What is the force F on the 1.0 nC charge at the bottom in FIGURE ...

Question

What is the force F on the 1.0 nC charge at the bottom in FIGURE P22.47? Give your answer in component form.

Accepted Answer

Hello, everyone. Let's take a look at this practice problem. Determine the force F acting on the two nano Coolum charge located at the bottom as shown in the figure below. Provide your response by providing individual components. Option A 2.6 multiplied by 10 to the power of negative five new ends with the J hat unit vector. Option B 4.0 multiplied by 10 to the power of negative five nuances. Option C 5.0 multiplied by 10 to the power of negative five nuances and option D 3.5 multiplied by 10 to the power of negative five nuances. This is a pretty typical Kums Law problem where we have a charge that we're analyzing. And specifically, we're trying to figure out the effect on it that multiple other charges have. So let's look at what's happening here and from what the diagram shows us and go through it charge by charge. So the bottom charge that we're asked to look at is a positive charge and recall that when you have two charges that have the opposite sign, they are going to be attracted towards one another. But if you have two charges that are the same sign they will be repulsed away from each other. So we can see from the diagram that charge two is positive. So that means that the electrostatic force between charges one and two will be repulsive. So due to charge two charge, one will experience a force pushing it down into the left. In this diagonal path, charge three is negative, which means that there is an attractive force between charge three and charge two. So due to charge three charge, one should actually be pulled upwards as I'm showing on the arrow diagram and lastly charge four is positive, which means they will have a repulsive force with charge one. So charge one will be pushed downwards and to the right thanks to the fourth charge. So in order to find the net force acting on charge one, we're going to have to go through this diagram charge by charge and use the Coolum law to find how each other particle will be affecting charge one. And recall that the Kums Law states that the electrostatic force between two charged particles is equal to the Coolum constant multiplied by the product of the two charges divided by the square of the distance between them. Let's start by looking at Q sub three charge three. The reason why I'm starting with this charge is because since it is directly above the bottom charge, it's the easiest one to look at because we don't need to worry about force components or, or diagonals or things to worry about there. So it's a, it's a simple, nice and easy place to start. So I'm going to write an expression a formula for the charge on particle one from particle three. So I'm writing this as F sub 31. And as we discuss with the Coolum law, this is going to be equal to the Coolum constant multiplied by the product of the two charges, which are in this case are Q sub one and Q sub three all divided by the square of the distance between them. So let's start plugging things in the Coolum constant is 8. multiplied by 10 to the power of nine newton meters squared per Coolum squared multiplied by the product of the charges. So that's two nano columns or two point or two multiplied by 10 to the power of negative nine cos and Q sub three has a magnitude of eight nano columns. So eight multiplied by 10 to the power of negative nine columns. OK. So that this is divided by the distance between them, which is not explicitly shown in the diagram. But we can see from this circular arc that the four centimeter number given for the other two distances can be thought of as a radius along this circular path, which means that the distance between charges one and three are also 0.04 m, that's four centimeters. So if we put this into a calculator, then we find a magnitude of about 9. multiplied by 10 to the power of negative five newtons. And because this is an attractive force that pulls particle one upwards along the y axis, there is AJ hat unit vector representing this. Also, since we discussed that the charge one is being pulled upwards, we're going to leave this component mark as positive. Now let's discuss the other two charged particles two and four, they're going to be a little bit trickier because you can see from the diagram that they are located at kind of diagonal positions from the bottom particle. Like there, there's gonna be a horizontal component here as well as a vertical component. But luckily for us, they actually that actually doesn't make it that much harder. It's it's actually a lot easier than it looks. And the reason is simple, both charge two and charge four have the exact same magnitude four nano columns. That means that if we were to draw out the individual components of these diagonal forces, we would realize that the horizontal components completely cancel each other out, the horizontal components contribute nothing to the overall net force on charge one. And also the way the problem is written also kind of gives us a hint about that because we can even see from the multiple choice options that all of our multiple choice options only give us options that are solely along the Y direction. So the X axis ends up proving to be irrelevant. But one thing we can do to simplify the amount of work that we have to do here is instead of trying to separately solve four Q two and Q four. And then using our sign and cosine components to break them both up into components. What we could do is instead since we already know that only the vertical component is relevant, we can just solve for the vertical component of one of these two charges. And since we know the magnitudes for both of them are the same, we could just multiply whatever we find for one of them by two. So that's all we really need to do here. And that takes out, that takes away a lot of work that we otherwise may have had to do. So the force from charge two, for example, you could go with four, it wouldn't make a difference either way, the force from charge two on charge one is going to be equal to uh it's the same Coolum law formula with one minor tweak. So the Coolum constant multiplied by the product of the two charges. So Q one and Q two divided by the square of R which is the exact same value. But the, but the one tweak that we're gonna have to make to this is we have to multiply this by the, by the sinusoidal function. That'll just give us the vertical component. And we're going to do this with respect to the 45 degree angle which we're given. And from what we know about our trigonometric functions, we know that it is the cosine function that relates to the adjacent side of a right triangle. And in this case, that adjacent side of, of this triangle is a, is the vertical component that we're looking for. So we multiply this by the cosine of 45 degrees. Now let's plug our numbers in and see what we get. So of course, the Coolum constant hasn't changed. It's just 8.99 multiplied by 10 to the power of nine newton meters squared per Coolum squared multiplied by the product of the charges. So once again, Q 71 is two nano columns multiplied by four nano columns for the other charges all divided by the same distance, the square of the same distance. So 0.04 m squared. And when we put this into a calculator, we find a magnitude of this charge of about 3.178 multiplied by 10 to the power of a negative five nuances. And since we discussed this is in the Y axis, there is AJ hat unit vector affecting this as well. And another important thing to realize is that since what we talked about earlier, the charges are the same, uh Q two has a positive charge and Q one is a positive charge. That means that they are repelling one another, which means that Q one is going to be directed downwards in the negative direction from Q two and Q three. So that is, it has a negative sign in front of it. So we could, if we really wanted to, we could do the exact same process. But solving for the uh for F sub 41, the force on particle one from particle four. But as we discussed earlier, there's not really much of a point if we already know it will be the exact same magnitude and direction as F sub 21. So what we can do here to find our net force is just simply add up the forces we have. But multiply the force we found for two sub one for sub 21 multiplied by two. So to show this in action, I'm gonna say that the net force on particle one is equal to the upwards directed F sub 31, the force from particle three plus two multiplied by the force from 211. While writing this out more numerically that is 9.0 multiplied by 10 to the power of negative five nuances in the J direction minus two, multiplied by 3.178 multiplied by 10 to the power of negative five newtons also in the J direction. And if we put this into a calculator, then we find a new force of about 2.6 multiplied by 10 to the power of negative five new MS along the J direction. And so this is our answer to the problem. And if you look at our multiple choice options, we can see that this agrees with option A. So option A is the correct answer to this problem. And that is all for this video. I hope this video helped you out and please consider checking out some of our other tutoring videos which will give you more experience with these types of problems, but that's all for now and I hope you all have a lovely day. Bye bye.

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What is the force F on the 1.0 nC charge at the bottom in FIGURE P22.47? Give your answer in component form.

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