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Chapter 28 Problem 0044canel Velrat a valocit maanttlde 460 mls throreh unliori macnetie fb %f maantttde 0.060 (4n loha ?aricahas chata" cfcharga cf +3.2 {039 ...

Question

Chapter 28 Problem 0044canel Velrat a valocit maanttlde 460 mls throreh unliori macnetie fb %f maantttde 0.060 (4n loha ?aricahas chata" cfcharga cf +3.2 {039 _ mar 6.6 ka) Ina batrean tha Partele direction motion and n8 Maanerie#e- 50*, WMt tht Tegnitude 08 (6) ihe Force akting (72 pattic - Jue Tad. Jnd (D) the ecceleratin &fiht particle dut Ihis force? (c) Does {he Jeeed 08 the particle increese- decre938 Feenha FannhNumbe(6) Number

chapter 28 Problem 0044 canel Velrat a valocit maanttlde 460 mls throreh unliori macnetie fb %f maantttde 0.060 (4n loha ?aricahas chata" cfcharga cf +3.2 {039 _ mar 6.6 ka) Ina batrean tha Partele direction motion and n8 Maanerie#e- 50*, WMt tht Tegnitude 08 (6) ihe Force akting (72 pattic - Jue Tad. Jnd (D) the ecceleratin &fiht particle dut Ihis force? (c) Does {he Jeeed 08 the particle increese- decre938 Feenha Fannh Numbe (6) Number



Answers

In Fig. $21-44,$ what are the (a) magnitude and (b) direction of the net electrostatic force on particle 4 due to the other three
particles? All four particles are fixed in the $x y$ plane, and $q_{1}=$
$-3.20 \times 10^{-19} \mathrm{C}, q_{2}=+3.20 \times 10^{-19} \mathrm{C}, q_{3}=+6.40 \times 10^{-19} \mathrm{C}, q_{4}=$
$+3.20 \times 10^{-19} \mathrm{C}, \theta_{1}=35.0^{\circ}, d_{1}=3.00 \mathrm{cm},$ and $d_{2}=d_{3}=2.00 \mathrm{cm} .$

Use the results of part being problems 65 for the motion along a single axis. In the following situation, frame A in figure 31 is attached to a particle that moves with velocity 310.5 C pass frame be Which moves past frames, see with the velocity of .5 c. What are M a C B and a C? And the velocity of the particle relatives of frame? See Okay, so first let's note that since n is one minus beta over one plus beta, that means that beta is a one minus M over one plus M. So from the problem we know that beta A B is 1/2. And since beta A B is one over to, that means M A B is one third, so M A C is M A B NBC, which is a third time's a third, Which is at 9th for part B. So beta A C is one minus M a c over one plus M a C. And that gives us 4/5 part C, the velocity of the particle relative to frame. See so by definition of beta, which is the speed parameter we have, that V a c is equal to For over 5 0.8 0.80 c.

Here about a in, but we can write emcee. So it is this m a C. We can write M A C, which will be equal to M A B Baby times, MBC BC that is equal to one or tree Diane's one or three, which gives us one or nine. Well, the one night in part B, we are asked to find a better. I see better I see the speed factor. Then we can write a better I see which is one minus and Macy I m a. C or one plus Macy. I'm a who see the value which we all tend here, which is 119 If we plug 119 here and one were nine here and do the simple mats. Then the better we get here he is four or five. Which musical to zero point age Joe. But see then, by using the definition off a better, we know that no one will last. The A C will be simply equal to zero point bait. Zil time Speed off light. End off the problem

