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Ratio of Electric Force to Gravitational force

  1. Aug 23, 2007 #1
    1. The problem statement, all variables and given/known data
    It is known that the electric force of repulsion between two protons is much stronger than their gravitational attraction. For two protons a distance R apart, calculate the ratio of the magnitude of the repulsion to that of the attraction.

    2. Relevant equations
    [tex] F_{g}=\frac{Gm_{1}m_{2}}{r^2}[/tex]
    [tex] F_{e}=\frac{Kq_{1}q_{2}}{r^2}[/tex]



    3. The attempt at a solution
    So the ratio [tex]\frac{F_{e}}{F_{g}}[/tex] =[tex] \frac{Kq^2}{Gm^2}[/tex]
    I get 1.3E28 which is wrong. Why?
     
  2. jcsd
  3. Aug 23, 2007 #2

    Dick

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    The equation is right. Hard to tell where you are going wrong with putting the numbers in.
     
  4. Aug 23, 2007 #3
    K=9.0E9
    G=6.67E-11
    Mass of proton aprox. 1.672E-27
    Charge of a proton 1.602E-19
    Right?
     
  5. Aug 23, 2007 #4

    Dick

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    Fine. But looking at your number, I think you are forgetting to square the masses and charges.
     
  6. Aug 23, 2007 #5
    ok, i get 1.2E36 with 2 sig figs and still wrong.
     
  7. Aug 23, 2007 #6

    Dick

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    I get 1.2355E36. So I disagree with who or whatever is telling you it's wrong.
     
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