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5) What is the total flux that now Dasses through the cylindrical surface? Enter positive number if the net flux leaves the cylinder and negative number if the net ...

Question

5) What is the total flux that now Dasses through the cylindrical surface? Enter positive number if the net flux leaves the cylinder and negative number if the net flux enters the cylnder.N-m-/C SubmitThe initial infinitie line charge is now moved that it is parallel to the Y-axis at * = What is the new value for Ex(P), the x-component of the electric field at point P?Scm;NIC SubmitWhat is the total flux that now Dasses through the cylindrical surface? Enter positive number if the net flux leave

5) What is the total flux that now Dasses through the cylindrical surface? Enter positive number if the net flux leaves the cylinder and negative number if the net flux enters the cylnder. N-m-/C Submit The initial infinitie line charge is now moved that it is parallel to the Y-axis at * = What is the new value for Ex(P), the x-component of the electric field at point P? Scm; NIC Submit What is the total flux that now Dasses through the cylindrical surface? Enter positive number if the net flux leaves the cylinder and negative number if the net flux enters the cylnder. N-m-/C Submit



Answers

The total electric flux through a closed surface in the shape of a cylinder is $8.60 \times 10^{4} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}$. (a) What is the net charge within the cylinder? (b) From the information given, what can you say about the charge within the cylinder? (c) How would your answers to parts (a) and (b) change if the net flux were $-8.60 \times 10^{4} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C} ?$

I haven't infinite line of charge. That's three micrograms per meter. Their electric field lines coming out of the Steiner charge you surrounded with on imaginary cylinder has a radius of Europe, 1 to 5 m and a height of deer on 4 m. You ask yourself, What's, ah, electric fields flux that's felt by the cylinder from this line of charge. Well, because this has cylindrical century, meaning that it's exactly the electric field. Lines are exactly the same. Regard loose what direction you're looking from. Around this line, you can use Scousers Law. Gasela states that the electric field flux for a symmetrical object is just equal to then close charge, divided by F salon, not electric, primitive iti and vacuum. So what is the charge in closed? You just great the Y in charge over the length of the cylinder. That's I'm a shards. You get so basically from these boundaries are over here and here, and you divide that by absolutely not. Since this is a constant of three micrograms per meter, he's a small supply. It advise your 0.4 m the height of the soldier as meters. Cancel out now when you divide by the illiterate primitive ity has value of 8.854 times 10 to the negative 12. But that has unit of Coghlan's per volts per meters. So you want to convert your cool ums to buy crow goals. So you multiply that by 10 to the six and now the micro Coolum on the top and bottom. Cancel app and you're left with units of fault meter, which just get because that is the unit of electric fields books. And that gives you a final answer of 1.36 times 10 to the five fault meters. And that's for a use of 0.25 m. It ask you again, what is it for? It is of 0.5 m and said, Well, you still get the same amount of electric field lines enclosed, so you're the radius of the cylinder share and make a difference. And you can confirm this by noting that there is no radius in the equation that I used. So is still the same number. Yeah, and ask you again, What if you change the length or height of the cylinder? Well, in that case, you doing close more electric field charges. These boundaries are gonna be farther away. What? I did this. I divided the original answer. I got by zero point for the original link. And multiply that by the new lengths to your 0.8. So it should be about double the flux since you have price as long as a cylinder and that gets you 2.7. Longtime send to the five fault meters for lines of 0.8 m.

Hi from 64 on the outward flux of field F goes X come a wife from a zero cylinder X squared plus y squared a square So for the parametric we get a co sign you comma a sign you comma v so we can find her And which is the partial respect to you across the park Shows like to be which is a cosign You signed you comma zero and we can set up our integral from 0 to 2 pi girl from like a l l a co sign you a sign you called zero dot a co sign you a sign you comma zero Do you d v giving us four pi l a squared? Yeah, Now it's from the outward fucks at the field So we have to find which is our over the magnitude of our rocky are to the peas Gonna be x squared plus y squared r p over to which just comes out to a to the p Therefore, our vector fields gonna be x over a p come a y over ap comma zero now setting a part integral. We go from negative l two l integral from 0 to 2 pi after N d a. Which is integral from l and I go from 0 to 2 pi of a co sign square to you over a p plus a sine squared you over a p do you devi and that comes down to four pi al over a 35 p minus one. See? Okay. So for part C with final values of P of the flux is finite. So when a approaches infinity we have the flux equals zero for p grand. The one when a bridge is a Finney We have flux approaching infinity for P lesson one. And then as l approaches infinity way have the flux l over aid the p minus one a is fixed at an Albertsons Vinny, there's there's no P that you could substitute where the flux is finite. So the flux is also going to get a penny

