Field due to infinite plane of charge (Gauss law application)

suppose we have a plate full of charge an infinitely big plate full of charges the question is what's the electric field going to be everywhere that's what we're going to figure out in this video so let me show you the same thing for from a side view so we have an infinitely big plate you have to imagine that even they have not drawn that and we need to figure out electric field everywhere so let's start with the specific point let's say we want to figure out what the electric field at some point at some distance r from the plate is going to be how do we do that the first question you might have is why do we want to care why do we care about infinitely big plates i mean is that practical well even in practice we may not have infinitely big plates we might have finitely big plate but then if you were to figure out electric field very close to it very very close to it we can assume the the plate is infinitely big so whatever we get over here we can use that values for very close distances so you can assume in practice what we are doing is finding the electric field very close to big plates okay if you go far away we can't use that but as long as we are close enough we can definitely use it so how do we do that well we can start with coulomb's law which you might be familiar with says electric field due to a point charge is q divided by 4 pi epsilon not r squared but by now you might you might appreciate that you can't directly do that but you'll have to break this up into tiny tiny pieces and then calculate electric field you to each piece and then add them all up and that's going to be a nasty integral which we're not going to do so we're going to go for coulomb's law but instead you know you might already guess we're going to use gauss's law and the whole idea behind why we can use gauss's law over here is because the electric field is going to be very symmetrical as we will see and because the electric field is symmetrical we can find a closed surface such that the electric field everywhere on that surface will be the same and so we can pull it out of the integral and then we can evaluate this expression without having to integrate and calculate what the electric field is going to be that's the whole idea behind it and if you're wondering wow that's amazing can we do that for every single problem no we can only do it for three special cases one is this one infinitely big plane the other one you may have already seen infinitely big line of charge and the other one is when we have a sphere of charge these are the only three cases where we can use this okay so this is one of them so where do we begin well we start by figuring out what the electric field looks like everywhere to to to apply you know to apply houses learn to choose a closed surface the first step is that so let's start over there how do we how do we calculate how do we figure out what the electric field looks like everywhere the steps are going to be very similar to what we did with the infinite line of charge so if you need a refresher of that credit to go back and watch that but what we do is because you want to use gauss's law the first step is to know what the direction of the electric field is everywhere figure out that based on symmetry and here's how we can do it let me first look at it from the side so i can see it nicely same thing i'm looking at from the side and what i'm going to do is i'm going to draw i'm going to divide this plane this sheet into two halves along this line okay along this line this one and i can say that the top part of this sheet is exactly equal to the bottom part of the sheet because it's uniformly charged it's infinitely big they are exactly same and so they are mirror images of each other okay what can we say based on that based on that we can guess what the electric field looks like over here how see here's how i like to do it first start with some arbitrary direction let's say electric field is over here this way now i can say that's wrong because why would the electric field point upwards because the the top part and the bottom part is exactly similar so why would the electric field point upwards there's no reason for that so for the same reason electric field can't point downwards electric field point cannot point this way so the only way electric field can be pointed is it's neither pointing upwards nor pointing down the only possibility there are only two now either it has to be towards the you know towards the right or towards the left and since we know this is positive charge we can guess that should be away from the plate and so it has to be the electric field over here needs to be towards the right what an amazing argument right just from the symmetry argument but we don't we don't just stop there remember point p was an arbitrary point i chosen that point could have been over here and i could have made the same argument i could have divided into two parts and remember this is infinitely big so whenever i divide it into two parts i will always get two halves the top half equal to the bottom half and so i could make that argument everywhere and therefore electric field everywhere at least over here somewhere on this line everywhere should be towards the right and over here everywhere towards left and not just that since the this point is very similar to this point there's no difference between these two points right i mean you can kind of say that every point is you know i'm i'm looking at the center of the sheet anywhere you go because the sheet is infinitely big i could say there is no difference between these two points and so the electric field here and here should also have no difference because absolutely no difference from these two perspectives so i could also say not just the direction but i can also say electric field everywhere over here must be exactly the same everywhere over here must also be exactly the same in fact if you go at a distance r anywhere you go top or bottom or or out of the screen or into the screen wherever you go the electric field must be the same at a distance r does that make sense that's our that's our symmetry argument so if i were to look from here just to make that more clear we could say that if i have to take a plane parallel to our given sheet anywhere on that plane the distance is the same from the sheet right i'm taking parallel and so everywhere on that plane the electric field must be the same any plane you take parallel to the sheath electric