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Magnitude of the E-Field at the center of curvature

  1. Apr 2, 2016 #1
    1. The problem statement, all variables and given/known data
    A circular rod has a radius of curvature R = 8.11 cm, and a uniformly distributed positive charge Q = 6.25 pC and subtends an angle theta = 2.40 rad. What is the magnitude of the electric field that Q produces at the center of curvature?

    2. Relevant equations
    E = kQ/r^2
    6.25 pC --> 6.25e-12 C

    3. The attempt at a solution
    [itex]dE\quad =\quad \frac { kdq }{ { r }^{ 2 } } \quad \quad dq=\lambda ds\\ E\quad =\quad k\lambda \int { \frac { ds }{ { r }^{ 2 } } cos\theta \quad \quad } ds=rd\theta \\ E\quad =\quad k\lambda \int { \frac { rd\theta }{ { r }^{ 2 } } cos\theta \quad \quad } \\ E\quad =\quad \frac { k\lambda }{ r } \int _{ 0 }^{ 2.4 }{ cos\theta } d\theta \quad \quad \lambda =\frac { q }{ s } =\frac { q }{ r\theta } \\ E\quad =\quad \frac { kq }{ \theta { r }^{ 2 } } sin(2.4)\\ E\quad =\quad 2.4e-4\quad N/C[/itex]

    Is this the right answer? What do you guys suggest?
     
  2. jcsd
  3. Apr 2, 2016 #2

    TSny

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    Your limits of integration do not look correct. Be sure you draw a diagram and indicate θ. Also, note that cm is not the SI unit for length.
     
  4. Apr 2, 2016 #3
    Untitled.png
    (theta = 2.4 rad)
    I can't visualize any other situation that would change the limits of integration
     
  5. Apr 2, 2016 #4

    TSny

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    You didn't indicate how θ is measured in your diagram. Also, you should draw the electric field ##d\vec{E}## at the center of curvature that is produced by a small element of charge and show the angle that the field makes to the central axis (the line that you drew and labeled r). Is the angle that ##d\vec{E}## makes to the central axis the same as the angle θ?
     
  6. Apr 8, 2016 #5
    Sorry for the late response, here is what I am getting,
    Untitled.png

    Seems like the angle is theta+pi
     
  7. Apr 8, 2016 #6

    TSny

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    I'm not following. A good place to set θ = 0 is the midpoint of the arc. See below.
     

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  8. Apr 8, 2016 #7
    Yep, was the dE right for my case?
     
  9. Apr 8, 2016 #8

    TSny

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    Yes, I believe so. And, yes, ##\vec{dE}## makes an angle of ##\theta + \pi## from the θ = 0 direction.
     
  10. Apr 9, 2016 #9
    Now, I'm getting the limits of integration from
    0+pi
    to
    2.4+pi
    is this right?

    (using the same integral equation from the original post)
     
  11. Apr 9, 2016 #10

    TSny

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    No. You want to integrate from one end of the charge distribution to the other end. What is the value of θ at each end of the charged rod?
     
  12. Apr 9, 2016 #11
    -1.2 to 1.2? from your picture
     
  13. Apr 9, 2016 #12

    TSny

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    Yes.
     
  14. Apr 9, 2016 #13
    I'm getting 6.64N/C, is this the right answer?
     
  15. Apr 9, 2016 #14

    TSny

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    Yes, I believe that's correct.
     
  16. Apr 10, 2016 #15
    Thanks!
     
  17. Apr 10, 2016 #16

    TSny

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    OK, good work. I hope you see why θ = 0 is chosen at the midpoint of the charged arc, rather than at one end. With θ = 0 at the midpoint, then θ also represents the angle that dE makes to the horizontal.
     

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  18. Apr 10, 2016 #17
    But isn't true even if it's chosen at one end?
    upload_2016-4-10_13-12-45.png
     
  19. Apr 10, 2016 #18

    TSny

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    Yes. Good point.

    We know E will point along the symmetry axis of the arc (the line that bisects the arc). In my picture, the symmetry axis was the horizontal x-axis that bisects the arc. In your picture, the line that bisects the arc would be at 1.2 radians (68.8o) with respect to your x-axis. (The charged arc should extend beyond the y axis in your picture.) In your picture, ##-\int_{\theta = 0}^{\theta = 2.4}dEcos \theta## would give just the horizontal component ##E_x##. There would also be a vertical component ##E_y = -\int_{\theta = 0}^{\theta = 2.4}dEsin \theta## that you would need to calculate. Then you could find the magnitude of E from these components.

    It is more convenient to define θ in such a way that θ represents the angle that dE makes to the symmetry axis. That way, you only need to do one integral,##\int_{\theta = -1.2}^{\theta = 1.2}dEcos \theta##.
     
    Last edited: Apr 10, 2016
  20. Apr 11, 2016 #19
    Ah, I didn't notice that 2.4 rad was actually past the 90 deg mark. That means that the integral from 0 to 2.4 would be missing the vertical component (the one at 90 deg). Also, there will be negative values of cosine for the angles in the 2nd quadrant which would subtract from the overall E. field right?
     
  21. Apr 11, 2016 #20

    TSny

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    I'm not sure which integral you are referring to here, the Ex = -∫dEcosθ integral or the Ey =-∫dEsinθ integral. Whether the arc goes past 90 degrees or not, the integrals will handle it. Even if the arc did not go past 90 degrees, you would still need to find both Ex and Ey for your picture.

    The cosine integral (in your picture), gives the x component of E. The part of the rod that is in the first quadrant will contribute a negative amount to Ex. The part that is in the 2nd quadrant will contribute a positive amount to Ex. But the integral ##E_x = -\int_0^{2.4}dE \cos \theta## will automatically take care of the signs in the two quadrants. The contributions to Ey from all parts of the rod in both quadrants will be negative.
     
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