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Homework Help: Electric Field Involving Gauss's Law Problem

  1. Feb 5, 2010 #1
    1. The problem statement, all variables and given/known data
    An atom that has Z protons, each with positive charge of magnitude e, reside in the nucleus at the center, surrounded by a uniformly distributed spherical cloud of radius R consisting of Z electrons, each with negative charge magnitude e, centered on the nucleus e. Use Gauss's Law to determine an expression for the electric field as a function of r, the distance from the center of the atom.
    (1) Find the value of electric field for r<R (magnitude + direction)
    (2) Find the value of electric field for R>r (mag + dir)
    (3) If the atom were ionized, such that it was deficient by one electron, what would be the new expressions for the regions r<R and r>R


    2. Relevant equations

    Gauss's Law: Integration of E dA = Q/eo
    E = kq1q2/r^2

    3. The attempt at a solution
    (1-2) I treated this problem as a hollow sphere
    The electric field is zero inside a conducting sphere.
    The electric field outside the sphere is given by: E = kQ/r2, just like a point charge.
    (3) This one I have no idea......
    Thank you for your help
     
  2. jcsd
  3. Feb 5, 2010 #2
    A conducting sphere alone has an electric field of zero within it. However, do you think there would be an electric field in a conducting sphere if there were charged particles inside of it - say, a positive nucleus with Z protons?

    Think of the nucleus as a point charge meaning regardless of how small your Gauss surface is, you still have the z*e Coulombs enclosed, so the E field is not zero within the electron cloud.

    Your second answer seems wrong, also. Once you've drawn a Gauss surface that encloses both the electron cloud and the nucleus, what does Q enclosed become?
     
  4. Feb 7, 2010 #3
    So for part one (r<R) would u use E=Q/4PiE0r^2 to give you that the electric field=Ze/4PiE0r^2
     
  5. Feb 7, 2010 #4
    Or would it be Integration EdA=Qenc/E0 and dA=2Pi*rL and Qenc=Ze so E=Ze/2Pi*rLE0
     
  6. Feb 7, 2010 #5
    yes that's right
    I'm unsure what rL is and i'm unsure how you got the integral to equal that.

    you can integrate to get the same answer as above. Your E field is constant at all points on your surface, so your integral becomes:
    [tex]E\int_A dA=\frac{Q_{enc}}{\epsilon_o}[/tex]

    That integral, I hope you see, becomes the total area of your Gauss surface. What is your surface? It is a sphere, so it becomes the surface area of a sphere:
    [tex]E(4\pi r^2)=\frac{Q_{enc}}{\epsilon_o}[/tex]
    [tex]E=\frac{Q_{enc}}{4\pi\epsilon_or^2}[/tex]
     
  7. Feb 7, 2010 #6
    Thank you very much.

    As for part B for r>R if it encloses both the electron cloud and the nucleus cloud then Qenc=0 therefore the Electric field=0?

    And for Part C: If it were deficient by one electron, does that mean the overall charge would be positive since there will be more protons then electrons so

    r<R (when deficient by one electron) I'm a little bit confused on this one.

    Now there are Z-1 electrons. Would you just plug that into the equation that we got in part a to give E=(Z-1)e/4Pi*E0r^2

    r>R (when deficient by one electron)

    Qenc=+1?? so would i just plug that in for the electric field?

    Thanks in advance.
     
  8. Feb 7, 2010 #7
    You are right about part B. For part C when r < R, it's the same problem as before! Does [tex]Q_{enc}[/tex] change? No, so the E-field doesn't either!

    For when r > R, [tex]Q_{enc}[/tex] becomes +e like you say. Just solve the integral the same way I solved it for part A
     
  9. Feb 7, 2010 #8
    So let me get this right,

    For when r>R (with one electron deficient), I solved the integral to get E=Z*+e/4Pi*E0r^2

    correct? the only thing that changes is the Qenc. The Qenc becomes Z*+e
     
  10. Feb 8, 2010 #9
    No, Qenc becomes +1e

    If there are Z - 1 electrons and Z protons, you get:
    Z(e) + (Z -1)(-e)
    Ze + (e - Ze)
    e

    So the total charge enclosed is 1 electron's worth but positive.
     
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