rabia 20121011 01:04:05  I have to solve a problem by using Coulumn' law given rho=rho not (1r/R) and find Q from given expression.  
physicsphysics 20111011 07:22:32  Does anybody have a problem by using Coulumn' law?
I used . The answer is . Can anybody explain what I am wrong?
3Danyon 20111013 08:44:10 
I made the same mistake initially. The thing is, one has to calculate the charge first and then divide by the . Otherwise you change the integrand and get the wrong numerical factor. To spell it out, the field is given by:

 
phoxdie 20101110 15:20:06  Sorry to get hung up on this, but it was very confusing the first time I looked at this solution. There is a small "typo" in the solution. A factor of epsilon is dropped for no reason.  
Kabuto Yakushi 20100919 10:28:26  Another method for solving this problem would be to take the volume integral of to solve for Q. Then to plug Q into Gauss' law.
where
plug Q into Gauss' law:
solving one gets choice B).
Yosun's solution is faster, but if using the shell method for integration doesn't come to mind, one must resort to the long method.  
tsharky87 20091106 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 20081105 15:43:14  in addition, if you set A=1, R=1, epsilon=1
it will be easier
then
E= 1/40
pam d 20110917 16:24:27 
nice

 
gliese876d 20081011 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?
flyboy621 20101102 21:12:58 
I think you are assuming constant charge density, which is not the case here.

 
hisperati 20071031 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 20080314 18:01:46 
Yea I'm confused too

blah22 20080314 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!

flyboy621 20101102 21:14:34 
You can't just multiply the charge density by the enclosed volume because the density is not constant. You have to integrate.

 
azot 20071023 06:04:01  the answer to this problem is wrong .
E shoud be equal to A*R^3/32*5
azot 20071023 06:05:00 
sorry I was wrong :) Everything is ok

kevinjay15 20071023 18:24:01 
ETS is never wrong.

 