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Nole; hesumattdumac dists epaced Ommn apan: ht is paralle-plate capueltor constrcted oftwo 20 cn charged potcntinl diftcrence of T0V_ Wnat Ihe clccirie field streng...

Question

Nole; hesumattdumac dists epaced Ommn apan: ht is paralle-plate capueltor constrcted oftwo 20 cn charged potcntinl diftcrence of T0V_ Wnat Ihe clccirie field strength inside?Hou:much ChancEmallolc ninc negative plale #ith _ spced of 2.0 x 10Sm/$.Does potun chctthrueh Feuch the othcz side? If not; whte Ine tuMine point ?Malc?

Nole; hesumatt dumac dists epaced Ommn apan: ht is paralle-plate capueltor constrcted oftwo 20 cn charged potcntinl diftcrence of T0V_ Wnat Ihe clccirie field strength inside? Hou:much Chanc Emallolc ninc negative plale #ith _ spced of 2.0 x 10Sm/$.Does potun chctthrueh Feuch the othcz side? If not; whte Ine tuMine point ? Malc?



Answers

Two Fixed Charges In Fig $23-29$, two fixed point charges $q_{A}=+1.0 \times$ $10^{-6} \mathrm{C}$ and $q_{B}=+3.0 \times 10^{-6} \mathrm{C}$ are separated by a distance $d=10 \mathrm{~cm} .$ Plot their net electric field $\vec{E}(x)$ as a function of $x$ for both positive and negative values of $x$, taking $\vec{E}$ to be positive when $\vec{E}$ points to the right and negative when $\vec{E}$ points to the left.

Okay, So this question we know that two point charges are held a certain distance apart, and we want to get the field at, ah, certain place and the 47 Electron at that point. So let's get down the Gibbons, and then we can discuss how to do the problem. So Q one is to tell them I seven and down cool arms. Um, and Q two is negative six times 10 to the minus. It groups at six point now point. Oh, okay. And they're held, um, 25 centimeters apart. So let's say are as 250.25 meters and then our goal is to get the electric field 0.5, um, centimeters from the negative charge and along mobile line between the two charges. So let's just draw it out here is plus minus. And then we want to be, um, five centimeters from the negative charge. So let's call this distance. Total distance are and let's just call this l just another, um, common special variable. And that's point of five meters. So our goal is to get the electric field here, So let's kind of do a free body diagram. Teoh on a test charge here to get the total electric field. So from the positive charge, um, there's a field to the right. That's positive. So we can call that you won. And then there's a field also to the right because positive charge should be attracted towards the negative direction come from, um, you two and I don't actually know the relative magnitudes. I'd have to figure it out. So now we just want to find the electric field, you know, either the charges and then add it up. And so, um, one thing to note the positive charges a distance of ar minus l from the from the, um from the point as it like point p or something. Okay, let's just not label then. But from the point it's are a distance of ar minus l. And then this is a distance of out from the point. So, um e then is going to be You won much of steel with magnitudes, so e one plus you too. So, um, each he is gonna be a K q over r squared. The both have K and Commons. I could take that out. So then we're gonna have Q one divided by the distance between, um, one and the point squared. So that's ar minus. L squared, and then this one's gonna be a Q two divided by the distance between one and two squared, which is l squared. Great. Now we can go ahead and plug all of this into a calculator. So we just need to sink fit, do nine times 10 to the nine times, um, 212 times 10 to the minus seven. Plus, I'm just going to the magnitude of Q two, Um, because I already took into account the sign of the charge by making the direction, um, to the right of you to So it's gonna be plus six times seven of my state, and I'll go ahead and communicate that after I'm done with my calculation. Okay. And then the two tens of my seven is divided by R minus. L. So that's 25 minus five. Um, some of us it's 20 centimeters, so it's 200.2 squared, and then we want to divide this 6 10 of minus C provided by, um l squared, Alice 100.5 squared. So you get 2.6 2.6 times Senate of five. Yeah, 2.6 times 10 of the five Newton's curriculum to the right. Um and yeah, let me hear. Great. And then we want to get the forest that an electron would feel if they replace at that point. So F is Q E. And then the charge of an electron is the elementary charge unit E. So that's 1.6 times 10 to the minus 19. So we just need to take our previous answer and multiply it by 1.6 times minus 19. I got 4.2 times, 10 to the minus 14 onions.

