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Four particles, a, b, c and d aresituated, respectively, at (+1.0 m, 0), (0, +1.0 m),(-1.0 m, 0), (0, -1.0 m).It is known that the charges of three of the particles...

Question

Four particles, a, b, c and d aresituated, respectively, at (+1.0 m, 0), (0, +1.0 m),(-1.0 m, 0), (0, -1.0 m).It is known that the charges of three of the particles areqa, qb and qd andthat these charges are equal in magnitude, but their signs areunknown. The chargeQc isunknown. What do we know about the signs of these charges if:In the y-direction, there is no net force on apositive charge situated at the origin.In the x-direction, a net force is experienced by a positivecharge situated at the

Four particles, a, b, c and d are situated, respectively, at (+1.0 m, 0), (0, +1.0 m), (-1.0 m, 0), (0, -1.0 m). It is known that the charges of three of the particles are qa, qb and qd and that these charges are equal in magnitude, but their signs are unknown. The charge Qc is unknown. What do we know about the signs of these charges if: In the y-direction, there is no net force on a positive charge situated at the origin. In the x-direction, a net force is experienced by a positive charge situated at the origin. This force acts in the positive x-direction. Which of the following is true? Charges qb and qd have the same sign: both are positive or both are negative. The magnitude of the charge of Qc is greater than that of qa. All charges are positive. If qb is positive, qd is also positive. If Qc < qa then the sign of Qc is negative. If qb is negative, qd is also negative. If Qc and qa are both negative, |qb| < |Qc|. If qb is positive, qd is negative. If |Qc|=|qa| then Qc is positive and qa is negative. If qb is positive, qd is also positive. If |Qc|=|qa| then Qc is positive and qa is negative.



Answers

Two objects carry initial charges that are $q_{1}$ and $q_{2},$ respectively, where $\left|q_{2}\right|>\left|q_{1}\right| .$ They are located $0.200 \mathrm{~m}$ apart and behave like point charges. They attract each other with a force that has a magnitude of $1.20 \mathrm{~N}$. The objects are then brought into contact, so the net charge is shared equally, and then they are returned to their initial positions. Now it is found that the objects repel one another with a force whose magnitude is equal to the magnitude of the initial attractive force. What are the magnitudes of the initial charges on the objects?

In this example, we're going to be working with point charges and specifically um working with force edition in columns law for point charges to calculate force, electrical force. Um So we're going to start with three point charges and they're going to be on the verge. Pieces of a rather nice right triangle. Um So the right triangle that we're going to draw is um that special right triangle where um your sides have an easy relationship to each other and maybe you'll recognize this triangle once I draw it. Okay. But the hypotenuse opposite the right angle has a length of five m. Um and then the longer side it's four m And the shorter side is three m. So you may have seen that triangle before. It makes calculations easy and up at the angles in there, even though we never get to actually worry about those angles. Um Okay, so what's true about this figure is that we know that the charge up at the top, so we'll call them Q one, Q two and Q. Three. What we know for sure about the charge up the top is both of its sign, it's positive and we know it's magnitude in columns. Um For the other two charges, all we know is there absolute values but we do not know what sign they have, they could be positive, they could be negative. Okay. And of course the question is um what sign do they have is the mystery that we are going to solve with our picture. Um But we have to have some more information. And what we're told is that those two charges create a net force that points to the left. So I'll call that f total um And it points to the left there. Um And the mystery is what sign to those two charges have. And the other question that we can address is how large is that total force? That's where we'll be using columns law. But to figure out the sign, we really just need a couple of pictures to consider. Um a couple cases there are only really two cases that we have to worry about And that's because if both of those charges Q1 and Q2 have the same sign um there's going to be an uncanny rustled why component either Pushing Q. three up or pulling it down. So what we know for sure is one of these has to be positive and the other has to be negative negative. Um And that will allow for the cancellation of the why component of force. Um and so we'll just kind of sort that through, we'll draw two pictures um of the forces that those charges exert On Q. one Q 3 And we'll first take the situation that Q one is um positive at Q. Two, it's negative and really only one scenario is needed to come up with the right picture. If sorry Q. one is positive then what's going to happen is that the force on Q three Is going to act along the line joining them and it's going to pull Q. 3 um push it upwards so that force would look uh upwards Along the line joining the two. And if Q two is negative, the force on it is going to pull attract towards you too and we immediately see that this is the wrong answer. Um Salt draw big frowny face there. Um Because the F total in this situation would then be pointing towards the right um Which we know is not the case. So we can rule this case out and it implies that Q. One as a minus sign and cute too has a positive sign. And we had a 50 chance of guessing it correct. Um and if you could imagine the picture of the forces in your head without drawing the picture, you might have avoided the step. Um But anyway, we now know the signs. Um and our next goal is to figure out uh the magnitude of F total. And for that, what we're going to have to do is calculate F one and F two using columns law for point charges. And remember there's a constant that you can go look up sometimes it's just called K. Uh huh. And has a value in S. I. Units of nine times 10 to the ninth. Newton's meters over Cool. M squared. Um And both of those forces are going to use that constant. And the product of the absolute values of two charges over there, distance squared. So Q one is 4 m away um And cute too is three m away. Yeah, Let me see, I've been using red for F2, so let's go go do that. And again, we don't have to worry about the signs when we're just figuring magnitudes of forces. And if you work that out, um plugging in for the K and the queues in in terms of their columns, let's see. You get 2.25 Times 10 to the -3 newtons and something a little bit smaller for the other. Okay, so those are the magnitudes of the forces. Um And even though I have them drawn incorrectly in this problem, because I had an initial guess for the signs of the charge and correct what we can see is that they are at right angles to each other. Kind of like an X and a Y component, so we can treat them. Uh Almost like we would for an X and Y component, and R F total is then going to be the sum of the squares of their magnitudes with a square root because F one is perpendicular to F two. Okay. And so we can easily work out that force magnitude, it's about 2.82 Times 10 to the -3 newton's. Um And one final question is, where would you put 1/4 charge in order to cancel the total force on Q three? And since Q three is positive? Um And if that fourth charges also positive, there's going to be repulsion along the line joining them. And so somewhere off to the left and it would depend How strong that positive Q four is. Um But it will create a 4th force one Q three that points entirely to the right. And by adjusting the distance and the amount of charge, uh you can get it to the point where it's identically canceling that Green Net force.

