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Flux due to a point charge

  1. Jan 24, 2009 #1
    1. The problem statement, all variables and given/known data
    A point charge q_1 = 3.45 nC is located on the x-axis at x = 1.90 m, and a second point charge q_2 = -6.95 nC is on the y-axis at y = 1.20 m.

    What is the total electric flux due to these two point charges through a spherical surface centered at the origin and with radius r_2 = 1.65 m?


    2. Relevant equations
    [tex]\Phi=\oint E_\bot dA[/tex]
    [tex]A=4\pi r^2[/tex]
    [tex]E=\frac{kq}{r^2}[/tex]


    3. The attempt at a solution
    I started with:
    [tex]\Phi=\oint \frac{kq}{r^2} dA[/tex]
    [tex]\Phi=A \oint \frac{kq}{r^2} dy[/tex]
    [tex]\Phi=4\pi r^2 \oint \frac{kq}{r^2} dy[/tex]

    To get r I did:
    [tex]x^2+y^2=r^2[/tex]
    [tex]x^2+y^2=1.65^2[/tex]
    [tex]x=\sqrt{1.65^2-y^2}[/tex]

    [tex]r=\sqrt{(\sqrt{1.65^2-y^2})^2+(y-1.20)^2}[/tex]

    So:
    [tex]\Phi=4\pi r^2 \oint \frac{kq}{1.65^2-y^2+(y-1.20)^2} dy[/tex]

    Evaluating this integral from -1.65 to 1.65 gives -1992.28 Nm^2/C

    I'm pretty sure I'm setting up this integral completely wrong. Any help on how to do it correctly would be greatly appreciated. Thanks in advanced for your help.
     
  2. jcsd
  3. Jan 24, 2009 #2

    Doc Al

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    Staff: Mentor

    Why mess around with integrals? Use Gauss's law.
     
  4. Jan 24, 2009 #3
    remember that the flux through a closed surface is equal to the charge enclosed divided by epsilon_not. aka gauss's law
     
  5. Jan 24, 2009 #4
    By use Gauss's law do you mean [tex]\Phi=\frac{q}{\epsilon_0}[/tex]? If so don't I need to know that permittivity of free space ([tex]\epsilon_0[/tex]), which isn't given in the problem?
     
  6. Jan 24, 2009 #5
    Alright I found that epsilon_0 = 8.854E-12. Thanks for your help.
     
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