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Three point charges are located In the xy-plane as in the figure. Given that Q = 33.4 /C , what is the magnitude of the total electric force ( in units of N) actin...

Question

Three point charges are located In the xy-plane as in the figure. Given that Q = 33.4 /C , what is the magnitude of the total electric force ( in units of N) acting on the charge Q located at the point (8, 6)?3Q(8, 6)40Select one_ A. 7.68B.3.38C.2.19C0 0.87E 1.21

Three point charges are located In the xy-plane as in the figure. Given that Q = 33.4 /C , what is the magnitude of the total electric force ( in units of N) acting on the charge Q located at the point (8, 6)? 3Q (8, 6) 40 Select one_ A. 7.68 B.3.38 C.2.19 C0 0.87 E 1.21



Answers

Charge $q_{1}=1.40 \cdot 10^{-8} \mathrm{C}$ is placed at the origin. Charges $q_{2}=-1.80 \cdot 10^{-8} \mathrm{C}$ and $q_{3}=2.10 \cdot 10^{-8} \mathrm{C}$ are placed at points $(0.180 \mathrm{~m}, 0.000 \mathrm{~m})$ and $(0.000 \mathrm{~m}, 0.240 \mathrm{~m})$ respectively, as shown in the figure. Determine the net electrostatic force (magnitude and direction) on charge $q_{3}$.

Here for the solution. Used Pythagorean Forum to solve for the distance from 2 to 3 here by using the Pythagorean theorem, we get sequel 200 0.18 m square plus 0.24 m square, and from here he gets equal to 0.3 0 m. Have electrical. Two k Q one Q two by our square now sold for the force by both particles on the third. So F one and three equal to K here by substituting the values of yeah, Cuban Q to end our and sold for this we get f for one and three. We get 4.54 multiplied by 10 to depart minus five of world. Now for I have 42 and three equal to K. Harris will develop you and killed three and ask Then, from here we get it. 3.74 multiplied by 10 to the power minus five and downward and rightward. Now, after when three has X and y force so sold for each in other mm. To do this though need to solve for theta value can do this with the given length values so cheetah can t take all to 0.24 m divided by 0.18 m. And from here we get to take her to 53.13 Larry now for the wife sign 53.13 degree equal to y. Divided by 3.74 multiplied by 24 minus five. And and from here we get y equal to 2.99 multiplied by 24 minus five. And for ex costs 53.13 degree equal to X, divided by 3.74 multiplied by 10 to the power minus five. And and from here we get X equal to the 2.24 multiplied by 10 to the power minus five. And now find the some of the vertical force is the force from particle to is downward. So it is subtracted from the force upward. So we get Force one and three, minus 42 and three. So it will be 4.54 multiplied by 10 to depart, minus five and minus 2.99 Multiplied by 10 to depart minus fine. From here we get 1.55 multiplied by two and three par minus five pen. Now the vertical and horizontal forces add up at an angle to write an upward. Using Pythagorean tore, um, one can solve for the value so have finally equal to underwrote 1.55 multiplied by 10 to depart minus five and whole script plus 2.44 2.24 multiplied by 10 to depart minus five and whole square. And from here we get difficult to 2.7 to multiply by 10 20 power minus five. And yeah, now use the values there to solve for the angle of the vector. So angles, Tita. So 20 take all 251.55 multiplied by 24 minus five and divided by 2.24 multiplied by 200 part minus five. And And from here, we get to take her to minus 34.7 degrees. So this is a complete solution. Step by step with the explanation in detail. Please go through this

Hi, friends. This is the problem. Baseball columns? No. In better form networks on Cuba. Good to be after. What? Unless f 23 plus I have to force it is to be back tradition. After what? Good week. Okay, Q Q d square. Jacob. Yeah, that is minus chicken. Force on to duty three repulsive force. So this could be Martin cape the chip Treat you is where? Minus I camp Force on cue to do to q four. Okay, good job. Food you upon rue de square direction If you see and community in this direction after 21 Just been weak if. Okay, 23 And that will be after four so it can be resolved in the two component minus cost. 45 I kept. Let's sign 45 j. Kevin. So you can write K. He was square by the square to be common. You will get minus two J cap minus six. I camp Mhm. My goodness. Eight by two. Yeah, One upon route, I kept notes. Eight right to one. Upon which Jacob So I'll simplifying it. This post could be recognized My next 4.2. Okay. +139 Jacob nutrition, you can give its magnitude as well as direction. So magnitude of net force on YouTube having the Meadows route off minus port while to square +139 square so it would be equal to four point to new trip. And it's the direction with me theater to return and worse off by component upon X component that is 10 and worse of while 39 divided by So you will get minus 5.4 degrees plus variety. Second is to me 1 74.60. That's all thanks for budging it.

