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Electric Potential and equilibrium

  1. Sep 24, 2008 #1
    1. The problem statement, all variables and given/known data

    Two spheres have a radius of 1cm each and carries 4.3nC and -9nC each. The two sphere are separated by an unknown distance. If both spheres are connected by a thin wire, what wil be the potential on each once equilibrium is reached? and how much charge must move between the sphere in order to achieve equilibrium?

    2. Relevant equations

    Potential of a point charge: V = kq/r

    3. The attempt at a solution
    The charge on each sphere when they are not connected is 38.7KV, and -8.1KV. I believe that if they are connected and are at equilibrium, the net charge will be zero hence charge at sphere 1 should be equal to charge at sphere 2. Am i right?
     
  2. jcsd
  3. Sep 25, 2008 #2
    You're right in saying that when they are connected, they'll quickly attain equilibrium- the potential difference across the wire will be zero, and the two spheres will carry the am charge. But it's not true that the net charge will be zero- net meaning the sum of the charges on the two spheres- and you've said that the charge on each sphere is a voltage!
     
  4. Sep 25, 2008 #3
    so, what should be the charges of the two spheres once at equilibrium knowing the charge of each when they are not connected?
     
  5. Sep 25, 2008 #4
    The two spheres will carry the same charge. The charge is also conserved- the total charge is the same before and after the spheres are connected. Does that help?
     
  6. Sep 25, 2008 #5
    By "total charge" i assume you mean net charge. Therefore, the charge carried by each at equilibrium will be the addition of their charges before equilibrium?
     
  7. Sep 25, 2008 #6
    Not quite. The point of the sentence "charge is conserved" is that the total (/net) charge is the same before and after you connect the two spheres- so all you're doing is redistributing it between the two of them, and what one loses the other must gain.

    By the way, I just noticed you're new here: welcome to PF :smile:
     
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