24. Electric Force & Field; Gauss' Law

Coulomb's Law (Electric Force)

# Electroscope (Find Charge)

Patrick Ford

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Welcome back, guys. We've got a fund example involving an Electra scope. If you haven't seen them before, maybe you you who are taking labs and physics, we'll see one. But basically, it's just a little device with these two conducting leaves on it. And what happens is you turn on the machine and these conducting leaves pick up charge and they because they're like charges they want to repel from each other. So the leaves start to separate, and that's why they don't fly off. So that's what Electra scope is. We're told the mass of each leaf in this Elektra scope, and we're also told, uh, the deflection angles to 30 degrees. We're supposed to be figuring what the charge on the end of each leaf is. So what I wanna point out first is that we're dealing with two identical charges, which means this Q one and Q two are going to be the same. That actually helps us simplify things because we know that Q. One If Q one is equal to Q two, then we could just replace them both with just queues. And we also know that these things are going to repel from each other because they're like charges. So that means that this force right here is gonna be out in this direction. That's the electric force and these things. I want to basically push off from each other, so they're just gonna point off in opposite directions now, the reason they don't fly off is because they're connected sort of bythe strings here. So we're gonna see that in a little bit how that plays out. So, basically, if we wanted to know what the queue of each leaf is, we have to use cool owns law, right? We have to use the force between two point charges. That's gonna be que que que over r squared, Remember? No. No Q one Q two because these things are the same thing. So basically, this just turns into K Q squared over r squared, and we're gonna be solving for these Q squared over here. So basically all have to do is move this r squared up to the top, move the k down to the bottom, and I'm just gonna raise that. Basically, what this becomes is our f e uh, sorry. R squared Effie, divided by K, is equal to Q squared. So all we have to do is just take the square roots. So we get Q is equal to the square roots of our sorry r squared F E divided by K. All right, so if we take a look at this equation, this is really all we need to solve. But if we look at this thing, we don't know what the center, We don't know what the distance between these two charges are. So in other words, that's this distance right over here which I'll call Little are. And we also don't know the force involved. So let's go ahead and take this step by step and solve each one of those separately. So how do I figure out our well, hopefully you guys realize that this is a 60 degree angle between these two sides and that these two sides are the same. So one way to think about this is that if this is 60 or this is sorry, if this is 30 and this is a right angle, this also has to be 60 degrees, right? And if this is 60 degrees and this is 60 degrees and this whole entire thing is just unequal lateral triangle. All the sides have to be equal to each other. So that means that this little are is gonna be 0.5. All right, let me know if that makes sense for you guys. So basically, we know now what this little our distance is. Now we have to just figure out what the electric force is now. How do we do that? Remember, there's multiple forces involved here, so let's go ahead and draw a free body diagram for each object. So the reason that these things don't just simply fly away from each other is because there's a tension in the string that connects these things. And there's a tension here, and there's a tension here, basically the free body diagram. So each of these things, they're just gonna be mirror opposites of each other. We also have a weight that pulls them straight down mg, and we're told the masses of each of these objects Now, just for sort of simplicity. I'm just gonna work with this free body diagram here. This one's gonna be the exact opposite. Just everything's gonna be reversed. Okay, so that's the free body diagram. And we've got the our distance. So how do we go about finding what this electric force is? Well, the reason that this electric force doesn't just simply push this thing off is because there is a component in the tension force that basically keeps it and perfectly balances it. This thing is an equilibrium. So remember that equilibrium condition When things are in equilibrium, it means that the sum of all forces in X and Y are equal to zero. All the forces basically cancel out. So you've got these components of these forces the tension in the X direction and the tension in the Y direction. And those things were going to cancel out with M G and the electric force. So let's go ahead and set up our conditions for equilibrium. Let me go ahead and make some room down here. Okay, So I've got to the in the X direction. Actually, we go ahead and do that in black. So I've got in the X direction I've got the sum of all forces should equal zero. What are the two forces acting on it? Well, if I just go ahead and pick this direction to be positive, then I've got the electric force minus the tension that in the story the tension component or the X component of the tension force which, if you guys remember, that's going to be tension times the cosine of the angle. Remember that T X is going to be t co sign of data. If our angle is relative to the X axis and t y is going to be t sign of data if it's relative to the X axis, Right, So we've got t cosine. Theta is equal to zero. Sorry, f e minus T coastline data is equal to zero. So if I figured out, basically the electric force is just going to be equal to the tension divided or sorry times the cosine of the angle theta. Now the problem is, I don't know what this tension force is. And whenever I have an unknown in the X direction, I have to go to the Y direction and solve for that. So in the Y direction Now I have to set all the force is equal to zero in the Y direction. Now I've got the tension of the Y direction is t sign of data and that's going to be minus M G is equal to zero. So if we move this sign of data over to the other side and also move the MG, what we're gonna get is that T is equal to m G divided by sine of theta. Right? So we have t equals that now he could actually go ahead and sulfur this. We've got the 50 g which we're told that's equal to 0. kg. Remember, Everything has to be an s I times 9.8 divided by the sine of the angle. And the angle here is just 60 degrees. So go ahead and plug that stuff and you shoot attention of zero point. Uh, let's see, I got 57 Newton's so now we can go ahead and Dio is we can plug this tension in that we figured out the reason we came to the Y direction is to solve for what? This tension was over here. So now we can just keep on going and figure out what the electric force is. So we've got the electric force is just equal to 0. times the cosine of 60 degrees and that's just equal to 0. 29. Great. So now that we have this theoretic force right here, we can go back to the original expression that we had up top and basically solve for what that charge is. And I could basically plug that in for the electric force. So finishing off and plugging everything in we've got to Q is equal to the square roots. This our distance, remember here is just 05 So you've got 05 squared and then you've got the electric force, which is 0.29 divided by this K constant 8.99 times 10 to the ninth. If you go ahead and work this out, you're gonna get Q is equal to 2.84 times 10 to the minus six columns where this 10 to the minus six just has a symbol, micro. Or that's that Greek letter mu. So you could also say this is to 84 micro columns and both of those things would be perfectly fine. You get full credit for both of those answers. Alright, guys, let me know if you guys have any questions with this

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