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A positively charged wire bent into a semi-circle

  1. Sep 24, 2015 #1
    1. The problem statement, all variables and given/known data
    A positively charged wire is bent into a semicircle of radius R, as shown in the figure below.
    The total charge on the semicircle is Q . However, the charge per unit length along the semicircle is non-uniform and given by λ=λocos(θ)

    Physics_2.png

    What is the relationship between λo, R and Q?

    2. Relevant equations
    Q = ∫ λ ds
    F = qE

    3. The attempt at a solution
    Q = ∫ λ ds = ∫(from θ=0 to π) λocos(θ) dθ.
    However, this yields evaluating the sin function from 0 to π which is just 0. I'm having a hard time believing that there is no relationship between λ, R, and Q. I was wondering if someone would be able to check my work.

    Any help would be greatly appreciated! :)
     
  2. jcsd
  3. Sep 24, 2015 #2
    It completely depends on where ##\theta = 0## is on the semicircle. Obviously, if it's on one end, Q = 0 because for a quarter of a circle the charge is the negative of the other. But, from your drawing, it seems you should actually use ##-\pi/2,\pi/2## as boundaries, which changes everything (the charge has the same sign throughout).
     
  4. Sep 24, 2015 #3

    Titan97

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    [Deleted]
     
  5. Sep 24, 2015 #4

    SammyS

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    Look at θ as it is shown in the figure.

    It looks like θ goes from -π/2 to π/2 .
     
  6. Sep 24, 2015 #5
    I'm confused as to why θ goes from -π/2 to π/2
     
  7. Sep 24, 2015 #6

    TSny

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    According to the figure, Θ is measured from the y-axis (not the x-axis).
     
  8. Sep 24, 2015 #7
    Oh, I understand what you're saying now!
    So now that I have the proper limits

    Q = ∫(from θ=-π/2 to π/2) λcos(θ) ds = 2λs = 2λr.
    Is this now correct?
     
  9. Sep 24, 2015 #8

    TSny

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    Why did you set s = r in the final step?
     
  10. Sep 24, 2015 #9
    Because I'm running off of 2 hours of sleep and making stupid mistakes like thinking that s is the distance when its not.

    So after I have the integral,

    ∫(from θ=-π/2 to π/2) λcos(θ) ds

    How can I implement R?
     
  11. Sep 24, 2015 #10
    Can I express it as the double integral: ∫∫ λcos(θ) r dr dθ?
     
  12. Sep 24, 2015 #11

    TSny

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    You need to express ds in terms of dθ in order to carry out the integration. (Yikes, only 2 hours of sleep? I feel for you.)
     
  13. Sep 24, 2015 #12
    But I think that's wrong because wouldn't that cover the whole circle as opposed to just the wire?
     
  14. Sep 24, 2015 #13

    TSny

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    Your setup of the single integral is fine. You do not want a double integral.

    In the single integral you are integrating the function cosΘ where Θ is the variable. If you want to integrate with respect to θ, you need to express ds in terms of dθ.

    Recall that if θ is measured in radians, then arc length on a circle of radius r is s = rθ.
     
  15. Sep 24, 2015 #14
    I just remembered the equation for arc length right when you posted that haha

    Which means that ds/dθ = r θ → ds = r θ dθ
     
  16. Sep 24, 2015 #15

    TSny

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    No. You did not solve for ds correctly.
     
  17. Sep 24, 2015 #16
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  18. Sep 24, 2015 #17
    s=rθ
    ds=rdθ
     
  19. Sep 24, 2015 #18

    TSny

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    Yes.
     
  20. Sep 24, 2015 #19
    I swear I'm not normally this stupid :)
    But this makes sense because for every little change in the arc length, there is a change in the angle, while of course R remains constant.
    Thank you!
     
  21. Sep 24, 2015 #20

    TSny

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    OK, good. Better get some :sleep:
     
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