1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electric potential and distribution of charge

  1. Jul 9, 2011 #1
    1. The problem statement, all variables and given/known data

    A conducting sphere of radius r1 = 10 cm and charge q1 = 2 µC is placed far apart from a second conducting sphere of radius r2 = 30 cm and charge q2 = 3 µC. The two spheres are then connected by a thin conducting wire. What is the potential on the surface of the first sphere after the two spheres are connected by the wire? Use the reference V = 0 for r at infinity.


    2. Relevant equations

    Meh....

    3. The attempt at a solution

    A total of 5mC(using m for micro, not not milli) is in this system. When the wire is connected, the charge would distrbute evenly according given a surface charge density of 5mC/(area of sphere one + area of sphere two). So the charge on one would be (5mC/(area of 1+area of 2))(area of 1). And the voltage would be the line integral from infinity to 10x10^-2 of q/(4pie0r) dl)

    I end up with 44,937.8V after all is said and done. But the "correct" answer is ~112,000V.

    Going backwards from 112,100V, you get that V*4*pi*e0*r = Q from the line integral to find potential, or that the Q on the first sphere is 1micro coloumb. Which would require that the first sphere is 1/4 the area of the second sphere given a uniform charge distribution. Where as the area ratio is 1/9.5 or something close.

    Any thoughts?
     
  2. jcsd
  3. Jul 9, 2011 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    No. The charge will distribute between the spheres so that the potential of both spheres is the same. Otherwise the potential difference causes a driving force for the charges from one sphere to the other one.

    ehild
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook