Problem

Find the electric field at the location of $q_{a}…

15:03

Problem 19 Medium Difficulty

(a) Using the symmetry of the arrangement, determine the direction of the electric field at the center of the square in Figure 18.52 , given that $q_{a}=q_{b}=-1.00 \mu \mathrm{C}$ and $q_{c}=q_{d}=+1.00 \mu \mathrm{C}$ (b) Calculate the magnitude of the electric field at the location of $q,$ given that the square is 5.00 $\mathrm{cm}$ on a side.

Answer

a)
Upward
b)
$E_{r}=2.04 \cdot 10^{7} \frac{\mathrm{N}}{\mathrm{C}}$

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Physics 102 Electricity and Magnetism

College Physics for AP® Courses

Chapter 18

Electric Charge and Electric Field

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Video Transcript

Welcome back to write. My name is Kevin Trucks. Looks consider here we have a much different charges here. We've got charge A. We have a charge be we have a C and we have a charge D We're considering the electric field that results here at some time in charge of top, which is going to be Excuse me, that's this charge, which I just called you. Now let's consider some the electric field lines. They're going to be omitted or or affected for resulting there we go from each of these charges. If you consider q A you A is a negative charge, so it's electric field lines are going to be pulling inwards so you can imagine that up in the center. The result is going to be in that direction. B is going to have a similar of similar situations, have a number of electric field lines that are all coming into the charges be so we can say that the resultant there or the effect on the first was going to be again up into the right those that I'm adding the S factories to detail because I'm doing vector addition here, let's consider the one at the bottom. But the bottom is going to be admitting because of the positive charge. So I can stack that right onto the composition of vectors that I have with you so far. And then he is gonna have a similar similar results. We're because it's positive it's would be admitting electric field lines. And so Q is going to be affected in such a way. I noticed that the resultant from all of this vector addition is just going to be straight upwards, which should make sense because this is pulling it up into the left. This is pulling it up into the right. This is pushing it up into the left. This is pushing it up to you to the right, so the effect should be just simply and up those lefts and rights or canceling out recertify nicely. Now, if we are looking to calculate the main to the electric field, we're gonna need some distances, and we're gonna need some equation. So let's go open type of the electric field here, which I'll make some e or signature e. Well, it's going to be K Times Q. Where Q is going to be. The magnitude the each of these charges in the quarter. So the first charge, for example, Miguel, choose to be charged a VK Hugh a over our and that be the radius between a and the middle squared. And we would do that for each and every one of these until we got all the way until que que de over the radius that's related to be But we defector k out front. And then we could just simply plug in all of the information that we know we know. For example, the I forgot one detail. Let's go ahead and add that there's one detail that we need happen. That's to treat each of these like vectors. So we actually also need to be including in here the angle that that is being affected here. So, for example, if the if I'm adding up all of these different vectors, we're going be adding up the X component and the wife opponent. So, for example, let's consider that this was just the sum of all of the electric fields in the white component. That would be times sign of the angle data, which would be right here. So time sign of data that was for D for this when it would be, uh there we are this day. Today we could say I'm sign, uh, they a And I'm using that that angle loosely. It's gonna be absolute value. And I'm just going to be using the direction that I got from my picture here. So because it's up into the left, I would say the Y value is positive because it's that up part and plus dot, dot, dot plus all the way until we get one for tea. And I needed against something to represent that the results of this white component is going to be positive because pushing it up into the left, but for just the white. But it's just up. Gets a time sign. I'll call that stated. Now, this may look quite intimidating, but realize that because this is a nice symmetric square, really, each of these is just being affected by a factor of rips you over to. So I could just as easily write all of this as being K times rip t over to and then multiply by que a over are a squared plus dot dot dot waas you d over R V squared just to keep my Alfred nice and simple. So essentially all of these signs of potatoes and these are all just going to be resulting in a value of route to over two for each one of them. Salafis back to that. Ah, Now, now, I do need to be a little bit careful with some of the signs I've used. Plus here, a little bit loosely, really would want to be again, considering what is happening with the direction of each of these. But in this situation, it doesn't quite matter, because each of these white components is going to be resulting in pushing it up. So there's going to be four different ups that are effective here. So I can again this recognize this has been before. Okay. Route to you. Over to where? The radius Where? Where? Where? I have the same charge for each of these. The same magnitude of the charge for each of these. So I can actually even back that out six and then the inside. I now have these these radio right, The radio I, which are one over and even the radio are all the same. So the radio. We're going to be 0.5 times or excuse me. Point. 025 Plans. Routes. Thank you. That's just trigonometry. We're calculating this size, right? This radius right here from this side right here, which I know to be 0.25 with half of my square. Okay, so now we finally have everything we need. We can just throw that nice and into the calculator and type later it all up. What you'll find is when you do that, the result is going to be 2.4 times 10 to the seventh. So the 2.4, any mission squared 2.4 times 10 to the seventh. Newton's cool. I don't enjoy this video. If he did is take a moment clicthat heart the bottom of screen to let me know as well.

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