Hi there. So for this problem, we have a situation that is shown in this figure, we have three positive charge particles particles B and C are so close to each other that they can be considered to be at the same distance from the particle A. We're also told that the force on the particle A due to the particles B N. C. It is a given information and that we're going to call forth A. And that is equal to 2.014 Times 10 to the -23 Newtons in the negative direction of the axis. Now in the figure in the figure part B. Of the starting we are given in here A particle B has been moved to the opposite side of the of the political A. But it is still at the same distance from it. Now the net force On eight is now Is now equal to we're going to call this 4th f. B. And that is negative And it is two 8:77 times 10 To the -20 for Newton's. So what we need to obtain is the ratio between the charges Q. C over cuba B. No. Yeah. Since the forces that are involved in this problem are proportional to Q. As the problem instead. And we see that the essential difference between the two situations is that in the 4th F. A. is proportional to the sum of the charges Q B plus Q C mm. And when these those two are charges are on the same side versus the other and versus than the other situation where B Is equal to the force B. Is equal to minus the charge be plus the charge Q. C. When they are on the opposite side. Now what we can do in here is to set up the ratios between these two. So we will have that the force F. A. Over the force F B. Is equal to the some of these charges overt minus Q B plus Qc. Now we substitute the values for the forces that we are given. So we substitute that. That is two .014 times 10 to the -23 Newtons. And the other one is 2 -2.8 77 times 10 to the minus 20 for needles. And this is also equal to Q B plus Q C over minus QB. Last Q C. So plugging this into the calculator, we obtain that this is approximately seven. That division at the left then what we can do now is to solve. And from this equation, well we can take out, we can take out for its samples QB. So From both the numerator and the denominator. So that QB Councils. So we were going to attain that this is equal to seven and that is one plus Q. C. Over QB overt minus one. Los que si over QB. Now what we need to do is to start solving for that ratio. So we will have, we pass this to the other side will have seven times minus one times that ratio And this is equal to 1-plus QC over QB. So we're starting here so we're obtaining here minus seven plus seven times the ratio, Which is equal to one plus the ratio. In here we pass the numbers to the right and the ratios to the left. So we will have six times QC over cuba over QB And this is equal to eight. So from here we obtain that this ratio of the charges is equal to eight over sets If I'm great. Yes, and this is approximately 1.333. So this is a solution for this problem.

So to start, I've drawn a diagram where I've labeled the particles A, B and C. Where is the one the left be used, the one in the middle and sees the one on the right. And then I've written out R constant K our values for the charges and our values were given for the separation. So the first thing I want to do for part A is to figure out the force between particles A and B. So to find that were to take a times a charge on a times a charge and be over the separation between A and B squared, we look in our values. For that, you find that this is equal to about 43 Newtons and on particle, be the direction of that is going to be to the left since it's an attractive force towards particle A. So next we want to find the force between particles B and C, and so for that we use the same equation. But we use the charge of particles B and C, a separation tune B and C squared, and that's equal to about 61 Newton's. It's also unattractive for so the force on particle be is going to be to the right then to find the net force. Since these are in opposite directions, all we need to do is subtract thes values. This is his equal to force when b and C minus the force between A and B, and that's equal to 18 Newtons to the right. The Net force on B is 18 Newton's to the right now for part B. We basically want to start the same way. We want to find the force between particles A and C, so that's gonna be equal to K times a charge on a times a charge on C or the separation between A and C squared and a separation to and see is just the separation between A and B plus the separation between B and C. So if we plug all that in, we find the forced to particles A and C is about 18.6 Newtons to the right, since this is now repulsive force, since A and C are the same sign, and so the force on C is gonna be to the right. Next we want to find the force between B and C, but we've done that already. Up here, you found that that was equal to 16 Newton's. That's gonna be the same. But the direction is going toe have changed now because it's an attractive force, and so C is going to feel a force towards the left. So the force on a particle see from particle be is 61 Nunes to the left. And then we're gonna find the net force in the same way, because again thes air in opposite directions. And so we can just subtract the two forces to find the net force. It's the net force on particle See is going to be before sweet B and C minus force between A and C, and that's equal to 42.4 Newtons to the left. So the net force on particle see is 42.4 Newtons to the left


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