Lecture 22 Gauss's Law

come back to physics 272 last time how did I feel the same way after a long weekend you know after a long weekend I've taken a break from physics or as the case may be not taking a break from physics but it's good to be back okay so motors and generators is what we talked about last time and the basic concept there is that because a magnetic field applies a force on a moving charge okay or because a magnetic field applies a force on a current carrying wire then we can use that idea to convert mechanical energy into electrical energy and vice versa by taking a wire and dragging it through a crossed magnetic field okay so that was the concept last time today is all about Gauss's law sometimes we have bullet points here but today it's just we will get to Gauss's law today everybody ready all right so for Gauss's law we're gonna have to think about flow and source and sink and drain and those kind of concepts and we're gonna have to define something called flux so let me try to give you an intuition about something called flux we'll talk about electric flux but first I want to talk about particle flux so I can think about water as it flows around and I can either think of it from kind of the big macroscopic level if I see this river going by and there's a certain amount of flow rate in the river or I could think very microscopically about it and I can look at a particular spot and look at individual molecules going by okay and if I were thinking and along the lines of individual molecules going by I might think well let me take a little area a little cross-sectional area and talk about the number of molecules that go by per second times that cross-sectional area that would be the idea of a particle flux so thinking along those lines let me give you a physical situation we're gonna make an analogy between water okay water flow and electric flux and so let me give you a physical situation here that I think you already have an intuition about and then we'll use that to develop intuition about electric fields and electric flux so here's a fountain kind of interesting alright I see water spilling out over the bowl and from this angle I all I see is that will I have a full bowl and I have water spilling out of the bowl all right now what should I conclude about what's going on inside the bowl with regard to the water I could think well maybe hidden water is being miraculously created inside the bowl fufufufu boof I've found the source of all water on earth you know it's just poof poof being created or I might think well there's probably a water faucet in there somewhere ok there's probably piping somewhere bringing water into the bowl so I observe water flowing out through some boundary in this case the boundary is the edge of the bowl and so I could take the edge of the bowl and I say oh there's all this water flowing over the edge of the bowl and I could count how much water how many water molecules per second are flowing out that boundary and that would tell me the flow rate of whatever faucet is coming into the bowl does that make sense this should be totally intuitive I hope right oK we've just got a lot of chatter going on in the classroom and it's hard for me to talk over that with a sticky throat thank you excellent ok so yeah oh we're gonna have the remote-control issue today ok then I'm going to have to go to here and actually push the button alright now this is a different case in this case this is actually a lovely fountain at the Getty Museum in Los Angeles and here you see water flowing down but if I look simply at this little pool right here I would measure well there's water flowing over the edge certain amount of water flowing over the edge there and I'd see over here that there's a certain amount of water flowing in and I could measure that what's flowing in that flow rate is equal to the flow rate going out and so I would conclude that inside this pool right here there is no source inside that pool ok water's flowing and waters flowing out there's not a source inside that that little surface area so if I see the same coming in as out I know there's not a source in this case I see a certain amount flowing out over the edge of the bowl and I conclude there must be a source inside so we're going to use that same kind of physical intuition and studying electric field lines and we'll define something like something analogous to flow which is electric flux and we'll be able to use the concept of electric flux to figure out if there's a charge inside of a closed surface or not so here and the case of the water fountain analogy I could look at the rim of the bowl and I count over the rim of the bowl and everywhere along the rim of the bowl I could say how many particles per second are passing over the edge here and I conclude there must be a source somewhere in the in the bowl and the source of the water is a water faucet not some magical pixie at the bottom creating water molecules but a water faucet in the case here of electric field I could do the same sort of thing I could say oh I have a point charge and it's putting out electric field lines and so you know what the shape of the electric field is coming off of a point charge all right it's 1 over 4 PI epsilon not Q over R squared all pointed away from the charge itself and now if I think of encapsulating this closed charge in an imaginary sphere so put an imaginary sphere over