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[-/3 Points]DETAILSSERCP11 15.3.OP.018.MY NOTESASK YOUR TEACHERPRACTICE ANOTHERThe flgure below shows three charged partlcles, Iylng alang the horlzantal Oyic Partl...

Question

[-/3 Points]DETAILSSERCP11 15.3.OP.018.MY NOTESASK YOUR TEACHERPRACTICE ANOTHERThe flgure below shows three charged partlcles, Iylng alang the horlzantal Oyic Partlcle 3,00 cm to the right of Particle has 2,15 nC charge and is 00 cm the right ofat left, has 6.25 nc charge: Particle has 1.00 nC charge and IsiWcm20cmWhat the magnitude (In NJC) of the clectric ficld at point D0 cm to the right of A? NCfourth particle wlth charge magnitude-D0 nc placed at thls point: What are the magnltude (In N} an

[-/3 Points] DETAILS SERCP11 15.3.OP.018. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER The flgure below shows three charged partlcles, Iylng alang the horlzantal Oyic Partlcle 3,00 cm to the right of Particle has 2,15 nC charge and is 00 cm the right of at left, has 6.25 nc charge: Particle has 1.00 nC charge and Is iWcm 20cm What the magnitude (In NJC) of the clectric ficld at point D0 cm to the right of A? NC fourth particle wlth charge magnitude -D0 nc placed at thls point: What are the magnltude (In N} and directlon of the net electrlc force on It? direction Select- Need Help? Raadut



Answers

In Fig. $21-26,$ particle 1 of charge $-80.0 \mu \mathrm{C}$ and particle 2 of charge $+40.0 \mu C$ are held at separation $L=20.0 \mathrm{cm}$ on an $x$ axis.
In unit-vector notation, what is the net electrostatic force on particle $3,$ of charge $q_{3}=20.0 \mu \mathrm{C},$ if particle 3 is placed at ( a $) x=40.0$ $\mathrm{cm}$ and $(\mathrm{b}) x=80.0 \mathrm{cm} ?$ What should be the $(\mathrm{c}) x$ and $(\mathrm{d})$ y coordinates of particle 3 if the net electrostatic force on it due to particles
1 and 2 is zero?

So from Sim entry, we see that the net force um come on the net force component along. Why access is zero now, the net force component along X axis will point right word Now, given Fada is 60 degrees. So if on the force would be twice off cue three times Kyu won plus component of Theodora or ah, prediction of fate Our protection off the component along, um X axis. That's why we have cool science data here. And then we have four pi unsettled, not discord. So that's the force Company lets a spread Excuse me. So that's the production of force along X axis. And since we know that co sign 60 degree is half so we can use that number on DA, we see that F three will be okay. Q three Q one divided by two. Sorry, uh, you wouldn't be here. It's a squid because we can get rid of this half with the two multiplied in front of the force. So this is equal to 8.99 times 10 to the bar. Nine Newton meters squared are Coolum SCA. Red Q three is five times into the bar. Native 12 Coolum and Q one is given us two times 10 to the power negative 12 column and then for its 0.9 five meter and then prescribed there, which finally gives us the force as 9.96 times tend to the bar negative 12 years. Thanks for watching as you

Problem Number 21. In this problem, there are three charges given. Judge Q one is equal to minus 1.59 No columns. Judge. You, too is equal to plus 3.2 Nano columns and charge Q three is equal. Do Plus five Nando columns. These charges Oh, God on the negative. Bye, exes. The charge. You too. Is that the origin with a positive side? The charge Q three is Boston and at 0.4 meters on the negative by direction. The charge you won, isn't it? Get there and is there 016 meters from the origin on the minus y line? No, we have to find the net force on the judge. Q. Three off plus five Nano columns so we know there done that. Force on Cue three is due to charges Que tu and Cue One. Since you two and you three are off the same side, which is positive so they will repel each other. The force on cue three by charge Yuto is after along minus y direction and Q three and Cuban are off. Oppose it sign so they will attract each other. The child Q one is negative and you three is posted. So the force on charge Q three by charge Cuban is attractive and towards Cuban or minus y direction. The scan with bitterness. Everyone So everyone and after our boat in the same direction, which is along the minus y direction. So we can write the net force on charge. You three is along minus y direction and magnitude. Off this net force Can we opened as equals everyone plus f too. This is a photo. Okay, you won you three bye. 0.2. It's square plus force after which is a cold. Okay, you three, you do by 0.4 is square now, substituting that given values K is equal to Coolum Constant and its value is nine into 10 to the power nine Newton meters squared for Columbus Square. The value off Cuban is 1.5 into 10 to the power minus nine. And well, you off you three is five into 10 to the power my nest nine. This is divided by 0.2. He's square No for force. After you can write lining to tend to the power nine indoor fight into 10 to the power minus nine into value off you two, which is 3.2 into 10 to the power minus nine. This is always rewarded by zero point four squared. Now, when we saw this, we get of is equal to one point 687 into 10 to the power minus six. Less zero for nine into 10 to the power minus six. This is a call to 2.5 beard, son into 10 to the power minus six. New does. This is the net charge on cue three. And it is along minus Why direction?

