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Lana straight melal rod nas 4 5 radius of 20 mm and & surlace charge of E densily = Mm Irom the axis, nClm"'. Detcrming the mennhvedh(reestsn150.7 NCO...

Question

Lana straight melal rod nas 4 5 radius of 20 mm and & surlace charge of E densily = Mm Irom the axis, nClm"'. Detcrming the mennhvedh(reestsn150.7 NCO N/C3013NCD. 47.98 N/CE 75.33 N/C

lana straight melal rod nas 4 5 radius of 20 mm and & surlace charge of E densily = Mm Irom the axis, nClm"'. Detcrming the mennhvedh(reestsn 150.7 NC O N/C 3013NC D. 47.98 N/C E 75.33 N/C



Answers

A rod $50 \mathrm{cm}$ long and $1.0 \mathrm{cm}$ in radius carries a 2.0-$\mu \mathrm{C}$ charge distributed uniformly over its length. Find the approximate magnitude of the electric field (a) $4.0 \mathrm{mm}$ from the rod surface, not near either end, and (b) 23 m from the rod.

For this problem. On the topic of gases law, we have a charge of uniform linear density to nana. Coolum zometa that is distributed along a non conducting rod. The road is co axial. Their long conducting cylindrical shell that has an inner radius of five cm and an outer radius of 10 cm within a charge of zero. We want to find the magnitude of the electric field 50 cents 15 cm from the axis of the shell. And we want to find the surface charge density on the inner and outer surfaces of the shell. Now we know that the linear density lambda is able to cue over L accused the net charge enclosed by a cylindrical Gaussian surface of radius R. The field is being measured outside the system so that the net enclosed charges only that which is on the rod. So consequently we have the magnitude of the field E. He is able to to lambda divided by four pi. Absolutely not our, which is two times the linear charge density too, Times 10 to the -9 columns, per meter divided by Yeah, four pi absolutely not Times a distance of 0.15 m. And so substituting the value for absolute not into this equation, We get the field strength to be 2.4 Times 10 to the power to Newton's Pakula. Now for part B. Since the field is zero inside the conductor, then there resides on the inner surface a charge of minus Q. And then the outer surface a charge of plus Q. And therefore with our I being 0.05 m, the surface density of charge on the inner surface, we'll call it sigma inna is equal to minus Q. Uh huh. two pi r. i. Times L. Which is minus λ over two pi ri. And so this is minus two times 10 to the -9 columns Permata divided by two pi times the inner radius 0.0 five m. So this gives the surface star density for the charge and the inner surface to be -6.4 Times 10 to the -9 columns per square meter. Now for the outer surface we have our oh equal to 0.1 m. And the surface charge density on the outer surface sigma outer his positive Q over two pi r. o. l. Which we can write as these linear density lambda of uh two pi r. o. And this gives us positive 3.2 Times 10 to the -9 columns per square meter.

For this problem. On the topic of gases law, we have a charge of uniform linear density to nana columns. For me to distributed along a long thin non conducting rod. The Lord is co axle with a long conducting cylindrical shell, which has an inner radius of five centimeters and an outrage outer radius of 10 centimeters, the net charge on the shallow zero. And using this information, we want to find the magnitude of the electric field 15 15 centimeters from the axis of the shell and the surface charge density on the inner and outer surface of the shell. Now we know that the linear density lambda is equal to Q. Of L. Accused the net charge enclosed by a cylindrical Gaussian surface, which has radius R. You feel is being measured outside the system so that the net includes charges only that which is on the rod. And so by gases law, we have the magnitude of the electric field. E. Is equal to two lambda of uh, four pi. Absolutely not are, which is two times the linear charge density, two times 10 to the minus nine columns commuter divided by four pi capsule or not times 0.15 m. Now in a 1/4 pi absolute note 8.99 times 10 to the nine in S. I. Units. And so using that value, we get the magnitude of the electric field to be 2.4 times 10 to the power to Newton's Pakula. Yeah. Now, since the field is zero inside the conductor, then there resides on the inner surface charge minus Q. And on the outer surface charge plus que where child accused the charge on the road at the center and therefore with the inner radius ri of 0.5 m, the surface charge density, we'll call it sigma una is able to minus Q over two pi R. I times L, which is minus lender over two pi R. I. And so this is minus two times 10 to the minus nine columns permit to divided by two pi times 0.5 meters. And so we get the surface charge density on the inner surface to be minus 6.4 times 10 to the minus nine columns per square meter. For the outer surface, we have our own equal to 0.1 m. And the surface charge density of the outer surface sigma. Outer is equal to positive Q, divided by two pi R. O. Times out, which is the linear density lambda of uh two pi R. O. And if we put the values in, we get this to be positive 3.2 times 10 to the minus nine columns per square meter.

