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Electromagnetism}Gauss Law

Recall Gauss Law \vec{E} \cdot d\vec{A} = q/\epsilon_0. Thus, E(4\pi r^2)=\int_0^{R/2}A r^2 4\pi r^2 dr = 4\pi A (R/2)^5/5. Solving for E, one has E=A/5(R/2)^3, as in choice (B).

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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. Help
2011-10-11 07:22:32
Does anybody have a problem by using Coulumn' law?
I used \vec{E}=\frac{1}{4\Pi\e0}\int_0^\R/2\frac{Ar^2}{r^2}r^2Sin\Theta dr d\Theta d\Phi. The answer is \frac{AT^4}{24e0}. Can anybody explain what I am wrong?
2011-10-13 08:44:10
I made the same mistake initially. The thing is, one has to calculate the charge first and then divide by the (R/2)^2. Otherwise you change the integrand and get the wrong numerical factor. To spell it out, the field is given by:

\vec{E}=\frac{1}{4\pi \epsilon_0} \frac{Q}{(R/2)^2} = \frac{1}{4\pi \epsilon_0 (R/2)^2} \cdot \int (A r^2) \cdot r^2 \sin{\theta} dr d\theta d\phi
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.Typo Alert!
Kabuto Yakushi
2010-09-19 10:28:26
Another method for solving this problem would be to take the volume integral of \rho to solve for Q. Then to plug Q into Gauss' law.


where d\tau=r^2sin(\theta)drd{\theta}d{\phi}

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.
2009-11-06 08:46:03
I was confused about the fact that I seemed to be getting a 1/160, 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 r=R and canceled away. The reason this is wrong is that the radius that you are taking the area with when multiplying by \vec{E} is actually R/2. So if you set r=R/2 you should arrive at the solution.
Common Pitfalls
2008-11-05 15:43:14
in addition, if you set A=1, R=1, epsilon=1

it will be easier

E*4pi*\frac{1}{2^2}=\int r^2*4pi*r^2 dr

E \frac{1}{2^2} = \frac{1}{5} \frac{1}{2^5}

E= 1/40
pam d
2011-09-17 16:24:27
2008-10-11 20:56:52
hmm. I'm just wondering... I had the following formula memorized for similar situations:


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 \frac{4piR^3rho}{3}

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?
2010-11-02 21:12:58
I think you are assuming constant charge density, which is not the case here.
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 \frac{4}{3}  pi  R^3  ?
2008-03-14 18:01:46
Yea I'm confused too
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 4 \Pi r^2.

Thanks so much Yosun!

2010-11-02 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.
2007-10-23 06:04:01
the answer to this problem is wrong .
E shoud be equal to A*R^3/32*5
2007-10-23 06:05:00
sorry I was wrong :) Everything is ok

2007-10-23 18:24:01
ETS is never wrong.

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