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FoncosInrcc-Cnan 0 SystcmCoulomb's Iaw for the magnitude of the force between two particles with charges Q and Q soparaled by distance d is Qq IFI=K whero K = ...

Question

FoncosInrcc-Cnan 0 SystcmCoulomb's Iaw for the magnitude of the force between two particles with charges Q and Q soparaled by distance d is Qq IFI=K whero K = and €0 8.854 x 10-12 C /(N - m?) is the permittivity of free space: T7co ` Consider two point charges located on the axis: one charge 416.5nC is located at 1,.715 In the second charge . 92 nC at the origin (x 0.0000) .Pan =What is Ihe net force exerted by Ihese two charges on third charge q 54.5 nC placed pehuaen 91 and % at Ij 065

Foncos Inrcc-Cnan 0 Systcm Coulomb's Iaw for the magnitude of the force between two particles with charges Q and Q soparaled by distance d is Qq IFI=K whero K = and €0 8.854 x 10-12 C /(N - m?) is the permittivity of free space: T7co ` Consider two point charges located on the axis: one charge 416.5nC is located at 1,.715 In the second charge . 92 nC at the origin (x 0.0000) . Pan = What is Ihe net force exerted by Ihese two charges on third charge q 54.5 nC placed pehuaen 91 and % at Ij 065 m Your answer may be posilive negative , depending on Ihe direction of the force. Express your answer numerically in nertons t0 three significant figures. Hints VuileUn" Submit My Answors Givo Up Fina= hetnte Caminu



Answers

Two point charges are lying on the $y$ axis as in the figure: $q_{1}=$ $-4.00 \mu \mathrm{C}$ and $q_{2}=+4.00 \mu C .$ They are equidistant from the point $P,$ which lies on the $x$ axis. Concepts: (i) There is no charge at $P$ in part $a$ of the figure. Is there an electric field at $P ?$ (ii) What is the direction of the electric field at point $P$ due to charge $q_{2} ?$ (iii) Is the magnitude of the electric field at $P$ equal to $E_{1}+E_{2},$ where $E_{1}$ and $E_{2}$ are the magnitudes of the electric fields produced by $q_{1}$ and $q_{2} ?$ Explain. Calculations: (a) What is the net electric field at $P ?$ (b) A small object of charge $q_{0}=+8.00 \mu \mathrm{C}$ and mass $m=1.2 \mathrm{g}$ is placed at $P .$ When it is released, what is its acceleration?

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.

