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Coulomb's Law and protons

  1. Jan 29, 2007 #1
    1. The problem statement, all variables and given/known data
    A point charge q = -0.80 nC is fixed at the origin. Where must a proton be placed in order for the electric force acting on the proton to be exactly opposite to its weight? (Let the y axis be vertical and the x axis be horizontal.)


    2. Relevant equations
    F=(kq1 q2)/r^2


    3. The attempt at a solution
    I first tried drawing this out and it seems like it should be a pretty straightforward problem, but I seem to be having more trouble with it than I probably should. I know that F would have to equal zero, but if I do that then I can't solve for r.
     
  2. jcsd
  3. Jan 29, 2007 #2

    berkeman

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    Staff: Mentor

    The other relevant equation you need to use it the force due to gravity. Assume that this experiment is being done in a vacuum chamber on the surface of the Earth. Draw a free body diagram assuming that the charge is physically held at the origin, and think about how you would balance the gravitational force and the Coloumb force on the proton.
     
  4. Jan 29, 2007 #3
    So if the proton is going to be placed below the point charge, then the gravitational force and the Coulomb force would be going in opposite directions, so you would have F(coulomb)-g...correct?
     
  5. Jan 29, 2007 #4

    berkeman

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    Correct concept, but a small oversight in your math. You are correct that you want to balance the forces, but "g" is the acceleration due to gravity. How do you write the force due to gravity on a mass m?
     
  6. Jan 29, 2007 #5
    Would you use the equation for the gravitational force? F=(Gm1 m2)/r^2
     
  7. Jan 29, 2007 #6

    berkeman

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    You could, but that's more complicated than necessary. Just use the appropriate form of the general F=ma, where you substitute the value of "g", the acceleration of gravity at the surface of the Earth. Then balance your forces, and you're done!
     
  8. Jan 29, 2007 #7
    So...would it work if I did F= coulomb - mg to take the weight into account?
     
  9. Jan 29, 2007 #8

    berkeman

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    In equilibrium (nothing is moving -- it's balanced), the sum of the forces on the proton must add to zero.

    I have to bail for a couple hours, but you are on the right track. Write the value of the two forces (careful about signs!), set the sum to zero, and solve.
     
  10. Jan 29, 2007 #9
    Would the charge of the proton just be equal to one? mg=(kq1 q2)/r^2
     
  11. Jan 29, 2007 #10

    berkeman

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    No. "Charge" is in units of Columbs in the mks system of units. Be sure to stay consistent in your use of units. What can you say about the charge of a proton compared to the charge of an electron? What is the charge of an electron in Columbs?
     
  12. Jan 29, 2007 #11
    The absolute value of the charges of protons and electrons are the same...so would I just use the charge of the electron in the equation?
     
  13. Jan 29, 2007 #12
    I just wanted to make sure that I was using the correct equation: mg=(kq1 q2)/r^2
     
  14. Jan 29, 2007 #13

    berkeman

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    Okie dokie. That's the correct force equation. What did you get for the final answer?
     
  15. Jan 29, 2007 #14
    My final answer was 8.31229692 x 10^24 m
     
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