Transcripts
1. Introduction (Episode 0): welcome to this series of videos that will introduce you to electric fields. A military charge generates an electric field around it. If you place another electric charging that field things charge will experience a force, and you know that a force triggers motion, right? Second Lord Newton Ethical there may. So the electric charge we just placed in that field will start moving. And what do we call the motion? Often electric charge? Well, an electric current. In other words, electricity. This is why the concept of electric fields. It's so central to all topics related to electricity, it is crucial to have a good understanding of electric fields when you study these topics, this is why I created these lessons and exercises to help you build a solid foundation. In the words electric field, there is the word field, So the first video focuses on defining what a field is and for their branches to the definition, often electric field. The two videos that follow describe in detail the cause and the effect of an electric field by introducing the notion off electric field strength. These videos are followed by three exercise videos where you can train your new knowledge as bonus. The last exercise would show you how Cools Law emerges naturally from the notion off electric fields, the four lesson and exercise videos. After that, we show you how to manipulate electric field strengths as victors. This will teach you to predict how a charge placed in one or more electric fields will move . This will naturally lead us to the concept off electric field lines. We will focus also on the two most common electric field geometries. Raid your fields on the uniforms fields, and then we will conclude with an exam level exercise that will centralize all the notions presented in this course. Follow this course diligently, work on the exercises, and you will gain a solid understanding off electric fields. Then you will be able to rely on this new understanding for the study of all other subjects related to electricity. So now head to the next video and let's get started
2. What is an electric field? (Episode 1): what is an electric field, and then that shell. An electric field is a region of space where an object carrying and the electric charge experiences Air Force off electric ology. Let's look at this definition in order to understand all of its facets. In electric field, there's a word field, so let's focus on this word first. What is a field? A. Field is a region of space for which each point is associated with a quantity. For example, Behind me, it represented a flat surface onto each point. Off this surface is assigned a number. The property associated with that quantity defines the nature off the field to understand what this means, let's look at this map. Every point is assigned with a number. This number represents the temperature of the point. So this map is a representation or that temperature field. You can try this with other type of quantities. Just check any map that provides some specific data, like for example, you need it. Population density, population average income, whatever. These are Oldfields. We could even imagine a field where the quantity is just either one or zero. For example, an electron field. So yes, imagine a region of space that we will present here with a surface for which every point is assigned. Either one was evil. If we have a one, it means that there is an electron present. At this point, if we have a zero, that means there is no electron at this point. If you look at the evolution of the field with time, you can witness the path of the electron. We can also imagine that instead of a number, the quantity associated with each point is a vector, which is a value with the direction there is a feel of this type you can see every day on your TV, this one here. The magnitude of the arrow represents the speed of the wind and the direction of the arrow , the direction of the wind. That field is a vector field. We could even go further and assigned to each point of a region of space, a function and operator, even a state. But looking into this would make us dive deep into quantum field theory, which goes beyond the scope of this course. So let's go back to our electric fields. The quantity associated with each point often electric field is the electric field strength . As its name suggests, this quantity will presents the strength of the effect of the field, which is the force that would experience an object carrying a charge off one. Cool. Let me pull out the board to illustrate this. Imagine this board is an empty space. An anti stays devoid of any electrical influence. Imagine now that the place that Charles Lee took, you've somewhere in the space a positive charge. Oh, yeah. When I say that I'm placing a charge at a given point. What I really mean is I'm placing an object which is carrying the charge. At that point, remember, charges the property, But in this course, I'll be placing lots of chargeable everywhere. So I'm not going to say each time I place an object which is carrying this chart, I'll just say I'm placing a church. But do remember that the charges a property not to think. Okay, So what happens to this Charles Lee took you? We're laughing. Okay, so this takes another charge in that space charge beak. You positive too. And let's pin it. Let's fix it. At that point, Little Q will experience repulsion force. Yeah, positive and positive repair. So that being forcibly took you in the perspective of little Q. What just happened? It was just They're not doing much right. And suddenly food, it feels accelerated is a force applied on it. And it's the state. If something changed fundamentally in the nature of space itself, something new happened. Now, the point where it was located has a new property. That is that if I place a charge, their discharge feels the force. This new property provided to space was provided by charge. Big Cube. They've actually modified the regional space around it. It gave it a new property, which is if I place a charge there, someone that stays this charge would feel a force charge Beak, You just created an electric field. And the electric field is a regional space where charge experiences a force. So if I placed a charge here, for example, and they get it one, it would feel force. A positive one here would feel force because it is in the electric field created by charge because okay, so what would be the value of this force? Let's go back to our Charlie took you here if would off course depend on the charging place there. If I place a big charge, where the force on it will be bigger with repulsion from Q would be bigger. So if it's proportional to, it's a charge of place at the point, but also proportional to the strength off the field. At that point, I the magnitude of the property that the point has because it is an electric field. This value is called the electric fear strength, the strength of the field. The electric field strength is a victim because you see it's proportional to the force. Forces affected has a direction. So does an electric field strength. We can also define the electric field shrinks that how much Force one cool off positive charge would experience one placed At that point, the electric field strength would be a property of the point. Right? Okay, I will discuss more about the electric field strength in the next video. For now, just remember what an electric field is. An electric field is a regional space where Charles placed in that space will experience a force
