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ReturnBlackboardS utelk Physls, 10e Help System Bnnouncements_ Nnread)PrINER VbreTo HnoANEtiChapter 18, Problem 33Four polnt charges have the same magnitude of 3.4 ...

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ReturnBlackboardS utelk Physls, 10e Help System Bnnouncements_ Nnread)PrINER VbreTo HnoANEtiChapter 18, Problem 33Four polnt charges have the same magnitude of 3.4 10-12 C and Mufo magnitude of the net electric field that exists at the center thc squarethc cornerssquare that 4.9 cmside. Three olthe chargespositive and on2 nedatit Determne &eNumberUnitsthe tolerance+/-29Click If you would Iike to Show Work for thls question: Qpen Show Workaccessing this Queston Assistance you will learn whlle

Return Blackboard S utelk Physls, 10e Help System Bnnouncements_ Nnread) PrINER VbreTo Hno ANEti Chapter 18, Problem 33 Four polnt charges have the same magnitude of 3.4 10-12 C and Mufo magnitude of the net electric field that exists at the center thc square thc corners square that 4.9 cm side. Three olthe charges positive and on2 nedatit Determne &e Number Units the tolerance +/-29 Click If you would Iike to Show Work for thls question: Qpen Show Work accessing this Queston Assistance you will learn whlle You earn points based Point Potential Policy set Instructon Questlon Attempts: Usnd aeeoneattee ubmit Ansiner



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Concept Questions The drawing shows a square, on two corners of which are fixed different positive charges. A third charge (negative) is brought to one of the empty corners. (a) At which corner, $A$ or $B$, is the potential greater? (b) Is the electric potential energy of the third charge positive or negative? (c) At which corner, $A$ or $B$, is the magnitude of the electric potential energy greater? Explain your answers. The length of a side of the square is $L=0.25 \mathrm{~m}$. Find the electric potential energy of a charge $q_{3}=-6.0 \times 10^{-9} \mathrm{C}$ placed at corner $A$ and then at corner $B$. Compare your answers for consistency with your answers to the Concept Questions.

Everyone here it is. Given three charges Cuban or to to fight my curriculum. Your group? Well, I think you'll find microphone. Um, you want to keep minus five? My curriculum at the vortices of equator cranking off site 2.95 centimeter are When do you go to nine? Five meter? We have to find a fare intensity and D. D. C is our and this value you can fight by protocol. Restaurant D squared minus. Do you square by foot? That is 0.0 95 Script Mhm, right. It's quite white food. So on solving it it is to me I am Gino. Fight, fight, Fight. Me too. You do. And if one can set out because they have insane magnitude and a party direction So that feel that we will be scarred to the three that escaped you three upon our square. Now substitute Travel K is given 9 to 10 to the power 9 to 3 is find my curricula and the stress We have measured 0.2555 So it is to be 6.92 And to the power seventh Newton Markkula, the sister party game now in Pardon me. The point are the location at this location electric fear intensity coming from Q two and Q trick and and the resulting field points. No. See when to watch 23 The field you to Cuban will have the same menu too. The feed due to Cuban will help. Yeah, the same magnitude? Yes, in part eight and will be perpendicular to combined Feed off Q 20 Children. So back to some of feeling vector some of things from all three charges. All right, will help. A magnitude? Yeah. Greater. Then them off. Part one or part eight. No. See that figure? This is Cuba just just killed them. This is guilty. Minus fight. Yeah, and that's fight Mhm. So, in a victory due to its midpoint, this midpoint here it is E two 83. And this is, um you won't. This anger will be 93. Oh, mhm. So net field at the sponge You were h he won, plus ito plus a tree as you mean. This is to me. Accepts is yeah. This is to me by access, so he will be okay. Cuban, upon our respect, keep you cool upon Invite me square say this is by Kevin. Okay, 23 appoint Dubai to spread. Exhale K will be covered. Cuban is find my popular. This is point Gino 555 square by cap Prosecutor to five. Into 10 to the department. Six. Yeah, two point 95 10 to depart Ministry upon to for his care. Oh, Uh huh. Yeah, this is X kept mhm. This is 19 to 10 to revival night five into 10 to 35 minus six. Oh, by cap 10 to 5 into 10 to the power six will be common. Yeah, to into five into 10. 24. Minus six until 42.95. But yeah, Uh huh. And finding the magnitude you will get X component is square to find components square on substituting the weather You will get 4.13 10 to the power end Nuclear parts Hula. That so? Thanks for watching it

