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Electricity electrostatic

  1. Aug 21, 2008 #1
    2 balls with a mass of 100kg are at a distance of 5m from one another, how many electrons need to be moved from one to the other in order to balance the gravitational force by an electric force,

    from what i understand, we have 2 neutral balls of the same mass, and i need to find out how much electric charge is needed to balance one of these balls with mg, i suppose the other is attached to the ceiling or something to prevent it from falling.

    whatever the ones charge is the others will be its negative

    mg=[k(q1*q2)]/r^2
    since q1 and q2 are the same
    mg=[k(q^2)]/r^2
    q^2=(mg)(r^2/k)
    q=[(mg)(r^2/k)]^0.5

    q^2=[(100*9.8)(25/9*10^9)]=2.72*10^-6
    q=1.65*10^-3 c

    now q is the charge of the balls, positive and negative,
    what i must do now is divide my total charge by the charge of a single electron (1.6*10^-19) and ill get the amount of electrons, only this doesnt seem to be working for me, where have i gone wrong??

    the final answer is 5.38*10^10 electrons
     
  2. jcsd
  3. Aug 21, 2008 #2

    berkeman

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    Staff: Mentor

    Are you sure about that wording? The gravitational force between them is attractive, and if you make one charged +q and one charged -q, that will also generate an attractive force. How can you balance two attractive forces. Maybe the question just asks how much charge has to be moved to make the two forces equal (but not opposite)?
     
  4. Aug 21, 2008 #3

    Dick

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    They don't want the electrical force to balance mg. Assume both of the balls are attached to the ceiling. They want the electrical force to cancel the gravitational force BETWEEN the balls. I.e. Gm*m/r^2.
     
  5. Aug 21, 2008 #4

    berkeman

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    Staff: Mentor

    Yeah, I wasn't referring to mg (that would be a huge charge!). But even just the attraction between the two balls isolated in space.... Seems like they would have to both have the same sign of excess charge, in order to generate a repulsive force. Sorry if I'm being dense here....
     
  6. Aug 21, 2008 #5

    Dick

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    Agreed. But I think the poster is picturing the problem as one ball suspending the other ball below it against the earth's gravity. In which case the signs are ok, but the poster is working with the wrong force.
     
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