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GR9677 #61
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Alternate Solutions |
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Comments |
tsharky87
2009-11-06 08:46:03 | I was confused about the fact that I seemed to be getting a  , so here's an explanation for those of you who were as confused as azot and I:
I was getting my r's confused in Yosun's solution, I just took  and canceled away. The reason this is wrong is that the radius that you are taking the area with when multiplying by  is actually  . So if you set  you should arrive at the solution. |  | jw111
2008-11-05 15:43:14 | in addition, if you set A=1, R=1, epsilon=1
it will be easier
then
E= 1/40 |  | gliese876d
2008-10-11 20:56:52 | hmm. I'm just wondering... I had the following formula memorized for similar situations:
E=
Where little r is the radius of the point from the center and R is the total radius of the sphere
Now using this formula I figured Q would equal 
so I used this in the numerator for Q and also plugged in r= R/2 and you get
which is *close* to the right answer, but not quite... where does this standard formula fall apart in reasoning for this particular situation? | | hisperati
2007-10-31 20:29:18 | To find the charge we need to integrate a charge density, that is per unit volume. The answers correspond to multipliying by a surface area ( I am not talking about the left side of Guass's Law, where it is obvious surface area is correctly used). Why isn't the charge density multiplied by  ?
blah22
2008-03-14 18:01:46 |
Yea I'm confused too
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blah22
2008-03-14 18:27:59 |
Ah I see, for anyone else confused by this in the future, it's just a method of getting the volume of a sphere through integration. Divide the sphere into lots of shells of thickness dr, each with surface area  .
Thanks so much Yosun!
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| | azot
2007-10-23 06:04:01 | the answer to this problem is wrong .
E shoud be equal to A*R^3/32*5
azot
2007-10-23 06:05:00 |
sorry I was wrong :) Everything is ok
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kevinjay15
2007-10-23 18:24:01 |
ETS is never wrong.
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