3. Electric field strength Part 1 (Episode 2): In the previous video, we saw that an electric field is a region of space where an electric charge would experience the force. We also introduced the quantity associated with each point of an electric field, the electric field strength. We can define the electric fear strength in two different ways. In this episode, we will look at the electric field strength as defined by the effect of the electric field on a positive charge. In the next episode, we will look at the electric field strength as defined about what causes the electric field in the first place. An electric field is a regional space where a charge placed in that space will experience a force. Let's go one imagine that the space within the circle is an electric field. So it means that if I place a charge little curia, the point off this space discharging took you will experience a force. This force. We depend on the charge they took, um, placing their, but also on how strong the effect of the field is. At that point, this quantity is called the electric fear strength. Okay, the electric fierce trance at a given point in the field will be the force experience by a point positive charge off one cool placed at that point. Well, the force experienced by a positive charge cool. The little fear strengths is thus Victor. It also has a direction units new terms for cool. Imagine that at this point off the electric field is associate ID and electric fear strength. I liked it downwards off, Say, free new terms Picolo. That means that if I place a charge off plus one cologne at this point, this charge will experience a force off three new terms If I place a for cool at this point , the force experienced by the charge of folk alone will be 12 new terms in the definition. Off the electric field strength, we say. And they fear strength at the given point is the force experience by a point positive charge per unit. Charge right, I said point positive charge. The positive is really important. Let me show you. You see that if Coop is one, then e is f. That's the definition. It's a force experience when Krys one if coup is minus one, then e is no f anymore. It's minus f. So it is very important that saying in an exam, you asked to give the definition of the electric fear strings. If you forget to say that they're considering a positive charge, he knows the mark, that's for sure. So remember the definition and remember to consider a positive charge. So the electric field strength at a given point in an electric field is a force experience by a positive charge placed at that point per unit charge. Imagine the whole bold is an electric field. Imagine also that this electric field is uniforms. That means that that every point of the board is associate ID, an electric fear strength, which is the same in direction and magnitude. The less taken Landon Point say whatever here and imagine in a tree fear strength directed that way and of magnitude, say, five new terms cool. That means that I have placed a child of one coherent feel, the force of five new terms and actually a charge off. One cool anywhere on the board will feel a force of five new terms selected that way because the field is uniforms, so that's place a charge of too cool here. What force with this child's experience. The force will be in the same direction as electric field and the magnitude of the force will be two times five. That is 10 Utahns. A charge of one cool place There would feel a force of five Newtons. A charge of too cool We should like twice one I feel a force which is double Imagine now that a place the child que off a minus two Cool! In that case, the force will be minus two by five Is my best in you. Don't you see? It is in the opposite direction. It is still calling here with E but in the opposite direction. Time for an exercise. Now a hydrogen atom consists of a proton with an electron in orbit around it. The radios off this orbit is five by 10 to the minus 11 meters. The electron is in circular motion around the proton because it is experiencing on the attractive force. Yes, electrons negatively charged Proton is positively charged. Well, they tracked the trouble so you would have air force here. Healthy Croton on the literal calculated. Find out what the magnitude of this force is using. Coombs Law. The proton is creating an electric field around it. That is why the electron is actually experiencing this force. Calculate the magnitude of this electric field. I e the electric field strength at the point where the electron is located. Good luck. An electron is a particle which is negatively charged. The charge of the electron is minus E. Where e is the elementary charge E is 1.6 by 10 to the minus 19. Cool. A photo tone is a particle which is positively tuft. Charge of a proton is plus e. There are opposite sign of charge. Therefore they attract each other. There is a force between them. On this force is what holds us together. It guarantees the integrity of the eso. This forces a Coolum Force Coolum Force which is proportional to the product of the two charges on inversely proportional to the square of the distance Between the charges Here it is a radius of the atom. Now we can replace this by writing down the elementary charges. So it's gonna be e by e squared, divided by the square off the distance between them. Note. I didn't consider the negative son because I'm interested in the magnitude here. I know the dilation of the fools. We like that. Okay, that's tracking the numbers. A for 99 by 10 to deny. Okay, by 1.6 by 10. 