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Homework Help: Need Help With Guass Laws and Amperes Law

  1. Dec 30, 2007 #1
    Hi, I am really stuck with these problems, is it possible you guys can help me out, by guiding me along this problem, I wil write the entire thing out, but would appreciate it if you guys help me step by step so that I can understand it, since I am having a VERY hard time with these problems.

    1. The problem statement, all variables and given/known data
    A point charge (+q) is located at the origin. A spherical conductor of inner radius "a" and outer radius "b" surrounds this point charge with its center at the origin. A total charge of =2q lies on this spherical conductor.
    {I put the diagram as an attachment}


    a) Find the magnitude of the electric field in region I, at a distance r<a away from the origin. Label on the diagram some E field line within this region, if necessary. [Hint: Use Guass Laws]

    b) Find the magnitude of the electric field in region II, at a distance r>1 (but <b), from the origin. Again label on the diagram some E fields if necessary.

    c) Find the magnitude of the electric field in region III, at a distance r>b from teh origin. Again label on the diagram E field lines if necesary.


    Consider 2 different designs for a printer cable. Design 1 has 4 cables each carrying a current I into the page. Design 2 has 4 cables all carrying current magnitude I but two have the current in one direction and two have in the opposite direction.

    d) Find the magnetic field at a distance r. (in the plane which is perpendicular to the wires), away from cable 1 by using Ampere's Law.

    e) Find the magnetic field a distance r away from cable 2 by using AMpere's LAw.

    f) Which design is better, why?

    2. Relevant equations

    Guass Law;
    [tex]\Sigma[/tex] Eperp * delta A = Qenclosed / [tex]\epsilon[/tex]0

    Ampere's Law;
    [tex]\Sigma[/tex] Bparallel * delta l = [tex]\mu[/tex]0 * Ienclosed

    3. The attempt at a solution

    Again, thank you for the help here.
     

    Attached Files:

  2. jcsd
  3. Dec 30, 2007 #2
    Ok, I 've been working on it and well, I got to this point, but am not too sure.

    a) sigma E perpendicular delta A = E * sigma delta A = E(4*pi*r^2) = Q / epsilon

    So wouldn't the charge in section I be +q / epsilon?

    c) sigma E perpendicular to delta A = E * sigma A = E * (4 pi r^2) = Q / epsilon

    so E = 1 / [4 pi r^2] * Q / b

    so, doesn't this tell us that the field outside the uniformly charged spherical shell is the same as if all the charge were concentrated at the center as a point charge??

    So that would mean that it is -2q?
    I am not too sure
     
  4. Dec 30, 2007 #3

    Doc Al

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    Staff: Mentor

    The charge contained within that surface is given as +q. So solve for E.

    That's correct. (The formula has a typo: You have b instead of epsilon.)

    What's the total charge within the Gaussian surface when r > b?
     
  5. Dec 30, 2007 #4
    The total charge is + q?

    Can we go over this a bit more? I am really feeling lost right now.
     
  6. Dec 30, 2007 #5

    Doc Al

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    To find the total charge within a surface with r > b, just add the charge at the center (+q) to the charge on the conductor. (What's the charge on the conductor?)
     
  7. Dec 30, 2007 #6
    The charge on the conductor is -2q, so +q + -2q will be -q so that is the charge in region I?
     
  8. Dec 30, 2007 #7

    Doc Al

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    No, that's the charge within a sphere of radius r > b. Region I is within r < a; the only charge in that region is the center charge of +q.

    Note: The regions are for the purpose of defining the field. For finding the charge, use a Gaussian sphere of the appropriate radius, counting all the charge within that radius.
     
  9. Dec 30, 2007 #8
    Ohh, ok, I think I understand now, so for region II the charge would be -2q since that is the charge of the conductor
     
  10. Dec 30, 2007 #9

    Doc Al

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    Careful here: What's the field in region II? (Hint: What's the field within a conductor?)

    While the total charge on the conductor is -2q, where is that charge located? What's the charge on its inner surface? Its outer surface?
     
  11. Dec 30, 2007 #10
    The E field in a conductor is zero right?
     
  12. Dec 30, 2007 #11

    Doc Al

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    Exactly. So part (b) is easy.
     
  13. Dec 30, 2007 #12
    Wow, thank you so much!!
     
  14. Dec 30, 2007 #13

    Doc Al

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    Just to prove to yourself that you fully understand it, try to answer my other questions in post #9. (Use Gauss's law.)
     
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