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The figure on the right shows sphere with radius a. It has a charge per volume ofp(r) Po r2Remember that the charge Q contained within a sphere of radius R for a sp...

Question

The figure on the right shows sphere with radius a. It has a charge per volume ofp(r) Po r2Remember that the charge Q contained within a sphere of radius R for a spherically symmetric charge density is4nr2 p(r )dr .What is the electric field for r < a? (5 pts)What is the electric field for r > a? (5 pts)

The figure on the right shows sphere with radius a. It has a charge per volume of p(r) Po r2 Remember that the charge Q contained within a sphere of radius R for a spherically symmetric charge density is 4nr2 p(r )dr . What is the electric field for r < a? (5 pts) What is the electric field for r > a? (5 pts)



Answers

ILW The volume charge denrigure $20=$ oblem $52 .$ sity of a solid nonconducting sphere of radius $R=5.60 \mathrm{~cm}$ varies with radial distance $r$ as given by $\rho=$ $\left(14.1 \mathrm{pC} / \mathrm{m}^{3}\right) r / R .$ (a) What is the sphere's total charge? What is the field magnitude $E$ at (b) $r=0$, (c) $r=R / 2.00$, and (d) $r=R ?$ (e) Graph $E$ versus

And we have a electrical equals to 3.269 manipulated by 10 to the power five volts per meter. And the electric potential we it is equals to 2.843 manipulated by 10. To the power five volt. At the distance D equals to 32.37 centimetres from the surface of despair. So we have to calculate the radius of the sphere. Capital art. Okay, So, uh we can write that potential. We this will be equals two KQ by R plus D. Which is where our plus D. Is the distance from the center of the sphere. And electrical E. This can be written as KQ by our is here because electrically is given at this surface. Okay, so from these two equations we can replace discharge you. So we can write that we it will be equal to E. R squared divided by R plus D. Okay, so from here, after rearranging we can write that E R squared minus. We are minus, really? It will be equal to zero. Okay so from here using formula so we can get that R equals two we plus minus. And the root of we square plus four E. V. D divided by 28 So now substituting values in this situation. So we get we which is equal to this value so 2.843 But player by 10 to the power five plus minus and the root of we square that is 2.843 multiplied by 10 to the power five. Holy square plus. Former player B. E. Which is 3.2 69. 10 to the power five. And again made clear by we which is 2.843 multiplied by 10 to the Power five. Molecular Biology which is 0.3237 And divided by total value to Molecular Biology which is 3.269 multiplied by 10 to the Power five World Park meter. Okay, so from here, after solving, we will get that radius R equals to 1.1 to 1 m. Okay, So this is the answer for this question. Okay, This is the radius of the matter is here. Okay.

Okay, so we're going Chapter 22 Problems 61. So we have a sphere here of radius are not, And it carries a volume charged incident of road geek. And it says a spherical cavity of radius are not over, too. Is that which is from a small circle here, and it's a scooped out left empty, as showed. So, partner, what is the magnitude and direction of the electric field at court A. Okay, so we want to consider this sphere as a combination of two spheres of a sphere of radius are not with row E the whole way. And then the small sphere of radius are not over too. Oops are not over to with the negative, Roby. Okay, so if we now consider this the combination of these two spheres, we can see that the electric field at point A will have zero combination from sphere one because it's in the very center sphere, one being positive. So the only contribution will be from listen mauler, negative spear. So let's write out. Dallas is a little here, which is its surface integral of ee dot d a. And this is equal to the charge in clothes over. Excellent. Not so Now. We can rewrite this as E times a equals Q over. Absolutely not. But this is also that's right. Cue enclosed in terms of the surface density. So we know the surface charged her that volume charging the city has given a negative road. E we just multiply this by the volume was 4/3 tie. 1/2 are not cute all over. Absolutely not. So now we have e equals Negative wrote e times 4/3 pi are not cubed over eight. We divide this by absolutely not times. What are surface area is of this small cube and the surface area is given by four pi times are not squared over four. See, these four is cancer out and we're left with negative Brody are not over six. Absolutely. This is the magnitude of our electric field at point a go. We also know that since the electric field is pointing, the electric field always points towards the source of a negative charge density. And since the court sources all to the right of point A, this is pointing to the right go. So now we know the magnitude in the direction um, part B now asks us what is the direction and magnitude of the electric field at point B? Okay, so at point B, we're gonna now have the sum of both spheres contributing. So we should know that he is now the sum of the one plus D, too. Let's take Eve one being or positive Ruiz of the bigger sphere and seeks you being the negative charge density or smaller spirit. So now let's do golf's long at each from each steer contribution. And this is charging closed slow. Not so we can rewrite this as e times four groups go back eaters. Four pi are not squared over. Four. That is the wrong one. So is four pi r squared And in this case, we're doing point point D. So they are Are in this case is the radius of the big circle which is our four pi are not squared in This equals the volume Just 4/3 pi are not cubed terms The volume charged density all over absolute nut. So now we know that our e one is given as a roadie are not over three absolute nut. Cool e too. Now I will be given similarly. But now we should see that way. Have e two tons four poor pie. And now they are for this situation. Now is gonna be three halfs our square because with our plus are not over to all the way to the center of the second sphere. So this is now three halfs are not squared. This equals 4/3 pi three halfs are not cubed times. Negative Row e. That's actually wrong. This isn't three halfs. And the reason that's not is because this fear stops right here. Right. So the radius for the charging city, the volume actually only goes out to our not over two. So this is 1/2 are not cube. And this is all over. Absolutely not. So we can immediately see something's cancel out. And we're left with E too. Equals we should have a pie over three. Now we have are not cubed over eight cars. Negative roadie. This is all over. Absolutely not. And we're left with 9/4 are not squared. And we we simplify this. Now we see this coming out to be Megan. Rudy are not for 54. Absolutely not. Yes. Equals two cool So now e total is just the one plus e to, and that comes out to being wrote. E are not over three. Absolute, not harness road Be are not over 54. Absolutely not. It's a weird fraction, but it comes out to being 17. Row e are not over 54. Absolute enough. Cool, that's it.

