GR9277 #12



Alternate Solutions 
Herminso 20090922 19:26:52  We have to solve the Laplace's equation in the region between the conducting plates , but since the potential difference between them is independenet of the cylindrical coordinates and , we have then , so . Now by BC's and . Thus   mikewofsey 20060210 14:55:32  An alternative way of looking at this is that as alpha gets large, the potential goes to zero, so alpha needs to be on the bottom or the solution will blow up. Likewise, this requires that phi be on top to make sense. Of the remaining solution, chose B, because of the 1/r relationship of potential.  

Comments 
istezamer 20091106 07:07:59  Actually I think it is much easier to eliminate choices... If phi equals to zero the potential must be Zero.. This Keeps choices (B) and (D)..
But if Phi equals alpha, the Potential must be Vo.. So only (B) works Out !!! :D
alemsalem 20100927 03:06:10 
yup just apply the boundary conditions, and you're there

  Herminso 20090922 19:26:52  We have to solve the Laplace's equation in the region between the conducting plates , but since the potential difference between them is independenet of the cylindrical coordinates and , we have then , so . Now by BC's and . Thus
mvgnzls 20110915 12:49:39 
I dont see where you get V=A*(phi) + B
where did you get this from?

mvgnzls 20110915 12:50:35 
where did you get V= Aphi+B from?

elxcats 20111031 12:01:13 
@mvgnzls
A(psi) + B is just the solution to the differential equation. Plug it in to the equation to see.

RusFortunat 20151022 11:01:48 
The best solution to this problem. Short and elegant.

  bkardon 20070923 13:21:19  Another way to approach this problem is as follows:
First narrow down the choices to those that have go from to in the space between the plates (that leaves B and D).
Since there are no charges in the region between the plates, the potential must solve Poisson's equation,
Even if you don't know what is in cylindrical coordinates, you can be confident that it includes a second derivative of \phi. Therefore, if the result of the second derivative in is to be zero, there can't be any terms in the potential.   pballjew 20061126 17:49:54  The way I found this is to note that the answer has to be zero when phi equals zero and V when phi = alpha. Then, by uniqueness, since boundary conditions are fit, you're done.
The only answer that fulfills this criteria is (B)
nyuko 20091030 09:43:08 
yes! this was what I did too! I think it is the easiest approach for this question.

  mikewofsey 20060210 14:55:32  An alternative way of looking at this is that as alpha gets large, the potential goes to zero, so alpha needs to be on the bottom or the solution will blow up. Likewise, this requires that phi be on top to make sense. Of the remaining solution, chose B, because of the 1/r relationship of potential.
hungrychemist 20070919 20:27:20 
1/r relationship of potential is not always true. Expecially when infinite (very large) charge distribution is involved.
B or D? B is correct since it's got correct units.

Poop Loops 20081005 13:15:53 
Also, if you only narrow it down between B, D, and A, you can see that the units only make sense for B.

sid 20081031 04:16:37 
dimensional analysis doesn't work in this question as all options have same unit of potential... angle in radians is already unitless..
angle is just a number....

neon37 20101110 10:51:39 
Can anyone think of anytime when dimensional analysis doesnt work for angle coordinates? I cant think of any. So as heres my hypothesis. Dimensional analysis works when you have angle coordinates eventhough angles technically lack units.

livieratos 20111108 10:55:00 
actually i would say it's easier to think of phi since V=0 when phi=0 and only choice (B) gives that result (at least from the ones that make sense...)

 

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