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Is this a correct mathematical solution?

  1. May 21, 2009 #1
    is this a correct mathematical solution??

    question:
    http://i40.tinypic.com/2hwcswm.gif

    solution:
    [tex]
    \sigma=\frac{7.5\mu c}{14}=0.53571428571428571428571428571429\mu c\\
    [/tex]
    [tex]
    14=2\pi r\\
    [/tex]
    [tex]
    r=\frac{14}{\pi}\\
    [/tex]
    [tex]
    dq=\sigma r d\theta \\
    [/tex]
    [tex]
    k\sigma r\int_{0}^{\pi}d\theta=k\sigma r (\pi-0)=...
    [/tex]

    is it a correct solution?
     
  2. jcsd
  3. May 21, 2009 #2

    dx

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    Re: is this a correct mathematical solution??

    Hi transgalactic,

    The rod is bent into a semicircle, not a circle, so you should have 14 = (2πr)/2 = πr. What's your reasoning behind the integral at the end?
     
  4. May 21, 2009 #3
    Re: is this a correct mathematical solution??

    ohh you are correct
    but thus is a minor mistake.
    the main issue is with the integral.
    i took a small part of a charge and calculated its contribution on the field of point p.
    and then i sum them..

    is it correct
     
  5. May 21, 2009 #4

    tiny-tim

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    Hi transgalactic! :smile:

    Your charge density σ = 7.5/14 µC and your radius r = 14/π are correct :smile:

    but I don't understand what τ is, nor your dq = στdθ equation …

    you need to integrate the field (a vector) from each dθ segment …

    what is it? :wink:

    EDIT: oops … I thought your r was a τ … sorry :redface:
     
    Last edited: May 21, 2009
  6. May 21, 2009 #5

    dx

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    Re: is this a correct mathematical solution??

    No, it doesn't look correct to me.

    Let's do it step by step. The small charge within dθ is (σ⋅r⋅dθ). You got that correct. This small charge is a distance r away from the point O. Now use Coulombs law to find the magnitude and direction of the electric field at O due to this small charge.
     
  7. May 21, 2009 #6
    Re: is this a correct mathematical solution??

    i forgot to use the formula of a field.
    [tex]

    \frac{k\sigma r}{r^2}\int_{0}^{\pi}d\theta=\frac{k\sigma }{r} (\pi-0)=
    [/tex]
    now its correct?
     
  8. May 21, 2009 #7

    dx

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    Re: is this a correct mathematical solution??

    You must take into account the fact that the vectors are not all pointing in the same direction. Your integral just gives you the sum of the magnitudes of the E field due to each small charge. That's not what you want. What you want is the vector sum. Draw a picture. What does the symmetry of the situation tell you about the final direction of the field? Using the symmetry, you will be able to ignore one of the components in the vector sum.
     
  9. May 21, 2009 #8
    Re: is this a correct mathematical solution??

    i know that the "y" components are being canceled
    i thought that the integral sorts that out.
    how to do it in a vectorized way so it will cancel the y components by itself,
    and not by the way of taking components of the feils of each part
    ??
     
  10. May 21, 2009 #9

    dx

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    Re: is this a correct mathematical solution??

    You used Coulom's law to find the magnitude of the field to to a small charge (σ⋅r⋅dθ) at the point P:

    [tex] |dE| = \frac{r d\theta \sigma}{4\pi\epsilon_{0} r^2} [/tex].

    What are the x and y components of this vector? If the x-component is dEx, then the x-component of the final vector will be

    [tex] \int dE_{x} [/tex]

    Similarly for the y-component. You can tell that the y-component of the final vector will be zero because of the symmetry, so you just have one integral to do.
     
  11. May 21, 2009 #10
    Re: is this a correct mathematical solution??

    i get the x component by multiplying by cos theta
    and because of the simitry i need to multiply by 2
    [tex]
    2cos \theta \frac{k\sigma r}{r^2}\int_{0}^{\pi}d\theta=2cos \theta\frac{k\sigma }{r} (\pi-0)
    [/tex]
    correct?
     
  12. May 21, 2009 #11

    dx

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    Re: is this a correct mathematical solution??

    Why is cos(θ) outside the integral? It should be inside. And there's no need to multiply by 2.

    You found that

    [tex] dE_{x} = \frac{k r \sigma d\theta}{r^2} \cos(\theta) [/tex].

    So x-component of the final vector is

    [tex] \int dE_{x} = \int_{0}^{\pi} \frac{k r \sigma d\theta}{r^2} \cos(\theta) [/tex]
     
  13. May 21, 2009 #12

    tiny-tim

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    unless you make the integral from 0 to π/2 :wink:
     
  14. May 21, 2009 #13
    Re: is this a correct mathematical solution??

    so we take the sum of all x components from 0 to \pi (no multiply by 2)
    [tex]
    \frac{k\sigma r}{r^2}\int_{0}^{\pi}cos \theta d\theta= ..
    [/tex]
     
  15. May 21, 2009 #14

    dx

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    Re: is this a correct mathematical solution??

    Now why didn't I think of that! You're so clever tiny-tim. :smile:
     
  16. May 21, 2009 #15
    Re: is this a correct mathematical solution??

    thanks
     
  17. May 21, 2009 #16

    dx

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    Re: is this a correct mathematical solution??

    Yes, that's correct.

    (BTW, you're taking k = 1/(4πε0) right?)
     
  18. May 21, 2009 #17

    tiny-tim

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    erm :redface: … with those limits, isn't it ∫ sinθ dθ?
     
  19. May 21, 2009 #18

    dx

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    Re: is this a correct mathematical solution??

    Oops, sorry! tiny-tim is right. It should be sin.
     
  20. May 21, 2009 #19
    Re: is this a correct mathematical solution??

    sin is the y components
    cos is the x
     
  21. May 21, 2009 #20

    dx

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    Re: is this a correct mathematical solution??

    It depends on where you're measuring theta from.
     
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