 Hi, I'm Zor. Welcome to Unisor Education. We continue talking about electrostatic field, and in this particular lecture we will talk about these field strengths, its intensity. Now this lecture is part of the course called Physics for Teens presented on Unisor.com. The website contains all these lectures as a course. So there is a logical connection between the lectures, and that's where I recommend you to watch this particular lecture from this website. The website also has for every lecture very detailed notes. There are problems and some problems we will solve today. And there are exams. The site is completely free by the way. There are no ads, no financial strings attached, etc. So just use it if you can. So we are talking about electric field intensity. Well first of all let me just talk about terminology a little bit. Sometimes it's called intensity, sometimes it's called strengths, sometimes it's called just electric field without any kind of additional words. In all these cases we are talking about the same thing. And that's what I'm talking right now about. So we have already learned the Coulomb's law. That's the main law of electrostatic. So if you have two charges there is also some kind of a force between them. One extorts on another and another extorts on the first one. So if there is a force and they are on the distance, we are talking about the field obviously. And since the force depends on the charge as we know, we can talk about quantitatively measure this force at any point. Now what's important is to divorce the second charge which is experiencing some kind of a force from the first charge and just leave the field around the first charge as a subject of study by itself. So there is one particular charge positive or negative. And there is a field around it. If you put anything into this field, this anything will if it has electric charge it will feel the force. But again the force actually does depend on both charges, the main one which is the source of the field and the second one which basically has the field of itself. It makes sense to talk about only one field, the field which is generated by one particular force. And measure it using a probe electrically charged object. So we have to agree about what is the probe. And if we will put this probe into one electrical field of one main object or in another based on the force which this particular probe object experiencing, we can make a judgment about which field is stronger and why. Or if it's the same field maybe in different places, closer to the main source of the field further. So we're talking about the source of the field which is I will call it main object electrically charged we have the field around it and we have a probe object which is experiencing the force if it's electrically charged of course. If it's placed into any place into the field of the main object. So first of all let's talk about the probe object. Probe object by definition is a point object with electrical charge equal to one positive one cologne. So this is the probe object it's by definition there is nothing to talk about here. And everybody agrees with this. Okay so this probe object can be positioned anywhere in the field of the main object and you can measure the force this force which this particular probe object is experiencing in that particular place of the field electrical field which is sourced in some kind of a main object is called the electrical field intensity or electric field strengths or sometimes plain electric fields. So if somebody says okay what's the electric field of this main object at this point how to answer this question. Well you take a probe object position it in this field and measure the strengths of the force which it experiences. Okay alright. So that's the definition of the intensity. Now let's talk about quantitative things. Now we all know that the force between two electrical charges according to the Cologne's law is some kind of coefficient which has certain dimension times the charge of the main object change the charge of another object divided by radius square divided by the distance square between them. So these are two charges I use the capital and lower case just to differentiate between the main object which is the source of the field and the probe object. Now if probe object is positive one Cologne that means that we can talk about this quantity as the characteristic of the field of this main object which has a charged capital Q at certain place at the distance r from it. Now when we are talking about the distance it means that we are assuming that this is the main object is a point object. If it's not a point object I'll talk about this just a little bit later in this lecture. But for the point object this is the definition of the intensity of the field created by this point object which has charged Q, electrical charge Q on the distance r from it. Okay that's done. Now this is the force which this particular probe object is experiencing. We know that if you have the main object and the probe object then the force is always directed along this line. If this is plus this is the probe object which is plus one Cologne so it's also plus then it's a repelling force. If this is minus then it's attracting force. So if it's a force it means it has magnitude and direction. So this is the magnitude of the electrical field intensity created by a point object charged with a Q Cologne and on the distance r from it. And the direction of the force is defined this way. So if my Q is negative then the direction is negative which means it goes towards the object. It's by the way similar to gravity when we're talking about gravitation it was also the same thing. Whatever the charge whatever the force was directed towards the source of the gravity we called it negative. Because of some other circumstances which we will talk about in the case of electrical field as well. So you will recall this. So if it's a negative then the intensity of the electrical field at this point is negative. Because the probe object is positive one Cologne. So that's why it's attracting so attracting is negative and repelling is