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Two conducting hollow spheres, a point charge and electric field prob

  1. Feb 5, 2009 #1
    1. The problem statement, all variables and given/known data
    A point charge is at the center of a conducting hollow sphere, with radius 0.011m, that is within another conducting hollow sphere of radius 0.041m. The point charge is Q0=+4.30e-6 C, the inner sphere has a net charge of Q1=-1.70e-6 C, and the outer sphere has a net charge of Q2=+6.50e-6 C.
    Calculate the magnitude of the electric field at a point A located 0.021m from the center.


    2. Relevant equations
    Gauss' Law [tex]\Phi[/tex]= Qin/ [tex]\epsilon[/tex] = E*Areas
    Areas=4[tex]\pi[/tex]r2
    [tex]\epsilon[/tex]=permittivity constant

    3. The attempt at a solution
    I've been stuck on this for awhile and don't really know where to begin. I know point A is between the two hollow spheres, one of which is positive, the other negative. But I don't know what to think of the conducting spheres when they aren't in electrostatic equil because they dont have net Q=0. Will the E of point A just be the sum of the electric fields of the spheres and point charge?
    EA= E0+E1+E2?
     
    Last edited: Feb 5, 2009
  2. jcsd
  3. Feb 5, 2009 #2

    LowlyPion

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    Think about the implications of Gauss Law.

    The total flux through a closed surface is going to be the net charge within.

    So at P the field you are looking at will be from the point charge and the inner sphere. The outer sphere since you are inside it ... what does that contribute?
     
  4. Feb 5, 2009 #3
    It won't contribute anything.

    So I tried KQ0/(distance to A)2 - KQ1/(distance to A)2 and it wasn't right.

    What's up?
     
  5. Feb 5, 2009 #4
    yes it would. you have a positive flux due the point charge and a negative flux due to the inner center. if you can find the net flux, you can find the net electric field. (also, think of A as being an invisible circle instead of a point.)
     
  6. Feb 6, 2009 #5
    alright i got it all figured out. s the net charge only the magnitudes or would you subtract the negative charge?
     
  7. Feb 6, 2009 #6

    LowlyPion

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    You would get a positive contribution from Q0 and a negative contribution from Q1.
     
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