is equal to the charge divided by a constant which is the permittivity of free space if there were another charge in this sphere then the total flux would be the flux due to this charge plus the flux due to that charge and that means that where you've got more than one charge the total flux will equal Sigma Q the total charge divided by epsilon and so we can say that Phi is equal to well it's equal to the integral of e dot Da integral of e dot Da over the entire surface of the sphere and we show that by putting a little ring there that means over the surface of the sphere the whole surface is equal to Sigma Q over epsilon and that is known as Gauss's law and it's also Maxwell's first law so Maxwell actually borrowed as we shall see three of his four laws are down to other people this is the first and it's Gauss's law one obvious point suppose you were to put a charge of plus Q and minus Q inside the sphere then according to Gauss's law the total flux would equal Sigma Q well plus Q and minus Q come to zero that means the total flux would be zero does that mean there's no e field no it simply means that the amount of flux going out from the positive charge will be balanced by the amount of flux coming in from the negative charge so the net flux which is what this is all about the net flux passing through the surface of the sphere would be zero now before we move on I want us to look at one or two interesting consequences of Gauss's law let's take a this time we're going to take a metal sphere and on that sphere we are going to place the charge Q and of course the charge will simply distribute itself right the way around that sphere evenly because the sphere is a metal sphere so electrons are free to move charge is free to move and consequently because it will be all this positive charge will be repelling itself it will try to get equidistant from all the other charges so that there is no potential to move and all the charge will then be equidistant and what I want to know is what is the electric field at a point a outside the sphere and what is the electric field at a point B inside the sphere well Gauss's law tells us that I can construct a sphere this is the sort of the imaginary sphere or if you like the plastic sphere around the charge in such a way that a the point that we are interested in is on that sphere and the total distance from the centre of the sphere is let's say ah now what does Gauss's law tell us Gauss's law tells us that the amount of flux which passes through the surface of this large sphere here is going to be equal to the integral of e dot Da that's the electric field through the total surface area of the sphere and that that will equal Sigma Q over epsilon that's Gauss's law well what is the Sigma Q for this sphere here it's the charge enclosed the charge within the sphere what is the charge within the sphere is plus Q so that is simply Q over epsilon EDA as before the integral over the whole of the surface of the sphere will simply give you e times the surface area of the sphere e times four PI R squared equals Q over epsilon which means that E is equal to we'll divide both sides by four PI R squared Q over 4 PI R squared epsilon and that is the value of the field at Point a but look that's rather interesting because that is exactly the same as the value of the field would have been if that entire charge had been at the center of the sphere so that if this charge instead of being distributed on a metal sphere itself had all been collected at a point at the center then the value of the field at a would have been this value here so what Gauss's law tells you is that if you've got charge distributed on a sphere outside that sphere beyond that sphere the electric field strength will be exactly the same as it would have been if all the charge had been accumulated in a single point at the center of that sphere now let's look at the consequence of being at point B in other words inside the sphere well once again Gauss's law tells me that I can construct a sphere and I've got exactly the same formula again that the flux is going to be the integral of e dot Da which is going to be Sigma Q over epsilon and Q has got to be the total charge within the sphere well how much charge is there inside this sphere the answer is none all the charges on the surface of the metal sphere here there's nothing inside this sphere here and that means that ETA is equal to the sum of the charges we zero so EDA you must be zero which means that II must be zero since the total area of the sphere is not zero so II must be zero so for point B in this sphere point B the electric field is zero and what that tells you Gauss's law is telling you that if you've got charge on a sphere inside that sphere there is no electric field at all which means that if you were to put say an electron there it would move because there's no a feel there's no force acting on it you might at first thought at first sight think that because the electron is much nearer this part of the sphere that the electron would be attracted to this bit of it but the point is that although there is a very close charge on this bit of the sphere here there is a much greater charge on the other side of the sphere but it's further away and the effect is they cancel each other out so that within the within the metal sphere there is no a field which means there's no force to act on any charged particles any charged particle inside a metal sphere on which are positive or an Indian