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Spherical Charge Ball, Gauss law

  1. Oct 2, 2012 #1
    1. The problem statement, all variables and given/known data

    A spherical charged ball of radius a has total charge Q; there is no charge outside the ball and no sheet-charge on its surface. The (radial) field inside the ball has the form
    Er(r) = constant x r2 for r between 0 and a.
    Use Gauss's Law in integral form to evaluate the constant in this formula in terms of Q and a, and then rewrite the formula in terms of these parameters.

    2. Relevant equations

    ∫E.dA=Q/εo


    3. The attempt at a solution

    Lets say constant is b and dA=4.pi.r^2

    ∫Er(r).dA=Q/εo

    from 0 to a ∫b.r^2(4.pi.r^2) dr=Q/εo
    (4.pi.b) from 0 to a ∫r^4 dr = Q/εo
    (4.pi.b)(a^5/5)=Q/εo
    b=(5Q)/(4.pi.a^5.εo)

    When i replace b (the constant) in Er(r) i get

    Er(r) = b x r2
    Er(r) = (5Q)/(4.pi.a^5.εo) x r2
    Er(r)= (5.Q.r^2)/(4.pi.a^5.εo)

    but the answer is Er(r)= (Q.r^2)/(4.pi.a^4.εo)

    can someone please tell me where am I miscalculating or i forgot to add something
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 2, 2012 #2
    You say that dA = 4.pi.r^2 , but where is the dr coming from in your integral?

    There is another way of expressing the differential area element. Look into that ;).
     
  4. Oct 2, 2012 #3
    I used the wrong dA, however even if I use dA=2.pi.r dr I'm still off

    from 0 to a ∫b.r^2(2.pi.r) dr=Q/εo
    (2.pi.b) from 0 to a ∫r^3 dr = Q/εo
    (2.pi.b)(a^4/4)=Q/εo
    b=(2Q)/(pi.a^5.εo)

    and when i replace b (the constant) in Er(r) i get

    Er(r)= (2.Q.r^2)/(pi.a^4.εo)

    I'm supposed to get 1/4 and not 2. I'm not familiar with other ways of expressing the differential area element
     
  5. Oct 2, 2012 #4
    The problem also says that there is no sheet-charge on its surface. Try using that.

    OK. nvm that. There is a simpler way. If you enclose the sphere with a gaussian surface, what can you say about the electric field flux going out.
     
    Last edited: Oct 2, 2012
  6. Oct 2, 2012 #5
    But wouldn't that just mean there is a magnetic field all around?

    If I enclose the sphere with a gaussian surface the electric flux would just be equal to the electric field times the area of the sphere.
    E.(4.pi.r^2)
     
  7. Oct 2, 2012 #6
    Exactly. And what does ∫E.dA represent? ;D
     
  8. Oct 2, 2012 #7
    I still can't figure out
    I mean..
    electric flux= EA=Q/εo
    EA=Q/εo
    (b.r^2)(4.pi.r^2)=Q/εo
    b.4pi.r^4=Q/εo
    b=Q/(4.pi.r^4.εo)
     
  9. Oct 2, 2012 #8
    Wait does that mean I can sub "a" for "r"? Therefore I get b=Q/(4.pi.a^4.εo)

    Am I right?
     
  10. Oct 2, 2012 #9
    Yes. You can make your gaussian surface as big (or small) as you want.
    Now, why does letting r = a work? (hint: look at the problem statement).
     
  11. Oct 2, 2012 #10
    Because there is no charge outside the ball and no sheet-charge on its surface

    Thankss a lot!

    So simple
     
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