30 miles 19 squared for the charges on the distance. Five by 10 to the minus 11. It's Quint. Let me grab my calculator. I find a force off 9.21 by 10 to the minus eight. Newtons, we've got two significant figures here, so I just put my point. That's really toe A It'll be is about the determination off the electric few strings at the position where the electron is located. Village is a charged particle, negatively charged and is experiencing a force because it is nearby. A positive charge. The photo the groom force would deter mined and little A. But we can see this situation The another perspective. The proton is a positively charged particle. Therefore it creates an electric field around it. The electron also charged, happens to be within this electric field. Therefore the electron We experience a force due to the electric field created by the photo . This force will be key here. Que is a charge on the electron said E T. The force experience is a cool in force so we can calculate the electric few strength at the position of the electron electric field strength. There will just be the force on the election divided by the charge off that this 9.2 by 10 to the minus eight, divided by one point safe by tens of miles. 19. Giving me 5.75 by 10 30 11 new terms. Cool, Cool No. 75. So here, 5.8 by 10 to 11. The definition of the electric fear strength we saw in this video expresses the consequence of the electric field on a positive chart placed in that field. Now, the next video, we'll look at how to define the electric field strength based on what causes the field. See you that
4. Electric field strength Part 2 (Episode 3): let's consider a positive charge. Big Cube, creating an electric field around it. The direction of the field is obvious. The electric field strength vectors. We point outwards regularly because that would be the direction a positive charge place in the field will take it. What about the magnitude off the electric fear strength at a given point of that field? Well, it will depend on the position of the point you are considering. Yes, it's obvious the further you are from the source off the field the week of the strength of the few. Actually, the fear strength at a given point in the field will be inversely proportional to the square of the distance between the charge creating the field. At that point, what about the magnitude of the charge creating the field? Will it have an impact on the field strength? Well, obviously, yes. If the magnitude of the charge creating the field is large, the strength of the field will also be large. The electric field strength is actually proportional to the magnitude of the charge, creating the field proportional. And what about the proportionality? Constant? It is written late. Okay. And it has a name the Coolum constant, but actually it is not really a constant. Yes, it depends on the medium in which the field is located, for example, in vacuum or when Air K is equal to 8.99 by 10 to the nine u turn meter square Bookworm square. But in water, it forced to have much lower value off about 0.11 by 10 to the nine new turn meter square per column square. So you see forgiven charge and distance from that charge. The strength of the electric field in water would be about 80 times weaker than in air. The medium can have a tremendous effect on the electric fields. We say that such a medium is a dialect. Okay, so now let's Sam allies the electric field. Franks at a position in an electric field is proportional to the charge, creating the field and inversely proportional to the square of the distance between that position. On that charge, I recommend that you learn this definition because by learning it, you actually truly understand what's going on. It can be a big service when you get a bit confused doing an exercise, for example. It's time for a short application. Now imagine a helium new clues. A helium nucleus consists of two neutrals and two protons, said carries to positive elementary charges. Position yourself and say, 190 meter away from the heat of Nicholas. So you at that point, can you determine the direction and can create the magnitude off the electric fear strength . At that point, give it a shot for the direction is pretty easy. It's just away from the healing Nicholas. For the magnitude, you can just apply the formula we discussed in this video. Electric fierce transfer at a point is proportional to the charge that creative field on inversely proportional to the square of the distance between that charge. On that point, if I plug in the numbers, I get E equals 8.99 by 10 to 9. Two is to elementary charges. So two times 1.6 by 10 to the minus 19 divided by the square of the distance between the helium nucleus. On the point I'm considering that is one centimeter so 10 minus nine meters squared. Let me calculate that. Where's my calculator? Seven for 88 by 10 39 neutrons per cool
5. Charged sphere in an electric field (Exercise 1 - Episode 4): In order for you to get family or with the notions of electric field strength that we have been discussing in the previous videos, I prepared some exercises. This video and the two next videos will be exercises and this one we have us feel 10 grams , which is carrying a charge of 0.3 Marco Cool. It is placed in a uniform field of electric fear strength 50 Newtons. Cool. The first question is determining the force on the sphere. Then calculate the acceleration of the sphere and finally find out how much distance it has covered in five seconds. Give it a shot. Let's look at the force on the sphere. The sphere carries a charge of the 0.3 Michael Cologne in this place in their field off strength 50 Newtons Pakula So we can calculate the force on the sphere, which is the charge of the sphere but applied by the electric field strength at the point where the sphere is located. The field is uniforms. I was going to be 50 neutral. Typical on everywhere. We can just plug in the numbers 0.3 by 10 to the minus six multiplied by 50 f will be cool . Therefore, to 15. Multiply that 10 to the minus six, which is 1.5 by 10 to the minus five new terms. The sphere is subjected to this electric force. It's the only force. So it is also the net force on the sphere. From that we can deduce its exploration. The Net force is equal to M a mass by suppression and this is 1.5 by 10 to the minus five Newtons. So we can find out the acceleration 1.5 by 10 to the minus five divided by the mass which is 10 grams in kilograms. So 0.1 giving you 1.5 by 10 to the minus sweet meters per second squared. We have the X elevation now, so it's pretty straightforward to figure out how much distance sphere covered in a certain amount of time. We can just use super at, um I down S u V A T s is the distance covered? That's what we're looking for you. The initial velocity Well, it's zero because of Sphere was at lest at the beginning V we don't know. A V is the final velocity and we don't know that a is 1.5 by 10 to the man sweet antes Five seconds. So what motion equation Contains s you A and T. It is s equals ut plus 1/2 a t squared. You're zero. So can kill this and I get s equals 1/2 a t squared. I plug in the numbers 1.5 by 10 to the minus three, but applied by five squared which is 25. And I need my calculator. For that I find 0.1 86 meters to see And if you configures so 1.9 centimeters.
