A Proton is orbiting a metal ball

  1. 1. The problem statement, all variables and given/known data

    A proton orbits a 1.0-cm-diameter metal ball 1.90 mm above the surface. The orbital period is 1.50 [tex]\mu s[/tex].

    What is the charge on the ball?

    2. Relevant equations

    [tex]F_{cp} = \frac{m v^2}{r}[/tex]

    [tex]F_c = \frac{k q_1 q_2}{r^2}[/tex]

    3. The attempt at a solution

    Since the proton is orbiting the metal ball, there is obviously a centripetal force involved, but first I need to velocity.

    [tex]v_{cp} = {\omega r} = \frac{2 \pi r}{T} = \frac{2 \pi (0.005)}{1.5*10^{-6}} = 2.1*10^4 m/s[/tex]

    I was wondering whether this statement is true [tex] F_{cp} = F_c[/tex]
     
  2. jcsd
  3. Kurdt

    Kurdt 4,941
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    You're on the right lines, just keep going. One issue is your value for r?
     
  4. so would q_1 = q_2 in this case, and for the r, I should use the 1.9mm instead?
     
  5. Kurdt

    Kurdt 4,941
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    You'll be solving for for one of the charges. The other will be the charge on the proton. The radius if measured from the centre of the metal sphere.
     
  6. so the radius of the metal ball would be .5 cm = 0.005m which is the value I used.
     
  7. Kurdt

    Kurdt 4,941
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    Plus the distance of the proton above the surface.
     
  8. to calculate the velocity needed for the centripetal force?
     
  9. Kurdt

    Kurdt 4,941
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    Yes.
     
  10. I don't see why I would need the distance of the proton from the surface.
     
  11. Kurdt

    Kurdt 4,941
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    The r is the distance of the proton from the centre of rotation. The distance is the radius of the sphere plus the height above the surface.
     
  12. so F_cp=(mv^2)/r

    and F_cp=F_c
    so F_cp=(kq_1q_2)/r^2
    also q_1=q_2 --> F_cp=(kq^2)/r^2

    Is this right?
     
  13. Kurdt

    Kurdt 4,941
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    If you're doing the same question you can't assume both charges are the same since it is one of the charges you are trying to find.
     
  14. r = 1.0cm + 1.9 mm = 1.19 cm = 0.0119 m
    m_proton = 1.67*10^-27 kg
    T = 1.5 us = 1.5*10^-6 s

    v_cp = omega*r = (2pi*r)/T = [2pi(0.0119 m]/(1.5*10^-6) = 4.98*10^4 m/s

    F_cp = (mv^2)/r = (1.67*10^-27 kg*(4.98*10^4 m/s)^2)/0.0119 m = 3.49*10^-16 N

    F_cp = F_c
    so F_cp=(kq_1q_2)/r^2
    F_cp=(kq^2)/r^2
    3.49*10^-16 N = (k*q^2)/r^2
    q^2 = [(3.49*10^-16 N)(r^2)]/(k)
    q^2 = [(3.49*10^-16 N)((0.0119m)^2)]/(k)
    q^2 = 4.937*10^-20 Nm^2 / 8.99*10^9 Nm^2/C^2
    q^2 = 5.49*10^-15 C^2
    q = 2.34*10^-15 C

    Is this the correct procedure and answer? Someone plz help
     
  15. I just saw your reply as I entered this post above, what am I doing wrong?
     
  16. radius would be r = 1/2(0.019 m) = 0.0095 m
     
  17. no wait, r = 1/2(1 cm) + 1.9 mm = 0.0069 m
     
  18. new calculation:

    r = r = 1/2(1 cm) + 1.9 mm = 0.0069 m
    m_proton = 1.67*10^-27 kg
    T = 1.5 us = 1.5*10^-6 s

    v_cp = omega*r = (2pi*r)/T = [2pi(0.0069 m]/(1.5*10^-6) = 2.89*10^4 m/s

    F_cp = (mv^2)/r = (1.67*10^-27 kg*(2.89*10^4 m/s)^2)/0.0069 m = 2.02*10^-16 N

    F_cp = F_c
    so F_cp=(kq_1q_2)/r^2
    F_cp=(kq^2)/r^2
    2.02*10^-16 N = (k*q^2)/r^2
    q^2 = [(2.02*10^-16 N)(r^2)]/(k)
    q^2 = [(2.02*10^-16 N)((0.0069m)^2)]/(k)
    q^2 = 9.62*10^-21 Nm^2 / 8.99*10^9 Nm^2/C^2
    q^2 = 1.07*10^-30 C^2
    q = 1.03*10^-15 C
     
  19. Dick

    Dick 25,735
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    Radius would be 0.005m+0.0019m.
     
  20. You're right, I corrected it and it comes to 0.0069 m. Do you know if this procedure I used is correct??
     
  21. the final answer should come to about -#*10^-12, but this is nowhere near that :(
     
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