negative charge has been placed will not move because there will be no e field acting on it okay let's move on and we're going to do exactly the same thing as we did four Maxwell's first law but this time we're going to consider not a electric charge of Q but a magnetic charge or a magnetic pole and what we want to say is that if we construct a surface and we have within that surface some kind of magnetic pole or magnetic charge if you like what is the equivalent formula to Gauss's law which we had up here Phi is equal to the integral of e dot Da which is equal to Sigma Q over epsilon would we expect to get a similar kind of formula well let's write down what that similar formula would be we would say that the magnetic flux which were call Phi B is equal to the integral over the whole of the surface of B dot Da where B is now the magnetic field and that's going to be Sigma the total amount of magnetic charge or the total amount of magnetic poles if you like divided by some constant K it won't necessarily be epsilon because epsilon applies to electric fields so we're looking for something along those kind of lines to equate to Gauss's law in relation to electric fields well the answer is that this term will always equal zero why is that because you can never have just a North Pole or just a South Pole in magnetic terms a North Pole is kind of equivalent to a positive charge and the South Pole is kind of equivalent to a negative charge but you can never have one without the other if you put a magnet in here it will have a North Pole and it will have a South Pole and consequently that's exactly the same in principle as having a plus Q and the minus q when you do the Sigma Q plus Q plus minus Q is zero you will always have the same strength North Pole as you have the opposite strength South Pole consequently this term will always be zero so now you've got that Phi B is equal to the integral of B dot Da which equals zero and that is Maxwell's second law and that says that the total flux the total magnetic flux passing through a surface of a sphere will always be zero because there's as much negative as there is positive the equivalent of is as much magnetic flux going out of the sphere as there is magnetic flux coming in there will be magnetic flux going out as a consequence of the North Pole but there will be magnetic flux coming in as a consequence of the South Pole and the two will balance one another out so the net flux will always be zero you might think that you can cheat you might think that if you take the surface of the sphere you could put a bar magnet like this where the North Pole is inside the sphere and the South Pole is outside the sphere and now you've got a North Pole with in by the South Pole outside and surely that will violate what we've just done well unfortunately of course it violates the very principle that the sphere has to be a closed sphere it has to be an entire sphere with no holes in it if you put a bar magnet across the sphere you have broken the sphere it's no longer a complete closed sphere and therefore it won't apply all sorts of funny things will happen so this is not allowed you have to play by the rules the magnet has to be holy within the sphere and if it is then you will have a North Pole and the South Pole both inside the sphere and that means that you will have a net flux and net magnetic flux passing through the surface of 0 that's Maxwell's second law we now move on to Maxwell's third law and for this we will consider a coil of wire okay it's just a plain coil of wire and we're going to take a bar magnet north/south and what we're going to do is we're going to push that bar magnet in and out of the coil so we're just going to change direction move it in move it out move it in move it out and what we know is that when you have a magnet movie inside a coil a current will flow the direction of that kind of course will keep changing it will go in one direction when you push the magnet in and it will come out in the other direction when you pull the magnet out but the current will flow when the magnet is moving if you simply hold the magnet steady inside the magnet inside the coil of wire nothing happens no current flows is only when you move the magnet that the current flows and that is because you've got an induced current and the induced current arises because of the change in the magnetic field and the faster you move the magnet in and out in other words the greater you change the field the greater the rate of change of field the greater the current will be and if there is a current flowing in this coil of wire it must be the case that there is a voltage difference a potential difference what we call an EMF the electro-motive force it's got nothing to do with force at all it's a voltage it's a potential difference causing the current to flow and so what we can say is that the electro-motive force which is of course the EMF which is a voltage which is driving the current is going to be what it's going to be proportional to the rate of change of the magnetic field the faster you move the magnet in and out by experiment you can also find that the EMF will be also proportional to the area of the surface that is this essentially this surface here but that is the open surface which is surrounded by the current loop and so you can also say that e the EMF is proportional to a the surface and so we can say that the total EMF is proportional to the change of the magnetic field times the area now definition