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JenniferSmart1

  • one year ago

What went wrong?

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  1. JenniferSmart1
    • one year ago
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    I'm going to write and draw the problem, give me a min

  2. tkhunny
    • one year ago
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    Sun spots. It's almost always sun spots!

  3. JenniferSmart1
    • one year ago
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    |dw:1360466298145:dw| the radius=a here is what is bothering me. I understand everything up to the point where we change \[\lambda(\theta) \] to \[\lambda\] ok here we go \[dE=\frac{kdq}{a^2}=\frac{k\lambda(\theta)}{a^2}ds=\frac{k\lambda(\theta)ad\theta}{a^2}=\frac{k\lambda(\theta)}{a}\] \[d\vec{E}=-dE_x\hat i-dE_y \hat j\] then all of a sudden \[\lambda(\theta) \rightarrow \lambda\] \[d\vec E=-k\lambda\frac{d\theta}acos\theta\hat i -k\lambda \frac{d\theta}asin\theta \hat j\]

  4. tkhunny
    • one year ago
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    Are you sure it's not just printing convenience?

  5. JenniferSmart1
    • one year ago
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    @JamesJ @phi

  6. JenniferSmart1
    • one year ago
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    hi @dumbcow :)

  7. dumbcow
    • one year ago
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    hi , not sure i can help you ... by the notation it seems lambda went from being a function of theta to being a constant within a function of theta

  8. JenniferSmart1
    • one year ago
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    here is the integral \[-\frac{k\lambda}a\int_0^\pi cos^2\theta d\theta \hat i-\frac{k\lambda}a\int_0^\pi sin \theta cos \theta d\theta \hat j \]

  9. JenniferSmart1
    • one year ago
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    i'm confused

  10. JenniferSmart1
    • one year ago
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    @zepdrix

  11. zepdrix
    • one year ago
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    \[\large d\vec{E}=<-dE_x,\quad -dE_y>\]\[\large d\vec{E}=<-k \frac{d \theta}{a}\lambda \cos \theta,\quad -k \frac{d \theta}{a}\lambda \sin \theta>\] This change is telling us: \[\large dE=k\frac{d \theta}{a}\lambda(\theta)\]\[\large dE_x=k\frac{d \theta}{a}\lambda \cos \theta\] Hmm sorry I can't quite seem to understand what's going on here :c Are the tiny x and y's supposed to be a partial derivatives?

  12. JenniferSmart1
    • one year ago
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    Sorry I was searching the internet....yeah that's probably what happened.

  13. JenniferSmart1
    • one year ago
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    I found it online!!!!

  14. JenniferSmart1
    • one year ago
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    they just have \[\lambda\] in stead of \[\lambda \theta\]

  15. zepdrix
    • one year ago
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    Jenn what class is this? This stuff looks really fun :D

  16. JenniferSmart1
    • one year ago
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    Physics II

  17. zepdrix
    • one year ago
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    An explanation seems warranted. I don't see how you're able to pull the \(\lambda\) outside of the integral if it's a function of theta...

  18. JenniferSmart1
    • one year ago
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    sorry the radius is "a"

  19. JenniferSmart1
    • one year ago
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    I forgot to draw that in

  20. JamesJ
    • one year ago
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    Oh I see. Lambda here is the linear charge density on the line. It is a constant. At exactly what step in the solutions you have linked to do you not understand?

  21. JenniferSmart1
    • one year ago
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    Yes! that's lambda... The document I linked is a little different from mine. Our professor wrote \[dq=\lambda(\theta)ds\] and half way through his solution he changed \[\lambda(\theta)\rightarrow\lambda\] anyhow, my question is in this step \[d\vec E=-k\lambda\frac{d\theta}acos\theta\hat i -k\lambda \frac{d\theta}asin\theta \hat j\] to \[dE= -\frac{k\lambda}a\int_0^\pi cos^2\theta d\theta \hat i-\frac{k\lambda}a\int_0^\pi sin \theta cos \theta d\theta \hat j\] how did we get an additional cosine?

  22. JamesJ
    • one year ago
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    The change from lambda(theta) to lambda, is because the charge density is constant, so that looks kosher to me. This part here at the end about multiplying each of the x and y components by cos theta just looks wrong and it certainly gives the wrong answer. For example, the integral for the E field in the x-direction--the integral before the i--is non-zero. But by symmetry, it must be zero.

  23. JenniferSmart1
    • one year ago
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    well his final answer is \[E=-\frac{k\lambda}{2a}\hat i\] and \[Q=\int_0^\pi a\lambda(\theta)d\theta\]

  24. JamesJ
    • one year ago
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    Definitely wrong. The net electric field at the origin must have NO x component: just look at the symmetry of the situation. So that simply cannot be correct. The solution you have linked to looks correct to me.

  25. phi
    • one year ago
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    Your problem is slightly different from the one you found on the internet You must have missed the critical fact that your charge density is not constant, but a function of theta. It looks like \[ \lambda(\theta) = \lambda \cos(\theta) \] If you start with \[d\vec E=-k\lambda(\theta)\frac{d\theta}a\cos\theta\hat i -k\lambda(\theta) \frac{d\theta}a\sin\theta \hat j\] (you still need to show lambda is a function of theta at this step), and now substitute in the definition of lambda, you get \[d\vec E=-k\lambda\cos(\theta)\frac{d\theta}a\cos\theta\hat i -k\lambda \cos(\theta) \frac{d\theta}a\sin\theta \hat j\] and simplify a bit, you get your professor's result.