For this problem we have two charges. Charge one with a total charge of three micro colognes and charged to with the charge of negative four micro combs and were given the X and Y. Positions of each of those charges. And first we want to figure out what is the magnitude F and direction data of the force between those two charges. So we'll start by plotting out where these charges are located on the xy plane just because that will give us a visual, which makes it easier to see what's going on. Um and how these charges are related to one another. So charge Q1 is going to be over here in the first quadrant And charge Q two will be over here in the second quadrant. In order to find the magnitude of the force, we need to use Q. Lam's law which is that The force will be won over four pi epsilon. Not Times the magnitude of key. one Times The magnitude of Q two all divided by the distance between them squared. In order to figure out the distance between them squared, we want to consider the triangle, you right triangle that's formed Using Q two and Q. one Where one side has a length of Delta Y which would in this case be one centimeter and the other has a length of delta X. Which in this case is negative 5.5 centimeters. Then the hypotenuse is D. Which would be equal to the square root of delta X squared plus delta Y squared by the pythagorean theorem. Therefore we can replace that in our expression for columns law and just simply divide the product of the charges by delta X squared plus delta y squared. Now we have all of the information that we need. You can go ahead and plug in all of the values that are known for these variables and you would get that thief magnitude of the force is 34.5 newtons. Next we want to figure out the direction and the direction is typically given by inverse tangent. So the data is going to be tangent inverse of delta Y over delta X. And so again, if you use a one centimeter for delta Y, negative 5.5 centimeters for delta X, You get an angle of negative 10.3°. So the magnitude of the force is 34.5 newtons, the angle is negative 10.3°. Now we want to consider adding a third charge Q three With a magnitude of four micro combs. And we want to place Q three such that the net charge felt by the, sorry, the net force felt by charge two is zero. We already know that Charge two is going to be attracted to charge one. And so in order to balance that out, it needs to feel a force in the opposite direction. Um from charge three. And because charge to have a negative charge and charge three has a positive charge, they will be attracted to each other. So we would expect charge three to be placed somewhere over here in the second quadrant. But we want to figure out the actual X and Y Positions Up Charge three. I'm gonna get you race the previous work to make a little bit more space for this. And there are a few different ways to go about figuring this out. I'm going to consider the X and Y components of the force separately because the some of the forces On charge to in the X direction must be equal to zero. And the some of the forces on charge to in the Y direction must be equal to zero. Excuse me if we are looking first in the X direction, Charge too will feel a force from charge one and from charge three. So we know that the magnitude of the force felt from charge one in the X direction Must be equal to the magnitude that it feels from charge three in the X direction. And again we'll use columns law to find the magnitude of those forces. So we would end up with 1/4 pi epsilon, not Q one Q 2. And in this case we're just considering the X direction. So this would be over delta X squared and then that is going to be equal to 1/4 pi epsilon not cute, three Q two. And the distance between them will be X three, the X component that we're looking for minus X two which is negative two centimeters squared and of course we can cancel a few things here. Um You would get that Q one times X three minus X two squared is equal to Q three times delta X squared. And we can go ahead and multiply to simplify this a little bit more. So I'm going to plug in the values for the charges as well. So we end up with three X three squared minus six times negative to you. Uh X three plus negative two squared is equal to for times And Delta X was negative 5.5. So this is negative 5.5 squared here And you can simplify, I'm going to skip that step. You would plug into the quadratic formula. I'm going to skip that step and just write down here are options for X three. So X three. Once you simplify this and solve using the quadratic formula Is either going to be 4.35 cm or negative 8.35 cm. And again looking at where X three is placed on our plot, we know that it's not going to be 4.35, it's going to be this negative 8.35 cm. So the X component Has to be negative 8.35 cm. I'm gonna go ahead and erase again to make more space but we're going to repeat that same process in the Y direction to figure out what the Y component is going to be equal to. So it looks like it should be a positive number and it should be larger than Why to which is 1.5 cm. Again Some of the forces is zero. So the magnitude of the force from particle one in the Y direction must equal the magnitude From Particle three in the Y Direction. I going to write out again from columns law 1/4 pi epsilon. Not Q one Q two over delta Y squared. And this will be equal to 1/4 pi epsilon. Not Q three times cute too over and again we would have here Why three minus Y two squared and you can cancel and cross multiply and you would have four times delta Y squared. Were delta? Why? Here is one centimeter is equal to U three micro cool. OEMs times And again you would have why three squared. Vitus two times. Excuse Me? times. Y three Plus 1.5 squared. And again you would go through the process of multiple, distributing and simplifying and plugging into the quadratic formula. And once you do that you'll again get two options for Y3, You'll get that. Why three can be 2.65 cm or that? Why three can be 0.34 cm And again looking at what it should be, it should be a positive number larger than 1.5. So we conclude that 2.65 is the correct answer and .34 is incorrect. So the Y component is two point or sorry. The Y position of charge three must be 2.565 centimeters.

