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Electric Field due to a Point Charge

Patrick Ford
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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