of flux we've already established this is the amount of the magnetic field or electric fields of the case maybe going through a particular area so flux is equal to BV in this case it's magnetic flux times the area and it's a dot product remember because of course it critically pens whether the window that is the area you're putting the flux through is as it one at 90 degrees to the field or not so if Phi the magnetic flux is equal to B dot da then D Phi by DT is going to be DB by DT dot da but we know that DB by DT times the area is equal to the EMF and so now you've got that the EMF is equal to D Phi by DT and that is equal to D by DT times V dot Da so what this is telling us is that the electro that the EMF electro-motive force is equal to the rate of change of flux which is itself equal to the rate of change of B dot da will park that formula we will need to come back to it but we'll park it we're not yet at the third law we now need to use Maxwell's first law which was Gauss's law to get us a little bit further along the road and what we're going to do is we're going to consider two large parallel plates metallic parallel plates if you like a very very large capacitor the reason it has to be large is that we're interested in what's happening away from the edges so there will be edge effects but if you get far enough away from the edges in the middle then the edges don't have very much impact and what we want to ask ourselves is let this let's just consider the surface of one of those plates on that plate there will be a charge and we're going to say that the density of that charge or the charge per unit area is Sigma so it's the amount of charge in a particular area is Sigma and the total charge will of course be Sigma times the entire area now gauss says i can construct a surface beforehand we've constructed a sphere it turns out it doesn't have to be a sphere it can be any surface so we're going to construct a surface which is essentially a cylinder which is straddles as it were the metal plate so that H is above the plate and H is below the plate and the radius of the cylinder will be R so the diameter will be 2 times R and Gauss's law your remember tells us that Y dot Da is equal to Sigma Q over epsilon that's Gauss's law the first of Maxwell's equations now then what will the electric field do from a plate which has got uniform charge on it well the field will always come out from a plate and at 90 degrees to the plate it can't go in that direction because there will be another if it did there be another charge going in that direction and the net resolution of the components of these fields would be out at 90 degrees so the field always leaves at 90 degrees so how much field goes through the sides of the cylinder answer none because it's all going parallel and it won't go through the side of the of the cylinder so the only flux goes through the top and the bottom of the cylinder so the flux that passes through which is e dot Da he's going to be the total charge within this cylinder and so now you've got that e times da well there's two areas that the flux will go out from this area at the top of the cylinder this area the bottom of the cylinder each of them is PI R squared and so the total area is 2 PI R squared top and bottom so that's y da and the flux of course is going out at 90 degrees to the surface so we don't really need to worry about the dot term what is the charge within that cylinder well it's going to be the charge per unit area times the area well the charge per unit area we've said is Sigma and the area is going to be PI R squared because that's the area of the charge on the plate inside always has to be inside the surface inside the cylinder divided by epsilon well if you now divide both sides by PI R squared you will get that E is equal to Sigma over 2 epsilon and that's a rather interesting result because it tells you that the field anywhere above or below the plate is going to be equal to Sigma divided by two epsilon and the fascinating thing about that is that that is completely independent of H notice H doesn't feature at all so the field above a very large plate remember this has to be a large plate if you get anywhere near the edges there's no longer applies but the field above a plate will always be Sigma over 2 Epsilon however far you are above the plate so now we consider the the position where you have two plates and a difference in charge let's say we put plus Q on that plate and minus Q on that plate what will the effect be well let's consider firstly just the top plate that will have an e field which goes out from the plate the e field always goes out from a positive charge and remember it's constant in space it doesn't vary so there will be an e field going in that direction there'll be an e field going in this direction and an e field in this direction all due to this plate here all of those fields and they will be constant the same in all points due to this field here sorry due to this plate here what about this plate this has got a negative charge on it so the fields will go into the plate which means that there will be a field going into the plate there there will be a field going into the plate there and a feel going into this plate here remember it doesn't matter how far you go above the plate the field strength remains the same it's what we just showed here so what does that tell us well if you look at the effect of the two plates outside this field will cancel this field so outside