Everyone. This is Question number 19 from Chapter 21 this problem were given to charges on the Y Axis and were asked to find the total force off those who charges on 1/3 charge. Also on the Y Axis. So Drew Quick picture. We have cue to cue three in Q one, and I didn't write the magnitudes down. But I did write the direction so we can go ahead and make a guess at what, Um, what direction this is gonna be acting in? So we're looking at a Charge three here. Q. Want is negative. Q three is positive, so that's gonna be in the negative y direction. And then Q two is positive. Q three is positive. That's gonna be repulsive. So that's also good. And push Q three in the negative y direction. So our net, um, Magnitude is gonna be in the night, the wind direction, and then we need to figure out force. And I went and wrote Cool arms law. Over here we have charges and we're asking for four so we can immediately think of that equation and because these are acting in the same direction, we just need to find the force of each charge to wanting que tu on Cue three and add those two together. So let's do one on three first. So F one on three equals K nine times 10 to the night Newton meters squared per Coolum squared. And I'm probably not see if I can squeeze um Kyu Wan minus five minus 1.5 times 10 to the where Nano's minus nine Cool arms and then Q three is five times 10 to the minus night. Cool moms kind of fit it. And then our radius is, um, let's seek you want is that mine is your 0.6 meters and Q three as at minus your 30.4. So that's gonna be a difference of 0.2 meters and you square that, and when you add that up, you get a total. You're when you put that in your calculator, you get a force of 1.685 comes 10 to the minus six Newton's and that's in the minus y direction. All right, so now we move on to force of two on three not skinny equal nine times 10 to the Ninth Newton meters squared per Coolum or charges charge. Que Tu is 3.2 times 10 to the minus knife, cool homes. And then, uh, that's five times 10 Q three to the minus ninth. Cool moms in our radius is minus 0.4 meters squared club out into your calculator and you get 8.988 times 10 to the minus seventh. So now our total force is just the addition of those two forces. There we go. Couldn't get another page to pop up. So forced total equals. They're acting in the same direction. So we just add them together. 1.685 times 10 to the minus sixth plus 8.988 times 10 to the minus seventh. Okay. And that gives us a total force of 2.58 times 10 to the minus 60 Newton's. And that is in the minus y direction, as we determined earlier

In this question we have three charges aligned in a two dimensional space here we have charged positive two Q. On the right, your charge positive queue up above and at the origin we have charge negative Q. There are three centimeters separating the negative Q. And the positive Q. Charge vertically. And four centimeters separating the negative Q. And the two or the two Q. Charge here. And we want to find the total force um on the charge Q. Here including the magnitude in the direction. So to do that we need to um look at the components of the force. So we will have to label these charges. So I'm going to call the top charge charge one the bottom left charge charge too. In the bottom right charge charge three. So the force is in the X. Direction. Um So are horizontal direction on it are only going to be due to the two Q. Charge. So here we have okay Q. Times two Q over the distance between them and the distance between them. Um We can calculate this will simply be a 345 triangle so be five centimeters. Um But this will be just distance 13 squared. Um But we need to only get the X. Component of this force now because this is the total force acting in that direction and we can find that yeah by looking at that X. Distance so divided by the total distance between the two charges. So sorry that should have been X 13 here. And this is simply the co sine of the angle here but we don't know the angle. So writing the co sign like this if we remember that the co sign of an angle well equal the adjacent side divided by the high partners. Writing it like this is just a convenient way to write it for us. And we also want the forces in the Y direction. So we'll get a very similar um force from that third charge where instead of the X 13 and the co sign of data will be using the sign of data, which will be why 13 divided by um the hypotenuse which is R. 13 But then we'll also have a force in the Y. Direction from charge to this one's only in the Y direction. So we won't need the co signer signed part. This will give us okay Q. Times Q. Where already took care of the minus sign by putting it out front, telling us what direction that forces acting in. And we divide that by our 12 squared. We are then told that the the charge Q equals three micro columns. We know the constant K. Is nine times 10 to the nine. And we want to use all of our ours um and excess and wise in meters to make sure we're using the correct units. Yeah, but we find a force in the X direction 51 0.8 newtons. And this force is to the left. So this should be a negative sign in front of these. Just saying that that force um points off to the left somewhere. Then we'll do the same thing and put the numbers in for the Y force and for for the force from charge three. That gives us a 38 point nine newton force up. And then we'll have to subtract the force from the negative Q. Charge. And this gives us a negative 90 newton force for a total force in the Y direction of negative 41 0.1 newtons. So we can find the total magnitude of the forest by using the by staggering theorem, which is the square root of F. X squared plus F. Y squared, which would give us a 66.1 newton force. And then we'll have to find the direction here. So I need to correct my answer here real quick. I'd put something in the calculator wrong this force in the Y direction here. When we had those together, we should end up with a negative 51.1 newtons. So I apologize for that. And then that gives us a total force of 73 newtons. Um So we just need to make sure we be careful and put things in the calculator there. And then we can find the angle of this force by looking at the tangent of theta and that will equal. Um And this will be the why force divided by the X force. So that say to our angle for the force will be the inverse tangent of fy over Fx. And if we naively put that into our calculator as is we get an answer of 44 0.6 degrees from the positive horizontal. But we know that can't be right because that would be implying that the force is pointing up into the right. So that can't be the case. So what went wrong or any time that you're taking an inverse tangent and ffx. So the denominator here is negative. You need to add 100 and 80 degrees to your answer. So if we add 180 degrees, that gives us a final answer here of 224.6 degrees from the horizontal, which will point down and to the left.


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