that point charge and I could stand anywhere on the sphere and I could make an analogy between the electric field lines that are poking through this sphere and make an analogy between that and water so if I think of the electric field lines acting a little bit like water flowing out and I walked all around the sphere and I measured all around the sphere and I saw electric field lines poking out of the sphere poking out of the sphere poking out of the sphere I would have to conclude there's some sort of charge inside in the same way that I concluded there was a water faucet inside of that bowl there ok so so in the two cases I'll draw some sort of a boundary around the system here water is flowing out of the bowl there must be a source inside the source of water is a water faucet in the same way when I draw a sphere around this point charge and I think about the electric field that's something that's kind of like flowing through the sphere poking the sphere and add up all that contribution I'd see that there must be a source inside there must be a charged particle inside so charges are the source of electric field to have any questions so far about the analogy or the physical intuition ok so the basic concept is that in the same way that you can diagnose whether there's a faucet somewhere based on the flow of water you can diagnose whether there's a charge somewhere based on the electric flux and we'll make that a mathematical concept so electric flux is very much like water flow so think of water flowing through a net okay water flows through a net at a certain rate and in the net I have little boxes okay little squares through which the water can flow and I could look at any individual square and I could simply count I could stand on the square and count one molecule to molecule three molecule however many molecules per second are flowing through that square and if I wanted to get the entire particle flux the entire water molecule flux through the net I would add up that contribution over all the squares and I would say okay there's a certain number of molecules per second times the area of this square and then go to the next square certain number of molecules per second times the area of this square and so forth and add up that whole contribution and I would get something that follows water molecules per second times area okay now I can take the same situation and let me take the same net so same exact net think of this net as being in a pipe so that the net is tied down okay all around the the where it meets the pipe and now as I start the water flowing I could think of this net deforming right the net itself may be a little bit stretchy I could deform it this way I could deform it that way I could deform it in a funny shape but when I go back to do that same calculation in order to measure the flow rate in the pipe even if the net deforms so maybe the net deforms and my squares get a little bit bigger I could still stand on a square and watch the molecules per second coming through that square times the area of that square and if I move to the next square in the net and did the same thing molecules per second coming through the square times the area of that square what I'd find is that even though they're not deformed and sum of the squares got bigger the angle of the squares with respect to the flow rate would would shift so that you know a square a square that's head-on the flow rate gets a lot of molecules per second through it but a square that's not head-on gets a smaller number of molecules through second per second through and so both cases whether the net deforms or not I would measure the same flow rate through the pipe okay so this gets us kind of thinking about we can make an analogy between this kind of concept and electric flux so in the case of electric flux the the the flow is going to be electric field lines so I'll think of electric field lines poking through surfaces and I'll stand on any square in the net okay and I'll measure the contribution of the electric field that's poking through that net alright times the area of that little square and then add it up for the next square and add it up for the next square and you can see where this is going anytime we take something where we're going to add up small contributions over a large system we're going to end up taking an integral do you have any questions so far okay either about the water flow analogy or where we're headed okay all right and so we're now ready to define electric flux itself so electric flux is defined as the three bars mean define as here's mathematically how I would write down all those ideas for an electric flux so I would say I need an electric field but I need to dot it into something the reason I need to take a dot product somewhere is again if I think of a flow rate coming by the front of the classroom and I'm thinking about a net trying to capture that flow rate if the net turns toward the side a little bit I need something in the mathematics that tells me a sideways net is going to catch less flowing through it so there must be a dot product somewhere okay and I can either think of taking a dot product of this sideways net the projection of it perpendicular to the flow or I can think of taking a projection of the flow onto the perpendicular of the net I'll show you what that means so here we write it down this way electric field dotted into some vector in hat