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Circular cylinder integral question

  1. May 22, 2009 #1
    the question is here:
    http://i42.tinypic.com/9j2alv.gif

    the charge on one small ring is
    [tex]
    dq=\frac{Qdx}{h}\\
    [/tex]
    the problem is with the distance squared part
    its constantly changes
    i dont know how to write in the distance square part
    ??
    [tex]
    E=\frac{kQ }{h} \int_{h+d}^{d}d x
    [/tex]

    and it a ring not a small partical
    so the normal formula of a field kq/r^2
    is not logical here
     
  2. jcsd
  3. May 22, 2009 #2

    tiny-tim

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    Just use Pythagoras :smile:

    (and the result of Example 3.8)
     
  4. May 22, 2009 #3
    this is the solution:
    http://i40.tinypic.com/2d14t2.gif

    regarding part a:
    it seems that they use one ring at a time.
    but the Pythagorean distance suggests that they use only a small part from that ring.

    if they use only a small part of a ring then the whole this should be a double integral
    one for the sum of every small part of a ring
    the other is for all rings.

    but if they use one ring at a time then they should take the distance from the center of a ring to the point,not the Pythagorean distance of a small part from a ring
    ??
     
  5. May 22, 2009 #4

    tiny-tim

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    No … they're using √(x2 + R2),

    which is the distance from every part of the ring (in other words, for every value of theta) …

    I agree if the ring wasn't circular, then those distances would be different, and you'd need a second integral (over theta) …

    but they're the same, so the field is exactly the same as if all the mass of the ring were concentrated in one part :smile:
     
  6. May 22, 2009 #5
    i dont know how to choose a small part of a ring:
    first way:
    dtheta is a small angle from from the center of a ring.
    i say that the whole ring is 2pi so a small part of that ring is dtheta/2pi
    [tex]
    dq=\frac{Qdx}{ h}\frac{d\theta}{2\pi}
    [/tex]

    the other way is saying that dx is the width and r*dtheta is the length of that small
    square.the density is [tex]\frac{Q}{2\pi r h}[/tex]
    so i get
    [tex]
    dq=\frac{Q}{ h2\pi r}dx Rd\theta
    [/tex]
    what way of choosing a small part is wrong?
     
    Last edited: May 22, 2009
  7. May 22, 2009 #6

    tiny-tim

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    They're both right!

    (I'm assuming you meant r not R in the second one :wink:)

    If you cancel the r/r, they're the same.

    So you can either do it that way, which gives you a ∫∫dr dθ, and the integral doesn't depend on θ, so you can immediately integrate wrt θ and get 2π times the ∫dr, which cancels out the 2π on the bottom

    or you can do what the book does, which is to take the whole ring together, which had no 2π on the bottom, so it comes out the same. :smile:
     
  8. May 22, 2009 #7
    in the second one:
    i meant R
    because the radius is R do its dR
    is it ok?
     
  9. May 22, 2009 #8
    i cant understand the whole ring method:
    why they do sqrt(R^2+X^2) if its a whole ring and not a small part of it
     
  10. May 22, 2009 #9

    tiny-tim

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    (hmm … just noticed I kept typing ∫dr when I meant ∫dx in my last post :redface:)
    Yup … I got confused between R and r :redface:

    but then you should have had 2πR on the bottom instead of 2πr, and the R/R still cancels. :wink:
    Perhaps I'd better write it out in full …

    ∫∫ (Qdx/2πRh)√(x2 + R2) dx Rdθ

    = ∫ [∫ (Qdx/2πh)√(x2 + R2) dx] dθ

    = 2π[∫ (Qdx/2πh)√(x2 + R2) dx]

    = ∫ (Qdx/h)√(x2 + R2) dx,

    which as you can see is what the book does …

    and it woks because the √(x2 + R2) is the same at all points on the ring :smile:
     
  11. May 22, 2009 #10
    this is my calculations.
    i got got a different double integral then you
    why??
    [tex]
    dq=\frac{Q}{ h2\pi R}dx Rd\theta\\
    [/tex]
    [tex]
    \cos \alpha=\frac{x}{\sqrt{x^2+R^2}}\\
    [/tex]
    [tex]
    k\int_{d}^{h+d}\int_{0}^{2\pi}\frac{dq}{(R^2+x^2)}\cos \alpha=
    k\int_{d}^{h+d}\int_{0}^{2\pi}\frac{x\frac{Q}{ h2\pi R}dx Rd\theta}{(R^2+x^2)^{\frac{3}{2}}}\\
    [/tex]
     
  12. May 22, 2009 #11

    tiny-tim

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    oh yes … I forgot about the cosine :biggrin:
     
  13. May 22, 2009 #12
    i solved the integral but i get an expression which differs alot from the books answer
    where is the mistake
    ?
    [tex]
    k\int_{d}^{h+d}\int_{0}^{2\pi}\frac{x\frac{Q}{ h2\pi R}dx Rd\theta}{(R^2+x^2)^{\frac{3}{2}}}=k\int_{d}^{h+d}\frac{x\frac{Q}{ h2\pi R}dx R(2\pi-0)}{(R^2+x^2)^{\frac{3}{2}}}=\frac{kQ2\pi}{2\pi h}\int_{d}^{h+d}\frac{xdx}{(R^2+x^2)^{\frac{3}{2}}}\\
    [/tex]
    [tex]
    R^2+x^2=t\\
    [/tex]
    [tex]
    2xdx=dt\\
    [/tex]
    [tex]
    \frac{kQ}{ 2h}\int_{R^2+d^2}^{R^2+h^2+2hd+d^2}\frac{dt}{t^{\frac{3}{2}}}=
    \frac{kQ}{ 2h}\int_{R^2+d^2}^{R^2+h^2+2hd+d^2}{-2t^{\frac{-1}{2}}}\\
    [/tex]
    the last part is not an integral i just dont know how to write this straight line with intervals on it in LATEX code.
    so its like that straight line on the right side.
     
    Last edited: May 22, 2009
  14. May 22, 2009 #13

    tiny-tim

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    Well, that looks ok so far.

    Maybe the ends of the cylinder are supposed to be solid also (it isn't clear from the question)?

    What answer does the book give? :smile:
     
  15. May 22, 2009 #14
  16. May 22, 2009 #15

    tiny-tim

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    But that is the same as your solution :smile:

    (assuming, in your last line, the ∫… is really a […] :wink:)
     
  17. May 22, 2009 #16
    wow i think i solved it i got identical expression.
    HIP HIP HOORAY.

    ok now in part b:
    its a solid cylinder .
    i need to take a 3d small part
    i divide the cylinder into small plates
    the density is [tex]\frac{Q}{\pi r^2 h}[/tex]
    [tex]dq=\frac{Q}{\pi r^2 h}rd\theta dr[/tex]
    but i am not sure regarding the dr part
    i know that our radius is temporaty
    why i cant write dR
    i cant see how the dr part changes from one radius to the other
    we can have a constant length that stays the same for every radius and get the same result

    correct
    ?
     
  18. May 22, 2009 #17

    tiny-tim

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    You're easily pleased! :biggrin:
    I'm not really following this :confused:

    The answer in the book uses the formula for the field at distance x from a solid disc …

    why don't you go back to Example 23.9, and see how they got it? :smile:
     
  19. May 22, 2009 #18
  20. May 22, 2009 #19

    tiny-tim

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    That's strange … part b clearly ends "Use the result of Example 23.9 to find the field it creates at the same point"
    Yes … in fact, you must, because R is fixed …

    you'll get the whole of one disk by integrating over dr from r = 0 to r = R. :smile:
     
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