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Flux through concentric spheres

  1. Aug 16, 2011 #1
    1. The problem statement, all variables and given/known data

    Charge of uniform surface density (4.0 nC/m2) is distributed on a spherical surface (radius = 2.0 cm). What is the total electric flux through a concentric spherical surface with a radius of 4 cm?

    2. Relevant equations

    [tex]\Phi = E 4 \pi r^2 = \frac{q}{\epsilon_0}[/tex]


    [tex]E=k_e \frac{Q}{r^2}[/tex]

    3. The attempt at a solution

    I tried finding the electric field due to the charge at the surface of the concentric spherical surface

    [tex]E = (8.9 \times 10^9) \frac{(4 \times 10^{-9})}{(4 \times 10^{-2})^2} = 22250[/tex]

    then used this to find the flux:

    [tex]22250 \times 4 \times \pi \times (4 \times 10^{-2})^2 =447.36[/tex]

    But what's wrong with my approach? This is very different from the correct answer (2.3 N.m2/C).
     
  2. jcsd
  3. Aug 16, 2011 #2

    gneill

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    Staff: Mentor

    You're given the charge density on the inner sphere, not the total charge. You should first work out the total charge on the surface of the sphere, then divide by [itex]\epsilon_0[/itex] as per your stated formula for [itex]\Phi[/itex].
     
  4. Aug 16, 2011 #3
    How do you know the charge density is in the inner surface and not inside the surface?

    [PLAIN]http://img707.imageshack.us/img707/6093/unledqqn.jpg [Broken]

    Like how do you know it is the green one and not the purple one?
     
    Last edited by a moderator: May 5, 2017
  5. Aug 16, 2011 #4

    gneill

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    Staff: Mentor

    The usual interpretation of the phrase, "distributed on a spherical surface" is that it is located on the outer surface unless otherwise indicated.

    Even so, it would not make a difference given Gauss' Law which proclaims that the flux through any closed surface surrounding a charge q is equal to [itex]q/\epsilon_0[/itex].
     
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