times the area contributions on each of the little squares and at the end of the day the summation over the tiny areas of every little square will end up turning into an integral in the limit of small squares so the N is a normal so here I'm thinking of the electric flux through a surface area here so here's a surface area that is a rectangular shape that's in the Y Z plane so your axes here are XY and Z this is a plane this is the Y Z plane but I've got a little piece of area there that's going to act like a net measuring flow through it and I have in this case a constant electric field just to have a nice simple case to get intuition about so the electric field in this case is along the x axis the magnitude is constant and here are the field lines I've drawn to represent that and what this n is is a unit vector that is perpendicular to the area that you're looking at so in this case where I have a small rectangle that is on the Y Z plane then the normal to that is this vector here ok or here when I've got when I've got the area tilted downward so that I have the area lying in the XZ plane the normal is perpendicular to that or here where I've got it tilted by some angle then the normal is always perpendicular to the plane so for any plane you can define there's always one and only one direction that's perpendicular to it all right and this normal defines that direction for you and the magnitude of that vector is set up to be one so it's a unit vector that tells you the normal to the plane do you have any questions about n ok all right so if I think then well what does this mathematically mean in these three cases so I'll think of a case here where I have an area that's perpendicular to the flow again think of this like a net that's perpendicular to the flow I should catch a lot of water so I'll get lots of flux in this case because II done it into that normal N will basically give me the full contribution right whereas here in this case the electric field is flowing past my net but I'm not really catching any because it's it's it's just flowing over the net and the way that's represented in the mathematics is that Y dot n in this case since n is perpendicular to the electric field e dot n hat in that case is zero and I'll catch no flux here so lots of folks here no flux here and in the case when I have it tilted a little bit I'll have an inner amount of flux okay so this is the equation that defines flux and these are kind of the the physical situations corresponding to different angles do you have any questions so far all right so this is the mathematical definition of electric flux but it's a flexible concept you could use it for different things so for example we thought at the beginning of the class about water flux we thought about water molecules per second times an area going through something you can also think in terms of of any other kind of flux you like so electric flux is right here but you can also think in terms of fish flux so here I brought Dory into the picture so here's Dory right there's a certain flux of Dory's going by and I could think of these areas as like nets that are gonna catch a certain number of fish so if the net is perpendicular to the direction that the school of fish is flowing I'll get a lot of fish if the net is tilted a little bit I'm gonna catch the fewer fish if the net is completely parallel to the the school of fish I'm not gonna catch any fish all right or Bruce here will go hungry in this case but Bruce will catch a lot of fish in this case all right do you have any questions about either fish flux or water flux or electric flux okay all right all right you might be asking yourself what does that stand for that stands for what the Fox right okay so we're going to calculate flux all right the flux is e dot n da and I have this situation here I want to say given that I have an electric field of five Newton's per Coulomb parallel to the X direction okay and I want to take this case right here I want I want things I want to take an area that's rotated by an amount theta theta is 30 degrees I'm gonna rotate from the Y Z plane around the z axis okay so the physical situation is this I had an area that was parallel to the Y Z plane this is the z axis right according to the diagram up there and I'm going to rotate just along this direction so it stays attached to the z axis but rotates around it by an amount 30 degrees all right and I'm gonna have a net here of 1 meter by 0.2 meters what's the flux what's the electric flux through the net okay all right we've got a lot of votes in so go ahead and take some time and discuss with your neighbor how to solve the problem all right you want a little bit more time are you ready ready to discuss all right so what's what's your step one here how do I go about solving this let's hear what's your step one yeah okay so you need to draw right okay so I hope everybody you know was scribbling diagrams and things like that so so one of the one of the ways to do this so we have our of the axes okay that we're given in the problem XY and Z alright and we have so so you're talking about one way to solve this problem is look it's about flux right so it's about catching things through a net so I you know in the top diagram I have this Y Z here the bottom diagram it's in the XZ plane but we want to calculate the middle point where it's rotated from the Y Z plane about the z axis by an amount