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Question 81.5m2.0m1.0 mThe electric field in a region is given by E-a/( cx)i where a= 200 _ b=1.0 , and c-2.0 What is the net charge enclosed by the shaded volume s...

Question

Question 81.5m2.0m1.0 mThe electric field in a region is given by E-a/( cx)i where a= 200 _ b=1.0 , and c-2.0 What is the net charge enclosed by the shaded volume shown above?Your answer:A)-8,8 x10-10 ,B) -70,5 x1O-1o €D) -21,2 xlO-10E) 15 xlO-10 €

Question 8 1.5m 2.0m 1.0 m The electric field in a region is given by E-a/( cx)i where a= 200 _ b=1.0 , and c-2.0 What is the net charge enclosed by the shaded volume shown above? Your answer: A)-8,8 x10-10 , B) -70,5 x1O-1o € D) -21,2 xlO-10 E) 15 xlO-10 €



Answers

The electric field in a region is given by 7. $\overrightarrow{\mathbf{E}}=a /(b+c x) \hat{\mathbf{i}}, \quad$ where $\quad a=200 \mathrm{N} \cdot \mathrm{m} / \mathrm{C}$ $b=2.0 \mathrm{m},$ and $c=2.0 .$ What is the net charge enclosed by the shaded volume shown below?

Fellow students in this question, we have given the electric field at X equals to 10 centimeter is even equals 200 world per meter and at the X equals two minus 10 centimeter. Electric fill it too is uh minus 100 volt meter. So we have to determine the magnitude of charge enclosed by the side equals to 20 centim of Cuba. Okay, so from the ghost here um goes here um we can write that the enclosing integral of E dot diaz. This is equal to the charge and close to barbie. Excellent. Not so this is the classic due to the electric field. So this is electric field which is outward and this is electric field which is also outward. So both flux will be outward. So they will add an E. Molecular bay area will become okay and both are outward and same in magnitude. So this becomes to E. A. And this is where the Excellent. Not So two more players. Electrically 100 area is silent square. So that is 20 cm from 0.20 m. And this Holy Square, this is equal to you'd ever be excellent. Not so you from here maybe equals to a excellent. Not so option A is the correct answer of the problem here. Okay, thank you

Hi. The given problem. There are two point charges. Cuban and U two, which are having the values. As for Cuban, this is 4.0 Nano Kurram. And for Q two, this is minus 6.0 Nando COLUMN These two charges are situated over the coordinate access as given in the figure this is X axis. This is why Axis and the first charge Cuban is located on the X axis at a distance off 2 m. So as young this Toby, the Cuban, the position of charge, Cuban from the origin. And this is 2.0 m and the second charge. Kyoto is on the Y axis at a distance off 1.0 m, so we can assume it to be here. This is the position off. Second charge at least dance off 1.0 meter. No, we have to find the electric flux due to these two point charges through the given spherical surfaces which are centered at the origin. No, in the first part off the problem, the spherical surfaces having a radius off our physical to 0.500 m as the radius is 0.500 m. So this is fear, obviously will not include either off that charge. This vertical surface will not include either off the charge within it. So using the expression for total flux as this is one upon Absalon, not the charge enclosed by the closed surface. And as the charge here is zero So the flux passing through the surface comes out Toby zero and it becomes answer for the first part of the problem. Now, in the third part of the problem, they are saying the spherical surface is having radius off 1.50 m needs to say this surface the second surface spherical surface is having a radius of 1.50 m. So it will include the charge Kyoto within it. But it will not include charge Cuban. So when R equals to 1.50 m this surface, this is very ical surface will include on Lee que two so expression for the flux Electric flux passing through it will become you too divided by epsilon. Not so the answer will be. Then we plug in the different values for Q two. This is minus six this is minus 6.0 in tow. 10. Dish par minus nine. Cool Um divided by 8.854 into 10. Dish par minus 12 column square per newton meters square So this answer comes out Toby approximately minus 678 Newton meters Square Park column So it becomes the answer for the second part off the problem. Now, in the third part off the problem, the radius has been given us 2.50 m. So obviously are spherical surface. Having this much radios will include all the two charges within it. So for this case, yeah, electric flux, the total electric flux will come out. Toby. The algebraic sum off the two charges means it will become five is equal to Q one. Bless you, too divided by absolute not. And when we put the values off these two charges here for Cuban, this is four for you to this is minus six means four minus six, multiplied by 10 dish one minus nine In order to convert Nana Colon toe column Divided by 8.854 in tow. 10 inch par minus 12. This is Colon, and this is Ghulam Square, part Newton, into Meters Square. And finally, this answer for the total flux. Passing through this surfaces, spherical surface comes out Toby minus 226 Newton meters Square Park column, and this becomes the answer for the third part off the problem. Thank you.

