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Central particle with charge Q -2 q is surrounded by two circular rings (with radii R and ZR) of charged particles, shown in the figure: What is the magnitude of th...

Question

Central particle with charge Q -2 q is surrounded by two circular rings (with radii R and ZR) of charged particles, shown in the figure: What is the magnitude of the net electrostatic force on the central particle due to the other particles?kqz /2R2 12 kqz/2R2 0 2 kqz /2R2 kqz /2R2 0 32 'kq /2R2-2q

central particle with charge Q -2 q is surrounded by two circular rings (with radii R and ZR) of charged particles, shown in the figure: What is the magnitude of the net electrostatic force on the central particle due to the other particles? kqz /2R2 12 kqz/2R2 0 2 kqz /2R2 kqz /2R2 0 32 'kq /2R2 -2q



Answers

In Fig. $21-25,$ four particles form a square. The charges are $q_{1}=+Q, q_{2}=q_{3}=q,$ and $q_{4}=-2.00 Q .$ What is $q / Q$ if the net
electrostatic force on particle 1 is zero?

Mhm. In this problem we have seven charges distributed. Has shown separation between any labor charges is D. So that between seven and four will be two D. And I've marked all of the appropriate charge values. Notice all positive. That tells us immediately and all forces between all Between any charge and seven are going to be repulsive. Same side. Repulsive. So I want to do Is kind of separate out charge seven. Just to give us some room. That the diagram would be just too too busy. So this is charged number seven here. And I will not I'll just be drawing arrows representing the forces. They are not intended to indicate any length relationship. Bank good relationship. But just give us directional information. Then we'll take care of the banquet of those forces in a minute. So we start with number one. As I said. All forces plus plus Everything's plus plus. So I'm repulsive. So 1 7 repulsive. That means it's going to move be a force to the right. F. 71 four on 7 2- one. Then we can do mother, let's just do the horizontal first three on one. That's going to be two left. Eft. I'm three on 7, I should say. So 73 four Also repulsive. How displaced them. So they again, don't read anything into the length of these indicate something is the same or longer or shorter. Just giving a sense of direction will take care of the magnitudes separately. So this is f. So that's everything. In the horizontal. Everything's along the line connecting the charges. So we don't get any angles out of these guys. Now let's look at two again are repulsive 2, 7. Repulsive. So that's going to be uh the negative Y direction. If you're thinking this this is our X. Why? Why negative Y direction. Then we have 57 repulsive And then we have 67 also repulsive cassette at all. All reports, it would be there will be nothing really to worry about if you had negatives in here. So you get some attractive ones. You draw an arrow, you know to the right or to the left when one of these would be the opposite are not anything of issues. So we have we have our six forces from the six charges on charge seven. And what we do is we know in general we look at athletics, We get the components of the vectors and then we would square each of those components to take the square root. That gives us the Mag two. So now let's let's just remember something about columns law the magnitude of the electric force, columns law the electric instead of constant K. Magnitude of one Charge Mag 2 to the second. Charge over the distance squared. That's the form of it. We've taken care of direction. This is the man to do. So we just got to put a plus or minus depending on if it's pointing in the plus X. Direction or the minus X. Direction or the plus Y or minus Y Direction. So let's let's start let's start with seven. Let's do with in order 134. And so on 71 is the pacific direction. So the ho vectors and the positive. So it's either plus or minus on that whole magnitude. So it's going to be a plus. Okay. And we can do it in any order you want 71 hears Q seven. He won over D squared. And then now for three for 37 that's in the negative direction. So minus K. Q seven Q three or D squared. Same way with four of the negatives -K. Q seven Q four. Now this one is 2 d. away so two D squares. Don't forget. That's not to D squared. That's in the end four. And then the race the second power because it's in the princess. Okay and we can put in we can put in our specific values here. It makes we can see it um one is to so this is going to be K. We have six e. And we have to eat, do you square minus K. Yeah 68. And three is E. Over the square minus K. Six. E. For is 48. And that's like I said over four D. Square. And we can then factor out we know it's going to be a K. E squared over D. Square. So let's factor that out K. E squared over D squared. And then when we have left 12 minus six minus 24 divided by four which is six. Well 12 -6 -6 zero. There is no that there is no X. Component of the net force and now we do the same thing and why And so we're going to have we're going to have We'll start out with 7 to minus K Q seven Q tube over D squared. Then we have we have the five trust Kay Q seven five over D squared plus K Q seven. You're six. And this one is again two D to D squared. So putting in our values out of this and it's K six E Q two is 4 E or D squared. Then we got plus K six E two eE over D squared to S K six E. Katie over two D squirt. And this comes out to be again we can factor out the K D squared K E squared or D squared. So we're going to get here -24 plus 12. And then we have here 48 Plus 48 divided by four. Well what do we have? 48 divided by four is 12, 12 plus 12. 24. zero. F net mm No. You just show you the general form F net X squared plus F. Now, why? That's what we would do if any of these, If these were 90, that's what we'd be doing taking the square root that give us the man, too. That's the way we do it for any factor. But in this case, Sarah, that's it. It's the whole problem.

