 Hi, I'm Zor. Welcome to a new Zor education. I would like to present to you a problem which, to tell you the truth, I was kind of surprised with the result of this problem. That was a long time ago and it kind of counterintuitive, if you wish. In any case, it's about electrostatics and this particular lecture is about this problem. It's presented as part of the course called Physics for Teens, presented on Unisor.com. Now, the whole subject is actually described also in writing as notes for this lecture. So, every lecture in this course has very detailed notes and I do suggest you to pay attention to these notes. All these notes are on Unisor.com. The lecture is basically, physically, as a recording. It's stored on YouTube and there are some descriptions on YouTube as well, but it's a short description. The real lengthy one is on Unisor.com. This course has a prerequisite which is Math for Teens. You do have to know math and in this particular problem I am using calculus, vector, algebra and stuff like this. So, you better brush up on your mathematics. And again, the course which is called Math for Teens is definitely sufficient to Physics for Teens. Okay, yes. And by the way, the site is completely free so no advertising, no financial strings. Okay, these are, it's kind of two problems but it's actually only one real problem. The second one is very trivial one. Here is what it says. Okay, let's consider you have an infinite plane, infinitesimally thin, so it's just the flat surface. Now, obviously you can tell that there are no infinite planes and no infinitesimally thin planes, etc. But again, we are talking about a model. Everything, whatever physics do, especially when it comes to a theory like this one, it's all about a particular model. Let's just bear with this from the mathematical standpoint. It's a plane, infinitesimally thin and it's charged with electricity. Well, it's infinite plane, right? So, the amount of electricity is infinite. However, we do have something which is called density of distribution of electricity. Which means that there is some kind of parameter which I call sigma, which is basically coulombs per square meter. That's how it is dimensioned. Which means that if you have any area on this plane, if you multiply the area by this sigma, you will get the amount of electricity in coulombs which this particular area contains. Now, what's necessary to do is to find out the characteristics of the electrostatic field around this plane produced by this particular distribution of electric charge. Now, in particular, let's just take some kind of a point P above this plane and let's find out the intensity of electric field at this particular point which this plane defines. So, how can I find out this particular intensity? Well, intensity is a vector, right? So, the vector must have first some kind of a direction. Now, for consideration of symmetry, this vector is supposed to be on a perpendicular from the point P to the plane. That's called the plane alpha. So, we have this plane and this is the perpendicular. Now, why is this the case? Well, it's very simple actually to explain why it's supposed to be on a perpendicular. Let's take any small domain here. Well, I didn't choose a point because point has no electricity. But any domain, however small, will have certain non-equal to zero amount of electricity. But let's consider that this particular domain is really very, very small. So, from every even smaller part of it, we can say that the vector of intensity is directed exactly on this line. Now, we know that vector of intensity, if you have two point objects, is always directed on a line which connects them. Because every charged point object has radial directions of the intensity vector. If this is the point object, then all the vectors of intensity at any point are always radial. So, we know that. Okay, so now we can consider that we have a probe object at, let's say, plus one cologne here at point P. And we have a very, very small area here. Well, this very, very small area has certain amount of electricity which is equal to sigma times the area, right? Now, this is the vector. I can always represent this vector as sum of two vectors. One is radial and another is perpendicular. Now, let's take the symmetrical relative to this. This is the perpendicular to the plane from the P, let's call it O. So, let's take symmetrical area here. Also, the same shape, the same area, just symmetrically positioned relatively to the point O. Well, obviously, since the area is exactly the same in size and because it's symmetry between these two segments are equal to each other, A, O and O would be. Now, A, P equal to B, P for obvious reasons because these are right triangles with the common catheters and the other two catheters equal to each other. So, the distance is the same, the charge is the same, so intensity should be the same in magnitude. But how about the direction? The direction will be this, right? Well, I put the arrow in this direction, it can be opposite direction. It depends on whether this is positive or negative, but it doesn't really matter. The matter is just the line along which the vector is stretched, right? And I can also represent this vector as a sum of this one and this one, all right? Now, it's the same magnitude, the same angle, right? Because the triangles are congruent, which means that my horizontal components of these two vectors are exactly the same. Because what is the horizontal component? It's