theta but it's about flux so one way to do that is that I want to take the projection of this area onto the normal to the electric field so that can be your step one okay and so that's that's definitely one one route to do it in which case you would need a cosine of theta in order to project this area on to something that's perpendicular to the field now the math is telling us a slightly different route but they're equivalent okay which is why both are gonna work so the math is telling us to take a dot and hat so let me show you both ways and hopefully then we'll see that they're equivalent okay so this is saying I need to find an in hat so let me draw out a geometric diagram and show you how this is going so this is a line that's parallel to the y-axis alright I also have the net that I need to find okay there's the theta this is the net so I need the area of that thing alright but the electric field in this case is parallel to the x-axis let me draw an electric field vector alright so that's the direction of the electric field parallel to X alright and what the math told me to do is that I need to figure out how to take you down it into N so I need to find the normal alright so here's that area and hat is a vector that's normal to the area right it's perpendicular to this plane of the area so if I'm very careful here I should be able to draw a vector that represents hat okay and now I hope from the geometry here we'll be able to figure out what the angles are that we need okay so I have an angle yet so I have some triangles that I've written down all right well one triangle here so there's an angle theta here what's this angle here back at the 90 degrees pi what yep okay so then what's the angle here okay I'm gonna have to yeah okay 60 degrees because that we were given theta equals 30 so in our in our specific example this is going to be a 30-60-90 triangle or in general I can write down over here PI over 2 minus theta okay can you kind of see that there all right and then the angle here since this is a normal is PI over 2 and then right here this angle right here has to be what I've got a PI over 2 minus theta I've got a PI over 2 what's gonna go right there it's got to be theta right or if you go back to thinking in terms of you a 30-60-90 triangle since theta was 30 here there's a 30-60-90 triangle here the angle here must be a gun 30 degrees was that too fast you got it okay all right so now I've found my normal that's what I needed to draw this diagram for and I need the electric field projected onto the normal because I need to solve the problem I need to take a dot n that's that's step one what's e dot it into n hat I needed to find the unhatched and know what the normal the angle was between them so this diagram shows me that the angle between the electric field and that normal to the area is Theta the same the same theta equals 30 degrees so then I'm going to get magnitude of electric field times magnitude of n sorry times cosine of in this case it's 30 degrees what's the magnitude of n yeah it's wine it's just set up that way so the magnitude of e and then times this guy is going to be 1 and so I need a cosine of 30 now we'll finish that in just a second but I want to show you that this this other way that was mentioned of I could go through this diagram and say the projection of Aeon end that's what the math formerly says it's actually equivalent to your physical intuition your physical intuition was well what if I do this a different way and I think of I've got this area and I've got an electric field coming in this way I could rather than taking the projection of the electric field onto the normal okay I could take the projection of this area onto something that's perpendicular to the field and it's geometrically equivalent so use whichever way you like okay and what you'll see is that if I take the projection of the area back onto here I'm going to get a cosine theta whereas if I take the projection of each I don't and I'm also gonna get a cosine theta okay so I get that I get the same cosine either way do you have any questions about that part about the geometry okay all right so now that we see where the normal is we see that we're going to get this guy let me show you a cool trick by the way if you haven't seen this trick before maybe it'll help you remember cosines I'm gonna set this aside for a second because I need cosine of 30 I know you just look it up on Wolfram Alpha but there's a cool trick so that you won't have to look it up anymore just in case you know you're stranded on a desert island and you have to calculate the cosine of 30 degrees and you don't have access to Wolfram Alpha until the satellite passes by all right so here's the unit circle there's only a few angles that that are worth memorizing on here okay so here I have this guy's PI over 2 up there right and then this guy is what PI well PI over 4 PI over 6 this guy is PI over 3 the PI over 6 corresponds to our 30 degrees and I can think about the cosines as the projections along the x axis right that's why we draw the unit circle so cosine of theta equals x component all right now I want to remember what these values are for the X components and it's all over - yes okay so you start off with square root of 0 over 2 square root of 1 over 2 square root of 2 over 2 square root of 3 over 2 square root of 4 over 2 you see the pattern square root of something over 2 let's go