Hi. The given problem here, this is the X axis. This is the origin at which we are placing a charge you three a charge you one has been kept here. It is having a negative charge. Let this point is a along the X axis at which this charge has been kept minus Q. One. The distance at this point from the origin is 0.200 meter. Then another charge positive in nature. Rescue has been kept here. Suppose that are going to be the distance from the origin is minus 0.300 meter. No, it is given that the net force acting at this charge U. Three due to these two charges charged you to put that be and charge minus Q. And put a day has given us that would be F at all. And that is equal to four points to zero micro newton. The value of human is minus 4.50 No problem. And value of kyoto is plus 2.50 No, no problem. So using this expression F. At all, there are two forces at this car's guilty. S Q three is a positive charge. It will be attracted by this minus Cuban by a force effort. Or do you do A. And it will be repelled by this positive. You too. If at all due to be both had in the same direction. And this net force fo will be given by the sum of F at all due to a yes effort to you to be so using cool. Um slow for the electrostatic force for F. O. A. This is A into Cuban into cutely you fight it by oh A. Which is little 0.200 m to the whole square plus for F. O. B. The game. And that is you two into Q three at all. This is three at B. This is due to the spend 0.300 the whole square. And this force come out to be 4.0 micro newton or 4.0 into english bar minus. Thanks new one. Now taking the common things out which includes the value of K. Which is 9 to $10 per nine. Then Q. Three. That can also be taken out as a common living inside for Cuban. This is 4.50 only the magnitude as the direction already. We have used in the diagram. 4.50 into 10 for minus nine column invited by 0.4 metre. Making us square off 0.2. And here this is 2.50 into 10 for minus nine divided by just squared off zero point three which comes out to 0.9 and it should be all in to 10 dish par minus six. Further taking this bandage for minus nine also has a comin out get nine into 10, Parts nine into 10 for minus nine into Q three bracket 4.50 divided by zero point or. And here it maybe written has 400 50 54 Similarly here this is 255 9. You see 24 into 10. H one minus six cancelling this. Finally this is nine into Q three else in 36. This is 450 multiplied by nine and 250 multiplied by four is equal to four into 10 days, part minus six. Then cancelling this 36. We've done nine if this is four and here it is U. Three into 4050 plus 1000 divided by four. This will to four into 10 for minus six. Finally this Q three will come out to be 3.2 into 10 dish par minus nine Poland. Or we can say Q three is 3.2. My broken lung politics is nano. And here it is the answer for the first part of this problem. No. In the second part of the problem we have to find the direction of this force and it is obvious this force is a king towards right direction of this force. Right now, we have to find a position at which if we put this charge Q three, the net force experienced by it may be zero. To suppose the position is here at point B. At a distance X from the origin at which the net force is zero. So here as the charges guilty, positive. This negative charge will attracted towards itself means this is effort. P. Due to A. And this positive charge will repel it. So effort I need you to be this shoe. This too should be equal and opposite F. F. T. Due to A. Should be equal and opposite to affected you to be So plugging in or non values here. Yeah into Q three. But at b into you charge at a Cuban divided by the distance which is this x minus 0.2 00 x minus 0.200 The whole square is equal to again U. Three. Again ask you to divided by the stairs and here it will be X plus 0.300 the whole square and slinky and Q. Three and plugging in all other values. What the value of Cuban, this is taking the magnitude only 4.50 into 10 for minus nine. Well um x minus 0.200 To the whole square is equal to 2.50 into 10% minus 9. 11. Divided by 0.300 plus. Acts to the hospital, cancelling this bandage for minus nine also. And then solving this equation. Making a cross multiplication. We get 2.50 divided by 4.50 is equal to 0.300 plus X. Invited by x minus 0.200 and then making a square root and making a cross multiplication. Again, we get the final answer to b minus 1.76 m is actually the position which we have assumed to be at positive X axis. Actually desert negative xx is somewhere there. This is a real position of the charge you three here. If I put it it will experience zero. Course answer for the third part of this problem. Thank you.


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