 ## Physics

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24. Electric Force & Field; Gauss' Law

# Electric Field

1
concept

## Intro to Electric Fields 3m
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Hey, guys. So in this video, I want to talk about electric fields. It's something that you'll see in your textbooks. You'll need to know how to solve problems with it. But it's also to answer a very important question, which is how to charges exert forces on one another. And the basic answer is that they do them by setting up these things called electric fields. So basically a single charge right which I'll call big Q is gonna produce an electric fields. You might see some confusing definitions in textbooks, but basically the idea is it's kind of like information this big. Q is setting up this thing called an electric fields right that exists in all directions. And then, if you place another charge at some distance, little are away from this produced electric field, then this charge will feel that field and actually have a force on it. So, in other words, it will have a repulsive or attractive force here, which we know from cool arms law. So basically what happens is this thing. Regardless of whether there's another test charge or another secondary charge to feel it, this thing produces an electric field in all directions. And then when you place a charge inside of that field, it feels of force. And that force that it's felt with the electric field is which is set up by big Q is just Q times E in which he has units of Newtons per Coolum, right? You probably don't need to know that, but there it is anyways, So he has units of Newtons per cool. Um, Okay, now which Q are we gonna use in these questions? Well, that Q is always the one that's feeling the force. So the words the secondary charge right here. Now, some of you might be wondering, Whoa, isn't the opposite also happening? Doesn't this Q also produced its own electric field and therefore this one is felt by it, and the answer is yes. So this little Q also produces an electric field that is either felt an attractive or repulsive force. And this is basically how to charges exert forces on one another. You have action reaction because of these things called electric fields. Okay, so again, I just want to reiterate in problems the queue that you're gonna use inside of this f equals qu equation is always gonna be the one that feels the charge or the the electric force. At some distance, little are okay. It's a simple is that Let's go ahead and take a look at an example. We've gotta to cool um, and three Coolum charge. They're separated by some distance. Are So I'm gonna go ahead and set these up. I've got to Cool. OEMs. Uh, got to cool. OEMs. Three columns. They're separated by a distance of little are in such a way that the electric field at this point is 10 Newtons per Coolum. In other words, we have an electric field here at this distance. Oh, sorry, Actually have it backwards. So this is actually gonna be the three Coolum charge, and this is gonna be the too cool in charge. Now, the electric field at this distance is just 10 Newtons per Coolum. And now we have to feel what we have to figure out. What is the force on this to Cool? Um, charge. What does that mean? That means that this is the Q, and this is the big Q. So it all depends on which one is the feeling charge and which one is the producing charge. So this one the too cool in charge is the feeling charge. So in order to figure out the force, we just need Q times E So the force is gonna be Q, which is too cool homes times the electric field at this particular point, which we're told is 10 Newtons per Coolum. And that's just equal to 20 Newtons. Okay, so that's just a simple example. Let's go ahead and get into some or examples involving electric field.
2
Problem

A 1.5μC charge, with a mass of 50g, is in the presence of an electric field that perfectly balances its gravity. What magnitude does the electric field need to be, and in what direction does it need to point?