For this problem. There are seven charges arranged as shown each separated by a distance D. That's equal to one centimeter. And we want to figure out what the net force on charge seven is in this case. So in order to figure out what the net charge is on seven, I'm going to figure out first what the X. And Y components are because the magnitude of the charges just going to be the X. Component squared plus the Y component squared. And then taking the square root. So that gives us the magnitude of the net force. So in the X. Direction, Particles seven will feel a force from particles 1 3 and four deep forced from Particle one will be in the positive direction because they have the same sign. Um And yeah so particle one will push Article 7 to the right The force felt from particles three and 4 will be in the negative direction again because they have the same sign. So they're going to push it away to the left. Um and so if we write it all out I'm gonna go ahead and factor out of four pi 1/4 pi epsilon. Not because that is the same between all of them. But between particles one and seven we would have Q one Q seven over D squared between three and seven it'll be Q three Q seven over D squared and between four and seven it's Q four Q seven over. Um And in this case the distance is two D. So in the denominator we have four D squared but if we go ahead and actually try and plug in values we would get 1/4 pi absalon not 12 E squared over D. Squared minus six E squared over D. Squared minus six E. Squared over D. Squared. Um And so we find that the force and the X. Direction is a big whopping zero. That's fine. We still can look in the Y. Direction. So in the Y direction we want to do the same thing. 1/4 pi epsilon not um Four star sorry particle seven will feel a force from particles to five and 6 with the force from to being in the negative direction and the force from five and six being in the positive direction. Uh So we would end up with a minus Q two Q seven over D. Squared Plus Q5 Q seven over D. Squared Plus Q. six Q 7. And again this is over four D squared. Because the distance is two D. When you plug in the numbers for the charges you would have minus 24 E squared over D. Squared plus 12 E. Squared over D. Squared plus 12 E. Squared over D. Squared. So again we find that the force is zero. And so the net force on charge seven is going to be a big whopping zero. Newton's, everything cancels out

For this problem we have two charges Q two and Q one seven and x axis. And we're going to put a charge Q three somewhere between those two charges and the X component of the Net charge on Q three is showed in the graph. And from this information we want to determine the sign of Q one As well as the ratio of Q 2 to Q one. So in order to think about the sign of Q one, we want to consider the fact that the problem tells us that Q three is a positive number. In fact it is equal to positive eight times 10 to the negative 19 cool albums. But we know that Q three is positive. And we also know that when Q3 gets closer to Q one the force becomes positive. If Q three is positive and when it's close to Q1 the force is positive then that must mean that Q1 is also positive because a positive and a positive are needed to make that force positive. Secondly We want to figure out the ratio of Q 2 to Q one. And in order to do this we're going to use this point where the force crosses over the X axis because at that point the some of the charges are sorry not some of the charges to some of the forces on charge three must be zero. And so if the some of the forces is zero then that tells us that the magnitude from force one must be equal to the magnitude from forced to and this point is located about one quarter of the way to the eight centimeter mark. So that's going to be two centimeters from charge one and six centimeters from charge to So we can go ahead and use columns law here. 1/4 pi epsilon, not magnitude of Q one magnitude of Q three. All over two cm squared is going to be equal to one over for pi epsilon, not magnitude of Q two magnitude of Q three. All over six cm squared. We can cancel a few things out and we're trying to find the ratio Q two to Q one so we can do a little bit of fancy algebra here and find that Q two over Q one Must be equal to six cm squared or 36 square centimeters over two centimeters squared or four square centimeters and 36 divided by four is going to be nine.

Hello students in this question we have to particles each of mass M. And charge Q. Are attached to the two and of light brought. So this is Q. Particle. This is Q. Having mass M. And this is mass M. Okay and this is the center and this is length L. And this is hell. The road is rotated about this center and perpendicular to the plane. With an angular speed. Who make up. Okay the ratio of magnetic movement to the angular momentum about the center of the Or we have to determine. So magnetic movement is given by I manipulate bay area A. So about this center this roar will rotate in a circle of really small part. Okay now the current I will be due to these two charges so that will be given by two human regular basis frequency F. And that area will be compiled manipulate our disappeared So to Cube and this frequency candidate and as Omega two pi marked by Ellis Square. So this too will be cancelled out. I will be canceled out. So this magnetic movement will become you omega. Let's square. Okay now angular momentum L. Will be equals two momentum nausea. I So this can be written as a dish multiplayer be angular speed omega. Okay so moment of inertia about the center of these two particle. This particle will be Emily square. And for this particle Emily square. So this will be Emily square plus Emily square mark. Larbi omega. So this is to and elsewhere Omega now we can take the ratio so M by L. Is equal stood that you omega elsewhere the verb to M. L. S. Former player B Omega. Okay, This omega will be cancelled out and elsewhere will be cancelled out. So ratio of M two L is equal to Q, the verb to em. Okay, so option A become correct answer here. Okay, thank you.


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