6. Electric field of a proton in an hydrogen atom (Exercise 2 - Episode 5): In this exercise, there is a hydrogen atom consisting of a proton and the electron in orbit around the photo , the radius off the orbit, this five by 10 to the minus 11 meters. We also know the kinetic energy off the electron. It is 2.3 by 10 to the minus 18 jewels. The proton carries one elementary charge, so it creates a feel around it. That is why the electron is sticking around. Can you figure out the strength off the electric feared created by the proton at the level off the electoral I E. Where the electron is located, an electronics is a chance to political. It is located within an electric field left field created by the proton so the electron would experience the force. The magnitude of the electric field strength at the location of the electron will be the force experienced by the electron, putting a charge that is divided by the charge of the electron. We know that we need to find that What is the force experienced by the ritual? Where's the Coolum Force force? Between two charges. It will be f equals K Q. One Q two of a D squared. So if we replace this by the relevant letters, well, the charge of the electron is e that of the proton. Is he also So it's gonna be e squared, divided by the distance between them square. This force causes a circular motion. It's a centripetal force, so it is also equal to and B squared off the kinetic energy of the electron can expressed as 1/2 and B squared or, if you prefer and b squared can be expressed, asked to k so I can't like this in then the force is therefore to k e off the electron divide by our that I can plug in here. They get the electric fear strength. Its magnitude is going to be equal to two k e of the electron divided by r and I just need to plug in the numbers So three times 2.3 by 10 to the minus 18 divided by the charge of the electron multiplied by the radius of the atom. So the magnitude of the electric field strength will be with my calculator arts here 5.75 by 10 to 11 year term Super cool. How many significant things to so he would be 5.8 by 10 to 11
7. Deriving Coulomb’s Law from electric fields (Exercise 3 - Episode 6): this exercise, we have to charges Q one and Q two separated by a distance deep. There will be a force between them. The Coolum force, knowing what you know about electric fields, prove it. They live the Coon Falls. That's considered the two charges. Each of these challenges will create a field, allowing them, for example, Q two will create a field to allow at the location off chance Q. One. There will be an electric fear strength that we can label e to. So if you want, isn't the field created by you two? It will feel a force. Yeah, you want isn't in the to field, so the force experience they won. Let's call it f one will be equal to Q one e two. Actually, let's go. It f off to on one because it's child Q two, which is acting on Cuban via the electric field it generates. Now Cuban also increase in a tree field that will affect Q two. There will be a fools if one of the two equals two Q two E one under chance Q. Two At the location of Chance Q. One. The electric field strength due to the Charles Q. Two. We'll be OK. Que two divided by the square. Same thing at the level of Q two. There'll be an illiterate fear. Strength E one caused by charge Q. One distance from a distance D. Well, I just need to plug in, listen to their I get F 21 equals Q one K crew to 70 square. Have you arrange a bit and you will recognize Coolum slow in the same way you have the same thing here. F one of two will be to que 21 of any squared. Therefore, F 12 I'm writing to Fast, easy Cool to K Q. One Q two over these quit. That's how we can derive columns low and more. You realize that the two forces have got the same magnitude. So in the same way you illustrate Newton's third law action reaction. Yes, supposes to charges are positive. That means that would be a propulsion f off to over one F off one of two, so two forces are the same magnitude, but a positive direction
8. Electric field strength vectors in 1D (Episode 7): Let's consider two positive charges. Q one and Q two separated by a distance. Little D Each charge generates at electric field in the space. Valiant. Let's place a charge. Q. Between Q one and Q two at a point located at a distance. X from child Cuban, the charge coup would feel a force f equals. Could be, I suppose, that I want to calculate that force. I would need first to know the electric field strength e at the point where coup is located , right? Let's remember that electric fuse traits are vectors, so the electric fear shrink at that point is a vector some of the to fear straits course by Q one and Q two so I can write as vectors. E equals e one plus e to. But before adding vectors, we must first define a reference way. Let's consider an axis aligned with the three charges so that we can work in one dimension only on that said the positive direction off this access to worlds to light. You see, within this restaurants frame E one would be in the positive direction Andi, to in the negative dimension the magnitude of the result in fear. Strength is therefore e one minus e to we can now substitute. You wanted me to buy their respective formulas, which I do here on the screen. The magnitude of the result and electric field strength is cake. Who? One of X squared minus K crew to of a D minus X squared. Suppose that two charges Q one equal to to cool and coo to equal to three. Cool a place that two meters from each other. Can you calculate what will be the magnitude and direction of the force experience by a charge off, plus one cool place at mid distance between them. Give it to try. So we have two charges, both positive to cool here, frequent there distance from each other by a distance of two meters. At midway, we would place another charge of one cool, and we want to know what will be the force on this judge. So for that, we need to find the electric field strength. At this point, they have to Charlie's. Therefore, there will be two electric fields, so we need to find a result and electric field strength. At this location. There will be an electric field strength due to the child off to school directed away from it. Because remember the direction of an electric field strength is the direction of positive charge things. Let's call it a two, and there will be another electric field strength do to the fecal charge. That's quite the three. These are vectors, so I need to define a direction to find a resultant because they do the damnation positive two worlds select, for example, the result and electric field strength at this location will be the song off the electric field strength as vectors because if the numbers we need to consider the directions, he too, is positive. E three is negative. Now I can plug in the numbers. He is a cool two K to cool. Divided by that's 1/2 of two is one. So one squared minus K by fecal. I once quit. That's basically to k minus three cases just minus came. That means that my electric fear strength, my result in people fear strength, just be equal. I would be directed towards the left because I've got a negative value Now remember the question we wanted to find the force that would act on the charge of one cooler place. At that point, we now know the electric field strength of the position, so it's quite straightforward to calculate the force that the child would experience if placed that position. Yes, this force will be f equals to keep now in the text. We were considering a charge of one. Cool. If I place one cool here, well, it will just be one by He is minus K. So that's my escape. The fools experienced by a charge of welcoming place. There will therefore be minus 8.99 by 10 to the nine New Times, and I've been represented tools the left because it's native. Now try this one. Imagine that I replaced the value of the child Q two by minus three. Cool instead of plus free cooler. What would be the magnitude and direction of the force on the one room? Now? What would happen if I replaced this charge here off equal by minus sequel, where that means E three would be to the light? Yeah, a positive charge would be attracted by my sequel. Positive Child would go that way. I'm going to Macy's, which are not relevant anymore. Therefore, here I have a plus. So do I have failed. Therefore, this is now five. Kate. So the force experience by my charge of one cool auditor place at midpoint would be off magnitude five K. That is I'm going to use 9 10 to the ninth. So five by nine by 10 to the nine, which is 45 to the married 4.5 to 10 and you talks. So my force now on the charge of one cooler place there would be to delight and of magnitude 4.5 by 10 to 10.