to 0 1 2 3 4 so it actually once you kind of get that trick in your head you don't have to look it up right so now I can kind of read off my diagram and say well I needed this guy our theta was 30 degrees so my cosine of 30 is going to be square root of 3 over 2 which I know you either already had memorized or what look up and Wolfram Alpha but that's kind of a fun trick in case you forget what it is so we're back to here cosine of 30 the square root of 3 over 2 so I'm ready to finish the problem I need the sum of e dot NDA so flux equals the sum over these en hat Delta a the area okay and so we've already got the e dot n okay so that's magnitude of e cosine of 30 and times the area and so our electric field was 5 Newton's per Coulomb cosine we said was square root of 3 over 2 in this case the area is 1 meter times 0.2 meters is 1/5 so 1/5 meter and then I get that the fives cancel and I get altogether square root of 3 over 2 Newton's meter squared per Coulomb that's the flux in that case okay do you have any questions about how that went all right and you see the way this cosine comes in you can either directly take u dot it into N or if you prefer you can think of projecting that area back on to something that's perpendicular to the electric field okay and different cases will make sense in different physical situations any questions all right so now that we've calculated a flux and have thought about what what a flux means Gauss's law is about a flux through a closed surface okay so Gauss's law says all right let me take that that flux we had defined flux as the sum of e dot n over Delta s and what we're gonna think about then is taking whatever area you like and dividing it up into little boxes ok those are the Delta A's and we're gonna sum over all the box and in the limit that all the boxes get teeny-tiny this becomes an integral we simply use a new symbol for it integral e dot NDA okay now the circle on the integral if you haven't seen that before that just denotes a closed surface when I'm thinking about fishing nets I'm usually thinking about something that's open okay but in the case of Gauss's law when I think about a surface that's closed and you can calculate flux through an open surface and you can calculate flux through a closed surface so in this case galsses law we're thinking of closed surfaces and all that means is that the surface comes back on itself okay it has no boundary so this will be the electric flux through some closed surface think of like a bag encapsulating something or a sphere or something like that or a box something that's closed and we find that the total electric flux poking through that closed surface tells you the charge inside so I get the sum over the charges inside of our epsilon naught so this is the statement of Gauss's law let's go through a little bit of the physical intuition of what that means and I hope you'll see that the statement of the law is actually very reasonable according to what you already know so let's say for example we take the first situation where let me have a point charge okay what does a point charge do well a point charge puts out an electric field all throughout space that's kind of a starburst kind of shape it goes like 1 over 4 PI epsilon naught Q over R squared pointing away from the charge if it's a positive point charge now let me enclose this thing in a box okay or maybe you want to enclose it in a sphere either way put some closed surface around it so what Gauss's law would tell you is that all right if I take the integral of the electric field the flux the electric flux through that closed surface and add it up all over the entire box it'll end up equaling Q over epsilon naught where Q is the total charge inside so I hope that's I hope that's not too far-fetched of an idea because for example in this case let's let's put a positive charge inside that box if all you had to measure was outside the box okay let's say you can't see in the box but the box is like some cardboard box so the electric field just pokes right through it and if you measured the electric field and you found it was a starburst shape that looked very like a much like a point charge you'd say look there's a point charge in the box or if you had it to do by Gauss's law Gauss's law would say well don't walk around and measure the electric field directly what are you gonna do is go sit on the you want to take this integral all over so what you're gonna do is go sit on the top of the box and walk along the surface at the top of the box and you're gonna take the box and divide it into squares so on any given square you're gonna measure electric field dot it under the normal of the square times the area of the square and set that aside and walk to the next square electric field dotted into the normal of that square times the area that's square you're just gonna sum it up all over the whole top of the box and you'll get a positive contribution poking out of the top you'll do that for all the surfaces of the box and what you'll find is that the entire flux coming out of the box equals 1 over epsilon not Q inside I'm not proving this to you by the way I'm just stating it for you and we're going to show that it actually makes sense in the case of a point charge and then we'll generalize that point charge case to other cases so this is very much like the case of that faucet or the source of of the fountain right in the case