 Hi, I'm Zor. Welcome to Indizor Education. I would like to continue leading you to a lecture which is dedicated to capacitors. Now the previous lecture was about permittivity, the property of the material to resist propagation of the field. This is the preparation number one. The preparation number two is the problem which I'm going to solve right now. And it's based on the problem which we have already solved. It was problem number two in this chapter dedicated to electric fields. And the problem number two, well there are two problems A and B there, but the A means to determine the intensity of the electric field of the infinite plane charged with certain density. And we have derived the formula for this particular intensity of this field which by the way did not depend on the height above the plane. So on any height considering the plane is infinite, on any height the intensity will be the same. So I will mention the formula. Now I will take the formula which was basically used during that solution of that problem, I will take it here. Now the problem right now is not the infinite plane charged with certain electric density, but a disk. So we have a disk which is flat, it's infinitely thin, but it has a finite radius. And we are talking about a point which is above it, at the height h above the center of that field, on the perpendicular to a disk itself. Now as before in the problem two I will use sigma as the density of electricity. So basically the electricity in any part, in any area of this disk is basically an area times sigma. Now during the derivation of the formula in the problem number two the technique which I was using was let's just consider a ring around the center from radius r to radius r plus gr infinitely thin in the horizontal dimension and obviously infinitely thin vertically because the whole disk is infinitely thin. Now I was arguing that from any however small area here is the intensity of the field here would have exactly the same magnitude as in any other area on this disk. Just because, but obviously if areas are the same, just because the distance from p to this to this to this to this will be the same, this is a ring around the center, right? So all these lines have the same lengths. So this r divided by r square remember this Coulomb's law? The same r. Now as far as direction here it is every particular vector of intensity has horizontal along the radius and the vertical along the z-axis, right? Now all the horizontal will nullify each other because for every point here there is a point there with an opposite horizontal component of the vector. But the vertical components are all together and we can actually add the whole charge of this ring as one particular source of the charge and use it to find out the intensity in this field. All we need to do is consider only the vertical component of this. Now I did all these calculations in the lecture which is called Problems 2 of this chapter and I will just write down the result. So this is the intensity produced by the charged ring and the formula is, it's differential obviously depends on the height h and the radius of this ring. It's equal to 2 pi k sigma h r dr. So dr is the thickness of the ring divided by h square plus r square to the power 3 second. So I'm using this formula as is because I have derived it in all the details in the lecture called Problems 2. Okay, now in the lectures on Problems 2 I just integrated the whole thing from radius r 0 to infinity and got some result. In this particular case I should integrate it from 0 to capital R. So if I will integrate it from 0 to capital R I will get a complete intensity of the whole disk at this point. Now again, only vertical components participate here because all the horizontal components are canceling each other. Now how can I calculate it? How can I integrate it? So there is a clever substitution r square divided by h square. Now using this clever substitution which is exactly the same which I was using in the previous Problems 2 so from this differential of y is equal to well one is a constant so I have 2 r dr divided by h square, right? So h square times d y is equal to 2 r dr. Now why did I do this? Because you see r dr in the 2 I will replace it with h square d y. So my integral would be now if r is equal to 0 then y is equal to 1 if r is equal to capital R y is equal to r square divided by h square. So these are, now for lower case r these are limits of integration but now we are integrating 4 by y so these are limits of integration 1 plus. Sorry. These are limits of integration by y. So now we can do this integral from 1 to 1 plus r square divided by h square and the function is this one but I will express everything in terms of y. Now 2 r and dr I will replace with h square d y so I will have pi k sigma then I have h and I have h square so I will have h cube d y. Now let me for a second leave this as is. Now I will convert the denominator. So here I will have this. Now what is h to the cube? If I will divide everything by h cube what will happen? So on the top I will have 1 on the bottom I will have now this is h cube and this is h square plus r square to the power of 3 seconds so what I can do is I can write it as h square plus r square divided by h square to the power of 3 seconds because if I will multiply h if I will raise h square to the power of 3 seconds it will be h cube in the denominator of the denominator and that will go to the up and it will get exactly this. Now this is 1 plus r square divided by h square if you divide you will have 1 and this is r square by h square so it's basically integral from 1 to 1 plus r square h square so this is basically y in the power of 3 seconds in the denominator so it's pi k sigma y to the power of minus 3 