9. Position of balance between 2 charges (Exercise 4 - Episode 8): in this exercise. Their two positive charges for Michael Cooler on nine. Michael Colon separated by a distance of one meter. If I place another positive charge between those two charges, there will be two forces. Where to repulsion forces. One portion forced you to this 11 repulsion force to through that one. Therefore, there'll be a point in between the two, where if I place a positive charge, it will be addressed because of to a portion forces will compensate each other. So can you figure out what that point is? By finding the distance X between that special point on the charge Q. One that's going to do any point between the two charges. Andi tried to find an expression off the result of electric history. Q One and Q two will generate to electric fields, so at this point, there will be two electric field strength. Let's look at them. You would have the one directed towards the light. Yeah, the direction of the electric fear strengths is the direction that the positive charge would take. This is a positive charge, So if I added a positive Charles there, it would just go away, find the electric field strength due to this one. Same thing about the other direction because this one is also positive. So e to on the resultant electric field strength at this point would be e equals the one plus two. If I want to work with magnitudes, I need to define a direction positive to the right. And I kind of like the magnitude of the electric field strength at that point is equal to the one which is positives that way, said minus E. To, because he, too, is the other way. Good. We are searching through a point between these two charges, which is special. It's a point where if I place another charge at that point, this charge would remain it fast. Ending the net force on the charge would be zero. If the Net force zero. It means that toe electric fear strength at that point needs to be zero. That needs to be zero. That would be the condition for that point. That means that the want equals the truth and you wanna we to there will be the notion of distance involved. Let's play in the formulas. Okay, 21 What will be the distance between the point and Cuba. It's X on between the point and Q two one minus six. We can cancel the case. Let's plug in some numbers. So we got full by tens of minus six. That's what you one X credit equals. Nine by 10 to the minus six. Other one minus X squared. Well, the 10,000,006. They couldn't go away. I can maybe we arrange things to make it a little sweeter. I prefer to work with linear equations, you know? So I got full by one minus X squared equals man X squared. And what I realized here because I'm trying to solve for X now, right? That's what I'm looking for. But I realize here is that this is an equation Where had a square one side in the square on the other congest Take this quote on both sides, giving me to buy one minus x equals three x and then so flex. So I developed this two months two x equals weeks so two equals five x giving me X equals to over five, which is point for meters problem. So if you have rigorous mathematician, you might have had a heart attack some. When this proof. Can you guess where? Yeah, When I have a squared equals B squared, there's no equals. B. It's a equals classman speak if there's two solutions. Therefore here. If I developed this by using the miners this time one minus X equals minus three X and up with X equals minus two, which is not relevant to our question. The question was, find a point between the two charges now outside of this segment. So this answer is not relevant. Hello. We have been working with electric fear strings as vectors in one dimension. Now it's time to have a look at what happens in two dimensions. See you in the next video.