at the top he will be zero similarly underneath the two plates this is going in this direction this is going in the other direction the consequence is that the e field will be zero outside at the bottom but inside the plate you've got two he fields both pulling in the same direction and since E is Sigma over to epsilon that means that the total field inside the plates will be equal to Sigma over epsilon just double this value here now I want to consider what voltage is voltage is the potential energy acting on a charge to move the charge from one point to another point technically if you have a charge Q then the voltage is the work done to bring a test charge little Q from infinity all the way in to a point which is a distance R away from the main charge Q so the voltage is the work done to bring this test charge from infinity to this point and that then becomes the essentially the electric potential per that unit charge well what is potential energy that is the work done which is equal to the force times the distance what is the potential energy at infinity the answer is zero if you look you ask yourself what is the force between these two charges at infinity will the force as we saw right at the beginning is going to be Q Q over 4 PI epsilon R squared where R is now infinite because it's the separation of the two charges and when Q is at infinity in other words infinitely far from capital Q then R squared infinity squared is going to mean that the force is zero because you've got an infinity squared term in the denominator so there's no force what is the force when you get to the point where Q is separated by little Q by a distance R that force will once again be equal to Q Q over 4 PI epsilon R squared where R now is just this distance here and the potential is going to be or the work done which is the force times the distance so we now have to multiply that by our and that gets you Q Q over 4 PI epsilon R and that's going to be the potential energy the force times the distance but voltage is potential energy per unit charge so it's potential energy divided by the unit charge and so voltage is going to be Q over 4 PI epsilon R but let me just remind you that the e field as we define right at the beginning of this video is Q over 4 PI epsilon R squared so that gives us a very easy relationship between E and V it's simply that V is equal to e times R or e equals V over R and just to notice that earlier on in the video we said that e the e field could be measured in terms of Newton's per Coulomb because it is force divided by charge now you'll see it's also volts divided by distance so E is not only Newton's per Coulomb it's also volts per meter you can express the e field in both ways so now if we go back to our parallel plates where you remember the e field between the plates is the same everywhere then what you can say is that there's a voltage difference V let's say that that voltage is zero and we've got a voltage V on this upper plate the electric field will be the same everywhere but the voltage will vary according to the distance from this plate so halfway up this will be V over 2 because E is V over R so the greater the distance the greater V will have to be in order for e to be a constant so we've established this general principle for large parallel plates but within those elected those plates the electric field is constant but the voltage has to change as the distance increases now as a general point you can say that even when you haven't got parallel plates even if you've got just a coil of wire if you break that coil up into very small distances which have a litter distance DL then the same principle that we've developed up here will apply which is that V is equal to the e field times the distance which means V which you can also write as the EMF is equal to the e field times the distance and once again to get it mathematically correct it's actually e dot DL because you have to take account of the difference between the e field and the direction in which DL is pointing and so we've now got that the EMF is e dot DL which means that the total EMF will be the integral around the wire so the total EMF is going to be the integral around the wire of e dot DL and now having done that excursion we can go back to the equation we derived before where we said that the total EMF is equal to the rate of change of flux and we said that that was the same as the rate of change of B dot Da but we've just said that he is also the integral of e dot DL and of course these terms also have to be integrated if you want to and integrate the full terms so the total EMF is the integral of the rate of trêpa of flux of change of flux which is also the integral of the rate of change of B dot Da which is also the integral of e dot DL and that is what is known as Faraday's law and is also Maxwell's third law and just to rehearse what it says the total EMF will be the rate of change of B dot Da over an open surface and that is equal to the integral of e dot DL over a closed loop just to remind what we're talking about remember we started off with the idea of a loop of wire and we have a magnet which we are moving in and out of that while that magnet creates a changing flux and the integral of that flux going through the area that is this closed so that open surface that's this surface here the integral going through that surface which is this term here is equal to the integral of the closed loop that is this loop here of e dot DL so the magnet