 Hi, I'm Zor. Welcome to a new Zor education. I'd like to solve a couple of problems for certain topics which we were studying before electrostatics. Now the previous lecture was about the Coulomb's law which is the most important law in electrostatics. So these problems are basically related to to this Coulomb's law and just to complicate the issue they are related to some other topics which we have learned before. I just combined them in the problems which are related to Coulomb's law. Now this lecture is part of the course Physics for Teens presented on Unisor.com This is a free website dedicated to advanced topics of mathematics and physics. There is a course Maths for Teens which is kind of prerequisite to this Physics for Teens course. I do suggest you to watch this lecture as part of the whole course which can be taken on Unisor.com because the course includes not only the lectures presented in certain logical order, but also detailed notes for all the lectures and there are problems like the one which we are talking today about and exams and again the site is completely free and there are no ads so do it as well. Anyway, so let's just solve these. I have four problems related to electrostatics. Now my first problem is you have an equilateral triangle and at each vertex there is a charge, a point object charged with one is positive and now there is the same and the third one is negative of the same magnitude. So this is point P, Q and R. Now this is equilateral triangle and the side is equal to D. Now, obviously these two objects are attracting the one which is at point R because these are two positive and this is negative. So they're opposite, so they're attracting each other. So my question is what is the force which this particular object experiences as a result of these two forces? It's a very simple problem and it basically includes only the Coulomb's law and the law of addition of vectors which represent the forces. So we have basically two forces of attraction. One from P, it goes this way and another is from Q, it goes this way. They are of the same magnitude and the magnitude of each force is this coefficient which is known from the Coulomb's law. The one charge, another charge minus, so minus would be here, Q and minus Q divided by D square which is the square of the distance between them. So this is one force and this is another force, exactly the same magnitude but directed slightly differently, right? So this is an angle of 60 degrees, right? So what is the result of addition of these two forces, these two vectors? Well, there is a parallelogram rule, right? You know, so this is, go this way, this way, this way. So what's the length of, let's put this point A, what's the length of RA? If you know the angle, you know, this is D, this is D, this is also D and D. Well, then obviously the half of this is equal to D times sine of 60, sine of 60 degrees. Sine of 60 degrees, this is 60 degrees or cosine of 30 degrees. So this is this piece, which is half of the force. So you have, the half of the force is equal to F times cosine of 30 or sine of 30, which is square root of 3 divided by 2, but the full force will be twice as big. So it's F times square root of 3, which is minus K, Q square, square root of 3 divided by D square. This is the result. So these two point objects will pull, well, on this particular drawing down, if you wish, but anyway, they will be pulling along the bisector of this angle with this force. Next. So this is basically a very simple exercise of the Coulomb's law. And the law of addition of mechanical forces. Next. Next you have, you have two objects, point objects. Each one has charge Q, but they originally, they are hanging on these threads. But considering they are both charged with the same amount of electricity, both, let's say, positive or both negative, then they will repel each other. So the thread will actually be slightly at angle from the vertical. Now, we obviously considering this on the surface of the Earth or any other planet. So there is a G, which is the acceleration of the free fall. They do have mass, let's consider the same mass and the same charge. So the question is, what is the angle from the vertical? They will deviate. Well, let's just think about this way. What kind of force acts on this one? Well, there is a one force is weight, which is mg. Another force is force of repelling. It goes this way. And the tension of the thread. Now the tension of the thread along the thread can be represented as some of two forces. One is vertical and another is horizontal. So this is the tension thread. Now, if this is electrical force of repelling, now, since this is the equilibrium state, because they're just hanging in the thin air, right? So this force of repelling should be equal to this force, which is the component of the tension. And the same thing with the weight. The weight should be equal and opposite in direction to the force, which is the component, another component, vertical component of the tension, right? So let's say the tension is t. Then this is t times sine phi, right? If this is the angle phi, then this is t times sine of phi. Now, this is the horizontal force and it should be equal to the electrical force. Now electrical force is kq square divided by, well, I didn't really tell you this. This is the given d, d square. Now, as far as the vertical component, this is t times cosine phi and it's supposed to be equal to mg. When I'm talking about equal, I'm talking about magnitude, because the direction is always opposite to each other. Now, we have to find the angle, right? So the simplest thing is just divide this by this and you will have sine divided by cosine, which is tangent of phi, is equal to kq square divided by mg d square. And from this we can define the angle. That's it. Problem number three. Problem number three is let's imagine an atom of hydrogen. So you have a proton plus e and you have electron, which is circulating on an orbit, which is minus e. And let's consider we know its mass. So we know the charge, charge of the proton is equal to the charge of electron, just opposite signs, negative electron, positive for proton. And let's say we know the mass. So my, and one more thing is we know the radius of the atom, atom of hydrogen. Now my question is what is the angular velocity of the electron as it circulates around the proton? And frequency of number of rotations per second, let's say. Okay, how can we determine that? Well, obviously we have to use the Coulomb's law to determine the force, which keeps the electron on its orbit, right? So again, this is f e, which is equal to I'm talking about magnitude. So forget about the sign k. Let's say this is q, which is equal to plus i and minus e is minus q, minus q. So that's the charge of electron, which is equal to the charge of proton. So the atom is neutral. So we have q square divided by r