10. Electric field strength vectors in 2D (Episode 9): when the test charges located inside to Morley Fields, we need to consider the resultant electric for your strength at the location off that charge. In the previous videos, we've seen this in one dimension. Now let's look at this process in two dimensions. We know now that the strength of an entity feel that given position off that field is a vector quantity. So to calculate the result of electric field strength, we need to have the fear strength off every field involved as vectors. Let me illustrate. Imagine you have two charges. One Q one here on another tree of charge, let's say Q two hit. That's consider them both positive. There will create two little feels wonder to this one wanted to that one. Suppose I place a charge anywhere? Let's say here has to travel to Cuba between color like and I want to know what would be the force on this time. So for that, I need to find the electric featuring that is occurring at this location. I need to find a result and electric field strength off the two fields. I have you won due to this judge, and I have you to due to this one. The result of field will be the vector sum of the two. So I can just Adam graphically and that would be the result in field. Therefore, the floors experienced by charging too cute, which was originally placed here, would be calling her With this resultant I feel strength and would be equal to F equals in order to calculate the magnitude and the direction of this result of village fear, strife and therefore the Force associate ID. We were working an example. In this example, we have a test charge. Little cute off two million cologne, three meters away from a child, Q one off nine Michael Coolen and one meter away from a charge Q two off to micro cool the I go between these two lines 30 degrees. The question is to find the force, which is only took you So when I mean find the force, I mean find the magnitude and the direction Q one and Q T creating electric fields around them. Therefore, there will both contribute to the electric fear strength at the location off the test charge. Yes, we have. He won due to Q one event. I would have to, due to co two. I can find the result and electric fear strength by adding the two as Victor's. I can do it graphically. That will be the result of electric field strength. Once I had the magnitudes and directions off result of military fear strain. It will be easy to find force on the took you That's ethical Street. The resultant electric fear strength would be the better song off the two contributors You want any to to do this vector. So my first need to find the components E X anyway, Therefore, I need to define from access so well True some convenient ones like this six and one. But before that, it would be very useful to find the magnitude off. One any to the one is okay. Do you want of a D squared plus B squared? It's like numbers nine by 10 to 9 by nine by 10 to minus six, divided by nine notes of nines and I can kill a path so good nine by 10 to the free. So that's 9009 kilo. New turns cool. You, too. Same thing Que que to over one squared at this nine by 10 to 9 countries to tens of my six giving me 18 kilo new terms for cool. Now we can go back to the store, mining the different components off the result of fear strength. So e x will be the one because he want is aligned with the X Axis Plaza contribution of Vetri on the X axis, which is plus you too. Course of 30. Yeah, The 30 angle here is found that too on the why will be equal to you to sign 30. So let's take a numbers. So I got nine. I'm going to use kill new tropical offers again for now, nine plus e true which is 18 cost of 30 discriminative for you too. And here I got 18 Sanders 30 which is 1/2. So that gives me nine cast nine. I went to three and this would be nine kilo new terms Pakula. So for this one, I calculated somewhere I found 24.6, 24.6 killing your terms Now that I have the magnitudes of the components e x n e y, it is pretty straightforward to find the magnitude off the resultant by using Peter goes formula and pledging numbers, I find 10.3 kill new terms. And for the direction I can define an angle theta this one which would be equal to the tangents minus one off the why component of the X component. So, like in numbers touching minus one off nine divided by 24.6. And I found 20.1 degrees. Right, So that's my resultant electric featuring Find the force test me easy f equals So to buy tens of my sweet to meet acquittal but applied by 10 0.3 I come to the feet So 20.6 new terms This worked example showed you how to determine the electric fear strength of the location where they are beautiful electric fields. Why don't you try it yourself? Check out the next video. There's a great exercise waiting
11. Multiple electric fields in 2 dimensions (Exercise 5 - Episode 10) : in this exercise, there are three charges, all identical, all positive. And that had for me and the correctable triangle trying was equal a jewel. That means that all the sides have got the same length l The question is, they will have an expression of the force that is occurring on any of the charges. Give it a try. Let's speak a charge that's picked this one. It will be under the influence of the electric fields created by this charge by this job. So there will be at this point and they fear strength like this and the need to fear strength like that There will have to say magnitude. What would be the magnitude of an egg? Fear? Strength. At that point, he would be okay, que of l squib. The disturbance in the charges is out and there will have the same magnitude, but not the same direction. That's why I will enable them differently. You want Andi to Let's call this one and two. Good. So the result of the electric field at this location would be the some of the to the vector sum of the two on That will be my resultant electric fish. If I can calculate the magnitude of this, I can find the force on the charge. Q. Most would be kooky. This is a vector some right? So I need to find the components on this one. Find the X and the Y. And for that I need to define stone axes, pictures positive that way and that way. So that would be except will be one. He X would be the contribution off the one on the X axis of the contribution of eating on the X axis. However, we know that have the same magnitude. We also know that we are in the liquid patrol triangle, meaning that here would go 60 degrees, 60 degrees, 60 degrees. Therefore, here we have also 60 to be 60 degrees. So I have to say magnitude they opposing direction. But they have the same angle with the x axis. They just cancel each other. That would be he won cause off 60 minus e to close off 60. But he want equals e to That's you using the Y axis. Now the contribution of you one on the Y axis here have 30 degrees B e one cost 30 and here e to cost 30. They both go in the direction of the Why access. Therefore, it's positive. But we know that you want equals e true that have same magnitude on their equal to k crew of l squared. So I get there two times. Okay. Two of us quit bike asserting, which is two times k of l squared price close to 32 I can kill the twos giving me the y equals case credit of three to all square e The result of the electric field strength is a some of its components, but the X's you so I can just Why That e is just k scary to three. Cooper was quit. I'm looking for the force, which will be called to buy E which is case primitive three to events Quit. I can square the cruise. Thank you. Okay. Square root of three. She's glad over those quick. We can also reflect on the donorship. Well, the force will be calling her with E so I could draw it here. That would be the force on charge. Q. What about the other charges? Well, the problem is exactly the same is just an imitation of 60 degrees. So perfectly symmetry. I can imagine the medium here, and you see that the forces aligned with the median off the triangle. I could do something here or here, giving the f that way and also f that way.