seconds I'm not using denominator but I'm using negative power dy and this is a very simple integral it's just a power function, right? So the answer is so minus 3 seconds so integral of x to the power of n dx is equal to x to the power of n plus 1 divided by n plus 1 plus c, right? This is the indefinite integral of the power function so in this case n is equal to minus 3 seconds so plus 1 is minus 1 second and divided by minus 1 second is multiplying by minus 2 pi k sigma y to the power of minus 1 second in the limits from 1 to 1 plus r square divided by h square is equal to minus 2 pi k sigma open parenthesis first we substitute the top one so I have 1 to the power of minus 1 half so it's 1 over square root of y, right? so it's 1 over square root of 1 plus r square divided by h square minus if I will substitute 1 will be 1, obviously and that's my answer so this is the intensity the electric field at point above the surface of the disk at the height h well, I think it's better to do it just slightly different minus and double reverse signs here so it's 2 pi k sigma 1 minus square root of 1 plus r square h square now, why is it better? because this thing is greater than 1 which means 1 over it would be from 0 to 1 so we are from 1 we subtract something which is less than 1 so it's positive and this is positive so everything seems to be fine as a formula now well, let me just make a slight change another slight change you remember from the previous lecture about permittivity I have replaced the Coulomb's constant with 1 over 4 pi epsilon 0 where epsilon 0 is permittivity of vacuum that was the previous lecture about permittivity I would prefer to do it the same way here so it's equal to so instead of k it's 1 root so it's sigma divided by epsilon, right? sigma divided by epsilon so sigma is the density of electricity epsilon 0 is permittivity of the vacuum times 1 minus 1 over square root 1 plus r square h square now basically that's it this is the formula which is the answer to my initial problem let's just think about this formula a little bit if r is increasing to infinity so our disk is bigger and bigger and bigger eventually becomes infinite plane then the whole thing disappears, right? because r goes to infinity so we will have 1 over infinity which is 0 and we will have sigma divided by epsilon 0 which is basically the same thing as we have derived for infinite plane in the previous in the lecture about the problem number 2 well in terms of epsilon 0 but it will be the same if you will use k it will be 2 pi k sigma which is the same thing which we have derived at problem 2 okay this is obvious but what's a little bit less obvious is the following what if h is getting smaller and smaller which means we are getting closer and closer to the center of the disk what happens? so h is smaller which means this is bigger which means this is smaller to 0 and again we have exactly the same type of result which is sigma divided by epsilon 0 so h is smaller so this one is bigger so this one is smaller goes down to 0 no, it goes down to 0 so we have only one left so the answer is sigma divided by epsilon 0 so that's very interesting observation but the further you are now basically it's kind of equivalent of the following if you have a point which is very very close to this disk it becomes much more important that this closeness kind of nullifies the fact that the disk actually is finite and the closer the point is to the surface of the disk the more from the point standpoint the disk looks like the infinite plane obviously if you go further from the disk that actually becomes if h is big then this is small so this is 1 and you will have 0 1 minus 1 so obviously the intensity is diminishing if you are going further and further from the disk but if you are going closer then it becomes from the point's viewpoint if you can imagine this it's so close to the surface of the disk that it doesn't really see what's on the periphery of the disk it's whatever is closer to this point which is more important and that's why the result is more and more resembles result of the infinite plane and there is one more important thing what if we are not talking about vacuum what if we are talking about some kind of a medium around our disk well then we have to substitute epsilon r here so it's not really anymore the constant it becomes dependent on the relative permittivity or dielectric constant so again go back to the previous lecture about permittivity in this case this coefficient of proportionality is not just vacuum it's vacuum times the dielectric constant which might be different for different materials and if material has a high level of dielectric constant which goes here as well and we will talk about this in the next lecture about capacitors so if epsilon r is big it diminishes the whole intensity of the field which means if you have two different charges let's say plus and minus and in between you have something which has a high epsilon r then the sparkle will be less probable even if you will charge them with I mean you can actually charge with a bigger amount of electricity without having any kind of a sparkle between them so that's basically why I'm talking about this problem because the capacitors will be the subject of the next lecture so remember this formula this is very important formula and I will use it obviously so it's obviously proportional to density of electricity but it's inversely proportional to dielectric constant which means if higher constant allows to put more electricity into these two objects without discharging electricity through the substance in between so that's basically it for today our purpose again was this formula remember it we will use it next time thank you very much and good luck