Gauss law of electricity

the goal of this video is to explore Gauss law of electricity we will start with something very simple but slowly and steadily we look at all the intricate details of this amazing amazing law so let's begin so let's imagine a situation let's say we have a sphere at the center of which we have kept a positive charge so that charge is going to create this nice little electric field everywhere right now the question we want to try and answer in and you know the question if you want to start with is what would be the flux through this sphere what is the electric flux through the sphere how do we calculate that well let me remind you how we calculate flux we define flux Phi as basically the integral of the electric field dot Da which means it's the integral of e da times cos theta and just to you know just to remind you because you've already talked about it but just to quickly recap the whole idea behind this is if we have some kind of an electric field flowing through some area let's call it as da then the flux is just the product of the electric field and the area alright that's how we calculate flux but then why there is a cos T Tower here well the idea is if you see from the fields point of view it can't see the entire area from the fields point of view the effective area that it's flowing through is only this much the perpendicular component and so if the angle between these two is Theta the note is the effective area through which the field is flowing is da cos theta from this right angle triangle is the hypotenuse this is the adjacent side this inside becomes the cost component and that's where the cos theta comes from and again why do we write it as e dot Da well we like to treat da the area as a vector by assigning an arrow mark perpendicular to it and if you did that notice if this angle is Theta this angle also becomes theta again this is not something new I'm just recapping what we all learned before okay and if you do that then we can say theta is the angle between electric field and the area vector and then we can say ah this is the dot product of electric field and the area vector but of course if you're not familiar with this we great idea to go back and watch this video on Elektra it's called vector form of electric flux or something like that okay so using this our first goal is to figure out what is the flux through this fear and so what I'll do is I'll do some of the work and then you'll do some of the work so let's start by zooming in so we can nicely see this so what I'm gonna do is I'm gonna take a very tiny piece of this particular sphere over here and let me call that tiny piece it's area to be da and let's imagine that the electric field at that point is e the question is what is the flux through the entire surface so let let me first look at what's the flux over here what is the angle between the electric field and D let's think about that first well because it's a sphere we might know one of the properties that the surface will always be perpendicular to the electric to the radius and since the electric field is along the radius this da is perpendicular to the electric field so what's the value of theta it's not 90 because remember theta is the angle between electric field vector and the DA vector da vector is this way right I mean you can draw a DA vector in words also but the convention is we always try to we always draw you know da vector outwards for a for a given closed surface so now what is the angle between the today in the same direction so theta becomes zero okay and that's not just here that would be true anywhere so if I were to take areas anywhere you will find that the theta will always be zero always the electric field will be in the same direction as that of the DA vector so that's nice that means that everywhere let me just write that down everywhere this is zero that's great because cos 0 is 1 that's one simplification another simplification that happens is notice that the electric field we'll be the same value everywhere on the surface why because it's the point on the surface is that the same distance from the charge R and therefore the electric field value should be a constant right so this is a constant and that's also great because constant numbers can be pulled out from the integral so let's see what happens to our integral now so if I were to calculate flux Phi will be the integral I'm going to pull out the electric field because it's a constant times da integral of DA times cost 0 which is 1 ok so now it's your turn I want you to try and find out what the final simplified version of this is because you already know what the electric field is you know how to calculate electric field due to a point charge at some distance R and I'm pretty sure you can try and calculate what the integral of D is going to be so please give this a shot and see what you get ok if you have tried let's see so the flux is gonna be e what is e electric field at any point you to a point charge is going to be from Coulomb's law Q divided by 4 PI epsilon naught R squared right and what's the integral of DA well what is the meaning of integral of DA you're basically adding up all these tiny pieces over the entire sphere so that'll be the total area a but do we know what that area is what the value of the area is yes the area of the sphere is 4 PI R squared and so now we can simplify we can say look R square cancels 4 pi cancels and what we now end up with is flux Phi equals Q divided by epsilon naught that's our answer so if the charge is positive we get a positive flux meaning it's an outward flow if the charge was negative we would have gotten negative flux meaning it's an inward flow now this party might say okay Mahesh so what's the big deal why this is just some problem that you have solved right so what's the big deal this isn't hair it gets really interesting okay I'm gonna start asking a bunch of questions and we'll go deeper into this concept so to begin with let's first zoom out and here's my first question what would have happened to this value of flux if our sphere had a size of two our radius of two are doubled can you pause and think about the answer alright if you're tried let's see notice that our flux is independent of the value R there is no R in our answer that means it does not depend on the size which means the flux value remains the same but why is that happening well you can actually see the answer over here the R squares are cancelling out and what does that mean because see flux is the product of electric field and DA right so even if the area is increasing why isn't the flux increasing well the main reason for that is that when the radius doubles sure the area increases it goes up as R square so when the radius doubles your area becomes four times as much area becomes four times as much but what happens to your electric field it goes down as one over R square so our electric field because now the distances are also doubled our electric field becomes 1 by 4 1 over R square right and so when you multiply notice the effects cancel