12. Electric field lines (Episode 11): on the electric field line represents the path that a positive charge would take when placed in an electric field. The presentation on the field using fear lines is very practical because it can describe the effect of the field on the charge at every point, like a map. To really understand how this works, let's put out the bold on Build some field lines together. Suppose that the board is a space in which a place two positive charges Q. One on cue to suppose also that I want to know what would be the direction taken by a positive charge placed at a given point. Say this one. Through that I would need to find the result of the electric field strength at that point, which is the vector sum off the two electric field strength through to Q one and Q two C. That would be the one, and that would be in two. I mean, the sum of the two, and I get the resultant electric fear string and wet. So let me remove the construct int victors. Joe, if it please the charge to here a positive charge, it would follow that direction, But wait if it moves along that line, say when it's here. You realize that the distance between the charges and that point I'm not the same than here . I would need to redraw the lifting fear strength at that point by considering the to individually tree fear strength he wanted me to and I would get something. Maybe it event. That's 1/2 electric featuring at that point like this. How did she feel strength at that point like that? And I could continue. What I'm doing here is I'm drawing the path off the particle, the path of the particle. We'll follow a line here for which all the fear strength would be tangents. That's a fear line I show here narrow to show the direction that the positive charge would take. If I placed it here, it would just follow that path. So I feel line is a path that a positive charge takes in the late field. And if we find out all the result of electric field strength of all the points in the field , we can draw the feelings. But you can do it intuitively too, right? You get something like this, don't forget to put the arrows. If at the test you are lasted through fear lines and you didn't put the arrows, you lose the mouth. So, for example, if I placed the charge at that point, it would be more like this. Actually, that place a charger. At that point, I know a positive charge. I know. Just follow this path. These are for two positive charges. Imagine now that I have one positive charge and one negative charge, let's say something like this. Plus here, on the minus that we can also draw the fear lines. Why don't you try? Let's consider this point. At this point, a positive child would be repelled by the plus. I am attracted by the negative, but the portion would be stronger. So we can imagine that the electric field strength would be maybe something like that for position midway between the two charges on the X axis. So here. Well, you see, I would have as much repulsion, then traction, so I probably would have any tree few strings which is or resentful here. I would have more traction to do negative standard portion from the positive. So maybe my fear strength would be a bit like that. And if I draw the line such as ALS, the fear strength attendant of this line, I get something. Magnets, only the arrow. To show the direction of the positive on, I can easily find tuition, figure out the others without forgetting potato the house. So the felines give you information about the direction that the positive charge would take . But it's not the only information you get when you're close to a charge. How do you expect it to fear? Strengths To be, well, large? Why you're close to the charge. So the effect of the charge will be strong. The electric field strength will be stronger. Look at the feelings. When you're close to charge, there are much more dense. So you see the connection here. The density of the felines is a measurement off the strength of the few to see if I take an area here. I think the same area there. Here, I have to feel lines in this area. He only have one. The density is higher there. Therefore, the strength of field is larger. There, the density of the felines inform you about the strength off
13. Radial & uniform fields (Episode 12): There are two types of electric feels which we have to pay special attention to Well, your fields on uniforms, fields. Let's explore these together. Imagine that place and you get in charge here. I want to draw the field lights. So imagine positioning a positive charge. Stay here. And I realized my electric fear strength will be the direction that the positive charge takes. So it would be like that for further away. It will be smaller, but it will have the same direction. Her further away, even smaller. Back with Samed Action. Second drawer line, which is tangent to all these actors, which is actually calling here with all the vectors which would be like that. That would be my field. So I need to put the arrow showing the direction of the politics of jobs would take Have appeared the operation. Well, I find my felines, all pointing tools. The charge You can imagine a circle around discharge and the fear land would be the radius of the circle. This is why we call it a radio field. Imagine that the charge here was positive instead. What should I move that Find my diagram? Yes, the direction of the hours because if it was positive a positive charge would go the other way would be repelled. Eso my fear lines need to have an error away from the positive charge. Let's consider two parallel place that are charged. One is positively charged and the other is negatively charged. There's also imagine that the legs of these plates is infinite. The charges within each plate repel each other so that distribution is such that the distances between them are all the same, basically maximum. What's important for us here is that the charges within the plates automatically distribute themselves uniformly. So now let's try to draw the electric fear, strength vectors and the fear lines in the space between the two plates I represented on the board to parental plates, which are charged one positively one negatively. Let's figure out the electric field strength within the plates. Take a point. I realize that I will have contribution off all the charges to marry from fear strength. Yes, each charge creates an aging field, so each charge will contribute on the part of the electric fear strength. At that point, for example, this one will have I need to feel strengths that way. This one that way. This one, that way. This one, that way, This one. That way, it's a trap. We can't do the same thing with the negatives. What you realize is that on the X direction, because the charges are uniformly distributed, you don't have that need. If she feels drinks, component only goes down. Let's the X component of all the electric field strength will cancel, leaving you only with the white composed on. If you consider the the plates like infinite, this would happen everywhere. The electric field strength will have the same magnitude in the same direction anywhere. Therefore, your field lines would actually be justly uniform and in the same direction. It's a uniform field that you have between the two power plants. If the power of place are not infinite, though, you can. I fi figure out the felines outside of the place, making kind of little ears. You can easily see this by trying to draw the electric field strength at the favors positions just at the outside off the two plates. So between two power place, you get a uniform field where the feelings are all uniformly distributed. on or in the same direction. In the next video, we will work on the family to exercise that occurs within a uniform field formed between two chart parallel plates. See you there.