All right, this is checked er twenty two problem for From this Sears and Lemansky. He's university physics textbook. It was shown an example twenty one point ten in section twenty one point five that the electric field, due to infinite line of charges perpendicular to the line and has magnitude e equals lambda over to Pi Epsilon, not times are consider an imaginary cylinder with radius r equals zero point two five zero meters and length l equals points zero point four zero zero meters. It has an infinite line of positive charge running along its axis. The charge for unit length on the line is Lambda equals three point zero zero micro columns per meter are a What is the electric flux through the cylinder due to this infinite line of charge? Part B. What is the flux through the cylinder if its radius is increased? Articles point five zero zero meters and see what is the flux of the cylinder if its length is increased to l equals zero point eight zero zero meters. All right, so we have three parts of this problem, but they're all really just based on the solving for the electric flux in general, and then just the part's A, B and C are just plugging in the numbers, so the strategy is going to be the same for each of these parts. So first thing is I've done is I've made a basic sketch of the cylinder cylinder is given his length. L So that's this. The length along here and the radius of the cylinder is our which I've drawn here just going from this access that goes to the bill of the cylinder. And then the line charges the noted with limp. So as we're given in the problem and shown in the example of the electric field is lambda over to Pi epsilon R efs on art. Eyes are universal. Constant eight point five four times, ten to the negative. Twelve Coolum Square for Newton Times Square meters. So I just made a note of the constant. We're looking at the the area of the of the cylinder so that total area is two pi r times l for the total area and then are the the angle for the electric field. So it's normal to the curve service of the cylinder, and it has the same magnitude at all points along the surface, so five is equal to zero degrees as we're looking at electric field, just going through the curved surface and standing out in all directions. So when we want to find electric flux, that's our basic definition for electric flux. Electric field times, the area times the co sign of the angle fire. But we already know the angle fires. Zero. So we just plug in the pieces that we have electric field, the area and the angle So the electric field is lambda over to Pi Epsilon. Not tens are times the area, which is two pi r L times. The Koh Sai of zero degree coastline of zero degrees, though, is just one. So that's an easy calculation to make. And then we have Lambda over two. Five times are times two pi R Over L. Well, we see we have two pie and then a nominee in both the denominator and the numerator. So that term cancels that term cancels out. So two pi term canceled out. The term are cancels out of focus is in the numerator denominator certain leaves us with the electric fucks as just lambda times elf over Absalon. So that's all we need to solve the problem. We just plug in the numbers well. In part, they were given. The radius is point two five zero meters in the length is point four zero zero meters. So we just plug that we see the radius term disappear. So we don't even need to know what that numerical values for the radio is. All we need is and our line charge Linda and we were given that value is three point zero zero micrograms per meter. I've already taken the liberty of writing this out, said Ah Mike Road. That's just times ten to the negative six. So I've already plugged that in because we need everything in standard units. So that's just three times ten to the negative six cool ums per meter. So it plugged that in for Lambda. Then we have l point four zero zero meters. So we plugged that in for l plug our universal constant upside not in and run that through a calculator and we get the electric flux is one point three six times ten to the five Newton Times Square leaders per Coolum in Part B were asked What is the flux of the Senate if the radius has increased the point five zero zero meters? So now we have a new radius, but our radios canceled out. So we have. The exact same calculation is part because the radius didn't even better. So we have the exact same answer from part A again one point three six times ten to the five noon times meter square two per kula and then part see asks us if the length is increased the point eight zero zero meters. That's just the exact same cock leash again. Exact same Linda exact. Same universal constant. That hasn't changed. All that's changed is the length that we've plugged in. So plugin Alec ALS point eight zero zero meters and that gives us then a final answer of two point seven one times ten to the five new meter square problem. As the final answer for part C.

So we have a cylinder off radius are the length of the cylinder is the same as the damn it off. It's circular pieces, which means this distances to our hands, how this distances are at the center. We have a lot of charge. Q. Now we want to find out the flux. We got service, so what we can actually do is find its flux. So these two circular surfaces on because in gossip still on, we confined the flux through the guard service. So now look ATT. Let's look at the card service, by the way. We're doing this because on the circular surfaces, the area victor is a constant, whereas on the road surface, the area vector changes depending on where you are on the circular surface. So the integration will be a lot easier if you do it on the circular surfaces. And hence we're doing this. No, let me draw a larger feature. No, we have this large circulars on this, so we want to integrate over small discs. Smaller rings off. Let's say three d a. R. On birth off the ring with video. No, the distance from any point on this ring. So the charge will be, let's say, smaller because we also already know that this distance are not. This is our day. We can see that are is equal to the square. Rode off are not squared plus r squared using fighter guardian here because we know this is 90 degrees other. We have this nothing off the electric field. The electric field obviously will be in this direction, which is obvious from the charge itself. So the angle between that and the area Victor would be I could deceive us this angle. Let's call this Tito has flux will be integral. Eve Arden Diva, which is innocuous. The elephant field will obviously be Q by four by its No, not our square area would be two pi r the, uh on Do you know that called the dot product Has a cost teetered on that baby cost. Now, one more thing to look at you got here is costing Isaac will do I just inside by high party New studies are not by smaller, which is are not by square root off r squared plus on our square which gives us in the girl off your way for my esplanade to bite. We have our idea. I was accosted a So there's another are not here. Derided by we have our square on Have another one. They are from Boston are giving us arse purpose are not square All part three will do Now we can cancel a few things here. But this integral is very easy. If you know if you consider our square is equal to some X. You will notice that this is the X bite you on by somebody doing that observation. We can write this Indigo Allies integral You are not by to excel or not we have in the denominator X plus are not square hole about three way too. I'm the new military help the X There is another by do so well liked that down here Now this is a simple integrate. This is off the form one by Effexor About three way too. So the integral big You are not by four extra Not you know that one by expert in integral will be off to form expert handling. I'm sorry. Integral. The XX parting will be off the phone expert in plus one by Empress one. So that's xB. Let's art are not square or are minus three by two less one divided by minus three wayto plus one which turns out to be You are not by two. Exit or not, there's a minus side inside, one by squared off our square. Let's are not square. Remember that the limits are from zero are not We started here we go all the way to are not no beverage concept before here you give us a No, that's it. You have cute are not by two Excellent minus one buddy a little tough to are not square plus one by squared off are not so great. So that is your way to a salon. Our times one minus one day. Now this is only through one face. The flux through the two circular faces will be twice this. We also know that the total flux is equal to the flux through the cylinder circular surface, plus two times this number. So that's Q wayto excellent. Not on minus one way with you, which is equal dough charge enclosed. Do everybody makes it or not? On, does it? Hence the field. You know this circular through the secular part will be we can simply cancel a few times. I get this nexus Hubei. Roto, I said I'm not that is that answer.


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