Hello everyone. And this problem we are asked to find out various things about a configuration of three charges that are given as shown in this diagram. So we have that we have Q. One and Q. Two lying on a horizontal line also although badly drawn in this diagram. Um And Q three is forming a completing a right handed triangle with size 45 and three centimeters. And we're also told two or three more things. We're told that the force, the total force on Q. Three coming from the other two charges Cuban and U. Two is pointing entirely in the negative X. Direction. So if we make this out to be the positive X. Direction then F. Three is pointing in the opposite direction. We're told that Q3 has a charge of plus four micro columns. And we're told that the size of Q. Two or Q. One is to tire centre of minus six. Cool. Um So it's two micro columns but we don't know its sign And Q2 we know absolutely nothing about. So, given these information, the first part of the question is asking us to sketch all the possible diagrams of the forces that you wanted to exert or can exert on charge Q three. So depending on the sizes or depending on the on the signs of these charges, the diagrams are going to be different. So we're going to have four possibilities. Right? So the first possibility is that both Q one and Q two are positive, in which case the D. Uh diagram I is the one that's applicable. So both of the forces are pointing above the horizontal and they're pointing kind of opposite each other. So that's the first possibility. Second part. So let's say that this is, you know, we could label this is Cuban is greater than zero and Q two is also greater than zero. That's the first possibility. Second possibility Q one is greater than zero but he too is smaller desert. If he too is negative then it's going to attract the charge Q three. So it's going to have a uh force factor that's pointing below the axis below the horizontal axis towards Q. Two. But the force coming from cuba is still going to remain the same above the horizontal to direct. Um then there's uh possibly number three where you have that Q2 is positive And he one is negative. So in this case key one exerts a force on Q. three which is pointing below the horizontal axis to the left. Where as Q two is exerting a force on you three which is pointing above the horizontal axis to the left. And then lastly we can have both of the churches be negative. So both Q one is smaller than zero and Q two is smaller than zero. In which case both of the forces are pointing below the axis, one is pointing to the right coming from Youtube and what is pointing to the left coming from cuba one. So then Part B is asking us to find. So these are the four possibilities, right? So this is one possible diagram with these configurations. This is another possible diagram with these church allocations or these signs. So the third possibility And the 4th possibility. All right, Good. Now, part B is asking us to use the information that we were given about the direction of the net force on Q3 to figure out which Diagram is applicable in this case. So, since we know that the net force is going to point to the left and this has to be equal to the sum of two forces coming from F. one and F. two. Right. So given that we actually know that the only if you look at the four possibilities in the uh in part a that you notice that the only way that the net forces can point purely in the negative direction or the negative X. Direction is going to be if you go with option A. Three. Right? So it's only the third possibility. So what that tells you then is that Q um It's the Q one is going to be negative Whereas Q two is going to be positive. Right? So that's what we that's what he tells us essentially. Now Part C. is asking us to find the magnitude and of course the signs of the charge in Q. two. So the way we do this is we analyzed the free body diagram, you three essentially. So this is what the diagram roughly looks like. Right? Where we know that the sides are here 5cm four cm 3 cm here. Uh huh. So this is three cm And this is four cm. All right. So we know we know this information. Um And so the figure diagram for church Q three is going to look something like this. So there's going to be force F. Coming from F from Q two, which is F two and it makes an angle alpha with the vertical and then there's going to be another force that's making an angle alpha, the sort of same angle With the horizontal, that's coming from F one or from Q one. Right? So the way, you know, you can convince yourself that this works is that we re label this Angle at Q 1 to be Alpha. And then by using the six o'clock, we know that that's you know, that's the angle that it's going to make with this dash line over here, so that's also gonna be alpha and then knowing that, So we know what's the angle, but that that F1 is going to make with the horizontal. And then the way you figure out what F2 makes with the protocol is to say that we know that there is a right angle between F two and F one, because we know that the sides opposite Q two and opposite you want are perpendicular as well. So we know that this is a right angle. You know that that is a right angle. That we know that this angle here, let me put that in red maybe. So that angle there is going to be 90 minus alpha, so that's 90 degrees minus alpha. And then since we know that, you know this angle over here, it's also a right angle and we know that part of that is that made up of 90 minus alpha, then the remaining angle has to be alpha. So that's how we know that. This is what the angles are doing. But so let's first figure out what alpha is or what Seinfeld franco's NFL for hard. Right? So going to say that sign of alpha is equal to let's see, So it's the opposite the angle, Right? So it's going to be three centimetres over the hypotenuse. So that's going to be um So that's going to be 5cm All right, so it's essentially just 3/5. Okay. All right, so if I read through this triangle then we would have that um these are decides so we have that this is the right angle, This is alpha, opposite death. So there's the three centimetres Next to it. We have four cm. Right? And up here we have the five cm. Okay. Right, so that's the that's the triangle. So knowing that and the definition, you can just figure out what sign Alpha is. We can also figure out what coastline of Alpha is. Co Santafe. Alfa is going to be the adjacent over hypotenuse. So it's four Centim over five cm. So that's four or 5. And now we're in a position where we can actually talk about the next forces. So first in C were only asked to find the management of Q. So what that is going to require us to do is to analyze this free body diagram over here, right? So if we analyze this free body diagram, we know that F three is purely pointing in the negative X. Head directions. So there's no there's no force acting in the Y direction. So that tells us that F net Y is equal to zero. And so F net Y is going to be equal to F two times sine alpha. You're sorry F2 times go sign alpha after Hussein, Alpha minus F one time. Sign off on. Okay, But what are the forces here? Right, so F two exerts a force of K times Q one times Q two. Sorry, Q two times to three. Right, Let's get the let's get the indices. Right? So what to exerts on three Over a distance of the distance between Q two and you three. And so that's getting to be three centimeters squared. So Over I'm gonna call this part 2 3 squared times. Cosine of alpha minus K times Q one Q 3 separation of 123 squared times the sine of alpha. Right? So that's F one sign alpha. Okay, so now we can put in our values here. So K is just um so that's we can factories KN Q three. Right? Just to simplify it a little bit. Then this is going to become Youtube ties, which you don't know. It's called ST alpha. It's actually let's rearrange this whole thing. Right? We know that is zero. So what we were interested in finding is Q two. So let's say that we have on the left hand side, we have K times Q two times Q three. Your sign of alpha over or 23 squared. This is going to be equal to K Times Q one Q three times sine alpha Over our 1 3 squared. So what Hansel's? So that is the K cancels Q three Cancell. Right? So now we have Q2 Over our 2 3. So this is between two and three. The distance between church too and church three is there is 3cm right? So that 0.0 two m So 0.02 m squared times co sign alpha, cosign alpha. We were adapted before over five. So this is 4/5. This is equal to Q one Over the distance between one and three. This is between one and three is four. So that's four cm. So that 0.004 0.0 point 04 squared sine alpha we said was 3/5. So the 1/5 also cancelled. So we have that Q two four, Q 2 over 0.02 squared is equal to three times Q one or 0.04 squared. So we're going to multiply through by 0.2 squared and divide before. So Q two is going to be equal to Um Let's see, so it's gonna be three over four. Q one Time 0.02 Over 0.04 squared. And of course that's just a half squared. So this is going to be 3/4 times Q one Times 1/2, 1/2 Squared. So cute too. So that's just a quarter. So it's going to be equal to 3/16 times. He went okay, so if Q one laws, what was Q one, let's go back to find out Q one. We were told what the size of Cuban was, It was too micro columns. So it was two Micro Cool Arms. So then we're going to have 3-16 tips to Just 10 to the -6, 3- six school arms. So then this is equal to three over eight Times 10 to the -6 school arms. So putting that into the calculator, so that's um 3.75 Times 10 to the -7 corns. Okay. And as we've established this is a positive charge, right? So this is boss, this is positive, it's a positive charge, right? Because Cuban had to be cuba had to be negative according to the diagram. B Um Q one had to be negative and Q two had to be positive. Okay, So there's that and the other thing we have to find is the magnitude, So this is not part D. You have to find the magnitude of f. So what is the magnitude of the force of the net force? Right? So the magnitude of the net force, it's just going to be uh since, you know, well, this would be the X component squared plus the white component squared square read it but we said that the white component is zero. Right? We specifically designed it to be zero. So then this is just the magnitude of F f f f X. And so going back to our free body diagram, we noticed that F X is just going to be F two times sine alpha Plus F one Times Coast Alpha. So F X. It's going to be F one times co sign of alpha Plus F two times sine of alpha. Sure. So we can work that out. So this is going to be K times Q one Q 3 over or 13 squared. There's co signs of the co sign. Going back up, the coastline was equal to four or 5. So this is four or 5 plus K Times Q two Times Q three For our 2 3 squared Times Sine Alpha which is for you or five. So glad we can put some numbers. So this is going to be um in fact er as K times Q three and that is going to be Q one Times 4/5 Over our 1 3 or 1 3 was 0.04 squared plus youtube Times 3/5 Over 0.0 uh Three I swear. Excuse me Actually, it looks like we made a mistake here. So the distance between two, that's right. So this was this was 0.3 squared. So the distance between two and three was 0.3 squared. So it's not true that it cancels out like that, It's not gonna be as nice. So we're going to have that, this is going to be um Actually maybe it's even better. This is gonna be a 0.03 Squared which is 3/4 squared. So it's going to be yeah right, this is going to end up Being 3/4 squared. Mhm um So it's going to be three or 4 cute all together. So this will be we just get rid of these wrong answers over here. So this is going to be instead. Um So three cubes over four cubes. Um Q one. So that's 27 over 64 Times 2.0. That's 10 to the -6. So that is 27/32. Alright. 27/32. Mm So that's 0.843. So let's let's say that this is eight 8.43. times 10 to the minus six. Cool Arms started 10 to minus seven columns. Yeah 8.4 times april 44 times Central minus seven problems. That's what you do is sorry I made a mistake in the distance between church Q two and Q three. So that's where that came from. So going back to our calculation of the force then, so we have the Q. K. Times Q three we can characterize and we had the sign and the crew sign Values and then we just have to add everything up. So now it's going to be equal to K. is roughly 9.00 times 10-9. Q three we were told was uh for micro Fulham's I believe. Right? Uh We just check. So Q three was 4 micro columns exactly. Mhm. So it's that times 4.00. I have sent through my six Q one, he said was to zero times 10 to the -6 Times 4/5, which is 3.8 Over 0.04 Squared plus. Need to add to that. Um 8.44 Times 10 to the -7 Times 0.6, Divided by 0.03 squared. So now we can just punch distance of the calculator and this will give us the size of the force. So this is going to be nine times 10,009 times four Stinted in -6 times cancer. This is going to be To two times there. About eight that's 10 30 negative 6/0 0.4 squared plus 8.44 times 0.6 times tent city negative 7/0 0.3 squared. And so this gives us 56.3. Yes. So this is the magnitude of the force, The magnitude of the net force that is being exerted on Q three by Q one and Q two.

