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Forces of gravity on protons and electrons

  1. Jan 17, 2009 #1
    The problem statement, all variables and given/known data

    1.Assuming that only gravity is acting on it, how far does an electron have to be from a proton so that its acceleration is the same as that of a freely falling object at the earth's surface

    2.Suppose the earth were made only of protons but had the same size and mass it presently has. What would be the acceleration of an electron released at the surface?

    3.Is it necessary to consider the gravitational attraction as well as the electrical force?
    yes or no?


    3. The attempt at a solution

    i have read through my chapter and none of the examples help me at all, I thought #1 could be zero but the online homework thing rejected that. I dont know if you could help me with the answer and also give a short explain why that would be greatly appreciated!! Thankyou!
     
  2. jcsd
  3. Jan 17, 2009 #2
    1) It's asking for when the acceleration = 9.81 m/s^2. You know about F = ma, right? This applies to all forces, including electrical ones. Mass, acceleration, and the two charges are known (proton's charge = electron's charge). Using what two formulas will give you everything but the distance (which can be easily solved for)?

    2) First, find the number of protons that can make up Earth. There will be a total charge here. The net force ( F = ma) will be the sum of TWO force vectors in the same direction, the eletrical force and the gravitational force. The distance in your equations will be the radius of the earth.

    3) Think. Everything with a charge has an electrical field. Everything with mass has a gravitational field.
     
  4. Jan 17, 2009 #3

    Dick

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    You need to get some equations and show us an attempt. For example, the force between two charges is F=k*q1*q2/r^2, F=ma, the acceleration at the earth's surface is 9.8*m/s^2. Try and put those together to answer the first one.
     
  5. Jan 17, 2009 #4
    for the first one I used f=ma and f=(Gm1m2)/r^2, im pretty sure the masses could be assumed they are close to zero so the answer would be 1?
     
  6. Jan 17, 2009 #5
    nevermind its wrong let me try again
     
  7. Jan 17, 2009 #6
    You assume the net force (m*a) is only comprised of the gravitational force. What other force could possible contribute to a nonzero net force?

    And zero divided by anything is always zero, so 1 would still be wrong.
     
  8. Jan 17, 2009 #7

    Dick

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    F=G*m1*m2/r^2 is the force due to gravity. You want the force due to the electric charge. And you do need to know the masses of the electron and proton. You can't assume them away because they are small. Better look them up.
     
  9. Jan 17, 2009 #8
    so would I do.... Fe +Fg = ma? and solve from there?
     
  10. Jan 17, 2009 #9

    Dick

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    You could. But the force due to gravity is completely insignificant compared with the electrical force. You should check this but ignore it for now.
     
  11. Jan 17, 2009 #10
    ok but with the electrical force they dont list either of the charge values, do I need them or are they the same value, ect???
     
  12. Jan 17, 2009 #11

    Dick

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    If they don't list them then you'll need to look them up. They are both the same but with opposite signs.
     
  13. Jan 17, 2009 #12
    With my online system, the website provides the constants that it wants us to use. Check to see if your system already has the constants just in case you use a number too exact or rounded (mine was in the Help section).
     
  14. Jan 17, 2009 #13
    i did....

    (k*q1*q2)/(m*9.8) = r^2

    still couldnt find the answer but is this the right direction it it f=ma and fe together, big thing also is you can take a negative square root... i am to confused
     
  15. Jan 17, 2009 #14
    What's wrong with that equation? In this case, positive and negative charges mean nothing, so don't make the force negative.
     
  16. Jan 17, 2009 #15

    Dick

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    Ignore the sign. That just tells you which direction the force is acting. It doesn't affect the magnitude of the acceleration. Take the square root of the magnitude.
     
  17. Jan 17, 2009 #16
    ok still isnt working... one thing for the mass, do i add both the mass of the electron and proton together for m cause that is what I am doing and is the only thing I can think of that Im doing wrong.
     
  18. Jan 17, 2009 #17
    No, only the mass of the electron needs to be calculated.

    However, maybe the system is so exact as to want you to factor in the force of gravity? I'm not sure how much it will affect your answer (if at all), so maybe you will have to use Fe + Fg = ma.
     
  19. Jan 18, 2009 #18
    nope did that this online system(my hw site not this site) is jacked up, Thanks all for helping though
     
  20. Jan 18, 2009 #19

    Dick

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    Fg would only affect your answer if you were carrying the results out to about 40 decimal places. Hope your hw site feels better soon.
     
  21. Jan 19, 2009 #20
    hi,1st post:smile: ..i m learning too

    izzit by taking the nett forces?
    Fg + Fe = ma

    i.e

    Gm1m2/r^2 + Q1Q2/4(pi)(epsilon zero)r^2 = ma

    epsilon zero = permittivity of vacuum
     
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