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If the distance between 92 (the charge that creates the electric field E2) and q is cut in half; what is the new value for the electric field E2?E2X...

Question

If the distance between 92 (the charge that creates the electric field E2) and q is cut in half; what is the new value for the electric field E2?E2X

If the distance between 92 (the charge that creates the electric field E2) and q is cut in half; what is the new value for the electric field E2? E2 X



Answers

If the electric field is $100 \mathrm{N} / \mathrm{C}$ at a distance of $50 \mathrm{cm}$ from a point charge $q,$ what is the value of $q ?$

This is Chapter 21. Problem number 36 were given this finger that consists of two point charges killing que tu and pee at a point P x away from Kyu won the next electric field. Due to these two charges air given to a stands zero and you're asked what X is so hit point P There are two contributions right due to Kyu Wan, the there is gonna be the electric field emanating from Q wanted point p. And then there's gonna be the electric field generated by cute at point people. So let's look at the direction of these electric fields and then we can calculate the magnitude. So since you want is a negative charge, the electric field emanating from a negative charge is always a radio towards the charge itself. So then, if you wanna show the direction of the one the electric field generated due to Q one, then it's gonna be towards Kyu won. Using the same logic if we have a positive tours, the electric field and arbitrary point away from this charge is going to be a way radio away from this chart. So here it goes R e to then, As you can see, E one is along positive. Positive extraction of the two is among negative X direction and the magnitudes. Let's see the magnitude of e want is gonna be in Touquet times the magnitude of charge divided by the distance between the charge and the point of interest. In this case, it's gonna be X squared. And then let's write a better early than we're gonna only put I had there because the one is in the positive extraction. If we write the victor for off E too, I'm again first. That's right. The magnitude que que chilled of a absolute value to divided by the distance between the charge and the pump that were interested in. So it's gonna do X plus 0.1 to converge. It'd two meters from centimeters square. Right now, since you're writing it in the vector form, a negative, I have better because the electric field you two is in the negative extraction. Now, when we add these two victories to each other, it's given to us a problem. Then that field that point gives zero. Then what's right? What you want is they want over X squared. I had plus que que tubes over exposed 0.12 squared negative by set for here. Since this equals to zero The K K one expert I had equals K severely a few too explosive 0.12 squared. I have rights seriously to get to the other side. So I had to cancel out case they're gonna cancel out. So we left Korea Magnitude of two on over X squared equals to the magnitude of few two over x los 0.1 to square. With that progressive important equation Now we're gonna d'oh! Then what's the problem? Expiry age. All I'm gonna do is I'm really divide the left inside by cute too, and multiply it by X squared. Well, the second page, that's what I'm gonna have is he was over you two than equals X squared over X plus 20.1 wanted to square right so we can just start calculating things a little. So a magnitude of cumin is 25 times temple burning off six schools, right, divided by cute. Who is given to us was 45. My purple on serves has so when we take the ratio, it's gonna look something like played 555 vehicles on forever. Something for this. Now, let's get here. If we take this fair route here, hand always a spirit of right hand side. This gives us X over X +12 equals to a skirt. A 0.556 Um, which is point 745 So then complete here is gonna be full points. Seven. Point him for five times expressed one point 12 So then, if you re distribute this, what we have is, um, points of him for 1745746 plus, um, Wayne +09 So if we, uh, subject the 0.7 for five from each side than have X minus points Simple, that's people's point whole night. Right? So here the war in the minus 0.745 he cool times X equals nine. From here. Some we're gonna get 0.2 55 X equals point line If you do. That dance is about two by five. Like to did we get 0.3 0.353 of meters? This is the distance. Wrong cue on to the point P

So in this question we have to point charges key one and cue tto onda distance between them. It's given by l on There is a point B at a distance X from Q one where the nettle actual field do Toki one on cue tto at this point is to go on, we have to find the distance x So we know that since Kyu won is negative and cuties positive So I electric field that point you two took you one will be two worlds Kyu won on and they looked at from that point Me to your door They'LL be away from Kyoto on since this At this point, the net electric field is zero. So the magnitude ofthe one should be called the magnitude ofthe that implies the electric field like Hee won which is given by Do you wanna know what ex Claire should be? Goingto littlefield I paid you to Kyoto to ski Q Do Maude over and you have six. Andi gives me my ex as is it going toe l A Times Q one Lord Bhutto, for what you're doing. Loved you don't write minus one mark for what? We don't have gained this question because gay just dance and sound here. So we have all the gallows now and plug them into this equation on DH. You get that to be sixty four point five talents and cynical. Thank you.