So for this problem, we want to find a key to using the various different point charges given to us. So first, want to find the force, Charge one to the charge of the organ. So using columns are McCall's cake thanks to you one. Missed you, too. Over our square. You just plug in the numbers that are given. You know that J is equal to a point on it until the ninth times. Negative 25 Micro cools Getting the problem for Q one, then times 8.4 Glencore kilns get for the third point charge at the origin. Take that divide by the distance doing it, which would be 0.22 square that square and you should get a force around 39.5 So that's the first charge. Uh, the force between the first charge and the charge at the origin. You also need the second charge from Q two. Mystery charge is forced to agree force. Would you do that? You use the same equation constant the same 8.9 times of the night, a needing Centimeter square or Coombs squared units for the cake constant. Multiply it by the mystery. That's when you kind of find just leave it as the Q two. For now, that's just the, uh, look variable for that on, uh, the charge of the origin is again cause it ain't look for Micro Koop's. They want abide by the divided by the distance. Between that, which would be was your 0.34 means that's where for that you should get 6.54 times 10 to the fifth times Q two. So we don't know the value for q T, but that's we're trying to find. They're two separate. Multiply together peacefully. So we have the two forces. Now that we don't know exactly what Q two is from this force, they give us the total force. So the Net Force, which is 27 unions. So if we just find the difference between the force the first force of the second force, um, we can just do excuse algebra toe isolate Q two. No, these are the two forces. And so, if you do, you subtract 39.5 we get they got 12.5 units, goes negative. The *** cancel out. You just divide so cute to divide 6.5, but you divide 12.5 by 6.5 times 10 to the fifth. Two new in units. Cancel out. You would get that. The cue to the charge of second charge is 1.85 times 10 to the negative five schools.


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