out and the flux does not change so the reason why the flux is independent of the size of our sphere is mainly because of the inverse square law if the electric field follows some other law let's say inverse cube law or inverse any other power this wouldn't be true so truly because of the inverse square law you're saying the flux is independent of the size okay now next question is what if I moved the sphere a little bit to the right or let's say I move the charge a little bit to the left off-center that's important now what will happen if you calculate the flux through the entire surface will the value remain the same now or increase or decrease this is a little bit more interesting so again all the video and give this a try okay now we need to read arrive it because it's no longer off center so this derivation will not work anymore okay does the first thing now if you had to read arrive see the problems first of all can we say the electric field is constant everywhere no because here we are closer to the charge so stronger electric field and as you go from here to here weaker electric field so electric field is no longer going to be a constant so this will not be true anymore what about this value of theta well if you were to draw again these tiny tiny DEA vectors notice that the angle is also no longer zero it will keep changing which means this is also not true so can you see how complicated this integral is going to become now we can't pull out that II that theta will not be zero oh man it's gonna become a little complicated now so we might think that the answer could be different right because we'll get a completely different derivation but what if I told you the answer remains the same and this is the mind-blowing part this is the most important part that even if I move the charge off-center anywhere the flux remains exactly the same but I'm pretty sure you'll be wondering how how is that possible can you prove that to me yes I can prove it to you there's a beautiful proof but I want to do that in a separate video because I'm gonna do it slowly it's a very logical proof I wanna do that separately but the flux remains the same now although I won't prove it to you but I want to give you some imagination ok and so the way to imagine this is instead of thinking of this as an electric field imagine that this charge is shooting out bullets like this and let's say when the sphere was kept this way when the charges up at the center let's say the number of bullets that are coming out 1 2 3 4 5 6 7 8 9 10 11 or 12 whatever that is let's say there are about 12 or 11 bullets are coming out for a minute through the sphere okay when the charge is at the center now here's my question what if I move that sphere a little bit to the right now how many bullets will come out for a second or sari per minute well it has to be the same isn't it it's still shooting 12 billet bullets every minute and those 2 al bullets have to come out through this fear I mean imagine is an imaginary listen I put it goes fear so the bullets will not get stuck anywhere ok put the answer should remain the same isn't it the floor has to remain the same of course you might say that over here the bullets will reach quicker and it will take longer time for the bullets to reach here sure but what a minute if I ask you how many bullets are moving out of the surface it has to remain the 2l isn't it so the bullet flux remains the same in a similar manner although not exactly this here but in a similar manner even the electric flux remains the same and you'll find it again due to inverse square law and now we can go even further we can ask why sphere what if I got rid of my sphere and took some other random surface what will be the flux of these bullets it should still remain the same so you see we derived this for a sphere but now we are saying that doesn't matter what kind of surface we have and doesn't matter where the charge we keep it inside that surface our flux will remain exactly the same here will be the case for the electric flux as well but of course there's a small detail over here what if this particular here you know this particular thing had a hole now notice some of the bullets can leak through that hole now the number of bullets passing through the surface will reduce which means there should be no holes in our surface in other words our surface must be closed so as long as we have closed surface the flux through it it doesn't matter what surface it's the most general case now has to equal the total charge sorry the charge inside divided by epsilon naught all right we can go even further and ask what if there are more than one charges what if there's another charge kept over here well think in terms of bullets you just add it's bullets as well so if this is charge Q 1 and this is Q 2 will be the value it'll be Q 1 plus Q 2 divided by epsilon not so if there three or four charges just add all the charges inside and divide them by epsilon not for a last question you might have is what if there's a charge outside will it also contribute to the flux so let's imagine a charge which is outside what do you think what will be the flux due to just discharge well let's see it's so exciting okay so if you again look at the bullet principle now you'll see not all the bullets will enter but whatever bullets enter this particular surface they will give you a negative flux and the same bullets will also exit that surface giving your positive flux notice that means the negative and the positive will cancel out that means no net flux due to this charge can you see that and now it doesn't matter where you keep this charge how strong that charge is even if it's shooting thousand bullets inside all those thousand bullets will exit some make sense meaning this only when you're calculating the flux you only calculate the flux due to the charges inside the external charges will not affect the flux bullets are there but when you're calculate or tell they just the inward and the output part contribution of it cancels out so now we can write this equation in all its grand glory so let me clear this let's write this in all its glory so this is Gauss law now so Gauss law states that if you calculate the flux through the closed surface and the way we write it in the integral is we put we put a circle over here to represent that we are calculating flux through a closed surface it doesn't matter what kind of surface that is doesn't matter at all that value should always always always equal the total charge inside that surface total charge if you want you can also write as Sigma if you want Sigma whatever total charge inside the surface divided by epsilon not I don't care about the charges outside even though they do contribute to the electric field they do give us bullets they do not contribute to the flux okay and so in all its glory this is the most general Gauss law for us so again the mind-numbing part for me is that we derive it for such a simple case the charge at the center of a sphere but look at this it is the most general case of electrostatics