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Homework Help: Magnitude of an Electric field on a point by 2 charged spheres

  1. Jan 24, 2010 #1
    1. The problem statement, all variables and given/known data

    The left-hand sphere has a positive charge Q and the right-hand sphere has a negative charge -Q . Charge is distibuted uniformly over each of two spherical volumes with radius R. One sphere of charge is centered at the origin and the other at x=2R .
    Find the magnitude of the net electric field at the point R/2 on the x-axis



    2. Relevant equations

    E=[tex]\frac{1}{4\pi\epsilon_{0}}[/tex] ([tex]\frac{Q}{R^{2}}[/tex]


    3. The attempt at a solution

    Since the point is located within the first sphere, I thought the electric field would be zero.
    Then I typed in:

    E=[tex]\frac{1}{4\pi\epsilon_{0}}[/tex] ([tex]\frac{Q}{\frac{3}{2}R^{2}}[/tex]

    but it said the answer was wrong. Can anyone please help?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 24, 2010 #2

    rl.bhat

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    Since the point is located within the first sphere, I thought the electric field would be zero.

    But the sphere is not a conductor. So the field inside a uniformly charge sphere the is not equal to zero.
     
  4. Jan 24, 2010 #3
    Oh ok, so then the E field for the first sphere would be:

    [tex]\frac{1}{4_{0}\epsilon\pi}[/tex][tex]\frac{Q}{\frac{1}{2}R^{2}}[/tex]

    And then since both fields of the spheres are pointing the same direction I will just add the E fields of the first sphere with the second sphere?
     
  5. Jan 24, 2010 #4

    ideasrule

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    Why? The shell theorem states that the part of the sphere from r=1/2R to R creates no net electric field. The sphere "under" that shell has radius 1/2R, so what charge must it have?
     
  6. Jan 24, 2010 #5
    Would it have half the charge it originally had? Q/2?
     
  7. Jan 24, 2010 #6

    ideasrule

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    Would it? How much of the original volume does the smaller sphere have?
     
  8. Jan 24, 2010 #7
    It would have 1/8 less volume...so then Q/8
     
  9. Jan 24, 2010 #8
    Ok, I finally understand the problem now. Thank you so much for your help!!!! I appreciate it =)
     
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