For this problem on the topic of castles law, we have the electric field in a particular space to be expressed to. I had Newton speculum and we want to consider a cylindrical Gaussian surface which has a radius of 20 centimeters co axial with the X. Axis one end of the cylinder is that X. Is equal to zero. And we want to find the magnetic. When I find the magnitude of the electric flux through the other end of the cylinder, which is X. Is equal to two m, we then want to find the net charge that is enclosed within the cylinder. Now there is no flux through the sides. So we have to contributions to the flux. One from the X is equal to zero end, which we'll call 50 and this is minus two times Pi times 0.2 squared and one from the X. Is equal to to end, which is what we want to calculate fi two, which is plus two plus two into pi times zero 0.2 squared. And so the flux through the end X. Is equal to two zero 0.5 newton meter squared. The cooler. Now the net charge enclosed by the cylinder can be found from God's law, and this is Q enclosed, which is absolutely not total flux. Fight to plus fire, not and substituting the values above. We get the total charge and close to be 2.2 Times 10 to the -12 columns.

In this problem, we have point charges that are fixed on the X axis of minus three and plus three. And we want to know what charge Creepers of the origin so that the field vanishes at X equals zero y equals three. So let's go ahead and draw this out. So and also right down that kid Givens Q equals Q two. You think it was for Micro Combs? Okay. And, um, we want it to vanish at X equals zero y equals three. And this was meter is not centimeters. Yes. Um, so here's our mystery charge Q. And then here is our Q one Q three. Let's say it's at a distance. Acts and then let's say that axis equal, Teoh three meters. And then, um, we want the field to vanish. Sure. And this is that. Why schools? I guess why you come up with a new name for it. Also call it X, even though it's in the vertical dimension. Hopefully, that isn't good. Too confusing. So, um, que want you to basically they're gonna create a field like this and like this, um and so what we need from the field charge here is we need to feel that, um, equally that opposes the sound so that some of these two fields let's think about that. First, the horizontal components would cancel, but the vertical components will add. So here is the vertical components, and so the some will look something like this. So we basically want the field from here to be equal. Teoh twice the twice the vertical component from one of these front from the Q one and Q three. So let's write down the electric field from Q one. At least the magnitude or he wanted you to. So we can say e as equal to K Q one over. And then we need this distance. So, um, between here and here, So bye, pat diary. And there, um, that's the square root of two times X. So then that's just gonna be so then if you square that will be two x squared, right? Because, um, it's 1/1 over this distance squared. Say that's distance are are equal, started to X, and, um, and then we want the vertical component. So let's call that you why? And so we need to basically multiply by. Um, we can find this angle, and we just need to multiply by the sign of that angle so we can say its times a sign of data. Um, what's this angle s strongest? The triangle created by this are So here's our This is a side of three. This is a site of three. So it's a 45 45 90 degree triangle. And so, um, this is gonna be the sign of 45 seconds. Uh, quickly. Oops. What? What happened? Oh, but okay, data equals 45 degrees. So then that's gonna be the sign of 45 which is one of a route to And then so that's you y of one of these charges. And remember, like I was saying that these two will sum up the total field from Cuba on a cue to it's just gonna be that double the y component of one right, cause he wanted Oh, and, ah, Q two will cause the same, uh, field actually write. That is cute, too. So then we know that the electric field from this one has to equal twice this. So then we can kind of transform this equation. We can say e from Q. So I'll put a subscript Q to the electric field from this. Q. Has to be double. Um, there's the to the vertical component of Cuba. Nick YouTube. So now we need to figure out um, but now you need to kind of right in the generic format of the electric field due to queue. So that's gonna be equal to que que and then, by the way, que needs to be negative. We'll just kind of pepper that at the end, right? If you're gonna make it an electric field that points down Q Q has to definitely be negative. Um, so I equal to cake you and then we wanted to buy that by the distance. Indeed, crew in this point and in that distance is just simply acts. So it's X word. So now, in a position to software Q. In terms of Q one, um, so let's do a bunch of cancellations so we can say goodbye to this K and say goodbye to this X squared. And then who did this to cancels as well. Eso Then we're left with the really nice, simple equation that Q is equal to Q one over a square to two, so it has to be sort of less than, um que Because, uh, it's only one charge. Um, you are? Well, yeah. I mean, I guess it it's kind of hard to tell on the surface of it needs to be less than or greater than Q. Because on one hand, it's closer to this point. And then, on the other hand, it's only one charge. So, um, but the mouth sort of balances that affect out, So you get this.


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