## anonymous 3 years ago What went wrong?

1. anonymous

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

2. tkhunny

Sun spots. It's almost always sun spots!

3. anonymous

|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

Are you sure it's not just printing convenience?

5. anonymous

@JamesJ @phi

6. anonymous

hi @dumbcow :)

7. anonymous

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. anonymous

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. anonymous

i'm confused

10. anonymous

@zepdrix

11. zepdrix

$\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. anonymous

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

13. anonymous

I found it online!!!!

14. anonymous
15. anonymous

they just have $\lambda$ in stead of $\lambda \theta$

16. zepdrix

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

17. anonymous

Physics II

18. zepdrix

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...

19. anonymous

20. anonymous

I forgot to draw that in

21. JamesJ

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?

22. anonymous

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?

23. JamesJ

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.

24. anonymous

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

25. JamesJ

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.

26. phi

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.

27. anonymous

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.... $\lambda=\frac{Q}{length}$?

28. phi

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

29. anonymous

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

30. JamesJ

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

31. anonymous

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

32. anonymous

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

33. phi

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

34. anonymous

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?

35. JamesJ

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

36. anonymous

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.

37. anonymous

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

38. anonymous

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

39. phi

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 $\lambda(\theta)= \lambda \cos(\theta)$

40. phi

I am guessing you missed that fact.

41. anonymous

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........

42. anonymous

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

43. phi

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

44. anonymous
45. anonymous

nevermind. that charge distribution is uniform....

46. phi

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

47. phi

*uniform charge density

48. anonymous
49. anonymous

page 4

50. anonymous

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

51. anonymous

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

52. anonymous

|dw:1360513618512:dw|

53. anonymous

MYSTERY SOLVED!!!!!!

54. anonymous

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

55. phi

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)

56. anonymous

I believe it's cosine theta

57. phi

confident is good, stubborn is bad

58. anonymous

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

59. phi

|dw:1360514118964:dw|

60. anonymous

you're right...sorry

61. phi

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

62. anonymous

Thanks so much everyone!!! That was fun!