the magnetic flux going through the surface which is essentially that term is equal to the electric field around this closed loop that's what's called Maxwell's third law and Faraday's law and it may look a bit odd but as many people point out that is the formula which runs the entire economy because that's the basic formula that describes how power stations work when steam drives turbines the turbines have effectively coils of wire which spin in a magnetic field it doesn't really matter whether it's the magnet that's moving or the coil of wire that's moving a moving coil in the presence of a stationary magnetic field or the other way around will cause a current to flow and that is how electricity is generated and that's the formula that governs the the generation of electric that is Maxwell's third law and now we move on to Maxwell's fourth and final law and here we are considering a current which is flowing in a wire which is going essentially into the paper and you demonstrate a current flowing into the paper by using a cross if it were coming out of the paper you would use adopt it's rather like an arrow an arrow going away from you you see the tail feathers the cross if you see an arrow coming towards you you just see the point of the yellow so this indicates a current flowing into the paper in the wire that's going into the paper and experimentally you find that around the wire you will get a magnetic field B and again experimentally it is shown that that value of the magnetic field will be proportional to the value of the current the greater the current the greater will be the magnetic field divided by R where R is the distance from the wire to the point at which you're measuring the magnetic field and that is an inverse term which means the further you get away from the wire these smaller the magnetic field will be whenever you have a proportional term you can always put a constant in and so what we say is that B is equal to a constant and the constant is called mu 0 over 2 pi times I divided by R mu 0 is what is called the permeability of free space not the permittivity that was epsilon this is the permeability of free space and this formula here is essentially a formula that was derived by biot-savart sometimes called the biot-savart law now once again much as we did in for the third law we can take this closed loop and we can break it down into lots of different segments of length DL and we can ask ourselves what is the value of B dot DL over the entire those loop well that sorry the value of B dot DL over the entire closed loop well that will equal B times the value of DL over the entire loop well the value of DL over the entire loop of course is 2 pi are the circumference of the circle so that's 2 PI R but B is mu 0 over 2 pi times I over R so this is equal to B which is mu 0 over 2 pi times I over R that's this term here times 2 PI R and you'll notice that the 2 PI R cancel out so now you get the interesting result that the integral of B dot DL that is around this closed loop is equal to the permeability of free space times the current passing through the wire and that is almost Maxwell's fourth equation is also called and PS law so you can see Maxwell got his first of all from Gauss his third law from Faraday and his final offer from ampere but we're not quite there because of a technical twist and it's this what we're saying is we've got a current in a wire which is going through this surface it's going into the paper through the surface and we're looking at a closed loop around the wire and we need a proper technical definition of what we mean by the current going through the surface the surface which is attached to this closed loop and the question is does it have to be a flat surface as it is on this piece of paper you know here is the closed loop and the flat surface is my piece of paper and the wire penetrates the paper penetrates the surface that's clear but does it have to be a flat surface and the answer is no it doesn't it could for example if this is the closed loop in other words that is this loop but a John you could attach a surface which represents essentially a bag so this is just a bag attached to this loop provided the current flows through and comes out of this surface there's the current I then this formula is applicable so the surface doesn't have to be a flat surface in other words it doesn't have to be just that surface there it can be a bag as long as the wire carrying the current penetrates the surface then this is the closed loop and this is the open surface and that's what gives us the problem because let's suppose we've got a wire fine so far it's got a current I and let's suppose that we're measuring the magnetic field around this closed loop here fine you would say still no problem this formula is going to work but let's suppose we have a capacitor here and let's suppose the capacitor is still charging which means the current is still flowing well what happens if I do this trick and put a bag in other words an open surface which doesn't have to be flat such that the bag does this now you say does the current penetrate the surface of the bag and the answer is no it doesn't because no current flows between the plates of a capacitor so your stein lead you'd have to say no can't flows and therefore the total B dot DL is going to be mu zero times naught which is not which means that you would not expect to find a magnetic field at this value at this point R on the closed loop which is madness