length times cosine of this angle. So these are oppositely directed and equal in magnitude, so all horizontal components will cancel each other. Vertical components are the same, and they are along the same direction and the same magnitude. So my point is that all components of all the vectors of intensity from all the points, horizontal components will all nullify each other. Vertical components are always directed towards the same direction and they are parallel to each other, and that's why they will all be summed together. So that's basically a description of, this is a qualitative description of the fact that our resulting vector must be vertical perpendicular to the plane. Direction will be either from the plane to the point or point to the plane, depending on whether this charge is positive or negative. If it's negative, it will be attraction, because this is positive. If this is positive, it will be repelling. Okay, now this is kind of an explanation, not much of a mess actually here. So let's calculate this vertical component, which is the only thing which we are interested, from all the elements on this plane and then add them up together. Well, I used the word add, basically it's integration. So we have to choose some kind of a small component, find the vertical component of the intensity vector and integrate along the same plane. Well, you can do it in many different ways. For instance, you can arrange the Cartesian system of coordinates. This will be some kind of a x, y point. We will make an increment dx and dy, have a little square here, and then integrate by x and y from minus infinity to plus infinity by x and from minus infinity to plus infinity by y. Yes, we can do that, but there is a kind of a better, a little bit simpler approach. And here is what I'm suggesting actually to do. Let me just forget about this piece and this piece. We will do it slightly differently. Okay, this is my base of the perpendicular. What I'm going to do right now is I will introduce, instead of Cartesian system of coordinates, I will introduce cylindrical one. Now, this will be the z-axis, okay? And on the plane, I will use the polar system. Let's consider, for instance, this is an x direction, whatever it is. So any point here will be characterized by angle and radius. Now, I will do a little bit even more in this particular case. Instead of using a small element with certain r with an increment gr and certain angle phi with increment d phi, I will use it slightly differently. I will use a ring. So the ring will have inner radius r, and outer radius r plus differential of r. So it's a very, very thin, infinitely thin ring. So it's very thin here. And obviously the whole plane is very thin as far as, as far as thin as in this direction, okay, infinitely thin. So what I would like to say that it's basically chosen for a purpose, this ring, y. For a very simple reason, because if you will take any small piece of it, it will have a certain amount of electricity and intensity vector would be directed from this piece toward this point p. Or this direction doesn't really matter. But anyway, it's along this line. So the length of this line is exactly the same for all pieces around this ring, right? Because it's the same radius. Here or here. So it's all right triangles with the common catatose, which is the distance from r to all. And the second catatose is the lower case r. So all these pieces will have exactly the same distance and the same angle here, which means that if they are all of the same area, infinitesimally small area, then the vertical component of the intensity vector will be exactly the same for all of these points. So when I will integrate them around the whole circle, the whole ring actually, they will basically be very simple result of this integration. I just have to take the whole area of this ring multiplied by sigma, which is the density of the electricity to get the amount of electricity. And knowing the amount of electricity, I will just use this formula for intensity, which we have learned before. So q would be the amount of electricity here. r would be the distance from any point on this ring to this point p. k is Coulomb's constant. And since the angle is exactly the same for all of them, I will just have to multiply by sine of this angle to get the vertical component. Okay, so how can we do it? Now this is actually a very simple thing. So the area of the ring is, well I have to take the area of a circle surrounded by r plus dr radius and subtract the area of a circle of the r radius, right? So it's pi r plus dr square minus pi r square, which is equal to pi r square plus 2 pi r dr plus dr pi dr square and minus pi r square. Now pi r square goes out. Now look at this. This is an infinitesimal variable of a higher order than this one. See, this is dr and this is dr square. And as we know from mathematics, whenever we will integrate, this will be a component which will be integrated and this will add an infinitesimal patch to the integration because we're integrating only once. So I can definitely get rid of this as well and say that my ring area is, area of the ring is 2 pi r times dr. Which is kind of obvious because 2 pi r is a circumference and dr is the thickness of this ring. And if dr is infinitesimally small, then the fact that it's curved instead of stretched doesn't really matter. The difference will be infinitesimals of higher order. So I have the area. If I have the area, I have the amount of electricity