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Proton between two plates

  1. Sep 13, 2009 #1
    1. The problem statement, all variables and given/known data

    problem.jpg

    2. Relevant equations

    Ki + Ui = Kf + Uf

    3. The attempt at a solution

    I need to find the velocity of the proton when it's 2.3mm's to the right of where it is now. I need to find the acceleration of the proton. I believe that I should use F=ma to do this. The only force acting on the proton is -Eq (to the left, hence the minus sign). I don't know the Electric field though.

    Any help?

    Edit: or, can I do this:

    Ki= 0
    Ui=qVi
    Kf=1/2mv^2
    Uf=qVf
     
  2. jcsd
  3. Sep 13, 2009 #2

    kuruman

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    You don't know the electric field but you know that it is uniform between the plates and you know the potential difference. How is E related to the potential difference in this case?
     
  4. Sep 13, 2009 #3
    E=-dV/ds isn't it?
     
  5. Sep 13, 2009 #4

    kuruman

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    That's it.
     
  6. Sep 13, 2009 #5
    I don't understand how to use that. I have a 20V potential difference. How does that relationship help me?
     
  7. Sep 13, 2009 #6

    kuruman

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    Integrate the expression.

    [tex]\Delta V=-\int Eds[/tex]

    You know ΔV = 20 Volts, you know that E is constant and you know the plate separation. Can you do the integral and solve for E?
     
  8. Sep 13, 2009 #7
    [tex]
    \Delta V=-\int Eds
    [/tex]

    [tex]
    20=-E\int ds
    [/tex]

    [tex]
    20=-E (2.3E-6 km)
    [/tex]

    E=-20/2.3E-6 = -8.695E6 V/km
     
  9. Sep 13, 2009 #8

    kuruman

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    The plate separation is 2.3 mm, not 6 km.
     
  10. Sep 13, 2009 #9
    That should read 2.3x10^-6 km since the problem will ultimately ask for KM/s velocities.
     
  11. Sep 13, 2009 #10

    kuruman

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    You can convert later. For the time being, calculate the field in Volts/meter, which the conventional unit for electric field. It is less confusing this way.
     
  12. Sep 13, 2009 #11
    It's all confusing no matter what units are used.

    E=-20/2.3E-3 = -8.695E3 V/m then
     
  13. Sep 13, 2009 #12
    The proton moves through a potential difference of 20V so therefore gain of PE=20e(e=charge of proton/electron).Now you can use PE gained =KE lost.This method may be a bit quicker.
     
  14. Sep 13, 2009 #13
    So how would i set that up? Something like this?

    [itex]U_i + K_i = U_f + K_f[/itex]

    [itex]U_i = qV_1[/tex]
    [itex]K_i = 0[/itex]
    [itex]U_f = qV_2[/itex]
    [itex]K_f = 1/2mv^2[/itex]

    [itex]qV_1 + 0 = qV_2 + 1/2mv^2[/itex]

    Something like this? Solve for v?
     
  15. Sep 13, 2009 #14
    K1 is not zero.
     
  16. Sep 13, 2009 #15
    Oh. Would it also be 1/2mv^2 (v being initial velocity?)
     
  17. Sep 13, 2009 #16
    Yes.Good luck with it.
     
  18. Sep 13, 2009 #17
    [itex]U_i + K_i = U_f + K_f[/itex]

    [itex]U_i = qV_1[/tex]
    [itex]K_i = 1/2 mv_i^2[/itex]
    [itex]U_f = qV_2[/itex]
    [itex]K_f = 1/2mv_f^2[/itex]

    [itex]qV_1 + 1/2 mv_i^2 = qV_2 + 1/2mv^2[/itex]

    The answer still doesn't make any sense. When you plug all the numbers in, the final velocity comes out to be WAY too fast.

    [itex]qV_1 + 1/2 mv_i^2 - qV_2 = 1/2mv^2[/itex]

    [tex](1.6x10^{-19})(-70) + (.5)(9.1x10^{-31})(94.5^2) - (1.6x10^{-19})(-50) = 1/2mv^2[/tex]

    [itex](-1.12x10^{-17} + (4.063x10^{-27}) - (-8x10^{-18}) = 1/2mv^2[/itex]

    [itex]-3.20x10^{-18} = 1/2mv^2[/itex]

    [itex]-6.40x10^{-18} = mv^2[/itex]

    [tex]\frac{(-6.40x10^{-18})}{(9.1x10^{-31})} = v^2[/tex]

    [itex]-7.033x10^{12} = v^2[/itex]

    First of all I have this errant negative sign that's killing me. I honestly have no idea whats going on.
     
  19. Sep 13, 2009 #18

    kuruman

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    Your potential energy is in Joules but your kinetic energy is not. You need to express the velocity in meters per second not kilometers per second then put it in the equation.
     
  20. Sep 13, 2009 #19
    yeah I did that too, but it still comes out way too high of a value and still attempting to square root a negative.
     
  21. Sep 13, 2009 #20

    kuruman

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    Yu say in part (a) that the particle is a proton. What is the mass of a proton?
     
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