field must be the same does that make sense that's our symmetry argument so now comes the question now that we know this what kind of gaussian surface would you use would you choose to use gauss law okay i want you to pause the video and think a little bit about this should be a surface such that that integral becomes nice like nice and easy here's gauss law again the integral should be nice and see nice and easy so what surface would you choose pause and give it a shot all right if you're giving this a shot let's see my first instinct is that whatever surface i choose needs to be flat in front of it or behind it why because we already saw such flat surfaces parallel surfaces will have same electric field all over it and that we can use to our advantage the second thing is whatever surface i choose it needs to go through the sheet it has to pass through the sheet only then i can enclose some charge so putting these two together the surface we can choose is a cylinder same thing if i show from the side view the cylinder would look somewhat like this should have the same length on both the sides are on the right and r on the left as well so now we can use gauss law we can equate the left hand side we can simplify the left hand side simplify the right hand side and go ahead and calculate it and so again before i do this great idea to pause and see if you can try this yourself because there's nothing new we've all studied about flux and we've done this for infinite line of chart so it'll be great idea to pause and really really really try yourself first all right if you've tried let's see so let's start with the flux what's the total flux through the entire cylindrical surface well i can find three distinct surfaces one is the front surface which let me draw that over here the back surface and the curved surface right let's start by drawing the let's calculating the flux to the curved surface how much would that be well notice the electric field lines everywhere over here is parallel to the curved surface right everywhere it's parallel and when you're calculating flux you're doing a dot product and so the d a vector wherever you go the d a vector is going to be outwards here it's going to be outward so here it's going to be downwards right so what's the angle between the da vector and the electric field vector it's 90 everywhere wherever you go even if i take a tiny piece over here the d vector is going to come out and that's going to be the angle would be 90 degrees and so that means wherever you go on the curved surface this value is going to be zero electric flux is going to be zero and that kind of makes sense nothing is flowing through the curved surface no electric field is passing through the surface so the curved surface gives me zero flux so the flux only gives me a value on the front surface and the back surface so what's the value over there so let's con let's come to the front surface let's assume that the electric field over here is i don't know some value e we already know it's going to be this direction and since the whole area is nice and flat what would be the direction of the area the area vector again normal outwards oh notice area vector and the electric field vector are in the same direction same direction so when you do the dot product cos zero would be one and so this dot product will be just e into d a and this entire d a is my a and so if you do this you just get the flux over here as e into a so flux here would be just e into a that's the flux through the front surface and the same flux to the back surface the story is the same which means the total flux the left hand side would be 2 times e into a a being the area of that front surface i'm just going to choose that as a let's use blue a so that's our left hand side so that should equal the total charge enclosed what's this charge enclosed we didn't say anything about the charge let me i just i totally forgot about the charge okay so first thing is total charge is infinity right because this is uniformly distributed and this is infinitely big and so whenever we have such cases you know what one thing we can mention is we can talk about how much crowded the charges are so we like to talk about the charge density and since the charges are distributed over the surface here we like to talk about surface charge density and so let's say the surface charge density provided to us in the question itself let's call it a sigma it's given to us and think in terms of units just imagine in standard units it'll be sigma coulombs per meter square so that means i'm saying that every meter square of this piece has sigma coulombs of charge let's say that's given to us okay now given that what would be the electric uh what would be the total charge enclosed over here well i know that each meter square encloses a charge sigma but we have a meter squares this is a meter squares how much would that enclose so one meter square includes a sigma two meter square encloses two sigma a meter square would enclose a times sigma so the charge enclosed would be a times sigma does that make sense divided by epsilon naught and so now we can do the algebra the a cancels out and so the electric field turns out to be sigma divided by 2 epsilon naught tada we are done couple of steps no integration done that is our answer now before we close can do you see something interesting in this formula i hope hopefully you see something interesting the interesting thing is there is no r in the formula it's independent of r what does that mean independent of r that means the electric field does not depend on the distance regardless of how far or how close you are to the sheet the value is the same and as we saw that means electric field everywhere should have the exact same value uniform field that means the sheet of charge unif in infinitely long sheet of charge produces a uniform field and what does this mean for our practical case level like just we saw before that means if you have a charged plate an actual you know finite plate then as long as you're close to it somewhere close to it you can say that hey electric field is pretty much uniform somewhere close to it near the center you have to be near the center close to it you can assume it to be infinitely big and you can use this value but of course if you go far away then of course electric field dies off and this will be useful for us in the future okay so electric field due to a metallic sheet close to it or infinitely big sheet would be sigma divided by 2 epsilon naught