Hi In the given problem, there is a combination of four charged particles. First of all, this is a positive charge Plus 5.0 Q. Then here this is another identical charge plus 5.0 Q. Then on the line joining them there is a midpoint of these two charges. And from this charge from this mid wind there is another charge that is also positive but Having a magnitude of 3.0 Q. And finally the fourth charge particles, this is negative minus 12 Q. And all these distances are given us B. Each. All these expenses are now as we know, electrical always goes away from the positive charge. So electrical, you do this Plus 5.0 Q will be downward and electrical due to this lower plus 5.0 Q will be upward and these two electric fields will be equal and opposite so we will cancel each other at and this observation point is marked as speak so at point B. Electric fields due to two Plus 5.0 Q charges are equal in magnitude as the distance the same. And the charges also saying so the magnet will be seen, so they are equal in magnitude but opposite in direction. So, net electric field at B. Due to these two charges will be zero. Now the remaining two charges Because of this plus 3.0 Q charge electric field, it is a way from the positive charge And you do this -12q. It is towards the negative charge. Like this. This will be so these are the two electric fields. Let it be even, Let it be E two. So finally, net electric field at this point B will be equal to as these two are opposite in direction, so they will be subtracted. So, first of all, due to this minus 12, you this is K. Q. The magnitude of charge. No need to use a sign as the electrical is a vector country. So this is K 12 Q by distance, which is too deep. The distance of -12 cue from this point, we will become deeply steeped, needs study and minus K trig. You divided by the stance. Again the square. So here it will be electric field at P E P is equal to Okay into 12. You buy for the square. Okay, Into 12. You by four. The square mine escape into three Q by T square. Now this is four. Threes are 12. So finally this is K three cube I. D square minus K three Q by the square. So this comes out to be zero. So there is no net electric field at observation point B. Thank you.

Problem. 24 points. 16. They have this arrangement of some charges, the mentions and he started units of charge has shown here we were told that get the convention that the potential zero at Infinity. We'd like to find what the potential at the center is, and we're told as a hint that thoughtful examination of the arrangement can reduce the calculation. So let's, uh well, let's take them at their word and see. See what we can think of about this rain. So one thing is that the corners are unequal distance away from the center, and we know that the potential, the contribution to the potential that some charge makes only depends on the distance between the point of interest in that charge. So these all have the same distance. And then if we look here two minus three plus two minus one adds up to zero. So at the senator, the contribution of these four charges all cancel out anywhere away from that and you're you're back to having to take all of them into account. But since this point has, uh, the symmetry of being equally distant from these, and they add up to zero it works out. So now we just have these two, four times. Q two charges the reach of distance of a over two away. So this is gonna be one over, or pi Absalon not, uh, work you two over a over to plus that the other charges identical. And then this works out to be 2.21 bolts.