square, right? So this is the Coulomb's law and this is the force which keeps electron on its orbit. Now from the mechanical part from kinematics of rotation we know that the force which is supposed to be the force which keeps the electron on its orbit, it's supposed to be equal to mass times angular acceleration, which is r omega square. If you don't remember this, I would suggest you to refer to the mechanics part of this course where I'm talking about rotation. That's a very simple thing. Basically, that's it. From here, we can determine the angular speed, which is, so k q square divided by m r square and r r cube. And this is omega square. So omega is equal to square root of this. Now in the notes for this lecture, in some cases, like in this one, I actually do some calculations and the calculations based on the numbers which I can get, the real values, what's the mass of electron, what's the charge of electron in Coulomb's, what's the radius of the atom of hydrogen, etc. And I do calculate the final result. What exactly is the number of, what exactly is the angular speed and the number of revolutions per second is omega divided by 2 pi. So if this is the speed, angles, regions per second, if you divide by 2p, it would be number of revolutions per second. That's how many times per second electron is circulating around the nucleus of hydrogen, which is a proton. I just have to make a note here. It's all based on classical Bohr's model of the atom as nucleus and electrons surrounding it. It's not the contemporary view which is based on quantum mechanics, etc. So I'm not going into that. So consider this as just not the reality, how exact how exact the frequency of electron in the hydrogen atom is. It's just this particular model with the numbers which are taken from internet somewhere, all right? So this is the orbital model as it was invented some time ago by Niels Bohr, the physicist. Okay, now this third, that's the force problem now. Force problem also sounds like a real science, but again, it's just our model. The same as the Bohr's model is a model of the atom, whatever I'm doing right now is another problem. It's also kind of a physical abstract, if you wish. So what we are doing is we are trying to bombard some atom with some elementary particle well for the purpose of, let's say, splitting this atom. Now, for instance, whenever we are bombarding uranium 235 with neutrons of certain intensity, it will split and it will be like atomic bomb or it will be more or less regulated reaction if we are regulating the number of neutrons and that can be power station, nuclear power station. Whatever it is, that's the practical side of it. In my case, I'm just kind of making up a problem which sounds similar, but it's not really similar at all. First of all, I'm not bombarding by neutrons but by protons just because I want to have this problem related to Coulomb's law. Why in reality, in the power station, for instance, atoms of uranium are bombarded with neutrons? Well, because neutron is electrically neutral and there is no resistance from the nucleus. Whenever the neutron comes into it, it just goes straight through without any kind of force which prevents to do this type of things. In my case, I want the force, so I want this proton to be bombarding the atom which has n protons. For instance, in case of uranium, you have 92 protons. What's interesting is that as we are bombarding this thing, according to the physics as we know it right now, according to Coulomb's law, the force between these two is inversely proportional to a square of the distance which means that it goes to infinity whenever the distance goes to zero. Now, obviously, this is not a point object. Atom has a certain size and to avoid this type of infinity things, what I'm actually asking is what if I would like this to be covering this distance from A to B. So, I would like to cover the distance from A to B. So, I have initial distance to the nucleus and they have an ending distance to the nucleus. I consider the nucleus to be a point object as well as this proton and what my interest right now is, I would like to know what's my initial speed I have to shoot this proton to cover this distance against the resistance of the nucleus and nucleus is resisting because this is a plus and this is the plus. So, there is a repelling force. So, nucleus is resisting and the force is variable and since the force is variable, I cannot just get to the final result of the calculation. What I do need, I have to divide this into infinitesimal number of little sevens from x to x plus dx. Now, we have even integration, right? So, you do have to know your math. So, we do have to integrate. Now, if it's on this particular place, the force is equal to, as it depends on the x, the force is equal to k times, let's say this is q, this is nq times n times q square, right? q times nq and divided by x square. That's the Coulomb's law. So, I have the force. Okay. Now, if I'm shooting this particular thing, which has certain mass, mass of the proton and certain initial speed, which I don't know yet, I have to find out what's the initial speed. Now, the work which is this particular force is doing when the proton is moving from distance a to distance b is equal to integral from a to b force of x times dx, right? Force times distance. So, this is the work and if I integrate it from a to b, I have the entire work. So, this is exactly the initial kinetic energy of my proton and this is the equation from which I can get the speed, which I don't know yet. So, now, this is equal to integral from a to b, k nq square divided by x square and dx. Now, obviously, this goes outside. So, I have k nq square, integral of x square is 1 over x, right? With a minus sign. So, we'll have 1 over x. So, it's 1 over b minus 1 over a. This is equal to, right? Because the derivative of 1 over x is minus 1 over x square. And so, that's it basically. Now, I have an equation. I have from here I get v is v equals to this thing, which is 2k. This 2 goes there times n times q square times 1 over b minus 1 over a and divided by m and square root of this. And again, I actually substitute certain real values, which I got from the net and I calculate the result of this. So, these are four really very, very simple problems. However, what's important is not only you have to know the Coulomb's law, but also some other things. For instance, how to add the forces, what's the angular acceleration whenever you're talking about the circulating. You have to know some mathematics like simple integrals, like in this particular case. And again, my purpose was not only the Coulomb's law, which is really very simple thing, but also to bring some other pieces from other parts of the course, which you really have to know by now. And if you don't, please go back and refresh your memory. That's it. Thank you very much and good luck.