In this problem we have given Ah uniformly distributed charge. Que is given 20 Nanda Kula It is distributed Yeah, along a straight line off length 4 m. And now it is bent into a circular are with a radius off 2 m Supports it is Bentley like this and the radius is given to us which is equal to 2 m and this angle Suppose he does. We know that the charge density Linear Children and Steve's equal toe Q I am and we know that he dies equal toe l Why are here This l is equal to four and radius is equal to 2 m so the theater will be equal to two Radian. Now we know that electric field, due to a circular arc formula, is given to us. Que Linda? They are science theater. We have theater using changing from minus theater I to two Plus he delayed. Suppose this is a diagram. So I take upper part. This is minus theater way too. And this is was duty delayed The total Analyst theater I do now we will put the values we know that electric field will be equal to nine in to 10 to the power nine. Lambda is equal to Cuba. L que y el divided by Are we just to equal do sign off, Tita? Were theater changes from minus one Radiant 21 radiant. Because theater, Why do equal toe one radiant? No. By putting the value off charge, which is equal do we have given the value off charges 29 Akula and the value off Ellis for this will be called to sign off one radiant minus sign off minus one Radium. Yeah. So after solving this, we will get the value off electric field, which is equal to 38 Newton for cola.

So in this problem we have ah broad, which is two meters in length on four centimeters where ineighty of cross section in the two cases, you know the charge distribution on it, and we think we want to find how many excess of electrons I was riding on it. So this is very simple. You know that the number of electrons is simply the total charge divided by the charge of each electron, and that isn't He used this to find a total charge with simply to integrate be charged density over the wall. You notice that because the age of cross section is constant, Devi can be written us a video of cross section times the X and that's it. So it's integral off a row. Yes, from zero to EL, which is two meters he wanted by E, which is an affront. Judge noticed that because the excess simply integrated over and there there is no other ex dependence. Integral off the X is simply X, so that's a Pro X by E and by simply plugging in the numbers. Age of cross section is four centimeter square, so that's four times two in four minus two well this way. Row is minus four micro columns. So that's empire minus six columns. We'll meet a cute times X, which is the length is two meters. Notice that there is a meters here you wanted by animatronic charge, which is 1.66 into 10. Fire minus 19 called him. This stands out to be two times 10 party elephants downstream party. Similarly, we need to do this So the number of lacrosse here also is integral of pro TV by e note is that row here is a function of X. So to be X squared the X by e. So that is be times excused by three bye, So we'll simply substitute. I'm sorry there's been a year so in some simply substitute our numbers. So that's 40 to 10 part minus two meters. What's where? I'm sorry. In bar minus two meters holds good. You in here? It's four centimeters squared, so it's only centimeters that is squared. Not the entire time be is minus two. Micro Colin permitted you so that's minus two into 10. Power minus six. Kulim Perimeter. Cute times X Q by three is two meters, all killed by three do. I did buy electronic charges. 1.6 interesting part. Minus 19. Who knows? This turns out to be 1.33 I'm Stan. Pardon Elephants. Simples that.


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