Okay, so in this question, we have to charges placed on meter apart, and we want to get the field midway between them and then a forest on a charge that situated there. So let's break down the Gibbons, Kiwanis 50 Micro Que Loans and Q two. Oops is equal to negative 25 times 10 of minus sex crew loans after, and the distance between them is one meter, and we want the field midway between them. So let's go ahead and draw this out. Here's one to positive two, which is negative, and we were in the field here, so we find the field by pretending there's a positive charge there and ask which force it would feel. So from the positive charge, it would feel repulsive force to the right and the negative charge. You would feel a attractive force also to the right. So here's let's call this he two and then this one z one. So they're in the same direction. So their magnitudes on together so we can say you total. At least the magnitude of it is the magnitude of B two, plus the magnitude of you one and my maturity to is gonna be K magnitude of Q two and then we want to divide by the distance between, um que two and the point, and that's gonna be our over to cause, right? R is the distance, this distance, and then that's situated 1/2 about halfway point. So it's just gonna be, are too squared. And then, um, e want is gonna be the same thing. Except for with Q two Cubans. Are you okay? Now? I can go ahead and pluck all of this into a calculator. So 9 10 to the nine time divided by, um, 0.5 squared, Um, and then Q two in Cuba, and they're gonna be 50 plus 25 7 minus six. So I'm adding their magnitudes on subtracting, prompting from 25 um, again, because they have field here in the same direction. Six. So I get 2.7 times 10 to the six, um, even spoke alone. And then for the next question, we want, um, what would be the forest on a charge situated there? A Q three is 23. Our story is 20 micro columns. So, um, so we know F is q E. So we're gonna just do Q three times try. Just call it G. Yeah, you total. So I'm just gonna take my previous answer and multiply it by 28 times 10 to the minus six. And I got 54 and kind of nice to know what kind of numbers in charge. And Dustin's will produce Newtons on the in the tens, so micro cruel OEMs and then meter scale distances.

No. In this problem, it could give it two point charges. Cuban and you do are placed on the accepted. This is Cuba, the Kyoto. Our place at the separation off. Three centimeters value on you, but is given three. My problem. That is three to minus six School. 302 minutes. Six full of you would do is give it milers to my problem that it minus two it to 10 to the power minus six Full of on that separation between them is five centimeters are Why did you don't pipe? Meet it. There are two parts of the question in first part and we have to find the position. We're electric police zero So white average electric field. This could be, you know, and in ticket part we have to find the point. Edwidge. What? They tell it to be a hero. Now be start solving for the first part. So here now we are discussing solution for Hebert soon Cuban and u two are like charges. The point average electorate. Really? You will be on the line joining the truth guys is outside of charge not been in the Dutch because that's actually quite electrically zero and three point it and the space of it from minus two to touch. So I see quite a prickly do do People is along busy and XY electric video Your job is a little CV. So electric video, Cuban attitude Oh, I see it in a positive direction. Support electric field, etc. Rubies, Udal even and you must be equal and opposite their magnitude should be the same. But directions would be uploaded. Now we will simply pay for it. The runner can be major. Okay? Cuban divided by is he could square. Okay, Uto divided by BC Chi Square So here against alone not a few bodies Give it three. My problem on a Z will be Are blessed eggs what is square? Similarly you do it through my group with him and it would be X squared. No, we have to simplify for eggs. Cross it by square. Good Approximately played Not so for finding the two up eggs but this time So where you off x will be a little too are divided by but all three miners photo Now that's a job. I who do is one quite for one foot Oddest given peace end. Jimmy doesn't starting audience given fights. Indeed, we did 1.732 By solving it, you will get Yeah, gonna do, centimeter. So you're uncertain. We don't get to send emitter to the right off my this judge. So it is going to to 20. Weaker. So the right up might it do? Judge? So this is the under off first park? No, we have to discuss the second part. The pickle part. We have to find the point average. What they tell it. Zero not supposed be considered, appointee average was Pencil it zero and we will consider it is extra spit from being so we have considered giving a V. It's going to our on we ever doom. Let we need to go to X. We have, as Jim let potential activities to zero. But angel activity will be put. Angel do a that is Hey, you won by And but But until act did you Toby Okay, you go divided by ready. This man who is zero if you want is given three. Micropal on 80 will be out of mine effects, huh? You do it minus two micro polar and B t X No, we have to simply fight for it's drugs would respect with the LAPD and say, Now we're simplifying it. No cross multiply it. But three exits Purdue who are minus two weeks. So here you will get X. It's panel who are divided by five. Our is given fight scent emitted. So your answer will be centimeter from minus you. Judge two bars. Look. So this is the answer for second part.

