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Equipotential Surfaces

  1. Aug 5, 2013 #1
    1. The problem statement, all variables and given/known data
    attachment.php?attachmentid=60700&stc=1&d=1375709755.png


    2. Relevant equations
    - Charge conservation
    - Equipotential surfaces


    3. The attempt at a solution
    Let Q1 be the amount of charge on the inner sphere with radius c, and Q2 be the amount of charge on the outer sphere with radius b.
    Using Gauss's law, I figured out that Q1=+2Q and Q2=-2Q
    1/ Charge conservation: Q1+Q2=0 (1)
    2/ Equipotential surfaces: [tex]\frac{kQ_{1}}{c}=\frac{kQ_{2}}{b}(2)[/tex]
    (1),(2)=>Q1=Q2=0
    Since my understanding on equipotential surfaces is not very good, please correct me if I am wrong, thank you!
     

    Attached Files:

  2. jcsd
  3. Aug 5, 2013 #2
    Do you even need calculation? When they are joined by a wire the two spheres can be said to be parts of the same conductor. We know that the total charge on the conductor is -5Q. There are many ways to distribute this charge but they will move so as to minimise the potential energy of the system and obey Gauss law. What will happen? (This situation is same as that of a solid spherical conductor with a cavity).
     
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