For problem 21 way have here our X and Y axes of the church, particularly a couple negative Q on the negative. Que small on positive Que small eso our negative US existence at X on the positive and negative molecule or distances off? Negative A and e. So for letter, a way will draw the everybody diagram. Must be capita negative, you charge particles Mhm. So here is just on the X Y plane off it. So here's our negative Pathetic. You, uh, small years before between the positive and the negative charge. And here is where the negative and negative deposited the negative petition attraction first and the department first. Let's call this death off some positive. U f off negative. Here's the way up by component off Positive Church. Here is the X component off Planet Church For the negative charge, Here's the my component directed upwards. Yeah, X component directed when the positive weeks. So that is the free body diagram off our I went in question particle. So three b were asked to find the net component accent that component by So for FX, we know that the ex components are in equal in magnitude. So they cancel out, leaving us a net component in next zero for the y component. Same magnitude. So they're edited. So we can say that it is two times d component off the F sub basket and potentially as a sub minus Qiwei. So it would be two times Okay, Capital Q smoke you divided by the distance between them, which is X squared, plus a squared the spirit off it. But it is squared. We can cancel the squared multiplied by the same data, so Saint data would be a divided by distance D. So this is data. So in the sign off, it will be a divided by distance. D So a divided by the spirit off X squared, less a square Simplify English would get over flight. Mhm. Okay, Cue cure a divided by expert less. He squared, raised to three hands. So this is our flight in the positive direction. Yeah. So first see it. That's as if I were church particle. Is that exist? Equals zero. So using the formula before you know that our next door should question though I component only. So it is OK, cute. Thanks. Smoke you times e invited by Seattle Square plus a squared these 23 hubs. So then canceled over to here so it would be squared only. So our net force to queue Small Q divided by he Screwed too, is just the answer deposited by direction. This is our net First for D tous asking us to grow power by component versus the excesses. So doing just our excellent biopsies. So what The point? XTC 40 our way component is the greatest and point negative foreign for eight our s y said each No, its value. So we out here values FBI for negative or alien for And it it's exponential and also symmetrical in the White House is so here's our draft.

Hi. Everyone the networks experienced by. Yeah, the more we really charge 8.8 in Jeju since it is Japan will or okay. Equal force by each of the mm. Fixed judge. Yeah. At one. To be the words per component. Mhm of forces and to produce non Gino force and vertical direction. What, in hora gentle direction. Yeah, Net force is zero, because origin it'll component and salute. Yeah, so Net force at the is a long vertical direction. No second part a spokesman. Second is correct. Okay, that that force at a can sit and had to be not non zero forces developed. So have B is greater than have pain. Uh huh.

