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Gauss's Law and positive charge

  1. Feb 10, 2006 #1
    A region in space contains a total positive charge Q that is distributed spherically such that the volume charge density rho (r) is given by:

    rho (r)=3 * alpha * r /(2R) for r ≤ R / 2

    rho (r)= alpha * [1-(r/R)^(2) ] for (R/2) ≤ r ≤ R

    rho (r)=0 for r ≥ R
    Here alpha is a positive constant having units of C/m^3


    What fraction of the total charge is contained within the region (R/2) ≤ r ≤ R?

    If anyone can help me with this problem it would be greatly appreciated. I am getting lost in all the integration with everything. Thanks
     
    Last edited: Feb 10, 2006
  2. jcsd
  3. Feb 10, 2006 #2

    jtbell

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    Please show us what you've done so far...
     
  4. Feb 10, 2006 #3
    This is what I got so far. It aint pretty.

    The integral from (r/2) to r of:

    Q/ (3R^3*pi/32)+(47R^3*pi/120) all this * (1 -(r/R)^2) * (4 * pi*r^2 dr)
    all divided by 4 * pi *r^2 * eplison not.
     
  5. Feb 10, 2006 #4

    jtbell

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    Please back up a few steps and explain how you got to that point.
     
  6. Feb 10, 2006 #5
    I am not really sure what I did. I used some integration formulas's, but I don't think they are right. Do you have a better starting way then what I did?
     
  7. Feb 10, 2006 #6
    For some reason, i don't know where to start off on this problem. There are so many specifications and I don't know if I'm doing the problem the right way.
     
  8. Feb 10, 2006 #7

    jtbell

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    OK, start with Gauss's Law. What does it say? How can you relate it to the conditions of your problem?
     
  9. Feb 10, 2006 #8
    I know that Gauss's Law is the integral of vector E * vector dA
     
  10. Feb 10, 2006 #9
    I could use Gauss's Law and use the second condition from the specs in the problem. I could take the integral from (R/2) to R of the second condition.
     
  11. Feb 10, 2006 #10

    Astronuc

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  12. Feb 10, 2006 #11
    I found the integral. It is ( a (1- (r^2/R^2) * R)/2
    Now I don't know how to use the Law to compute anything?
     
  13. Feb 10, 2006 #12

    jtbell

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    That's the definition of the flux of [itex]\vec E[/itex] through a surface.

    Look up Gauss's Law in your textbook. It says that the total flux of [itex]\vec E[/itex] through a closed surface equals the total charge enclosed inside that surface divided by [itex]\epsilon_0[/itex]. Both of these can be written as integrals:

    [tex]\int{\vec E \cdot d \vec A} = \frac{1}{\epsilon_0} \int {\rho dV}[/tex]

    where [itex]\rho[/itex] is of course the charge density.
     
    Last edited: Feb 10, 2006
  14. Feb 10, 2006 #13

    jtbell

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    Bleah, the server must be overloaded. My equations haven't shown up in my preceding posting yet. [oops, I had to fix a bug... now they're OK]

    Anyway, to use Gauss's Law, imagine a spherical surface of radius r (the radius that you want to find E at). You need to find (a) the flux of E through that suface, in terms of r and E and constants, and (b) the total charge enclosed inside the surface, using the densities that you've been given. Then substitute into Gauss's Law and solve for E.

    Note that integrating the charge density between R/2 and R (as you proposed) gives you the total charge contained inside that spherical shell. The r that you want to find E at, is < R, so your imaginary spherical surface doesn't contain all of that charge.

    Also, don't forget to include the charge that lies between radius 0 and R/2!
     
  15. Feb 10, 2006 #14
    I don't know how to relate electric flux to this problem.
     
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