Okay, so this one is a calm, a little bit of a complex problem. We have three point charges arranged. A such in space. We have to charges that are 2.5 times 10 to the negative. Dying cool looms there. Positive one. Is that negative? 10.5. The other ones at 0.5 this year is the origin zero. Um, and then we have one on the X axis, which is 10.7 meters away from the origin. And that one is three 0.0 times. Tend to negative nine. And the question is, what is the total force acting on this particular charge? Now, we're only focused on what's happening on this charge. So we're going. I'm just gonna draw quick force diagram. They're all positive charges, which means they're going to repel each other. So, um, this charge is going to feel a force of that way from this charge. We'll call this one, will call this, too, and we'll call this three. So one on three is going to make it go that way. And then 213 is going to make it go that way at the same angle. Um and so, um that's gonna be kind of a set up for later. Now the first thing we have to do is we have to. We have to use Cools Law, which is force is equal to K a constant times. Q. One We'll do one in three, first divided by the separation distance between one and three squared. So we're looking for this distance, which we don't know off the top of our heads. But we know this is 0.5, and we know this is point seven, so we can actually use Pythagorean theorem to figure this out. So 70.5 square causes a right triangle. Um, so 0.5 squared plus 0.7 squared is equal to C squared where r squared. I guess I should call it, um, and that will tell us what are is when you plug that in the calculator, you end up getting 0.86 meters. So that's why we're going to use for our are. So let's go ahead and calculate this now that the magnitude of the force So K is nine times 10 to the nine. That's a constant that's just given to you. Um, the first charges 2.5 times 10 to the negative nine. And the second charge is 3.0 times 10 to the negative nine. And that's all over that distance that we just calculated, which is 0.86 squared. Now when you plug that into the calculator, what you end up getting is 9.1 time or, I should say, 9.1 to give a couple more digits 9.1 to 7 times 10 to the negative eight. And so that would be the magnitude of this force. Right here. We'll call this F 13 now F 23 is going to be the same magnitude. Why? Because we have the same charges. The same distance, right? That's still also gonna be 230.86 meters. Everything is gonna be the same. The only difference is the direction. So both of these are going to be the same magnitude. But here's the issue. We can't just add them together and say, Well, that's the total force because we have to include the direction. And if we look at this, I'm going to redraw this over here. Um, here's that charge one forces going like this, the other forces going like this. What we have to do is we have to break them up into X and Y components. So here's the X component for the 1st 1 And then, um, here's the don't sees a different color. Here's the why components and then the X component for the other one is the same. And then the y component is this one like that. So what you're going to see is these two X components are gonna add up to cause a force going that way like that. But these two forces are actually gonna cancel out. So we actually don't even care about the why components, Because they're just going to cancel each other out, and we're just gonna have a net force in the right direction. So the answer is for the direction, at least is going to be going to the right. Now let's find the magnitude. So what we're gonna have to do is we're gonna have to figure out what is the angle here so that we can then figure out, um, how to get this x component. So the way I I found the angle is I looked for this angle because that's gonna equal to that angle. Um, and this angle can be determined by using trigonometry. So if we look at co sign of an angle, we know that CO sign is equal to the adjacent side, divided by the high pot news. So co sign of the angle, which is what we're looking for is the adjacent side is the side that's adjacent or next to. So if this is the angle here, the adjacent side is this 0.7 and then the high pot news was that point groups 0.0.86 To get data by itself, you have to do the co sign in verse on both sides, which cancels that out. But you have to do it to both sides. When you plug that in the calculator, you get 35 points, five degrees, and that's gonna be our angle here. So this both angles, we're gonna be the same again because they're at the same dimensions. So what we're gonna do now is we're going to, um, solved for the, uh, adjacent sides. If I just look at one of these, let's look at this one. This angle here is 35.5 degrees. The magnitude of the force we just calculated to be want 9.127 times 10 to the negative eight. And we're interested in this X component right here. So to figure that out, if I look at it as a right triangle, I can use trigonometry again to find the adjacent side. And so I would just plug in again this co sign, um Fada equation So co sign now theta is going to be 35.5 is equal to the adjacent side, divided by the high pot news, which is 9.127 times 10th and negative eight. Multiply both sides by that value, and I get that the adjacent side is, um, equal to 7.4 three times 10 to the negative eight Newtons. Now that is just one of the X components. But we said it was gonna be two because there's two charges. So all we have to do now is just multiply that by two for both charges. And when we do that, we get 1.4, uh nine times 10 to the negative eighth. Sorry, negative seven Newtons. And that is our final answer. And also we have to add it's going to the right


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