positive. Well you can always see the system of coordinate where this is at the origin of coordinates. Which means this goes along the X axis. And that's why it's considered to be positive. And if the Q is negative it goes against the X direction. The traditional X direction. So it's negative. Now this is all about the point objects. So the probe object is always the point object by definition. The main object in case of the point object this is basically the quantitative value of intensity and this is the direction so completely solved the problem of what exactly is the intensity. If the charged object is more complex than the point object then we have to deal with different story. Why? Because this is the force. The force has a direction. Force is a vector. And if this particular probe object experiences the force from more than one point object then all these forces must be combined together according to the rules of addition of vectors. The force is a vector vectors must be combined together using the regular addition of the vectors. By the way that's one of the reasons why I was telling you many times that you have to be proficient in mass especially vector algebra and calculus. And I will use both of them in this particular lecture. So let's talk about a couple of examples where I would like to demonstrate what to do to define what exactly the intensity of the field is in case we have more complex source of the electrostatic field than just a point object. So my first example is as simple as it can be in case it's not a point. So what's more complex but really very simple than a point object well two point object right? So if you have two point objects okay now each of them has a charge which is the same thing just for simplicity. Now this creates certain field and this creates certain field. Both fields exist and they are interposed on each other and what I'm interested is the intensity of the field at any point on the perpendicular through the midpoint. It's as again as simple as it can be probably the most simple problem which is not involving the one point object. So two point object and we're talking about the perpendicular to the midpoint of the segment between them. So what's given? Given this is the distance so distance is known and charge is known and I would like to have the intensity at the distance x from the midpoint right? So this is plus one C, Coulomb. This is a probe object and it has two different forces. One force is along this line and one force along this line. So let's talk about two forces and then we will combine them together. So one force along this line is let's call this point A and this point B this point M. I'm not sure we need M but anyway this is P. Of the object this is electrical intensity of the object at point A. So the value is divided by AP square. What is AP square? It's D square plus X square right? By Pythagorean theorem right? Now AB this is electrical intensity at this point by the charge Q at point B. Obviously it's equal to the same quantitative value because it's the same distance and the same charge. But the problem is that these two vectors are not collinear, they are at angle to each other. So let's assume that these are negative for instance doesn't really matter whether negative or positive. In which case this is attraction force. How do I determine the resulting force? Well the easiest way is to represent this vector as a sum of horizontal component and vertical component. So this vector is equal to sum of this plus this. And this vector is also represents this plus this. These two vectors for obvious reason they are opposite in direction and equal in magnitude so they will nullify each other. But the vector which is component of this one and is exactly the same as the vector which is vertical component of this one, they will add together and they will be doubled. Because the collinear vectors are just expanding to each other along the same direction just adding the distances. So what I have to do I have to find the vertical component of this and this and just double it basically. And that would be the resulting force which this particular probe object will experience. So what is it? Well obviously it's the magnitude of this vector times cosine of this angle. So this is phi and this is phi. So I have to multiply EA times cosine of phi and that would be my vertical component. Now what is cosine of phi? Well I can use this triangle. So it's this divided by this. The radius which is participating in this angle and hypotenuse. So it would be x divided by square root of d square plus x. Same thing with eb. Angle is the same because triangles are the same and that's eb times x divided by square root of d square plus x square. So what would be my final result? Well I have to substitute instead of EA and eb I have to substitute this which is the same thing. So I'll just use one of them and use the coefficient too. So it will be 2 times k q x and divided by d square plus x square. The whole d square plus x square plus to the power of one half so it will be 3 halves. This is the resulting electrical intensity, electrical field intensity at this particular point. That's it. This is as simple as it can be. And now, so you don't think that everything is simple I will present a little bit more complicated problem where the calculus will be involved. Now consider you have a thin rod. A thin rod. Now, thin means it's infinitely thin. Now let's consider that the length is from minus d, this is 0 this is coordinates y and x to plus 0, to plus d. Now if it's charged with certain amount of electricity, let's say q, we can always say that this amount of electricity is evenly distributed along the lengths. And we are talking about the density of electricity in this case which is q divided by L the length of this thing which is q divided by 2d. Now this is lambda which is called a density of electricity which means that any particular segment, let's say little segment of some lengths has amount of electricity equal to this lambda which is density times the length of this. If it's the entire thing it will be entire q. If it's half it will be half