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Using electric force to find the force of gravity

  1. Jul 25, 2008 #1
    1. The problem statement, all variables and given/known data

    A plastic sphere with a positive charge of 4.8 x 10^-19 C is held stationary in a gravitational field of strength 9.8 m/s^2 by an electric field of strength 1.2 x 10^5 N/C. What is the force of gravity on the sphere?

    2. Relevant equations

    F_e=qE=kQq/r^2
    F_g=mg=GMm/r^2
    g=GM/r^2

    3. The attempt at a solution

    I think the only way to find the force of gravity is to find the mass of the sphere first. Since r^2=kQ/E, g=GM/(kQ/E)=EGM/kQ. So M=gkQ/EG

    M=(9.8 m/s^2)(9.0 x 10^9 N x m^2/C^2)(4.8 x 10^-19 C)/(1.2 x 10^5 N/C)(6.67 x 10^-11 N x m^2/kg^2)=5.3 x 10^-3 m

    If this mass is the mass of the sphere and not some other random object, then the number can be substituted into F_g=mg

    F_g=(5.3 x 10^-3 m)(9.8 m/s^2)=5.2 x 10^-2 N

    If M isn't the mass of the sphere, then I'm completely wrong, but that's the only solution I've been able to come up with. Can anyone tell me if I'm right?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 25, 2008 #2
    if i understand the question correctly... the sphere is floating so like one force acts upwards and one acts downwards. If it is floating, what does that say about the magnitudes of the electric and gravitational forces?

    I'm not sure if what you did is right, but if I understand the problem correctly, you just need to do one calculation...
     
  4. Jul 25, 2008 #3
    I guess I was needlessly complicating the problem. The strength of the gravitational field must have been a red herring.

    If the sphere is indeed floating, then the upward electric force must balance the downward gravitational force, so that their magnitudes are equal. F_g=F_e=QE
    So F_g=(4.8 x 10^-19 C)(1.2 x 10^5 N/C)=5.8 x 10^-14 N

    That probably is the right answer. Thanks for your help.
     
  5. Jul 25, 2008 #4
    yea that's what I was thinking, glad to help
     
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