Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

JenniferSmart1 Group Title

What went wrong?

  • one year ago
  • one year ago

  • This Question is Closed
  1. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    I'm going to write and draw the problem, give me a min

    • one year ago
  2. tkhunny Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Sun spots. It's almost always sun spots!

    • one year ago
  3. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    |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 \[\lambda(\theta) \rightarrow \lambda\] \[d\vec E=-k\lambda\frac{d\theta}acos\theta\hat i -k\lambda \frac{d\theta}asin\theta \hat j\]

    • one year ago
  4. tkhunny Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Are you sure it's not just printing convenience?

    • one year ago
  5. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    @JamesJ @phi

    • one year ago
  6. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    hi @dumbcow :)

    • one year ago
  7. dumbcow Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    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

    • one year ago
  8. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    here is the integral \[-\frac{k\lambda}a\int_0^\pi cos^2\theta d\theta \hat i-\frac{k\lambda}a\int_0^\pi sin \theta cos \theta d\theta \hat j \]

    • one year ago
  9. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    i'm confused

    • one year ago
  10. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    @zepdrix

    • one year ago
  11. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\large d\vec{E}=<-dE_x,\quad -dE_y>\]\[\large d\vec{E}=<-k \frac{d \theta}{a}\lambda \cos \theta,\quad -k \frac{d \theta}{a}\lambda \sin \theta>\] 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?

    • one year ago
  12. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    Sorry I was searching the internet....yeah that's probably what happened.

    • one year ago
  13. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    I found it online!!!!

    • one year ago
  14. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    they just have \[\lambda\] in stead of \[\lambda \theta\]

    • one year ago
  15. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Jenn what class is this? This stuff looks really fun :D

    • one year ago
  16. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    Physics II

    • one year ago
  17. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    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...

    • one year ago
  18. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    sorry the radius is "a"

    • one year ago
  19. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    I forgot to draw that in

    • one year ago
  20. JamesJ Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    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?

    • one year ago
  21. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    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?

    • one year ago
  22. JamesJ Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

    • one year ago
  23. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

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

    • one year ago
  24. JamesJ Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

    • one year ago
  25. phi Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

    • one year ago
  26. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    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.... \[\lambda=\frac{Q}{length}\]?

    • one year ago
  27. phi Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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

    • one year ago
  28. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    How did you know that it is max at theta=0 degrees and zero at theta=90 degrees?

    • one year ago
  29. JamesJ Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Ah ... if the charge density is NOT constant, then yes indeed: this is a different problem!

    • one year ago
  30. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    well I wrote \lambda(theta)ds in the picture....sorry I didn't make it clear.

    • one year ago
  31. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    How did you know that it is max at theta=0 degrees and zero at theta=90 degrees?

    • one year ago
  32. phi Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  33. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    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?

    • one year ago
  34. JamesJ Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    What exactly is the wording of the problem as it was given to you?

    • one year ago
  35. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    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.

    • one year ago
  36. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    so the image states: \[dq=\lambda(\theta)ds\]

    • one year ago
  37. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    \[s=r\theta\] therefore \[ds=a\;d\theta\] this would give us \[dq=\lambda(\theta)\;a \;d\theta\]

    • one year ago
  38. phi Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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 \[ \lambda(\theta)= \lambda \cos(\theta) \]

    • one year ago
  39. phi Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    I am guessing you missed that fact.

    • one year ago
  40. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    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........

    • one year ago
  41. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    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

    • one year ago
  42. phi Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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

    • one year ago
  43. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    http://online.cctt.org/physicslab/content/phyapc/lessonnotes/Efields/EchargedRings.asp

    • one year ago
  44. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    nevermind. that charge distribution is uniform....

    • one year ago
  45. phi Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    I was going to say, "those are for a uniform charge. you're problem is different"

    • one year ago
  46. phi Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    *uniform charge density

    • one year ago
  47. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    page 4

    • one year ago
  48. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    \[\lambda(\theta)=\lambda cos(\theta)\] was probably given.....oh my....i feel dumb

    • one year ago
  49. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    why are we integrating form \[-\frac \pi 2\; to\; \frac \pi 2 \]?

    • one year ago
  50. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1360513618512:dw|

    • one year ago
  51. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    MYSTERY SOLVED!!!!!!

    • one year ago
  52. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    <-----slow but determined to learn physics

    • one year ago
  53. phi Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    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)

    • one year ago
  54. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    I believe it's cosine theta

    • one year ago
  55. phi Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    confident is good, stubborn is bad

    • one year ago
  56. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1360514047313:dw| honestly....

    • one year ago
  57. phi Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1360514118964:dw|

    • one year ago
  58. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    you're right...sorry

    • one year ago
  59. phi Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    unless you are changing your definition of theta. (Allowed, but you should say so)

    • one year ago
  60. JenniferSmart1 Group Title
    Best Response
    You've already chosen the best response.
    Medals 2

    Thanks so much everyone!!! That was fun!

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.