3
concept

## Electric Field due to a Point Charge 6m
Play a video:
Hey, guys. So now that we've seen what the electric field is and how charges respond to electric fields, we're gonna look a little bit more closely at the electric fields of point charges. Okay, you don't need to know that. So basically, the idea is we know that charges produce electric fields, and in previous videos, we were sort of like drawing squiggly lines here. But the reality is, instead of squiggly lines, thes positive charges, positive point charge is actually produce an electric field that points radial e outward from it. The way I like to remember this is that positive people are very outward going people. So this is the electric field produced by a positive charge, like one cool, um, or something like that. Whereas negative charges are just the opposite. And we just chose this convention, by the way, hundreds of years ago s Oh, this is just what we established as positive and negative electric field. So these things produce fields that point inwards. That's the main difference between them. We've got outward going and inward e fields like this, but you'll definitely need to know that now We saw how a single charge where it was like this Q. Or this Q over here was basically producing an electric field. And that that when you brought a secondary charge closer to it. So, for instance, if I brought a Q like this, then it would feel a force, whether it's in this direction or this direction, depending on the magnitude of that. Well, the magnitude of the electric field now has an equation, and it's k times big. You divided by r squared. Notice how this kind of looks like Coolum Law, except it's missing one of the little cues. So the main thing is that remember, the electric field is set up on Lee just by one charge. The producing charge Big Q. And the other thing is that this little our distance, instead of being the distance between two charges like we had in columns law. Now it's the distance to the point of interest. So what that means is, that problem will ask you what is the electric field at a certain number, or sometimes you'll be given a diagram of charges, you'll need to figure out an electric field at a specific point. So that's how you're going to get that little our distance anyway, So we have this equation of the electric field generated by a producing charge. Q. And we said that a second charge, if it's moved closer to it that little. Q. It felt a force from that electric field. That equation was given by F equals Q E. What we could do is basically plug in this expression. Now we can actually come up with something interesting. We know that e is just k Q. Divided by r squared. So if you multiply times this Q. If you recognize this formula, all this is is just cool ums law. So this really brings the relationship between the electric field and the electric force. So a producing charge Q produces an electric field. And then when you bring a smaller charge or doesn't have to be smaller, a second charge key. Little. Q. They both feel a force, but the opposite is also true, So you could actually flip little Q. And big Q. One could be the producing. The other one could be the feeling charge, and they're gonna be the same. That's why we have action reaction pairs between two charges because they both exert that mutual forces on each other through their electric fields. So now we get the complete picture between the relationship between Coghlan's law and the electric field. Let's go ahead and take a look at some examples right here. We've got a distance X and the electric field is some number at another distance. Here we have 10 columns and now what is the ratio of between X and Y? So here's what I want to dio. I basically want to draw this little producing charge, which I'll call Q right here Now at some distance away so that I'm gonna call this X. I'm told that the E field here E X is equal to Newtons per Coolum, right, And then at some other distance. So you keep going over here this other distance right here the electric field is going to be equal to 10. By the way, this doesn't necessarily have to be half. It's not to scale or whatever. So this whole entire distance here is equal to why. And now I need to figure out what is the ratio of X and y given those electric field intensities. Okay, so basically what happens is I know that, uh, the electric field at a certain points at, for instance, point x. The equation for that is gonna be K times big Q divided by the distance between X squared. Whereas this guy over here, the electric field at point why is K big Q divided by Y squared. Now what happens is we actually don't know what the producing charge, what the magnitude of the producing charge is. All we need to do is just get the relationship between X and Y. So to do that, you have to divide. We're gonna need a ratio. Right? So what I'm gonna do is I'm gonna divide e X by ey so you're gonna get K big Q over X squared all over K big Q over. Why squared now? The reason this is important is because we can actually cancel out the case and the cues that appear on both sides. Right. If you do copy dot flip. If you manipulate this equation, you're just gonna get the same thing on the numerator and the denominator. So what we end up with when you do this fraction is you're gonna get e X over e y is equal to we have one over x squared, divided by one over y squared. So this is actually gonna go all the way up to the top, And it's gonna be why squared over X squared. OK, so now what we can do is we want to get rid of this X squared over here. So we want to square root both sides. So we're just gonna get this square roots. We make some room for that. We've got the square roots of e X over E Y is equal to the square. Roots are actually when you when you take the square root of these y squares and X squares, they're just gonna turn in tow Y over x. Okay, so now we've got something that kind of looks like our ratio here. We've got y over X. All we need to do is we need to flip it because we need X over y. So all we have to do is just flipped both sides and I'm gonna move this over here, actually, so I get the square root of e y over E X is gonna equal X over y right. So all I did was I just flip that this was the X over e y. Now it's ey over e x, and I would just flip. This is well, right. So you could do that. You could just flip both sides of them cool. So I know I have to plug this stuff in, so I know that e y over here is 10. And I know the X is 20 Newtons per Coolum, and that's just in e equals X over y. And if you plug that and it's just gonna be a ratio, and you should get I get 0.71. So, in other words, that is how far E X is relative T. Y. So this is about 0.71 of the distance between this whole entire line segment here. Okay, So let me know if you guys have any questions, let's keep moving on
4
example

