Gauss law logical proof (any closed surface)

if you keep a charge at the center of a sphere then we've seen that the electric flux the total flux to the sphere happens to be q divided by epsilon naught but what's amazing is that even if this was not a sphere and you had kept a charge in some random shaped surface as long as it's closed the flux through that would still remain the same so the flux through any closed surface would still be q divided by epsilon not regardless of where the charge is kept and in general if there are multiple charges then the total flux will be just the sum of all the charges inside the closed surface divided by epsilon naught and this is what we called the gauss's law so the goal of this video is to mathematically prove that this is true for not just spheres but for any random shape so where do we begin i think we can begin by recapping how we proved it for a sphere we said hey flux through the entire sphere can be calculated as flux phi equals the product of electric field times the perpendicular area at that point and of course we have to integrate this over the entire surface but because the electric field due to a point charge everywhere over here is the same and because the area everywhere is the perpendicular i can just multiply them and if i do that the electric field is 1 by 4 pipes not q by r square so q divided by 4 pi epsilon naught r square where r is this distance or you can imagine the radius of the sphere and the perpendicular area is just the area of the sphere the entire sphere is perpendicular so the entire area of the sphere is 4 pi r squared and as a result this cancels and guala the flux through the sphere happens to be q divided by epsilon naught and this is a quick and dirty derivation but if you need a more detailed derivation we've talked about all of that in all these nuances in our previous video on gauss's law so feel free to go back and check that out but now how do we prove this for any random shape because now electric field everywhere would be different and the area is not nice and area is all crooked so it's going to be a nightmare to calculate it right so how do we do it there are probably multiple ways to prove it but one of the ways i love is what i found in richard feynman's book and the method was as follows even though we're now calculating the flux through this big surface let's still imagine a tiny sphere of any radius you want which is centered at that charge now we know the flux through this sphere is q by epsilon naught we know that now all we have to prove is that the flux through the big surface is exactly the same as the flux through the sphere see what i'm saying if we can prove that the flux through this big closed surface is the same thing as the flux through this sphere then we are done then we can say that should equal q divided by epsilon naught so how do we prove that for that we'll divide the sphere and the big surface into teeny tiny pieces like this so i'm going to show you how we'll divide it so let me draw a couple of dotted lines over here so let's say this is one of the lines i'm dividing this into tiny piece now and this is another line so we'll find one tiny piece over here remember this is in three dimension this is a surface so you'll see a circle over here and there's another tiny piece over here okay and you imagine that these two lines are very close to each other this angle is almost zero so you imagine these are like almost point you know it's like very tiny surfaces now all we have to do is prove that the flux through this piece is exactly equal to flux through this piece if we can do that we are done because then i can do the same thing over here and i could say hey flux through this piece by the same argument would be flux through this piece then i can say hey flux through this piece will be the same as flux through this piece and as a result i can then say hey flux to the sphere is the same thing as the flux through this whole surface and as a result the flux through the closed surface will be q divided by epsilon naught okay so you see how we are changing our proof it all finally boils down to proving that the flux through this piece is exactly equal to flux to this piece and so the question now is how do we do that well let's go ahead and try and calculate so what's the flux through this tiny piece let's call that as phi 1 only through this piece well that would be the electric field at that point which we know how much it is it's q by 4 epsilon not r square i'm just going to write it as e for now so that's going to be e times the perpendicular area over here now since this is a sphere we know that spheres are always perpendicular to the radius electric field is along the radius so this area automatically becomes perpendicular so this entire area if you call that as a the entire area becomes perpendicular and as a result the flux through this piece is just e times a now let's figure out what the flux through this piece is and let's prove that it will also be e times a okay all right how do we do that so flux through this piece let's call that as phi 2 that's going to be the electric field here multiplied by the area perpendicular over here but what's the electric field over there we know electric field is going to be radial this way but that electric field is going to be less than e right because we have gone farther away but how much less i want to compare this if i call this electric field as e dash i want to know e dash i want to compare e dash and e so how do i write e dash in terms of e well one way to do that is we can look at the distances so let's say this distance is r this distance is r and then let's say this distance from here to here is n times r and remember because i'm because you know this is a very tiny piece this distance and this distance and this distance is all same they're all just n times r okay so here's my