  26. JenniferSmart1
    • one year ago
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    I'm sorry....but what's the definition of lambda? I thought it was dq? Here is where I'm getting confused. I guess I don't know the definition of lambda.... \[\lambda=\frac{Q}{length}\]?

  27. phi
    • one year ago
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    I think lambda is the "charge density per unit length" if you have a string of length 1, it will have a total charge of lambda. But in this case, the charge density is changing with theta. it is a max at theta=0º and zero at theta = 90º

  28. JenniferSmart1
    • one year ago
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    How did you know that it is max at theta=0 degrees and zero at theta=90 degrees?

  29. JamesJ
    • one year ago
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    Ah ... if the charge density is NOT constant, then yes indeed: this is a different problem!

  30. JenniferSmart1
    • one year ago
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    well I wrote \lambda(theta)ds in the picture....sorry I didn't make it clear.

  31. JenniferSmart1
    • one year ago
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    How did you know that it is max at theta=0 degrees and zero at theta=90 degrees?

  32. phi
    • one year ago
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    because, based on the equations you posted, I figured out that the charge density as a function of theta must be \[ \lambda \cos(\theta) \]

  33. JenniferSmart1
    • one year ago
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    how did you figure cos(\theta)....sorry I'm trying really hard to understand it. How did you know that \[\lambda(\theta) \rightarrow \rightarrow \rightarrow\rightarrow \rightarrow \rightarrow \lambda cos(\theta)\] what was your thought process?

  34. JamesJ
    • one year ago
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    What exactly is the wording of the problem as it was given to you?

  35. JenniferSmart1
    • one year ago
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    Shoot! I left it at school in my locker D: I'm going back to school at 2pm again today 4hrs from now :S but the image was: |dw:1360511966796:dw| The first question asked which direction the Electric field would point. And the second question was to find the magnitude of the Electric field. Since the Electric field is pointing down, the final answer should only consist of the y component of the electric field.

  36. JenniferSmart1
    • one year ago
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    so the image states: \[dq=\lambda(\theta)ds\]

  37. JenniferSmart1
    • one year ago
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    \[s=r\theta\] therefore \[ds=a\;d\theta\] this would give us \[dq=\lambda(\theta)\;a \;d\theta\]

  38. phi
    • one year ago
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    First, I doubt the professor would make a blunder when presenting an explanatory example. second, explicitly writing \( \lambda(\theta) \) signals that the charge density is a function of theta. Third, based on the final answer, it must be \[ \lambda(\theta)= \lambda \cos(\theta) \]

  39. phi
    • one year ago
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    I am guessing you missed that fact.

  40. JenniferSmart1
    • one year ago
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    Here is what I don't understand. How did you know to write \[\lambda(\theta)= \lambda \cos(\theta)\] why not \[\lambda(\theta)= \lambda \sin(\theta)\] or \[\lambda(\theta)= \lambda \tan(\theta)\] or \[\lambda(\theta)= \lambda \csc(\theta)\] or \[\lambda(\theta)= \lambda \sec(\theta)\] or \[\lambda(\theta)= \lambda \cot(\theta)\] how did you come to the conclusion that it's \[\lambda(\theta)= \lambda \cos(\theta)\] what was your thought process? you said "based on the final answer it must be...." There must have been some thought process.... sigh........

  41. JenniferSmart1
    • one year ago
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    LOL oh sorry...is it because he substituted \[\lambda cos(\theta) \; for\:\lambda(\theta)\] in the integral haha....I see what you mean now.....sorry

  42. phi
    • one year ago
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    yes, that is what I thought. you end up with a cos^2 and a cos sin when you integrate, one goes to zero, the other to your result

  43. JenniferSmart1
    • one year ago
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    http://online.cctt.org/physicslab/content/phyapc/lessonnotes/Efields/EchargedRings.asp

  44. JenniferSmart1
    • one year ago
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    nevermind. that charge distribution is uniform....

  45. phi
    • one year ago
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    I was going to say, "those are for a uniform charge. you're problem is different"

  46. phi
    • one year ago
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    *uniform charge density

  47. JenniferSmart1
    • one year ago
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    page 4

  48. JenniferSmart1
    • one year ago
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    \[\lambda(\theta)=\lambda cos(\theta)\] was probably given.....oh my....i feel dumb

  49. JenniferSmart1
    • one year ago
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    why are we integrating form \[-\frac \pi 2\; to\; \frac \pi 2 \]?

  50. JenniferSmart1
    • one year ago
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    |dw:1360513618512:dw|

  51. JenniferSmart1
    • one year ago
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    MYSTERY SOLVED!!!!!!

  52. JenniferSmart1
    • one year ago
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    <-----slow but determined to learn physics

  53. phi
    • one year ago
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    Good question on the limits. The integral comes out the same for 0 to pi versus -pi/2 to +pi/2 however, I believe the physics is different (where and how the charge is distributed is different for the two ways...) btw, in you picture, that should be dE sin(theta) not dE cos(theta)

  54. JenniferSmart1
    • one year ago
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    I believe it's cosine theta

  55. phi
    • one year ago
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    confident is good, stubborn is bad

  56. JenniferSmart1
    • one year ago
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    |dw:1360514047313:dw| honestly....

  57. phi
    • one year ago
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    |dw:1360514118964:dw|

  58. JenniferSmart1
    • one year ago
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    you're right...sorry

  59. phi
    • one year ago
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    unless you are changing your definition of theta. (Allowed, but you should say so)

  60. JenniferSmart1
    • one year ago
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    Thanks so much everyone!!! That was fun!

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