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Using Gauss' Law on a Solid Annular Sphere

  • Thread starter deathsink
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Homework Statement



Imagine a solid, annular sphere. At the center of the hollow is a point charge +Q. The inner radius of the sphere is r0, and the outer radius is R. Assume the charge density p = p0/r (for r0 < r < R). Calculate using the integral form of Gauss's Law the electric field in all three regions:

0< r < r0
r0< r < R
R < r < infinity

Homework Equations



[PLAIN]http://www.forkosh.dreamhost.com/mimetex.cgi?%5Coint_S%5Cmathbf%7BE%7D%5Ccdot%7Bd%7D%5Cmathbf%7BA%7D=%5Cfrac%7BQ%7D%7B%5Cepsilon_0%7D=%5Cfrac%7B1%7D%7B%5Cepsilon_0%7D%5Cint_V%5Crho(%5Cmathbf%7Br%7D)%5C,dV [Broken]

The Attempt at a Solution



for 0 < r < r0

E(4 Pi r^2) = 1/e0 (p0/r)(4/3 pi r^3)
=> E = (1/3e0)p0

But this can't be right for because all of the r's cancel out. And since I can't do this one, I know I can't do the rest.

Thank you.
 
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Answers and Replies

  • #2
gneill
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For this first interval, where 0 < r < r0, the enclosed charge is just +Q, the charge at the center of the hollow.
 
  • #3
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For this first interval, where 0 < r < r0, the enclosed charge is just +Q, the charge at the center of the hollow.
When you take the surface integral of E*dA, do you get E*(area of sphere) which you then need to divide through by?

I have not had Vector Calculus, so a lot of the notation is odd to me.

Edit: I got [tex]E=\frac{Q}{4\pi \epsilon _0r_0{}^2} [/tex] by dividing through by the area of a sphere.
 
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  • #4
gneill
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Your E dot DA in the surface integral turns into E da, because the field is radially directed and so the dot product is just the product of the magnitudes of the vectors E and dA (which is normal to the surface over which you're integrating).

So, at radius r you end up with

E(4 π r2) = Q/ε0

Or, rearranged,

E = Q/(4 π ε0 r2)

Which should look familiar.
 

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