of the fountain we saw that water was spilling over the edge of the bowl and so if we think of a net being along the edge of the bowl catching that water spilling out we'd have to conclude there's a source somewhere inside so we'd have a net flux out of the bowl and we conclude there's a source inside here in the same way that if I have a net electric field flux poking out of a closed container I must conclude that there's a charge inside of the container does that make sense given what you know about the shapes of electric fields coming off of point particles okay you have any questions so far all right okay so I can think of a negatively charged particle so let's say that I had a negatively charged particle inside the box then you know that the electric field lines would point towards the negatively charged particle and I could do the same calculation walk along the surface of the box on any given tiny square on the surface of the box I'll calculate e dot n in this case e dot n is going to be negative okay how do I know that what I have a closed surface then everywhere on that surface n is a normal that points away from the inside okay so so points outside so when I take the electric field here dotted into n if n is pointing towards the outside and E is pointing towards the inside Edina and there's a negative number so all along this box here I'll get negative contributions summed up and I'll find that there's a negative charge inside the box okay this is an in the water analogy this would be very much like like observing a sink where you can see that the water is draining out of it okay so if the water is draining out then I would have net O net water flux going into into that sink there going into the drain so this tells you you know it's this analogy that is the origin of the source and drain or source and sink language that you often see in electricity and magnetism in this case here let's think of of having no charge in the box so now I've got this cardboard box with no charge in it but let's say there's an electric field around to measure anyway and the electric field just goes straight through the box so let's say I have a constant electric field all pointing along the x axis and I just put the box somewhere in space where there's no charge inside what I would measure then is okay in this case of a constant electric field going to the right any of those areas that are parallel to the field will give me no contribution alright but this side to the right and this side to the left those will give me contributions and on this side here I'd have that the electric field is poking out of the box where the electric fields poking out I get a positive contribution on the other side where the electric field is pointing in I get a negative contribution the reason I get a negative contribution when the electric fields poking into a box is remember all along a closed surface and points out so endpoints out of the box over here it's in the opposite direction to the e so e dot n is a negative number and so the contribution here is positive the contribution there is negative they cancel so what I'd find if I take Gauss's law the sum of the flux over this cardboard box in this case I get zero and that would tell me that there's no charge inside which is the same situation as watching this fountain here where I can see the water flowing I see the water flow out so if I measure the flow rate through you know an imaginary bag inside here I'd see that it flows in it flows out and there must be any source or sink inside that spot do you have any questions about that or the analogy okay all right does no questions mean it's clear or does no questions mean it's clear as mud and we should spend more time in it okay feel good go like this if you want me to slow down go like that okay all right getting a lot of thumbs up all right so what's apply Gauss's law notice I did not prove Gauss's law to you okay I'm just telling you what it is giving you some physical intuition about it and I'll show you how to use it so Gauss's law we already stated Gauss's law but we want to look at how it works for a point charge okay so in the point charge case what I would think about is well I know what the electric field is for a point charge it's 1 over 4 PI epsilon naught Q over R squared times R hat okay if Q is positive the electric field points outward if Q is negative the electric field points inward and Gauss's law would say I need to now measure the flux I need electric flux integrated over a surface so I need the electric flux dotted into the normal of a surface and then integrate that all over the surface so let me think of a point charge and let me think of drawing a sphere around it I've drawn several spheres here but let's think of the outer sphere okay so think of an outer sphere of radius R centered on that positively charged particle in that case if I think of taking this integral surface integral of e dot NDA all over that sphere well what do I know if I've got a sphere that's centered on the charge okay so let the sphere be of radius a then anywhere I look along the sphere the magnitude of the electric field is going to be the same because I'm all at the same distance R away from the from the point charge okay so the magnitude anywhere I look is going to be according to this equation it'll be 1 over 4 PI epsilon naught Q over R squared that's just the magnitude and you see