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Two spheres with a conducting wire

  1. Sep 14, 2015 #1
    1. The problem statement, all variables and given/known data

    1. Two solid metal spheres are very far apart: sphere A (radius 44.1 cm) is charged to potential +2,346 Volts; sphere B (radius 63.6 cm) is charged to potential +8,848. Now, a long conducting wire connects the two spheres. When the charge has stopped moving, find the new charge on sphere A, in μC.


    2. Relevant equations
    v1=kq1/r1 v2=kq/r2
    3. The attempt at a solution
    since there are connected by a conducting wire both of the spheres will have the same potential thus :
    v1+v2 / 2 = 5597 V
    then:
    5597= kq1/r1
    q1= .275 μc
    im not sure about this so any help will be appreciated !!
     
  2. jcsd
  3. Sep 15, 2015 #2

    haruspex

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    How do you justify that? Is potential some conserved quantity?
    What equation do you know relating the charge and radius of a conducting sphere to its surface potential?
     
  4. Sep 15, 2015 #3

    rude man

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    You have 4 equations and 4 unknowns: qAi, qBi, qAf and qBf.
    where i = initial and f = final. You can solve for qAi and qBi by your 'relevant' equation'. Now invoke equality of potential for the final state, and charge conservation.
     
  5. Sep 16, 2015 #4
    I'm working on the same problem and this is what I got from following this thread:
    - the equality of final potential means: VAf=VBf
    ⇒ kqAf/rA=kqBf/rB
    ⇒qAfrB=qBfrA
    ⇒qAf=qBfrB/rA and qBf=qAfrB/rA (1)

    and then from conservation of charge:
    qAi+qBi=qAf+qBf
    ⇒qtotal=(qBfrA/rB) + qBf → solve for qBf
    and the same for qAf with substitution from (1).

    Is that right?
     
    Last edited: Sep 16, 2015
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