Find the distance between the two electrons

  • Thread starter rcs9390
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  • #1
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I have this problem that I'm stuck on



An electron is released above the Earths surface. A second electron directly below it exerts just enough of an electric force on the first electron to cancel the gravitational force on it. Find the distance between the two electrons.

The distance is 5.07m......I need to know how to get that. Thanks
 

Answers and Replies

  • #2
rock.freak667
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The electric force between two charges ,Q[itex]_1[/itex] and Q[itex]_2[/itex] is given by
[tex]F=\frac{Q_1 Q_2}{4\pi \epsilon_0r^2}[/tex]

the gravitational force of attraction between 2 masses M[itex]_1[/itex] and M[itex]_2[/itex] is given by
[tex]F=\frac{GM_1 M_2}{r^2}[/tex]

you will need to use these equations...try something and tell me what you did
 
Last edited:
  • #3
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Yes but I do not have the force or the distance....so i dont think that equation can allow me to find r....I'm figuring I have to do something with the gravitational force formula also
 
  • #4
rock.freak667
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Assume the distance between the first election and the earth is the radius of th earth...which is 6x10[itex]^24[/itex]kg
 
  • #5
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thats all i did...i pluged in the numbers that we have such as Kc Q1 Q2 G M1 and M2 and got (8.99X10^9) (-1.6X10^-19 * -1.6X10^-19)/(r^2) and the gravitational one is (6.6732X10^-11) (9.109X10^-31 * 9.109X10^-31) /(r^2)...thats as far as I got......I tried setting them equal...and tried finding r that way....but it obviously doesn't work since the r's are the same......I also tired using r=5.07 and working backwards somehow but all of those attempts didn't work out
 
  • #6
rock.freak667
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For the gravitational force put r as the radius of the earth
 
  • #7
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can you explain why?
 
  • #8
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because to me that doesnt make sense
 
  • #9
rock.freak667
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Well I can only guess that above the Earth's surface means that the electron is more or less near to the surface of the earth and if you were to add the radius of the earth and the height above the earth's surface, then the distance between the masses(i.e. 'r') would still be 6x10[itex]^6[/itex]m. The addition of some small distance say 1m added to 6x10[itex]^6[/itex]m would still be 6x10[itex]^6[/itex]m in standard form and wouldnt affect the output very much...
 
  • #10
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well i put that in and it still doesnt work
 
  • #11
rock.freak667
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Are you sure? because i did and got 5.08m but i didnt use the permittivity of air...which the question didnt state if this was done in a vacuum or not
 
  • #12
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wait how did you do it then?
 
  • #13
rock.freak667
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[tex]F_G=F_E[/tex]

[tex]\frac{GM_1 M_2}{R^2}=\frac{Q_1 Q_2}{4\pi \epsilon_0 r^2}[/tex]

[tex]\frac{(6.67*10^{-11})*(6*10^24)(9.11*10^{-31}}{(6.4*10^6)^2}=\frac{(1.60*10^{-19})^2}{4\pi(8.85*10^{-12})r^2}[/tex]

rearrange for r^2
 
  • #14
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i think you used the wrong equation for electric force because i used Kc(q1 q2/r^2)....and Kc is 8.99x10^9....and thats what you are supose to use and if you do its not the same answer
 
  • #15
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k is just 4pi * 8.85x10^-12 inversed rcs9390.
 

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