that's here 1 over 4 PI epsilon naught Q over R squared that's the magnitude I need to think now about dotting that into the normal so think of this sphere alright and when I'm standing at the top of the sphere which direction does the normal point if I'm right here at the North Pole X Direction the normal point yeah okay and when I'm over here the normal points out and when I'm down here at the South Pole the normal points down and it turns out that that's the exact same direction as the electric field right so I've set up the what's what's nice about the mathematics here is that I've set up the the Nets of Gauss's law to match the physical situation okay I've put a sphere on it and everywhere on this sphere the electric field is poking straight out and the electric field is normal to the sphere anywhere I look okay so that means that anywhere I look the magnitude of the electric field is the same and anywhere I look on that sphere a dot n is the same number does that make sense this is actually the crux of how to do Gauss's law how to apply Gauss's law to a real problem the the entire well kind of the fun physics problem there is trying to find a shape where that's the case trying to find a shape over which you can calculate Gauss's law in a in a pretty simple manner so I have them that edad n is the same everywhere on the sphere and in fact that it's a constant okay when I'm thinking of what does this integral mean the integral over the area the integral over the surface area literally means take your area divide it up into little boxes so tile it with boxes and add up the contribution in one box go to the next box and add up the contribution and so forth so I'm going to move along the along the area that way and all along that surface again we already said that a dot and is a constant number all right and that this is it's magnitude 1 over 4 PI epsilon naught Q over R squared and now as I walk around the entire surface area the sphere the only thing left to add up is the areas in all the boxes and if I add up the areas on all the boxes and sum up the entire area what's the surface area of a sphere it's not PI R squared and it's not 4/3 PI R squared all right it's ok it's 4 4 PI R squared is the surface area of a sphere thank you front row ok so 4 PI R squared that's the total area that I'm going to need and you see what happened in this case we were thinking about this largest sphere here notice that the R squared here that's the same as the R squared there right I was sitting on this particular sphere the magnitude of the electric field at that sphere is down by one over R squared but then the area itself puts back in a contribution of R squared so R squared cancels R squared and the four pies cancel and I get altogether Q over epsilon naught so this shows you what Gauss's law means for the point charge and when I take an area that's spherical okay and notice that because the R Squared's cancel it didn't matter which sphere I was pointing to I was thinking of the largest sphere but the math applies equally well to the next fear and it applies equally well to the next fear and so forth so for any sized sphere centered on this this is all being very reasonable do you have any questions so far okay yep all right now here's what's kind of fun to think about this cancellation of the R Squared's that's kind of I don't know when you see something simple like that happened in physics we always always look for a deep reason like if you do a really long calculation and you find out that the answer is 1 you go asking well was there a way to do this without the long calculation okay so in this case the R is canceled and we think there's something very deep going on there we think that in fact the fact that as I get away if I think about the magnitude of the electric field as I get away from a point charge that electric field falls off like 1 over R squared okay we think that's very much related to the fact that those electric field lines get distributed over an entire sphere of surface area 4 PI R squared that those are squares in fact are related ok and we think that this is probably why gravity itself follows a 1 over R squared law as well so we think there's probably something very deep there associated with the geometry of this universe ok as far as why the electric field falls off like 1 over R squared and you can define analogously gravitational fields and see that they also fall off like 1 over R squared that's just kind of a fun fact there any questions so far all right so on each sphere the electric field went like 1 over R squared and on each sphere the surface area went like R squared those cancel out to give me Q over epsilon naught all right so now I'd like though to see if we can they about this in a shape that's not a sphere right I showed you it works for the sphere but at the same time Gauss's law is much more general than just the easy case the sphere that I gave you Gauss's law says this will work for any surface area that encloses that charge so we did the sphere all the spheres were easy but what if I had a box or what if I had some funny Wiggly shape okay ya'll says law says it doesn't matter think of this as like your fishing net remember we started the class by talking about a fishing net and how much water was flowing through it and even if the net deformed one way or the other the flow rate through the pipe was the same so in the same way the