of course there will be a magnetic field there so we've got to have some kind of corrective term which we add to this equation here and how are we going to do that well here we've got the key term is mu zero times I so let's write that down that's the term that we want inside this capacitor here because that's what's causing us all the trouble and what can we say about mu zero i well current is simply the rate of flow of charge that's its definition so mu zero sorry I is equal to DQ by DT the rate of flow of charge that's what current is now you can always not multiply a formula by the same values as long as you do top and bottom because essentially that just means you're multiplying by one so I am going to multiply by I'm going to take mu zero and then I'm going to multiply by epsilon D a divided by epsilon D a that's just 1 so I've done no damage to the formula times DQ by DT let me just show you what I've done I've taken mu 0 DQ by DT at some Yuki 0 DQ by DT and then multiplied it by 1 epsilon da divided by epsilon da that's one I've done no damage to this term at all but now I can rearrange that that mu zero times I is equal to MU 0 epsilon da times D by DT Q divided by epsilon da but let me just remind you of Gauss's law which is Maxwell's first law which says that eta is equal to Q Sigma Q divided by epsilon which tells us that E is equal to Q divided by epsilon D a and Q divided by epsilon da is this term here so now we can write this term as mu 0 epsilon D a times D by DT times e the rate of change of field with time but I can go further I can take this D a term and bring it here so we now have mu 0 epsilon D by DT e da and what is e dot da or e da that of course is flux and so we now get that this term mu 0 I is equal to this term here which is mu 0 epsilon D Phi by DT rate of change of flux and just to remind you that since I came from an e field it is effectively electric flux not magnetic flux and this is often called the displacement current that's its technical term it's when you've got this situation where you've got a capacitor no current is flowing there is a kind of an implied displacement current and that term has to be added to this term and that gives you the grand final term that the integral of B dot DL around the closed loop of wire is equal to MU 0 I which is the formula we had up here and apply is if you haven't got this capacitor in the way but if you've got this capacitor problem here you also have to add this displacement current here which is mu 0 epsilon D Phi and just put me there to tell you that it's an electric field D Phi by DT and that is Maxwell's fourth law there's just one final point to mention which is a point of intrigue and that was this term here the permeability of free space multiplied by the permittivity of reach free space Maxwell discovered that mu 0 epsilon equals 1 over C squared where C is the speed of light and so he discovered that there were two terms associated with electromagnetism which when multiplied together related to the speed of light and that's what gave Maxwell the concept that there was an eternal speed limit or a universal speed limit the speed of light C squared which is now related to electromagnetic terms from this Maxwell was able to deduce that light was electromagnetic radiation but that's for another video so for completeness now let's just write out Maxwell's four equations so we know what they say the first equation says that the total flux is the amount of electric field going through the area which is the integral over a surface it doesn't have to be a spherical surface it's just the integral over a surface and that will always be the total charge contained within the surface divided by the permittivity of free space that's the first law the second law does the same thing for magnetic flux so this was electric flux this is magnetic flux and that is the integral of B dot Da the magnetic field going through an area over an entire surface is equal to zero and we said it was zero because you can't have a North Pole without having a sample then we came to Maxwell's third law which is also Faraday's law which says that the total integral of the change of flux rate of change of flux is equal to the integral of e dot DL over a closed loop that's the field at times the distance which is equal to D by DT times the integral over the open surface of B dot Da and that's Maxwell's third law and then as we've just derived Maxwell's fourth law says that the integral over a closed loop of B dot DL is equal to MU 0 times I which for most at times is sufficient that will work but where you've got that capacitor problem you also have to recognize that there will be a displacement current mu 0 epsilon times D Phi e by DT so when that term is 0 which it will be if if the surface goes through between the capacitor plates this is the term which will enable you to calculate the magnetic at the the B dot DL term that is the changing flux and those are four basic derivations of Maxwell's equations they can be put together in a more vector form but I just wanted to do a basic derivation for the purposes of this video

Channel: DrPhysicsA

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Keywords: Maxwell, equations, law, gauss, faraday, ampere, permittivity, permeability, coulomb, electric, field, force, electricity, magnetism, charge, displacement, current, flux, volts, emf, rate, of, change

Id: AWI70HXrbG0

Length: 54min 31sec (3271 seconds)

Published: Fri Sep 07 2012