14. Motion of a charge between two charged parallel plates (Exercise 6 - Episode 13): to conclude this lesson on electric fears. Let's work on their fund little exam type of exercise. They have to payroll plays here, which are charged creating in a field inside the plates, which is uniform on and off strength. 20 killer neutrals. Pakula. The length of the plates are five centimeters on the distance between the plates two centimeters. Midway through the place, there was a proton that is going to enter the two plates. The proton has a speed off. To buy 10 to the five meters per second on the mass of the proton is also given. The first question is about drawing the past of the proton when it is between the place we know roughly. The second question asks you if the proton will hit any off the plates. On the final question, ask you what should be the maximum electric fear strength so that the proton party through the plates without hitting them. Give it a try. The first question consists, enjoying the parts of the photo, but first, while it goes in a straight line. But here, when it enters the plates, it was under the influence of an electric field. The electric field strength will be downwards. Therefore, positive charge would feel forced downwards. The photo is a positive charge, so fear the force downwards. So the poor tone accelerates downwards when it enters the planes. So the path will look like this. And this representation. I see that my proton hits the plates. I don't know that yet. That's the next question. So what is the force on the photo? It's to be. Q. Is the charge of the Proton is electric fear strengths Off the field in which Proton is located? This is the only force on the photo. Therefore, it's also equal to May. From that, I could calculate the exclamation. Two e. R. M. Let's take in numbers is 1.6 by 10 to the minus 19. It applied by 20,000 divided by the masses, 1.67 by 10 to the managed 27. And on my notes I found 1.92 by 10 to the 12th meters per second squared. Good. So what can I do with this acceleration? Well, I could calculate the time it takes for the proton to reach the plank and compare it to the time it takes for the photo to pass. Sue. Five centimeters. Yes, on the x axis. Just be this constant. So how long would it take for the proton to past five centimeters at this speed? Well, that's pretty simple. I just two vehicles to have a tea giving me t equals D of a V. At this point, you your 55 centimeters divided by the speed to 10 to the five. And I think this as being 2.5 by 10 0 minus seven seconds. That is a time the proton would take to cover five centimeters horizontally. If the time it takes for the proton to hit the plate is less than this time, then it hits a plate. So let's calculate this. Just do. That s one centimeter you? Well, it is in our office rental velocity over generally. So on the 30 Corps direction, it has a velocity of zero B. I don't care a I just calculated it here. 1.92. I tend to 12 anti. That's what I'm looking for. Circled use s equals UT plus 1/2. It is quiet beauty councils because it's it doesn't cancel this Jesse course zero So now you've got 1/2 a T squared T equals skirt of two s of A. That is 0.2 divided by this number 1.92 by 10 to 12. I find for tea 1.0 by 10 to the minus seven seconds. Well, this is much less than that. They're full. The proton will hit the plate. It doesn't have enough time to pass through the plates. It hits of plates in 10 minutes, seven seconds. But it needs 2.5 by 10 months. Seven seconds to cross the whole frame. So it does hit the plate. There is an alternative way to see this. If you prefer. It depends on the people you could consider. Okay, this is the time I have available. How much distance vertically can I cover in this amount of time? So I do the suit battle. So But this time I'm looking for the distance covered in this amount of time. Vertically. So they steal zero. This is I don't care. This is 1.92 by 10 to 12 on the time would be this amount of time. So I could use also s equals UT plus 1/2 a T square. So you're is you're so 1/2 a T squared. Plug in the numbers and the answer I find is to your point to six meters which is more than 0.1 meters. So in this amount of time vertically, the proton would cover six centimeters, so it would hit the plate. 10. Conclusion, Of course, in the next question were requested to find the maximum value off the electric field strength so that the proton does not hit the plate. If I decrease the value of the electric beer strength, I will decrease the force on the proton right because equals three. But this is also equal to you may so I can write down that e equals and may have acute. So if I decrease the force on the bottom, I will decrease also its acceleration. The past would look maybe something like this. Now if I could find the acceleration so that my path is exactly like that, meaning that when the proton covers five centimeters, it covers exactly one centimetre vertically. If I can find this acceleration, I would find my electric field strength. So let's do that vertically. I want to sue. That s would be 0.1 you zero the I don't care a will be, um that's what we're looking for. And t where t will be the time it takes for the proton to carry out five centimeters off is entirely that we calculated last time it was 2.5 by tens of minus seven seconds. So I get s U 18 s equals UT plus 1/2 a T squared zero So s equals 1/2 a t squared. I can find a two s of a t squid. So 0.2 divided by 2.5 10 to the minus seven squared. So a equals What did I find? 3.2. I tend to the 11 meters per second Quit. So how the acceleration for which the proton will cover five centimeters horizontally and one centimetre vertically. Exactly. Now I can just plug in this in here. My fear strengths would be So the masses put on 1 67 by 10 minutes. 27 by this accident. Asian 3.2 by 10 to 11. If I did by the elementary charge 1.6 by 10 to the minus 19. But she's 3340 new terms