Charges in a Plane

by Patrick Ford
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Welcome back, guys. Let's try to solve this one together. So we've got these three point charges arranged in this triangle right here, and we're trying to find out what the net force on this three Coolum charge is. So let's go ahead and first draw. What? The forces that are acting on this three cologne charge. So we got We're focusing on this guy right here. And the first is that he feels a attractive force from this negative too cool in charge. Why is it attractive? Because these things are unlike charges, ones positive ones. Negative. So this is gonna point in this direction, and I'm gonna call this F two because it comes from the too cool in charge. Now, there's also this cool, this one cool in charge here on the bottom left corner, that is some distance are away from this other charge, and it's gonna exert a repulsive force on the three cool in charge because these things are like charges. They're both positive. I'm going to call that F one, and what we're supposed to do is take thes forces, which is the only force acting on it and figure out what the net force is on this object. So in order to do that, we have these forces there pointing in different directions. We have to use vector addition. So with vector addition, we always have these steps right here. First we get a label and calculate these forces, which we've labeled already. So I'm gonna take that over here, and then we have to decompose them, pick our pick our directions and add them in this all for the, uh, net force. All right, so let's go ahead and calculate the forces. So I got f two and I'm gonna be using columns. Law write columns. Law is just k times the Q one Q two divided by the distance between them squared. So that means that F two, which is the force between the negative two columns and the three columns, is gonna be K which is that constant? 8.99 times 10 to the ninth. And then I've got the two charges involved. So I've got to two and three divided by the distance between them square 0.6 square. Now, some of you might be wondering why I haven't chosen that negative sign and it's because remember, whatever we're finding, the Coolum force were always just going to plug in positive numbers. We're gonna worry about the direction later. So if you work this out, you should get The F two is equal to 1.5 times 10 to the 13th. And that's a Newton right here. Right? So you don't need the units for that if we try to work out F one. So in order to work out F one, we need 8.99 times 10 to the ninth. We've got the two charges, right? You've got this three Coolum and the one cool. Um, but we don't have the center of mass or sorry. We don't have the distance between them. That little are. So hopefully you guys realize that this is a 68 10 triangle. Right? So this should be 10. But if you didn't, you could always just use the Pythagorean theorem. So you've got six squared plus eight squared, and that equals 10 centimeters. All right, so we want this in s eyes. This actually just be 100.1 m. So first we have to go. We have to do three columns, One column and then divided by zero. We got 0.1 square. And if you work this out, you should get 2.7 times 10 to the 12th. Alright, I'm trying, right? That's small enough just so we don't run into this little thing right here. But anyways, now we've labeled and calculated both of these forces. So we're done with that step that first step. Now we just have to go ahead and decompose this thing. Right? So you've got these forces that are pointing in opposite directions at once pointing to the left one is pointing upwards, so we got to decompose it. So I've got to decompose this F one vector by basically moving it and projecting it down to the X and y axis. And I do this by needing by using my trig and my co signs and things like that. Right? So, in other words, I need my f one X components and my F one. Why components. Now I've got this. Um, I know if I wanted to break up this F one into its X and Y components, I know I relate that using the sine and cosine. So F one is gonna be Kasey in F one X is gonna be a F one co sign of data and F one. Why is gonna be f one of sign of data? So I know what these forces are I'll have to do is just figure out how to get Thetas. So I use that by going ahead, looking at my triangle, and usually I'm told what the angles of these things are so I can go ahead and figure what this angle is. Or I could just use the relationship of sine and cosine in the triangle itself. What I mean by that is that I've got this angle Fada right here, which, by the way, is the same as this angle theta right. These things are parallel lines. This thing is a parallel parallel horizontal lines, so could basically complete this little triangle. And I know that this is eight centimeters. So I've got 68 and I know the hypothesis of this triangle is 10. So that means Well, let's use our sine and cosine rules. Right? Co sign of anything is just adjacent over high pot news. So, in other words, co sign of this angle right here is the adjacent side over there. Hi, pot news. That's just 6 10 6/10. And if I did the same thing for sign using so gotta I've got opposite over iPod news. So I've got 8/10. So, basically, instead of having to figure out the angle theta and then having to use that, which is normally what we do here, all we have to do is just replace thes co signs and signs with these fractions right here that we figured out it's the same exact thing. So let's keep going. We've got f one X is just gonna be F one, which is 2.7 times 10 to the 12th, and I have to multiply by co sign of fatal, which is just 6/10. So if you do that and you work it out, you should get Let's see, I've got 1.62 times 10 to the 12th. Now, the white component is gonna be very similar to 0.7 times 10 of the 12 Now, instead of 8/10 we just do 10. Uh, sorry. Instead of 6/10 you do 8/10 for a sign. So if you work this out, you're gonna get 2.16 times 10 to the 12th. All right, so now you've got the components. So now you have to basically just pick our X and Y sorry Are positive or negative directions. So we've been working with up into the right. Usually that's what we picked to be positive. So let's go ahead and just stick with that up into the right is gonna be positive. Which means left and downwards gonna be negative. And now all we have to do is at all of our components together. So I'm gonna go down here and make some room for myself. So I got Let's see, that should be enough room, That little box right here. I'm gonna have the F two and F one, and I add them together to figure out what the Net Force is thes air my X and Y components. Now, if you remember, F two pointed purely along the X direction and to the left. So that means that X component is going to be negative. So that's negative. 1.5 times 10 to the 13th and a white component of zero. And now the F one pointed in this direction, which means both of the components gonna be positive up into the right. So we got positive components for that. We've got 1.62 times 10 to the 12th and then 2.16 times 10 to the 12th. So, again, this is why we don't wanna concern ourselves with the negative directions by plugging in the charges first. Because all these things are gonna change anyways, So we're gonna go ahead and pick our directions after all of that stuff. That's why this is that protest is really important. And if you go ahead and add everything straight down, we're gonna get that the F net components, there's gonna be negative 1.34 times 10 to the 13 because it's actually one power higher. Let me just go ahead and make sure you can, As you can see, that that's gonna be 13 and then this is just gonna be 2.16 times 10 to the 12th. So if I actually go back into the diagram, I can actually draw out what that vector is going to look like. It's gonna look like this is gonna have a high negative x component, and then it's gonna have a positive Why components gonna look something like that? Because this F two is actually really, really powerful compared to the white components or start compared to the X component of this guy. This one's actually relatively weak. Um, okay, so I've got my components all added up, So that means that the magnitude of the net Force is just going to be using the pie, the Pythagorean theorem. So I've got negative 1.34 times 10 to the squared. Plus, Now I've got 2.16 times 10 to the 12th, and then I've got to square that Let me just make sure that's written properly. So 10 of the 12 square that when you get the magnitude of F net is equal to what I got was 1.36 times tend to the 13th, because again, this is like one power higher. So it's actually really strong. And that's Newton's. Now the question didn't ask us to find out what the direction is, but when you know, you could do that using tangents and things like that. All right, so that's how you find the net force of these kinds of problems. Let me know if you guys have any questions