by multiplying by q. So my differential q at the radius r. Now why I'm using differential? Well, because the whole thing is infinitesimally small. It's differential only for this ring. It's 2 pi sigma r dr. Okay, now what's the distance from p to any point on this ring? Well, let's just take the distance from p to the plane alpha as h and then my r would be equal to the radius square. r square would be equal to radius square plus h square, right? This is the Pythagorean theorem. Okay, I've got that. And so my dE of r equals 2. This is my intensity according to this formula. It's the k times k times 2 pi k sigma r dr divided by square root distance which is r square plus h square, right? Now, I would like to know not just the intensity but the vertical component of the intensity because the horizontal components will all be canceling each other. However, only the vertical components will be calculated in the same direction, so I can just add them up. So instead of vector addition, because these are all vectors differently directed, I can't really add them up. But if I will represent each vector as sum of horizontal and vertical component, all horizontal will be wiping out each other but the vertical components will be adding to each other. Okay, so I have to multiply by sine of this angle, right? Sine of paO. So my vertical component which I will use ez as a suffix z. It's along the z coordinate. As a function of r, it will be equal to 2 pi k sigma r dr divided by r square plus h square. And I have to multiply by sine of this angle. What is the sine of this angle? It's h divided by the lengths. Now, we know this length. It's this one. Well, actually, it's the square root of this one. So I have to multiply it by h divided by the square root of r square plus h square. Okay, let me write it down slightly differently. 2 pi k sigma. These are all constants, right? h is also constant. What's not constant is r dr. r. dr I will put at the end. And this one is r square plus h square to the power of minus 3 second times dr. Right? Square root is power of 1 half. This is power of 1. So it's 1 and 1 half would be 3 seconds because it's in the denominator. And I will not use the denominator here. Everything will be in numerator. So this is my final amount of intensity vector, magnitude of vertical component of intensity vector of one ring. So the whole ring contributes that much to the magnitude of the entire intensity of the plane. Now, what I have to do is I have to integrate it for all the rings, obviously, concentric to each other. So which means I have to integrate it from 0 to intensity by r. Right? So it would be 2 pi k sigma h. And what I have here is r times r square plus h square to the power of minus 3 second dr. So this is basically the answer. This is the magnitude of the entire vector produced, entire intensity vector produced by the plane. Okay, so let's concentrate on doing something about this integral. Now it's, as we are saying, it's a pure technicality. That's why you have to know math before actually attempting any kind of a serious calculations in physics. So let me do it in two steps. I can do it in one step, but I can really prefer to do it in two steps. You see, I don't like that this is r square plus h square. I prefer to have something like 1 plus x square to some power which is more palatable from the integration standpoint. So what I will do is I will introduce a new variable called r divided by h. Now, what happens here now is my r would be equal to xh. So my integral, if r is from 0 to infinity, x also will be from 0 to infinity. Now my dr is equal to obviously h dx. I'm just differentiating, x is a variable, h is a constant. So differentiating just multiplier will be outside. Okay, in terms of x, the whole thing would look like this. But the integral also from 0 to infinity by x now. And I have this unpleasant multiplier in front of it. Which is 2 pi k sigma h. Now, instead I will put here r is xh. And here I will have x square h square plus h square to the power of minus 3 second. I replaced r with xh. And dr would be h dx, right? Now that's equal to, obviously h square would be outside. So it's 2 pi k sigma h integral 0 to infinity xh. Now it would be h square times x square plus 1 and all of this in the power of minus 3 second, right? dx, no h, one more h, sorry. Well I can put h square here, or h square here. Actually I can put h cube. Okay, I put all h's together. Now equals to dx equals to 2 pi k sigma h cube integral from 0 to infinity x times. So h square times minus, to the power of minus 3 second would be what? H to the power of 2 times minus 3 second, right? Powers are multiplied. n times x square plus 1 minus 3 second dx. Now what is 2 times minus 3 second would be minus 3. And this is the plus 3, which means, and this is something which is again for me was absolutely remarkable. Look at this again. There is no h anymore. Let's take a graph from 0 to infinity x, x square plus 1 minus 3 second dx. There is no h here. So this is something which I think is the whole purpose of this lecture. What does it mean that there is no h in this formula? It means that the intensity of the field does not depend on the distance of the point to the plane. So no matter how far from the plane you are, intensity of the electric field will be exactly the same. This is a remarkable stuff. I mean, intensity, intuitively you feel that the further you are from the source of electricity, the weaker should be the field. So