A solid conducting sphere of radius, capital R equals to 1.351 m. And it has potential we equals to 3.618 manipulated by 10 to the power five volt volt. Okay. And at the distance D equals to 34.95 70 m. So we have to calculate the magnitude of electric potential at the surface of the spear. Okay, So we know that the electric filler the surface, it will be given by take you by our square and potential at the distance deep from the surface it will be given by KQ by R plus D. Okay, so we can replace from these two equations this child skills. So we it will be given by E R squared divided by R plus D. So from here after rearranging we get equals two. We are plus D divided by capital are scarce. So now substituting values here. So we get electrical equals two. We Which is 3.618 molecular by 10 to the Power five World and our which is 1.351 m plus D, which is 0.3495 m developed by our square, so 1.351 m Holy Spirit. So from here after solving we get electrical er the surface of conducting spear equals to 3.371 But car by 10 to the power five volt meter. Okay, so this is the answer for this question. Okay, this is the electric field at the surface of connecting spear. Okay.

In this problem, we're going to be calculating the enclosed charge into using God's Law, calculate from that the magnitude of the electric field for a sphere where the volume charge density is given by this equation. And we're going to be doing that for the regions R. Is less than big are and R is greater than bigger. So that will be inside the sphere and outside of the sphere. All right. To start. We're just gonna want to go ahead and calculate the enclosed charge. So cute. Uh given the radius R from the center of the sphere when we're starting this calculation for the region R is less in our And using that we're going to move on to our is greater than our Okay, so you are is just equal to the volume integral of row of our times the volume element. Okay, we're going to do this in spherical coordinates. So setting up the integral. We have Go from 0 to 2 Pi. The girl is here at a pie and integral from 0 to our of row of our prime. We're just changing our symbol of integration here. Just so we don't confuse ourselves. Times are prime squared stein data. The our prime if they did they feel okay, we're going to go ahead and gather some terms here. So nick Kroll of 0 to 2 pi D. Theta and general, Oh sorry, let's defeat Integral of 0 to Pi signed data. D Theta and Federal Reserve are our prime. Our prime squared D. R. Prime. So this is just a little bit of an abuse of notation that we're going to use. Uh specifically just to calculate out these two integral really quickly um, calculating those out we get equal to or pie times the integral. There are and I'm also going to go ahead and expand out row of our prime. Now it's going to be beta over our prime times sine of pie are prime over to big are times are prime squared the our prime canceling some terms and factoring is equal to four pi beta, you go from zero to our our prime sign of omega, our prime D our prime. And what I've done here is I've just assigned a constant omega is equal to pi over two big are, I've just done that for the sake of convenience here for a little bit. We'll go ahead and see why that is the case here in a second. So in order to solve this integral, that's going to go ahead and require that use integration by parts. So once we've done integration by parts, I'm not going to show that here, you should know how to do integration, but parts if not, just go ahead and look at any first year calculus textbook, but given that we can go ahead and find that this The whole thing is equal to four pi beta times negative two our big are over hi times co sign of pie little or Over two big are minus for pipe, R squared, sorry, just four R squared big R squared over high squared times sine of hi over uh sorry times to our the car. Right. So now we have our charge as a function or in close charge as a function of radius. So for our greater than sorry little are greater than big are ah the sphere has stopped so we can just find that Q of our is actually just equal to Q of big are so if we plug in big art too, this equation here we get that that is equal to negative or pi beta times for big R squared over pi squared. So you can go ahead and distribute out these terms. I've gone ahead and not done that for now. Just because it'll be more compute for the next step. You can go ahead and do that if you want. We're gonna move on to calculating the electric field. Just putting out we now have the electric or the enclosed charge inside as well as its outside the sphere. These two expressions perfect us then our All right. So moving into causes low we have for the spherical case, The magnitude of eight times for high R squared is equal to ah coun closed that radius over epsilon? Not so that implies for R is less than big. Are we now have that? Yes, is equal to Q of our over four pi epsilon? Not R squared. And that gives us beta over epsilon. Not Times -2 big are over our co sign of pie are over too big. Are minus for big. R squared over pi squared, R squared times sine of hi are over to our and this is for inside of the inside of the sphere of charge. I'm just gonna go and given a subscript of the less than sign to indicate that this is inside the sphere. So for are greater than big are that gives us that. Yeah, on the outside is equal to Q of big are over or pi epsilon not little R squared. Which if we plug in all of our factors here we have that that is equal to negative four big R squared. Better over pi squared epsilon, not R squared. And the reason I didn't factor out these things here is to make this calculation easier. Afghan hadn't skipped through the actual algebra of it, but it shouldn't be too much. Okay, so we now want to check that E inside our, sorry Big are at the boundary is equal to E outside the boundary. And if we plug Big are into this and into here, we find that that is the case.


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