of the q. So whatever it is. Anyway, the amount of electricity inside a little segment is proportional to its lengths and the coefficient of proportionality is lambda which is density of electricity. So the dimension of this is cologne divided by meter. Amount of electricity divided by unit of lengths. So this is given. This is the amount of electricity. The rod is given. This is the dimension. We positioned it right in the center of the system of co-ordinates. And what we are interested in we are interested in the intensity at some point p with co-ordinates AB. Now, how can it be approached? Let's start from the beginning. As we see we don't have a point object. We don't have even the finite number of points. We have infinite number of points basically. Whatever number of points in the rod. Now how can we basically solve this problem? Well, this is a typical for the entire physics. We divide the big thing into small chunks. So let's consider that I have somewhere at the distance x from the zero a small chunk of this small piece of the rod differential of x is its lengths. So from x to x plus dx that's the segment which I am considering. And it's so small infinitesimally small that I can actually assume that this is a point object. And do whatever I need with this as a point object. And then I will integrate. If it's a finite segment I will summarize. But if it's an infinitely small segment I will integrate. Which is the same thing as optimization. But on an infinite scale. So first I have to find out what is my intensity based on this particular piece of the rod. This is the force. This is the vector. I can find its magnitude and direction. And then since it's a function of x I will add these factors together, summarize them, integrate them basically to get the final point. So far so good, right? Okay, so let's find out e as a function of x. So this is plus one c. Now again it's k times what is the charge of this particular thing. Well again its lengths gx, the density of electricity charge is lambda. So it's lambda times times gx. Now divided by square of the distance, right? Okay, what is the square of the distance? Let's just use this particular drawing. Now this is a this is b, right? So we will use this right triangle this category is b and this category is a minus x. So the square of the distance which is the hypotenuse is a minus x square plus b square. This is my magnitude of the force. Now can I integrate this by x from minus d to plus d? No. Why not? Well come on this is the magnitude, there is also a direction and this direction is different from this direction or from this direction. I cannot integrate this because all the vectors have different directions so what should I do? I should represent this a vector e of x as sum of two vectors. One is horizontal and one is vertical. So let's assume our vector is this way. So I have to represent it as a sum of these. So this is my e of x this is my e y at x and this is my e x of x. So I have to represent this vector as a sum of this plus this. This plus this. And then I will integrate separately all the x's and all the y's. Okay, now how to do that? Well that's basically kind of simple because you know this angle. This angle is equal to it's sin phi is equal to b over b over a minus x. And the cosine equals to this which is this a minus x no I'm sorry not a minus x. a minus x square plus b square. This is b square. This is a minus x square. So this is square root of a minus. Okay, now and the cosine is equal to this. So we know sin and cosine and that's why multiplying by cosine and the sin this vector, this vector we will get components this and this one. So x of x is equal to that's the projection to here. That's the cosine right? So it's hypotenuse times cosine. So which is this one k lambda dx divided by a minus x square plus b square times cosine a minus x divided by square root. So in which case I will do this into the power of three seconds. One and a half, right? This is my px and e y of x is equal to the beginning is the same and now I have to multiply this by sin right? Sin of this angle to get the vertical component which means it's by b and by three seconds. What should I do next? Well, next let me just slightly change my retching in this case. This is not the entire ex. It's only a differential which is of this particular thing and this is also differential. So now I have to integrate it. I have to integrate this from minus d to d and this from minus d to d to get complete vertical and horizontal components of the resulting intensity of the electrical field. And that is basically the end of my story because from here on it's just technicality as we are saying it's a pure mass. Take these integrals and done with this. You will have two components vertical and horizontal which means you have an entire vector. Now I don't want to do right now this derivation of these integrals. I did it here in the notes to this lecture. It's relatively simple integrals. You can definitely take them yourself. You can try it yourself. If you know go to the notes and notes contain complete derivation and there where you will find basically the final expression for horizontal and vertical components of this vector. Now this is I would say more complex example of intensity of the electrical field. I will probably try to use one more. Now this is the rod, right? Now I will probably in one of the problems which I will solve I will also use the rectangle a flat kind of an object as the source of electrical field and also it will have its own charge and I will try to make calculations related to this. Other than that basically that's all you really need to know about the intensity of the electrical field or just simply electrical field as sometimes people are saying. And again don't forget that in all the more complex than the single point object cases you need some mass. There is nothing you can do without it. So there is a mass for teens course on the same site Unisor.com and I kind of suggest you to refresh your mass skills using that particular course. Also free without any ads etc. That's it for today. Thank you very much and good luck.