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Insulating spherical shell - E Force on a rod

  1. Sep 23, 2007 #1
    An insulating spherical shell of the external radius b and the internal radius a is charged uniformly with volume charge density p. What force F does it produce on a uniformly charged thin rod, as shown in figure2? Rod's total charge is Q and length 2a. The rod is arranged radially from the center of the sphere, the distance from the rod's center to the center of the spherical system is r.

    [​IMG]

    E[tex]\bot[/tex] = [tex]\frac{\sigma}{\epsilon_{o}}[/tex]

    E = [tex]\frac{q}{r^{2}}[/tex](Ke)

    The goal here is to find the net force acting upon the rod. It's obvious the rod lies parallel with the radial electric field produced by the sphere, so the field the rod produces will not affect the field from the sphere (and thereby the force acting upon it). Though the fields may interact outside the radial line of the rod, but I'm not sure if that will determine the force (hunch says yes...).

    The force at any point along the rod will differ dependent upon its distance from the outer surface of the sphere where the charge is maintained.

    Any advice in regards to this problem?
     
  2. jcsd
  3. Sep 23, 2007 #2

    Doc Al

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    Set up an expression for the force on an element of the rod's length and integrate. First you'll need to figure out the field from the charged spherical shell as a function of distance from its center.
     
  4. Sep 23, 2007 #3
    So first the electric field of a hollow sphere is E = (q/r[tex]^{2}[/tex])Ke where q is the charge of the sphere. I'm guessing I need to use the surface charge density to obtain q?

    The equation I'm finding for the force is F = E(q) where q is a point charge. I have Q which is the total charge for the rod, so dQ will be the infinitesimal charge at any point along its length. The integral will be from r-a to r+a..
     
  5. Sep 23, 2007 #4
    I'm given the volume charge density (Charge / unit[tex]^{3}[/tex]), so I use a and b to get the volume and multiply this by [tex]\sigma[/tex] to get the total charge on the sphere?
     
  6. Sep 23, 2007 #5

    Doc Al

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    Exactly. Then use Gauss's law to get the electric field at any point outside the sphere.

    The use that field expression to find the force on an element of the rod's charge, expressed in terms of its length. The element of charge should look like [itex]dQ = \lambda dx[/itex], where [itex]\lambda[/itex] is the charge per unit length.
     
  7. Sep 23, 2007 #6
    Putting all that rambling nonsense together, I get F = E(dQ), so I will end up integrating: Ke[tex]\int[/tex][tex]^{r+a}_{r-a}[/tex](q/r[tex]^{2}[/tex])dQ

    Is this correct...? It's been a long weekend.
     
  8. Sep 23, 2007 #7
    Okay great, so I am getting closer. :) Let me work this out...
     
  9. Sep 23, 2007 #8

    Doc Al

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    Yes, you're almost there.
     
  10. Sep 23, 2007 #9
    With my shipshod integration I managed to find F = Ke*q*Q[[tex]\frac{1}{r - a}[/tex] - [tex]\frac{1}{r + a}[/tex]]. That's got to be close at least, though I'm not sure about Q.
     
  11. Feb 16, 2008 #10
    so it seems a few years later, another lucky physics student (that's me) gets a chance at solving this problem. Let me explain my thought process in the hope of getting some feedback because the explanation is not making much sense to me.

    This problem should be treated no differently than the F= keqq/r^2 right? And if I know rho = Q / V and shouldn't I be using V = 4pi(b-a)^2 since there is no charge on the inside? Why are the limits on the integral from r-a to r+a? And why do I need the charge density of the rod? If the formula is just charges over r squared and I am given Q isn't that sufficient? Sorry for all of the questions but I am sleep-deprived and desperate. Maybe some sleep will help me gain some perspective.
     
  12. Feb 17, 2008 #11

    Doc Al

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    Since that's the force between two point charges, I'd have to say no. But since the charge distribution of the insulating shell is spherically symmetric, you can treat its field as equivalent to a point charge at its center.
    Yes, use the volume of the shell to find the total charge of the shell. Edit: But that volume should be 4/3 pi (b^3 - a^3). (Thanks to V2Viper25 for bringing that to my attention.)
    While you can treat the shell as a point charge, you can't treat the rod as one. Find the force on each element of charge on the rod ([itex]\lambda dr[/itex]) and integrate to find the net force.
     
    Last edited: Feb 17, 2008
  13. Feb 17, 2008 #12
    Shouldn't the volume being used be 4/3pi(b-a)^3 because it's an insulating sphere? and after your force integration, what happened to the 2a? Shouldn't it be F = ke*(Q of sphere)*(Q of rod)/2a[1/r-a - 1/r+a]
     
  14. Feb 17, 2008 #13

    Doc Al

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    You're almost right. Volume = 4/3 pi (b^3 - a^3). I must have been napping before. :redface:
     
  15. Feb 17, 2008 #14
    If r >> a, wouldn't we be able to treat the rod as a point charge? And in that case, wouldn't F = ke*((Q of sphere)/r^2)*(Q of the rod)? because the charge wouldn't have to be distributed over the rod if it's being considered as a point charge. Or am I looking at this all wrong?
     
  16. Feb 17, 2008 #15

    Doc Al

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    It's true that if r >> a, you could treat them both as point charges. But in that case the problem isn't very interesting. They want you to integrate.
     
  17. Feb 18, 2008 #16
    Okay, sorry for the delayed response. First, thanks for the clarification - I guess it's a good think I have this and other problems to solve if I wasn't able to recognize that by looking at the problem. It means I def. need the practice. I forgot that coulomb's law was just for point charges. And lack of sleep def. manifested itself in getting that equation wrong so thanks for catching that error.

    Okay, so now I understand the integral mitleid set up on post: 09.23.07, 19:20 (sry I don't know how to use the graphics on this thing yet). Is the q in that equation the q on the shell that we're essentially solving for by multiplying rho by volume? (w/ the correct formula for volume of course)

    Also, where our dQ = lambda * dr should I be substituting (Q/2a) for lambda? Will that make it more confusing? Should I integrate and THEN substitute my values in? It's been a few years since calc so I just want to make sure I'm setting everything up correctly.

    Thanks again for the help!! :)
     
  18. Feb 18, 2008 #17
    Never mind, I got it! :)
     
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