question now if this is r and this is n times r if this electric field is e what will be e dash can you pause the video and think about that what will be e dash in terms of e it's going to be e divided by something can you pause and think about what what what that will be all right so e dash is going to be q divided by 4 pi epsilon naught n r squared and so there will be an n square and so the whole thing since this whole thing is e it's going to be e divided by n squared does that make sense so electric field over here e dash is going to be e divided by n square if you think about it that kind of makes sense because electric field goes down as one over r square if you increase the distance by n the electric field decreases by n square this is the whole one over r squared law so electric field has reduced one n square now the final question is what's the area over here how do we calculate that so at first we'll be like okay how do i calculate this area how do i know how much this area is but remember i don't care about this area i care about the area perpendicular and this area in fact is not perpendicular i know it's a little hard to see but you can see it's kind of like slanted and so this is really not perpendicular to the electric field so i need to calculate and let me try and draw that now i need to calculate the area that is perpendicular so you can imagine it's a component of the area it's a little hard to see let me use a little lighter color so let me use pink in fact let me zoom in a little bit okay so let's say this is the component of this area which is perpendicular to our electric field all right so this is our area perpendicular the question now is how much is that area compared to this area at first it looks like things have become more complicated now we have a perpendicular component and all of that but if you think about it it's actually become easier it's easier to calculate this area because if you look at if you see carefully you can see this area and this area and this charge together form a nice cone and we know this distance is r and we know this is n times r so using geometry can you figure out if this area is a how much will this area be it's going to be bigger than a but how much how much will it be can you compare again this is the last time i want you to pause the video and see if you can try this on your own again what would be this area perpendicular all right let's see let me redraw that somewhere down over here big let me zoom out so that we can see properly so to calculate this area let's first concentrate on the diameters because if i know the diameters of the radius then i can calculate the area so there's one diameter over here let's consider the diameter over here and now actually i see two triangles and if you look at it they're similar triangles because they have one common angle they have this corresponding angle and this corresponding angle oh this means the sides are in the same ratio so if this is n times this side this diameter should be n times this diameter or this radius should be n times this radius so if the radius is n times more than this what can you say about the area well area is pi r square so this will be pi r squared this area would be pi n r whole square or will be n squared times pi r square or will be n square times a so what we find is that this area this area is n square times this area and now you can see right in front of your eyes something absolutely beautiful you're seeing that the n square cancels out i've never been happier about canceling things in my life before and as a result the flux through this is exactly equal to the flux through this piece why is it happening that way because when i go farther away the electric field goes down as one over n square but the area increases by n square and so the product stays the same and since the flux here is exactly the same as flux here this means the flux through this entire sphere should be the same as the flux through this surface and that means the flux through that entire closed surface is also q divided by epsilon naught hence proved but wait before we celebrate remember gauss law works for multiple charges um how can we prove that what if there was another charge over here how can we prove that the flux now is q1 plus q2 by epsilon how can how can we prove that well for that what we can do is we can individually calculate the flux due to q2 and for that we can draw another tiny sphere over here and we can prove same way that the flux through this surface has to be q2 by epsilon naught and then we can use superposition principle and you can say hey flux through due to both the charges must be q1 by epsilon naught plus q2 y epsilon naught so it'll be q1 plus q2 epsilon and we can do the same thing for all the charges regardless of where they are but finally remember gauss law says that charges outside do not contribute to the flux can we prove that as well so what if there is a charge q3 over here can we prove that the flux due to this charge is zero yes we can in fact using the same technique so what we can do is again we can divide this entire surface into tiny pieces this way let's say it divided like this and now you have two surfaces over here and now using the same idea we can prove that the flux entering here is exactly the same as the flux exiting here and we can do that for the entire surface same way by considering tiny tiny pieces and we would have proved that the flux entering this entire piece entire blue surface is exactly equal to the flux exiting it and therefore the total flux contribution from this charge would be zero and so even if there are 100 charges outside their contribution to the flux would be zero and now we have completely proved gauss's law that the flux through a closed surface indeed equals the total charge inside divided by epsilon naught this is easily one of my favorite proofs in physics