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JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.2I'm going to write and draw the problem, give me a min

tkhunny
 one year ago
Best ResponseYou've already chosen the best response.0Sun spots. It's almost always sun spots!

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.2dw: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 \[dE=\frac{kdq}{a^2}=\frac{k\lambda(\theta)}{a^2}ds=\frac{k\lambda(\theta)ad\theta}{a^2}=\frac{k\lambda(\theta)}{a}\] \[d\vec{E}=dE_x\hat idE_y \hat j\] then all of a sudden \[\lambda(\theta) \rightarrow \lambda\] \[d\vec E=k\lambda\frac{d\theta}acos\theta\hat i k\lambda \frac{d\theta}asin\theta \hat j\]

tkhunny
 one year ago
Best ResponseYou've already chosen the best response.0Are you sure it's not just printing convenience?

dumbcow
 one year ago
Best ResponseYou've already chosen the best response.0hi , 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

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.2here is the integral \[\frac{k\lambda}a\int_0^\pi cos^2\theta d\theta \hat i\frac{k\lambda}a\int_0^\pi sin \theta cos \theta d\theta \hat j \]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.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?

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.2Sorry I was searching the internet....yeah that's probably what happened.

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.2I found it online!!!!

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.2they just have \[\lambda\] in stead of \[\lambda \theta\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Jenn what class is this? This stuff looks really fun :D

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0An 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...

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.2sorry the radius is "a"

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.2I forgot to draw that in

JamesJ
 one year ago
Best ResponseYou've already chosen the best response.0Oh 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?

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.2Yes! 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?

JamesJ
 one year ago
Best ResponseYou've already chosen the best response.0The 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 xdirectionthe integral before the iis nonzero. But by symmetry, it must be zero.

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.2well his final answer is \[E=\frac{k\lambda}{2a}\hat i\] and \[Q=\int_0^\pi a\lambda(\theta)d\theta\]

JamesJ
 one year ago
Best ResponseYou've already chosen the best response.0Definitely 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.

phi
 one year ago
Best ResponseYou've already chosen the best response.1Your 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.

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.2I'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.... \[\lambda=\frac{Q}{length}\]?

phi
 one year ago
Best ResponseYou've already chosen the best response.1I 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

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.2How did you know that it is max at theta=0 degrees and zero at theta=90 degrees?

JamesJ
 one year ago
Best ResponseYou've already chosen the best response.0Ah ... if the charge density is NOT constant, then yes indeed: this is a different problem!

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.2well I wrote \lambda(theta)ds in the picture....sorry I didn't make it clear.

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.2How did you know that it is max at theta=0 degrees and zero at theta=90 degrees?

phi
 one year ago
Best ResponseYou've already chosen the best response.1because, based on the equations you posted, I figured out that the charge density as a function of theta must be \[ \lambda \cos(\theta) \]

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.2how 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?

JamesJ
 one year ago
Best ResponseYou've already chosen the best response.0What exactly is the wording of the problem as it was given to you?

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.2Shoot! 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.

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.2so the image states: \[dq=\lambda(\theta)ds\]

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.2\[s=r\theta\] therefore \[ds=a\;d\theta\] this would give us \[dq=\lambda(\theta)\;a \;d\theta\]

phi
 one year ago
Best ResponseYou've already chosen the best response.1First, 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 \[ \lambda(\theta)= \lambda \cos(\theta) \]

phi
 one year ago
Best ResponseYou've already chosen the best response.1I am guessing you missed that fact.

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.2Here 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........

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.2LOL oh sorry...is it because he substituted \[\lambda cos(\theta) \; for\:\lambda(\theta)\] in the integral haha....I see what you mean now.....sorry

phi
 one year ago
Best ResponseYou've already chosen the best response.1yes, 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

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.2http://online.cctt.org/physicslab/content/phyapc/lessonnotes/Efields/EchargedRings.asp

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.2nevermind. that charge distribution is uniform....

phi
 one year ago
Best ResponseYou've already chosen the best response.1I was going to say, "those are for a uniform charge. you're problem is different"

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.2\[\lambda(\theta)=\lambda cos(\theta)\] was probably given.....oh my....i feel dumb

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.2why are we integrating form \[\frac \pi 2\; to\; \frac \pi 2 \]?

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.2dw:1360513618512:dw

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.2MYSTERY SOLVED!!!!!!

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.2<slow but determined to learn physics

phi
 one year ago
Best ResponseYou've already chosen the best response.1Good 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)

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.2I believe it's cosine theta

phi
 one year ago
Best ResponseYou've already chosen the best response.1confident is good, stubborn is bad

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.2dw:1360514047313:dw honestly....

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.2you're right...sorry

phi
 one year ago
Best ResponseYou've already chosen the best response.1unless you are changing your definition of theta. (Allowed, but you should say so)

JenniferSmart1
 one year ago
Best ResponseYou've already chosen the best response.2Thanks so much everyone!!! That was fun!
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