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Coulomb's Law

  1. Nov 25, 2003 #1
    Yeah, i'm having difficulties with this one question...

    A stationary proton holds an electron in suspension underneath it against the force of gravity. How far below the proton would the electron be suspended?

    I understand that Fe = Fg

    so i would have __ = mg...
    but i dunno what to fill the __ to get an answer... i've tried using
    Eq = mg... but that doesn't seem to get me anywhere.. can anyone show me my error and point me on the right path?
     
  2. jcsd
  3. Nov 25, 2003 #2

    chroot

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    The force due to electromagnetic force is Coulomb's law,

    [tex]F_e = \frac{q_1 q_2}{4 \pi \epsilon_0 r^2}[/tex]

    where [tex]\inline{q_1}[/tex] and [tex]\inline{q_2}[/tex] are the magnitude of the charges in coulombs, and [tex]\inline{r}[/tex] is their separation in meters.

    The force due to gravity you already have:

    [tex]F_g = m g[/tex]

    The force pulling the electron up is given by Coulomb's law. The force pulling the electron down is given by the previous equation. Set them equal to each other, and solve for r.

    Does this help?

    - Warren
     
  4. Nov 25, 2003 #3
    we didn't have the same formula for coloumb's law... we had

    Fe= kq1q2 / d^2
     
  5. Nov 25, 2003 #4

    chroot

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    That's actually the same formula. Someone along the line decided to make a new variable,

    [tex]k = \frac{1}{4 \pi \epsilon_o}[/tex]

    just so Columb's law would look more similar to Newton's law of gravitation,

    [tex]F_g = \frac{G m_1 m_2}{r^2}[/tex]

    Also, they called the separation d instead of r, but it's the same quantity.

    - Warren
     
  6. Nov 25, 2003 #5

    ShawnD

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    I'm going to use Fe for electric force and Fg for gravity force.

    Fe = Fg
    kq^2/d^2 = mg
    (9x10^9)(1.6x10^-19)^2 / d^2 = (9.11x10^-31)(9.81)

    To solve I just graphed it and looked for the intersection.
    d = 5.0774m
    d = 5.08m
     
  7. Nov 25, 2003 #6
    why did you square the 1.602x10^-19... like i would understand that to substitute that into q1, but why q2 also?? This is basically why i was having problems with the question.. if i knew you could use the same value for both q1 and q2 i would have never asked this...

    BTW that is the right answer...
     
    Last edited: Nov 25, 2003
  8. Nov 25, 2003 #7

    ShawnD

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    protons and electrons have the same charge, that makes it squared.
     
  9. Nov 25, 2003 #8
    well that makes sense now... thanks a lot all of you guys... man, i'm soo happy i found this website...
     
  10. Nov 25, 2003 #9

    chroot

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    Take it step by step.

    [tex]-\frac{k q_1 q_2}{d^2} = m g[/tex]

    Solve for d:

    [tex]-k q_1 q_2 = m g d^2[/tex]

    [tex]d = \sqrt{\frac{-k q_1 q_2}{m g}[/tex]

    Since your problem deals with a proton and an electron, which have opposite but equal charges, the quantity

    [tex]q_1 q_2 = -e \cdot e = -e^2[/tex]

    where [tex]\inline{e}[/tex] is the charge on the electron (and proton).

    Plug that into the equation above to get

    [tex]d = \sqrt{\frac{k e^2}{m g}[/tex]

    - Warren
     
    Last edited: Nov 25, 2003
  11. Nov 25, 2003 #10
    sweet.. i got the answer, thx guys...
     
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