Hello for problems, 16.36 We are asked to find the distance of a charge from a point p. We know that the electric field at Point P is just equal to zero. Because of this, we then can say that the electric field, due to charge one, is going to be equal to the electric field due to charge to charge Q one charge Q. Two. In addition, we are told that we we are not told. We can see that Charge one is equal to charge q two divided by two. Thus, we can start figuring out what our electric fields are. To do this, we can use the equation. Electric field is equal to k que over r squared. Thus it is going to be equal to que que won over x squared that must equal que que to all over x plus our quantity squared. Simplifying this equation out, we can come out and see that X squared must equal Q one over Q two times X plus our quantity squared. Substituting in that queue that Q one is equal to Q 2/2 week Now can come up that X squared is equal to X plus our quantity squared all divided by two. When we multiply and simplify this this out we come up that X squared minus to X r minus R squared equals zero. We can use the quadratic equation minus B plus or minus the square root of B squared minus four a. C all over to two. Factor this equation out. We're then told once we used the quadratic equation that X is going to be equal to are times the quantity one plus or minus the square root of two. Since the X in this direction here cannot be negative, we know we just have to look at the positive value of this. Thus, X is equal to our times. One plus the square root of two or X is equal to 28.9 centimeters.

So in this problem, we're looking at a point charge Q. And we're some distance away from it on. We're told that the potential and the magnitude of the electric field is at this point eso We want to find out how far away we are from this point here, what the magnitude of the charges for Q and what direction the electric field is pointing. So in order to do this, we need to establish a couple things first. So first we need to know is how to find the potential due to a point charge. And we can do this using this'd equation right here. So we have the potential equal to K, which is cool, um, is constant times Q. The charge divided by our so the distance away from that point charge in order to find the magnitude of the electric fields, what you do is, uh, pretty much it looks almost the same. So we have Q onda the absolute value of er k rather times Q. The absolute value of Q divided by r squared. So the only difference here is that r squared and this absolute value sign on dso just to remember how we define electric field on electric fields usually directs towards the negative charge and away from a positive charge. So can write that field direction, if we have a negative charge, will be pointing toward it and a positive charge will be pointing away. So now that we have that established, we can start by finding the distance. How far away are we when we have thes specific, uh, this specific potential and electric field magnitude. So we're gonna use the two formulas one potential and one for the electric field to sort of cancel out our unknown, which at this moment is cute. So if we take V divided by a yeah, what we're gonna have is cake divided by our times R squared over K keep. So we get que que to cancel out and what we're just left with is our or well, you can re label that are so we know that we're talking about this distance, and so what are is essentially to just, uh, just going to be the potential divided by the electric field magnitude. So we take these values here 4.98 divided by 16.2, and that leaves us with 0.31 m. So that's how far away we are from the charge. And now what we want to dio is we want to solve for Q. Because we now have our distance on. We know our potential difference. What we can do is we can use this equation on DSO for cute so rearranging to get que will give us The times are over K and case Coolum is constant, just in case you forgot. So we'll have 0.31 times 4.98 tonight, about nine times 10 to the negative ninth. And that's the value of okay here. And this is going to give us 1.72 times 10 to the negative 10th columns. So now we know and what the charge the magnitude of the charge is. And now finally, what we want to find out is, um what point? Which way? The electric fields pointing. So if we just look above at our ah, at our solution for Q here we see that Q is actually a positive value. It's greater than zero eso remember again, if we have a positive charge, the electric field will actually be pointing away from it. So we have the electric field in this case, uh, going to the right. If that's how we define our coordinate system, Just directing away from the charge. And when we have a charge like this, the, um, electric field is Obviously I started taking a radio shape. Eso it'll generally just be away from uh huh away from the charge.


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