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Charges In A Line (Find Zero Force)

Patrick Ford
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Hey, guys, how's it going? So I wanted to work this problem out together. We're supposed to figure out where we need to put a one Coolum charge so that the force on it is equal to zero. So what does that mean? We're just gonna assume that the net force So what's asking for is that the net force needs to be equal to zero, because we have to cool OEMs, or we're sorry. We have to charges on the outsides. So what this problem is saying is that we need to put a one Coolum charge somewhere around here in order for the forces on it to cancel out. So before I actually start plugging anything in, I just want to figure out qualitatively meaning no numbers where this thing needs to be in order for those horses to cancel out. So let's take a look at the left side. So imagine I dropped a one Coolum charge here and I want to figure out if those forces would cancel. So all I have to do is just figure out the directions of the forces on it from the two Coolum charge. Remember, all of these things are positive. So they're all going to repel. They're gonna repulsive forces on each other. This thing has to move to the left. Right? So this is the force from the two Coolum charge that's gonna point to the left. But the force from the three Coolum charge also has to point to the left. It might be stronger or weaker, we don't know, but it's going to point to the left. So in other words, there's never going to be a situation where these forces are going to cancel out. They're always going to add together and point to the left so it can't be here, right? And so I'm right here that these things never cancel. There's never gonna be a situation in which one will point in one way one will point in the other. So it can't be in this area in this region right here. So let's look at the same. Let's look at the situation on the rights. Who got this one cool in charge here and you don't do the same thing from the three Coolum charge. You have a four set points in this direction. The too cool own points points in this direction. So for the same exact logic this thing will never cancel out. These forces will never cancel out. So that means it has to be somewhere in the middle. And the reason for that is if you work out the forces here, the force from the too cool on charges gonna point in this direction and the fourth from the three column charge needs to point in this direction. So that means that there is some magical distance in which these things will cancel out. Now we have to be careful because it won't exactly be in the center. If this thing were in the center than this stronger charge here will produce a stronger force. So this thing actually needs to be slightly closer to the left so these things will balance out. So that means that the condition that we're trying to solve for the rest of the problem is that we need the magnitude of the two Coolum force to be equal to the magnitude of that three Coolum force or the forces from those two charges. In other words, this is the condition that we need to solve, right? They're gonna be pointing in opposite directions. But if their magnitudes of the same, they're going to cancel out. So let's go ahead and write out the expressions for those two. So the two Coolum charge right here I need a distance between these two charges. By the way, this is a one Coolum test charge that we're gonna drop right here. So I'm gonna call this distance here, X. And that means that if this distance between the two and the three Coolum charge is are so this is our equals 10 centimeters, then this is just our minus X rights. That's the distance between eyes just the whole entire distance minus that little chunk of X R minus X. So I've got the K constant times the two charges that are involved. In other words, this is gonna be the too cool in charge and the one Coolum charge divided by the distance between them. That's X squared. Now I set up the equation for F three. So that's this guy right here between the one Coolum charge. So okay, and then I've got the three and one. So that's three columns one cool arms divided by the distance between them squared R minus X right and then squared, so I don't know what those distances are. Now again, what we're supposed to solve is that these two things are supposed to be equal to each other. So in other words, these expressions right here that I've just figured out have to be equal to each other. So let's just go ahead and actually said those things equal. So we've got K to one divided by X squared equals que three one divided by R minus X squared. So this thing is an equal sign, so we can actually cancel out anything that appears on both sides of the equation. We're gonna get the case will cancel, and also the ones will cancel. I mean, one doesn't really do anything, so I mean it. Just cancel it anyways. And so we come up with an easier expression. Now, what I'm trying to solve is where I need to put this one cool in charge. So that's a distance. So in other words, I'm gonna go ahead and solve for X. So X is my target variable here. So just because of this, the easier one to work out like I could actually go ahead and solve for what ar minus X is and refined. But it's just that X is an easier variable to solve. So let's go ahead and do that. If I wanted to solve for X, I had you to get all the things involving X to the one side. So I'm gonna move this ar minus X over to the other side, cross multiply, and then this to needs to come down and basically trade places with it. Right, So you just have to cross multiply. Okay, so we get our minus X squared, divided by X squared is equal to three over to right. We have the three and the two on the outside. So all you have to do now is just take the square roots. So you might think you have to foil this thing our our might X squared, and you're gonna get these all these ugly terms. But the thing is that both of the terms in this expression R squared so you can actually just take the square root of both of the whole entire thing. So if you take the square root of both sides where you're gonna get his ar minus X just one term and over X is equal to the square root of three over to. So now how do we get How do we start isolating this X? I just have to move this stuff to the other side. So I have to bring this X over. And this becomes ar minus X equals the square root of 3/ X. And now all I have to do is this a subtraction? So I could move this X to the other side and and add it right? So I got r equals the square root of 3/2 X plus X, right? So and then I have this X as the greatest common factor inside of both of these terms. So God's r equals that about X parentheses, the square root of 3/2 plus one. Right. So if you distribute this X back inside this expression, you should get back to what the thing on the top is. All right, so now all we have to do Is this our distance? Remember, it was the 10 centimeters. So that means if I wanted to figure out in meters, that's just gonna be 0.1 So I got 0.1, and now I always to do is just divide this over to the other side in order to get rid of it. So I've got this expression here on the bottom is gonna be square root of 3/2 plus one, and that's gonna equal X. If you plug all of this stuff into your calculator carefully making sure that you have a parentheses here in the denominator, you should get an X distance of 0.45 That's in meters. So in other words, that's 4.5 centimeters. And that's our answer for X. So, in other words, going back to the diagram, this thing, this one Coolum charge has to be 4.5 centimeters away from the left. Coolum are from the left charge, and that means that this remaining distance over here is gonna be 5.5 centimeters. So in other words, it confirms we were saying before, this charge has to be slightly closer to the weaker charge so that smaller distance can balance out the larger charge on the right. Alright, guys, that's basically it. Let me know if you have any questions with this