flux coming off of this charged particle is the same and it doesn't matter if I take my Gauss's sphere think of it like a net if I deform that net a little bit I'm gonna catch the same amount of electric flux no matter what the shape of the net okay so that's kind of weird let me show you mathematically how that's going to work out this slide got a star by the way because it's really important so the star means you know I post the lecture notes online and so when you see the stars on one of the slides that's a really really important slide to go back and study so I'd like to build up to the case let me show you where I'm heading I'd like to build up to the case where I'm gonna be able to take any funny shape okay I want to surround this charge by any shape I like and and see how Gauss's law is gonna work so here's how we're gonna get there we saw that we could take any sized sphere and that worked but let me think about a segment okay so let me think about this angle right here it's kind of like pac-man now eating a ghost that's over here and so I've got a ray coming out in a ray coming out and I just want to think about these little segments of spherical shells okay as long as these are all in that same wedge the flux through that wedge is the same no matter where you look the flux through this wedge is the same if I measure it here or here or here or here does that make sense okay because as the electric field lines going as the electric field is decreasing in magnitude the area I'm sweeping out is also increasing in such a way that they cancel well that was a weird delay okay so in any segment contribution from any shell is the same kind of like a flashlight right so think of this is like shining a flashlight and if I'm shining a flashlight it doesn't matter whether I look up close or far away if I think of how many photons hit a particular spot okay and I move that spot away and so forth as long as I'm catching the same angle of the flashlight I'll catch all the photons right the photons coming out of the flashlight of the photons coming out and whether they hit a piece of paper that's close by or hit a piece of paper that's farther away I'll still get the same number of photons catching it does that analogy make sense okay so same thing going on here now so the flux through any of these segments of the shell is the same what about the flux along this lot instead of this radial line coming out okay what's the flux along that radial line yeah you have a question no just resting that's okay all right so if I think of having an area that is that's parallel to the electric field lines that's like trying to take a net and catch fish but I put the net the wrong direction okay everything just passes it by so if I think of some area that's along here I'm not gonna catch anything all right so we can think now okay so in any segment in any segment the contribution from any shell is the same kinda like a flashlight so I'm gonna break this up into several segments we have the red orange yellow green blue and so on all those different segments around the sphere and now what if I do this I'm gonna surround the charged with any shape by breaking it up into these segments what I'm gonna do is I'm gonna follow the spheres at different places okay so you see I've drawn this complicated blue line but there are places with a blue line follows one sphere and then comes back down and follows another sphere and then comes back down and fellows another sphere so as long as I set this up carefully all of my areas will either be lying on a sphere or the next area will be something that's completely perpendicular to that and catches no flux and goes back to the next fear and now I'm on a sphere again and then the next segment is perpendicular and catches no flux so I've set everything up in this geometry such that all the flux contributions are lying on the surface of the spheres all the other ones have zero contribution and now just think about making everything very finely grained right so in the same way that I can take any surface area I want and tile it with squares as long as I make the squares small enough or maybe you're used to telling things with triangles and said I can make any shape I like by making these spheres close together and just going to very small bits okay this is just the idea of calculus that I'm going to make any shape I like by adding up all these small contributions so really I could make any shape I like right and I would just have to take the limit then that I have spherical shells that are very close to each other and I'm gonna hop on and off each shell in a very fine-grained manner I don't know why that advanced again all right so in the limit then gives me the case rec and consider any geometry I want the limit of small segments works for any smooth shape and then the flux through the outer surface the trucks to the surface is the same so it works for any shape I like alright so Gauss's law then says take that flux over any enclosed surface and count up the flux and you'll always get the sum of the charges inside over epsilon not any any questions about that okay all right that's actually as far as I wanted to get today so we're done for today and I'll see you guys on Wednesday

Channel: Prof. Carlson

Views: 1,627

Rating: 4.8571429 out of 5

Keywords: iMovie Physics

Id: ULWIQvF-muo

Length: 44min 23sec (2663 seconds)

Published: Sun Nov 13 2016