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Balancing a Pendulum in Electric Field

Patrick Ford
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Hey, guys, let's get more practice involving electric fields. So we have a charge that's at the end of a pendulum. And this pendulum, we're told, is equal in equilibrium and suspended inside of a uniformed electric field With that E over here. Now we're told, with the electric field, Magnitude is and our job is to figure out what is the charge on the end of this pendulum? Que so there's a couple of important things to pick out from this problem. The fact that it's an equilibrium, right? This condition right here, equilibrium means that the sum of all forces in the X and Y directions is equal to zero. Right. All the forces basically cancel out. We're going to see how we can use the forces in order to solve this problem. But if we're solving for the charge. Q. There's basically one of two equations that we can use. We can use cool OEMs law that EFC over our streak of r squared. But the thing is, we don't have another charge, right? There's no second charge over here. So instead, what we're gonna do is we're gonna use the electric force is equal to Q e right, The charge times, the electric field, which we actually have over here. All right, so we're gonna go and use this formula right here to solve for Q. And if we go and rearrange to this, the Q is just getting equal f divided by electric fields. So I have with the electric field is if I can figure out what this electric force is, then I could basically just plug that in and software the charge. All right, so how do I go about solving that force? Well, there's a bunch of forces that are acting on this, So let's go ahead and draw a free body diagram for this. We have an electric field on a point charge over here, so we know that the force is going to be in the same direction. If this is positive now, the reason that thing doesn't just go flying off is because it's held in place partly by this string that it's attached to, which has some tension. And there's also another force due to gravity, right, because this thing has some mass 30 g. So we get We said that this object was an equilibrium, which means that all the forces were gonna cancel out. So in order to set this thing up first we have to break up this tension into its components. So we have an X and A Y component, and we get those components T y and T X by relating the tension with the CO side or sorry, the angle theta that it makes with the X axis. Now what is this angle right here? So right, this over here, this angle is equal to Well, if you sort of, like, project out this sort of horizontal line, this is a 90 degree angle. This is a 15 degree angle, which means this has to make up the difference between 180. So this angle is actually 75 degrees, right? So it has to basically, uh, this angle this angle has to add up to make 90. So we've got the angle right here. So that means that the tension in the X direction is going to be t co sign of data and the T in the Y direction is gonna be t sign of data. So now we have these components. Let's go ahead and set up our equilibrium conditions in the X direction I have that the sum of all forces in the X direction is equal to zero. So what I have to do now is I have to pick a positive and a negative direction. So I'm just gonna go ahead and assume that the top and right directions are positive. If you guys chose to do something else, your answer is still going to come out the same. You're just gonna have some negative numbers first, but the physics still works out. So I'm just gonna choose this to be the positive directions. Now, starting off my forces in the X direction, I have the electric force, which is really what I'm looking for, minus the tension or the X component of the tension. In other words, the electric force here is just equal to the tension times the cosine of the angle. So I'm like, Great. If I have the coastline of the angle and I have the tension, I could just go ahead and figure this out. The problem is, I know what the co sign of data is. I don't know what the tension is. So remember, Think back to Newton's laws. Whenever we had a situation where we needed one variable in the X direction, we usually went to the Y direction to solve for it. So let's go ahead and do that. We're gonna go over to the Y direction. We have the same exact condition we have. The sum of all forces in the Y direction is equal to zero. So you start from the positives, which is gonna be your t y. And then that's gonna equal your minus M G equals zero. In other words, your tea sign of data is equal to M. G once you move this to the other side. So now tea is actually equal to M G, divided by the sine of theta, and I actually have what all of these numbers are. I have the mass I have G and I also have sign of data. So in other words, I get 30 g, which is 300.3 kg, 9.8 for G. We're gonna move down, and then I've got divided by the sine of the angle, which is 75 degrees. If you work this out, you're gonna get attention of zero point through 30 Newton's. And now you can just take this number and then plug it back inside of this formula right here. So I know that the electric force is just equal to 0.30 Newton's times the cosine of the angle, which is 75 degrees. So that means that my electric force is equal to I got 2. 78. So now what I can dio is I can take this electric force and I just have to plug it all the way back into my initial equation right here to solve for the charge. Remember, that's my final variable. So I've got two. Q is equal to this is gonna be 0.78 only highlight that's you guys makes you understand what that connection is. And I've got 0.78 divided by 100 Newtons per cool. Um, so you should get 7.8 times 10 to the minus four columns. And this is, by the way, our final answer. So let me know if you guys think about this problem, let me know if you guys have any questions. I'll happily explain any steps and I'll see you guys the next one