the forces should be weaker. But in this particular case, it looks like no matter how far we are from this infinite plane, the intensity of the force will be exactly the same. Well, why? Because it's an infinite plane. You see, it's infinite and all these infinite tails on all the sides are combining together, are making this intensity independent of the height. No matter how high you are above the plane, it will still reach you with the same strength. Strange, but the fact is the fact. Now, how can we find out what this particular integral x relative to x is? Well, that's kind of simple. Now the best thing is to say that we will have y is equal to x square, right? What does it mean? We have 2x dx would be dy, right? dy equals 2x dx. So the whole thing is equal to 2 pi k sigma. Now x is from 0 to infinity and that's why y will also be from 0 to infinity. Now we are using this 2 already, 2 and x. So we will have y square plus 1 to the power of minus 3 seconds dy. Now I'm sorry, not y square plus 1, y plus 1. x square is y, 2x dx is dy, right? So I have this. Now let me make it a little simpler. I will put x plus x minus 1, right? Then a plus 1 second, y is equal x square plus 1, plus 1. dy will be exactly the same thing, but now instead of y plus 1 I will have y here. But now integration should be not from 0 to infinity but from 1 to infinity, right? If x is equal to 0, y is equal to 1. If x goes to infinity, y goes to infinity. Now 2x dx is my dy and x square plus 1 is just y. Now this is just a plain power function and we can very easily integrate it. So what's the integral of this? You can wipe out this. Now you know that y to the power of n derivative is n y to the power n minus 1, right? You know that. That's the derivative from the power function. So if I will integrate, indefinite integral of y to the power of minus 1 dy, it will be equal to y to the n divided by n, right? Plus constant which we don't care about because we are talking about definite integrals. Alright, so my n minus 1 is minus 3 second. So n is equal to minus 3 second plus 1 which is minus 1 second. So this integral is equal to pi k sigma. So indefinite integral would be equal to y to the power of minus 1 second divided by minus 1 second, right? According to this. n is equal to minus 1 second. Okay, and now what should I do with this? I should really put it in limits from 1 to infinity. Okay, first we go to the... Well, let me just write it a little bit differently. Divided by minus 1 half is minus 2 pi k sigma and 1 over square root of y, right? If you don't mind. Now, if you substitute infinity, obviously we will have 0. So this is equal to 0 minus. And now you substitute 1 here. So minus and minus would be plus. Square root of 1 is 1 and this would be just plain 2 pi k sigma. So this is the intensity of the vector of electric field, electrostatic field produced by infinite plane where sigma is density of electricity which signifies how much electricity every unit of area will have. Well, this is the end of the first part of this. Which basically is 99% of the whole lecture. And another 1% is the question, how much work do we need to move any charge from one place above this plane to another? Well, that's obviously very simple thing to do. Look, we do have, this is the plane and I have 2 points. P1 and P2 and I have to take some kind of a charge q from this place to this place. Now, obviously I know this sigma is density. Now, look, this is projections onto this plane. Now, the intensity is exactly the same, here and there. And it's all directed vertically. Which means that my horizontal movement parallel to the plane needs no energy. And the only energy I need when I'm acting, when I'm moving against or along the vector of intensity. Since all vectors of intensity are directed vertically. And so the horizontal is perpendicular to the force. Perpendicular to the force requires no work, right? So only when I'm going along the force, with the force or against the force, I have to spend energy. I have to, my work will be positive or negative depending on the direction. So it's very simple. So first I move this horizontally to be on the same vertical. And then I'm moving along the vertical from whatever the height h1 to height h2. And my force is constant, because as we have seen, it does not depend on the h. So it's just the movement with a constant force, which is this one. From point which has height h1 to height h2. Which means my work would be the force times the distance, right? So the work would be equal to the force, which is 2 pi k sigma. By difference h2 minus h1. Now, obviously there is one more little thing about the direction of this force. We were just talking about magnitude. Direction depends obviously on what kind of charges, what kind of a charge q is. And what kind of a charge is in the plane. If they are of the same, then if h2 is greater than h1, it would be repelling. So it will be one sign of work. If it will be positive or negative, it would be another sign. So it all depends. So let's not talk about signs. Sign is easy, but the magnitude of the work is this one. And that basically completes this particular lecture. This is, as I was saying, 1% of the whole thing. Okay, I do suggest you to read the notes for this lecture. They are maybe in a little bit more detail than whatever I was just talking. And it's just a good reading anyway. It's like a textbook, as was probably every other lecture. So that's it. Thanks very much and good luck.