 GR 8677927796770177 | # Login | Register

GR9677 #61
Problem
 GREPhysics.NET Official Solution Alternate Solutions
This problem is still being typed.
Electromagnetism}Gauss Law

Recall Gauss Law . Thus, . Solving for E, one has , as in choice (B).  Alternate Solutions
 There are no Alternate Solutions for this problem. Be the first to post one!
rabia
2012-10-11 01:04:05
I have to solve a problem by using Coulumn' law given rho=rho not (1-r/R) and find Q from given expression. physicsphysics
2011-10-11 07:22:32
Does anybody have a problem by using Coulumn' law?
I used . The answer is . Can anybody explain what I am wrong?
 3Danyon2011-10-13 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
2010-11-10 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
2010-09-19 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
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
 pam d2011-09-17 16:24:27 nice 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?
 flyboy6212010-11-02 21:12:58 I think you are assuming constant charge density, which is not the case here. 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 ?
 blah222008-03-14 18:01:46 Yea I'm confused too
 blah222008-03-14 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!
 flyboy6212010-11-02 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
2007-10-23 06:04:01
the answer to this problem is wrong .
E shoud be equal to A*R^3/32*5
 azot2007-10-23 06:05:00 sorry I was wrong :) Everything is ok
 kevinjay152007-10-23 18:24:01 ETS is never wrong.      LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces . type this... to get...$\int_0^\infty\partial\Rightarrow\ddot{x},\dot{x}\sqrt{z}\langle my \rangle\left( abacadabra \right)_{me}\vec{E}\frac{a}{b}\$