In rehab Cuban equals to my curriculum without its charge nature and acutely is equal to plus four point double zero micro cooler. Its nature is no. Okay so and the neck forces towards the left that is in negative X axis. For the part of the question We have to sketch all the possible outcomes depending on the sign of the charges. So this is charged Cuban, this is charge Q2 and this is our security, This is Cuban Q2 and this is cute. Okay so now we can take it as positive and It is also positive and this is given as positive. So the force due to this charge on this will be in this direction and due to this charge, the force will be in this direction. Okay so say this is a fun force and this is f to force. So this is situation one. Okay, no the situation will be this is cute recharge, this is Cuban charge and this is Q. To charge. So let's this sign is negative and its sign is negative and it is already positive. So the force will be in these directions. So this is force F. One and this is force F. Okay. No, 40. This is situation. So now for the third situation we can draw that this is Q3 charge, this is Cuban charge and this is Q to charge. So let us take it as positive and it's negative and it is given as positive. So the force due to this charge on this will be in this direction away. Okay, and this force will be in this direction. So this is a force F. Two and this is a force F three F four. Okay, so this is situation third. Okay, now the situation fourth will be this is Q three Cuban and Cuban. So let us take it is negative, it is positive and this is positive as further questions. So the force due to this negative charge on this will be along this direction and the force due to this two positive charges were repulsive. So this is forced have to and this is a force have fun. Okay, so these are the four possible outcomes of the question. Okay, So this is the 4th 4 possible outcomes of the question. Okay, now Moving to the part B in which we have to determine the nature of the charge Q two according to the problem. So for the part we can say that the force will be the force will be in the upward direction and towards right direction because the forces are mm in the left and right direction and for the situation to the force may be in the downward direction. And also indeed, the horizontal forces may cancel out may cancel out. Okay? So for the situation third, the net force may come out in the right direction and the vertical force may cancel out. Okay, Because if the magnitude somehow changes or rearrange itself so may cancel. So for the fourth situation, the force maybe towards left and the vertical force maybe cancel out. Okay, so the fourth situation is the required situation. So we can say that Cuban should be negative. Cuban is given as negative charge. Okay? And You two should be positive. Okay? Uh this is the outcome of that question. Okay. Q three is given as positive already and Cubans should be negative and Q two should be positive so that we get net force in the left the direction. Okay, So we can say that Cuban is equal to as we have magnitude of Cuban. So we can write that Cuban is equals two minus two point double zero micro column and Q three is equal to plus four point double zero micro column. And we can say that you two is greater than so this is the answer for the but be of the problem. Okay Cuban is negative and Q2 is positive. Okay, so now moving to the part C in which we will determine the magnitude of charge Q two. Okay, so in this part we have determined only it's signed. So the case fourth is the right one. So the vertical force must be equal in magnitude and opposite in direction. So we can write that angle cheetah one of force F one and horizontal is because in the words of four by five from the geometry so it is equals to 36.86 degrees. And angle theta two between F two and horizontal is equals two causing words of three by five. This is equal to 53.13 degrees. Okay, the vertical course must be Equal so we can write that Afghan costs 36.86°.. This is equal to f to sign 53.13 degrees. Okay, say this is a very question number one. So now the fourth F one can be written as F one equals two Cuban cutely deal whereby are 13 square. So substituting values, we get nine multiplied by 9 19 to 10 to the power nine multiplied by Cuban charges to multiply by 10 to the power minus six And Q three is 4-10 to the power -6. So now we will consider magnitudes only and urban three is 0.04. So square. So from here we get F one equals to 45 minute. Okay? And From here substituting this value in previous one equation. So we get F2 equals to 45, multiplied by sign 36.86 divided by Sign 53. Okay, so from here after solving we get F2 equals to 33.8 Newton. No, if you can also be returnees have two equals 2 K Q two Q three divided by R 23 square. So now substituting values, K is 19 to 10 to the power nine and Q two multiplied by Q three is equal to 14 to 10 to the power minus six. They were by our 23 is 0.3 m square. Okay. And F two has magnitude of 33.8. Newton. Okay, so from here after solving we get to two equals to 8.43 multiplied by 10 to the power minus seven column, Which is equals two 0.843 Micro column. Okay, so this is the answer for the question. Okay, as Q2 is positive as you do is positive. So its magnitude will be in the positive. So this is the answer fully part C of the problem. No. Yeah. No. For the party The magnitude of net force will be equals two. F will be equal to F one cause 36.86 degree plus F two. Because 53.13 degrees. Okay, so now after substituting values of F one and F two from here is 45 after 33.8 we get 45 Because 36.86 degree plus 33.8 Cause 53.13°.. So from here after solving we get f equals to 56.2 Newton. Okay, so this is the answer for the party of the problem. This is the net force on the charge. Okay.


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