Question 45 states that the figures shown here, with three charges listed on the left hand side. But the distance D provided part asks, Find the magnitude of the electric field at eight point halfway between Chartres one and two. But at the midway point indicated by the green dot in this figure on the left hand side, uh, we'll do. We'll talk with the rest of the question later. Old Selves when first. So we want to find the electric field at this point. So when you look at how the forces, or how the charges, I should say affect the military field at that point, so let's do that. A little is green, so I, my electric field from Q one will point towards it from Q to also points towards it, however, from Q three points away from it. So if I know that my neck out for this black black Sorry. So I know my let electric field at a given point is the sum of the electric fields produced by in Chartres near it. I consult for that. These for the magnitude of what I'm looking for. However, one thing we can know because each of these charges have the same magnitude, Um, primarily between Q one and Q two. The sector of the vector. That's you know you're kills for charge. One is exactly equal. And opposite that of Q two. I mean, the electric field component do one equals the exact opposite electrical component. Do do it, um, charge to So therefore, these two will cancel each other out, and I don't need to compute it in such I only into calculate he don't feel do could point to the charge three as my solution. So easy, not frame. So again, I feel that this well, this is it. This is true. You would need to say why you don't need to look at these two electric fuels when solving this type of problem so we could do this with K Q three over the distance squared between the midway point, um, and Q three. So if you look at what that would be so in our if you make a triangle here, que three is here, this is a midway point. This is night length on medical arm just for a second. This is like D, and this is D over two. We're getting value for you. You're not giving a value for our arms We want to solve for this near this is a right triangle. So you see that d squared equals r squared plus de over to all squared. So the link for the distance between these two here again is given my arm. So I need to rearrange the equation to solve for R. And by doing that, I find my distance term years D squared minus d squared over four is my expression for the net electric field the main engine of it. Of the successful again, Um parte a asks for is magnitude and my solving this was playing my values. Um, for Katie. No Q three vinos Negative 5.0, Jimmy 0.0 Micro Coombs the distance thes given some students enemy Find that then that electric field the centers Magnitude 23 significant figures. 6.89 I tend to the Seven Newtons per cool, of course. Great. That's party are being as a man and do the electric field halfway between Chartres Q two and Q three greater than less than or the same as bone part A. We're looking for? If you were yourself electric, you'll at this point would be greater. Listen, well, I can say right away that it will be greater than right greater than state. Why momentarily? Because looking at Electric Fuel's for that perspective for Q two up top, it's pointing down que three is also pointing down. So instead of these components cancelling out like they did in the first round, they'll actually two times the electric fuel to do one of the charges. So there's a net. Well, I should kill twice are, too. I mean, more than zero. Those two contributions rather than zero plus the contributions. Cute one. So because of this, because the cancellation between Q one and Q two the electric field at this point must be greater and part C basically asks to calculate what it is trying to get, at least the electric field at our new point indicated here in the circle again, it's the sum of all the contributions from 12 and three. I like to put on the same page if I can fit it just because that's where the other work were shown. So from Q uh, electric field due to charge one. So if we look at this is now so again where the point is here. Just this distance between Q one and the midpoint was the same as you found for in part main sowing death K Q. One over this distance D squared minus D squared over four. So now what I can do. It's good. Like you mentioned because we have the same magnitude. Charge Q two and Q three. They're both in the same direction. I only need to look at one of the two terms and multiply it by two to get the total electric you'll for that. Those two components to go to times eat to, let's say, because we can't eat, too. The manager t two equals a magnitude e three. It's a very place se e three with e to energy two times e to this equation. And so I consists a k que two over the distances, dear. Over to square. It's a D squared over four. And so, by calculating this Oh ah, my polities. This is not quite true because they end. There are different reactions. I just didn't count for that. So I can say, um not means That's Adam. That's interesting. Uh, yeah. Okay, this is E one. This is E to 83 but when they make a note, is that looking at our figure here at least 82 83 or this line? He wanted to this line there perpendicular each other. Think of the net on the trip field. Because these two terms are perpendicular. You just take the square root like a right triangle rate. So it's e one squared, plus, essentially, remember, it's I want to eat you plus the three squared or Yeah, I'm cynical. That's right. I could I can just do this two e two square in this whole time squared. So that's again. This is my e one term. This is my to be to term. I substitute in this equation here, and I find that my net electric field in the scenario 4.19 times 10 to the eight. Okay, neatness per cool. So that's our final answer. It is larger than we found party, which is what we hope for my plan for, um So the tricks, tips and tricks of this question is that you to a party at least you want me to? They do cancel out. Because remember, we could need to consider the electric field at vectors. Um, and this near because the magnitude of the charges are the same. They cancel out. If Q on was larger than Q two, then we couldn't do this or vice versa. So have to be the same. Charge the same distance away in order for those two to cancel, um, number three, we found the components for each one and eat two nd three. And because they're perpendicular, we can apply our Pythagorean rule to solve for Annette Electric Field in which we did and found again it was larger than party.


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