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Electrostatics question

  1. Feb 1, 2004 #1
    Two conducting spheres (of the same size), one negatively charged, one positively charged are at a finite distance apart. Magnitude of neg charged sphere < pos charged sphere.

    When the spheres are brought together and touched, and brought back to initial distance, what is the net force between the two spheres?

    The first question [unwritten here] was computed by finding the force charge between q1 and q1

    F = (8.99E9 N*m^2/C^2)*(q1*q2)/r^2

    -I know when the two spheres are physically juxtaposed, the charged is then evenly distributed:

    Qt = q1 + q2

    So, after the distribution, each charge would be Qt/2.

    My question is, would the charges then become neutral, therefore there will result in no repulsion or attraction? Or, would the according spheres retain their sign charge but the magnitudes will become equal: Qt/2? Or, would the charge with the larger magnitude be the pervasive charge and both will become positive when touched? Therefore, the two spheres would repulse instead of attract. Since the positive charge was the dominant charge in terms of magnitude, they would become positive.

    In any case, the resultant electric force between the would be:

    F = (8.99x10^9 N*m^2/C^2)*(Qt/2)*(Qt/2)/r^2

    Can anyone verify any of my assumptions?

  2. jcsd
  3. Feb 1, 2004 #2
    The magnitude of the positive ball is greater, so both charges are left with a slightly positive charge. I think they would then repel each other.
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