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Onducung sphere 0/ jadius 60 m has sunuce charge dcnsity 628 00 10*-6 Clmn"2 denosirod on It The magrilude &t (na clectne jahl NE st outsido Ine surace ol ...

Question

Onducung sphere 0/ jadius 60 m has sunuce charge dcnsity 628 00 10*-6 Clmn"2 denosirod on It The magrilude &t (na clectne jahl NE st outsido Ine surace ol Ihe spheto 2 402+11 Ho2 07 778-05 ena Viese Hee20y

onducung sphere 0/ jadius 60 m has sunuce charge dcnsity 628 00 10*-6 Clmn"2 denosirod on It The magrilude &t (na clectne jahl NE st outsido Ine surace ol Ihe spheto 2 402+11 Ho2 07 778-05 ena Viese Hee20y



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22.16. A solid metal sphere with radius 0.450 $\mathrm{m}$ carries a net charge of 0.250 $\mathrm{nC}$ . Find the magnitude of the electric field (a) at a point 0.100 $\mathrm{m}$ outside the surface of the sphere and $(\mathrm{b})$ at a point inside the sphere, 0.100 $\mathrm{m}$ below the surface.

This is Chapter twenty two. Problem fourteen from Sears ends man skis, University Physics Ah, solid metal sphere with a radius point four five zero meters carries a net charge of point two five zero Netto, poop, find the magnitude of electric field A at a point point one zero zero meters outside the service of the spear and be at a point inside the sphere point one zero zero meters below the surface. The first thing I've done is I have drawn a model of this year. So we're given that the radiance of the spheres point four five zero meters, which I've denoted with capital are for the radius. And then I'm looking at some distance point one zero zero meters outside of the sphere represented by D for that distance and were given that the total charge is point two five zero nano columns. But at the same thing as point two five zero times ten to the negative nine. Cool. Um, so I just substituted in for Nano. So that way I am in standard s I units and the basic way of solving this expression is to use the equation. That's the definition for the electric field due to a point particle cake over arse where Kay is our electro static constant eight point nine nine times ten to the ninth meter squared per Coolum from cool ums Law Q is going to be the total charge that total charge of this fear. Well, as I just said, we were given that and lower case R is the distance from the center, so we can assume since we're outside of the sphere, we can model this as if the entire charge of the spear was concentrated at the center. So that means the total distance from the centre of the sphere is simply the radius of the sphere Capital R plus the distance away from the sphere D. It's our distance away. Little R is the same thing as big R plus D. So I made this substitution into the expression, and then I've just plugged in the numbers. So the constant K plum betting and then plug in the charge of the spear, plugged it in and then plug in. The numbers were given for the radius of the spear and the distance away from the spear. And then we just enter this into the calculator and we get that the electric field at the point that we're interested in a seven point for three Newtons per Kula and that is part a of the problem, Part B is we want to know inside this here. So as I've modeled here, we're going inside the sphere. However, we know that the sphere is metal, metal is a conductor and because its conductor, all of the charges on the metal sphere are going to be on the very surface. Because if doesn't matter if it's all positive charges are all negative charges. All the charges push each other as far away as possible, so they all get pushed towards the surface of the conductor. So over assuming we have electricity IX, the charges aren't moving. Therefore, inside this sphere right here inside the surface, there are no charges in here. So by KAOS is law. If there's no charges and closed than the electric field at at, the surface must be equal to zero. So the electric field is zero inside of the metal conductor and all metal objects are going to have this exact same phenomenon, so equal zero. There is no electric field inside of the window

Correct. So what? We have a solid metal sphere off radius to your point full is your 0.45 millions, which is 45 7 minutes. Everyone flying t electric, few at the point to appoint wine outside and inside. Right? And we didn't We did this view main point. You want to take notice that it's ah medals few. So it's conducted. So the charges itself, who be only on the surface because they tried to repel each other silly. We all on the surface off this year. So any point we didn't dispute itself is surrounded by discharges evenly, uniformly, and as a result, it experiences no electric field. So any point in citing meadows fear itself is zero with on the outside because the electric huh charges are spread out. Really? What experiences? As if the charges are originating from the center off this fear. As a result, we can simply use the Coolum Stone for a point judge which is stated that ah, 1/4 pi absolutely not. Register permitted tee off free space divided by r squared. That is your distance, right? From the point to the center off the off your point church with cute atop richest each church amount. Now take note that all of this has to be sa units. So based on what we were given, Q is 0.25 No, no columns. So you gotta convert that to clubs, right? Not playtime. Stent off my last night and you get absolute notes. Ah, primitive Dios Free space which is can be searched online. You are right. Are it's actually your distance, right? From the points to go center. So what's your point? One plus 2.45 Just the radius. Meet this so it skips us actually 0.55 meters and putting it all together. Ah, we should have all the values over here. We should be able to get, uh, seven point five. Sorry. It's certain point for 144 nutrients. Quill

Solid conducting sphere of radius capital R one equals to 1.206 m. And has charged Q equals to 1.953 micro color evenly distributed over its surface. A second conducting sphere of radius capital R 20 of radius zero R two equals to 0.6 development five m. And initially uncharged at a distance B equals to 10 point double zero m from the first state spared enough to. Spears are connected with the wire which is then removed. So we have to calculate the charge on the white second. Is there? Okay. QQ. So when the both spills are connected, the potential will become same. So we can write that Cuban by Arvin. This will be equal to take you to buy our two. So from here we can write that Cuban by Q. Two. This will be equal to R. One by our to and from the charge conservation. We can write that you will be equals two Cuban plus Q. Two. Okay, so from these two equations this equation one and the situation to we can write that Q two. It will be equals two. R two divided by R one plus R two manipulated by you. So substituting values here so are two is 0.6. Development five. They were the R. One which is 1.206 plus plus 0.6. Development five charge Q. Which is this value so 1.953 micro column. So from here after solving we will get you to equals two six point 57 multiplied by 10 to the power minus seven column. Okay so this will be the answer for the charge on the second is here. Okay.

For this problem on the topic of course is law. We are told that the volume charge density of a solid, non conducting sphere varies with radial distance are and we are given a function of the charge density rho R as a function of our. We want to use this information to find these fears total charge. The field magnitude E at R is equal to zero. R. Is equal to big are over two and R is equal to big are. Lastly, we want to graph E versus our. Now we can integrate the volume charge density over the volume and the result will be equal to the total charge. So the integral of the X. And the integral of Dy and the integral of disease of role is equal to four pi times the integral From 0 to our of the R. R squared throw, which is equal to the total charge Q. Now, if we substitute the expression row is equal to rho s little are of big are wear OS is 14.1 bigger columns per cubic meter. And perform the integration. We get four pi into the OS over are times R. To the power for over four is equal to the total charge cube. And so rearranging we can solve for this charge Q. And we get Q. Is equal to pi times row S times the radius are cute. This is pie into 14.1 Times 10 to the minus 12 columns per cubic meter. Times zero point zero 56 meters cubed. This gives the charge to be 7.7.8 Times 10 to the -15 Cool Arms. Yeah, for part B At R is equal to zero. The electric field is equal to zero since the enclosed charge is equal to zero. Yeah. For part C. Or before we get to party at a certain point within the sphere at some distance R. From the center we can use gases. Law to find the field and we know the field E. is equal to one over four pi. Absolutely not times the enclosed charge over R squared. And the enclosed charge can be found similar to the total charge Q enclosed is equal to four pi times the integral From 0 to our of D. R. Times R squared rho which is four pi times the charge density. Your Os over our times R to the power four over four. And so this implies that the electric field at any point are is one over four pi. Absolutely not times pi rho S Or to the 4th of a big are times little R squared. And so this simplifies to one over four pi epsilon not pie row S, little R squared, divided by big are. And so for see we have R. is equal to the radius big are divided by two Where big. RS 5.6 cm. And we get the electric field E to be one over four pi. Absolutely not times pi times row S. Times are over two squared or divided by our. So this becomes 1/4 pi. Absolutely not pi times row S Times are divided by four. Yeah. And so if we substitute values into this equation, this becomes one of the four pipes alone, not is 8.99 times 10 to the power nine near to meet a squared McCullum squared times pi Times Row S. Which is 14.1 Times 10 to the -12 columns per cubic meter Times are which is 0.05, six m. All of this divided by four and so at R is equal to our over two, we get the electric field strength E to B. Five 0.58 Times 10 to the -3 Newton's Pakula. Yeah. At the position of the radius are equal to big. Are we have the electric field E equal to 1/4 pi absolute or not times Pie Rue S big R squared over big are which is pyro S. R. Yeah. Yeah. So it's simply times pi rho S. R. And so if we substitute our values again into this, this becomes 8.99 Times 10 to the Power nine. Again suppressing the units. Here times pi times 14.1 times 10 to the -12 columns per cubic meter. Time zero point zero 56 m. This gives the electric field and are to be two 23 Times 10 to the -2 newtons Pakula. Yeah. And lastly, we are asked to graph the electric field E as a function of our, which looks as follows. And here we have the electric field E versus our


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