## Zero Electric Field due to Two Charges 6m
Play a video:
Hey, guys, lets you got this problem together. So we've got these two, these two charges which lie on the X axis, and we're supposed to figure out what distance or a point on the X axis Is the electric field going to be zero? But I want you to remember that for this problem, we're gonna be using the fact that again outward or positive charges produce outward going electric field lines, they always wanna go away. Whereas negative charges produced in we're going electric field lines. This is gonna be useful because we actually have a positive charge right here and a negative charge right here. So we're gonna need to remember which directions those electric fields go in. And we've done these kinds of problems before when we were looking with columns law. So in order to figure out basically before we start plugging in numbers, we have to figure out which region is this possible for the electric fields to cancel out. So I've got three basic choices I've got on the left I've got in the middle and I've got on the right. Now, if you look the middle, what's gonna happen is at this point of interest. Some of the words this p right here, the electric field from this positive to cool, um, charge all the electric field lines, we're gonna point away. So that means that at this position right here, the electric field from that positive charge is gonna be towards the right. You see, You see, that always wants to go away. Whereas the electric field from the negative charge always wants to go towards that charge. So that means that at this point, it wants to go to the right as well. So, basically, this is the electric field from the negative charge. And there's never gonna be a situation which these things will cancel out. They're always going to add together, no matter where I am along along this line between them. So that means that it can't be in this area because they never cancel out. So let's take a look at the right region now. So we have the inward going electric field lines, so that's gonna be e minus. And from this positive charge, we have a an outward going electric field line. Now, you might be saying you might be tempted to say Hell we actually might be ableto you know, find a place where these things cancel out. What I want you to notice is that the electric fields for any charges is just k times big Q divided by little r squared. So the words it's the magnitude of the charge divided by the distance between them squared. So this three Coolum charge is always gonna be at a smaller distance from this point of interest remembers to some random point here than this entire distance to the weaker charge. Now, what this means is that at this region here, the magnitude of the net of the negative electric field is always going to be bigger than this positive electric field because it's a larger magnitude of a charge, and it's always going to be a smaller distance because this distance right here is always going to be larger, and it's a weaker charge. So, in other words, the magnitude of this is always gonna be bigger than the magnitude of the electric field. So it's never going to cancel, so it can't be this, and instead we're gonna have to use our last scenario, which is on the right. The magnitude of the electric force. It goes in this direction, their direction of the electric field. Wow. And I've got the inward going electric field from the negative three Coolum charge. Now there is possibly a scenario where I could find a distance in which these will cancel out the weaker charges at a smaller distance. The larger charges that a larger distance so there might be a way than for them to balance out the variable I'm looking for is this distance right here, which is X at what point on the X axis is the electric field equal to zero? And what I'm gonna do is I'm just gonna set the origin over here to be zero. So this is just gonna be X equals zero, which means I'm probably gonna get a negative number or I'm just gonna have to specify to the left of the cuticle, Um, charge something like that. And the condition that I'm trying to solve for is that these things should be equal but opposite in direction to cancel out. So in other words, I should have that the magnitude of the e of the electric field from the positive charge should be equal and opposite to the magnitude of the negative electric field. All right, so we know what the magnitude the electric fields are for This guy over here, it's gonna be K. We've got the producing charge, which is the two Coolum charge. And this is gonna be at a distance of X squared now that has to equal the inward going electric field. That's gonna be K times the three. Cool. OEMs. Notice how I didn't put the negative sign there because I'm just working with the magnitudes of them, right? So just set their magnitudes equal to each other. And now this distance right here is actually not going to be X. But it's gonna be X plus the seven centimeters square, so actually could just write that. So I've got X plus 0.7 square. That's seven centimeters. Just written in s I Right. So now what I can do is I can try to solve for this x variable over here. So I got this equation and I could go ahead and cancel some stuff out. I got the K's. And now, in order to get this X over one side, I have to move this thing over and I have to move the two down, so basically, they just have to trade places. And if I do that, I'm gonna just continue writing all this stuff in black because it's gonna be, ah, pain if I try to switch colors. So I've got X plus 0.7 squared, divided by X squared equals And then I've got three halves. Great. Now, rather than starting to foil all of this this whole, this whole mess out and getting a whole bunch of quadratic terms, things like that, what you can do is realize that both of these terms on the top and bottom r squared, which means that we can take the square root of each side. So we're gonna take the square root, take the square root, and then it just becomes X plus 0.7 over X equals the square root of three halves. Okay, so now all we have to dio is to start isolating for X. I'm just gonna move it over to the other side. This becomes X plus 0.7 equals square root of 3/2 times X. And now all I have to dio is I have to subtract this X to the other side. So by up to minus X, and I have this 0.7 equals the square root of three halves X minus X. Now I can do is I can pull this X out as a common factor because it's like, you know, it appears in both of these terms, so basically have 30.7 is equal to x times the square root of three halves minus one. Notice how if you distribute this X back inside of both of these terms, you're just gonna get back to the expression that you had right above right? So it's just it's a good shortcuts is basically just pull out that greatest common factor. Okay, and now what we could do is we could basically just move that over and divide. So we have 2.7 divided by the square root of three halves minus one, and if you just plug that stuff in, you should get the X distance, and you should get 0 31 meters or that's just 31 centimeters. But remember we said that it's going to be to the left of that too cool in charge. So we just have to specify its to the left off the too cool, um, charge. And this is the answer. So if there was, there was, like, an actual coordinate system here, this would be, like minus 31 centimeters anyway, but this is actually perfectly fine. If you just left it like that, let me know if you guys have any questions. And if not, we're gonna keep going with some more examples.
5
Problem

If two equal charges are separated by some distance, they form an electric dipole. Find the electric field at the center of an electric dipole, given by the point P in the following figure, formed by a 1C and a −1C charge separated by 1 cm. 6
example

## Electric Field Above Two Charges (Triangle) 4m
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7
Problem

4 charges are arranged as shown in the following figure. Find the magnitude of the electric field at the center of the arrangement, indicated by the point P. 8
example

## Balancing a Pendulum in Electric Field 5m
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