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A proton is released from rest at the positive plate of a parallel-plate capacitor.

  1. Sep 14, 2007 #1
    A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the negative plate with a speed of 50,000 m/s. What will be the proton's final speed if the experiment is repeated with double the amount of charge on each capacitor?

    I seem to be stuck on this one. I initially tried to use
    1/2 m[tex]^{}_{i}[/tex]v[tex]^{2}[/tex] + V[tex]^{}_{i}[/tex]q = 1/2 m[tex]^{}_{f}[/tex]v[tex]^{2}[/tex] + V[tex]^{}_{f}[/tex]q
    but i ran into a wall. Help please.
     
  2. jcsd
  3. Sep 14, 2007 #2

    Chi Meson

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    That's the wrong way to think of it.

    What is the same about the capacitor? (Hint, it's the same capacitor)
    What does doubling the charge of a capacitor do to the voltage across the capacitor?
    What effect does this voltage increase have on the electric potential energy of the electron?
     
  4. Sep 14, 2007 #3
    the capacitor plates are still the same distance apart. same size. same signs(one is positive and one is still negative)

    by doubling the the charge of the plates it would seem that the voltage would increase by a factor of 2 as well.

    increasing the voltage would increase the electrical potential of the proton.

    Are these asumtions correct?
     
  5. Sep 14, 2007 #4

    Chi Meson

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    Doubling the voltage will increase the potential energy of the proton by how much?

    And then, how much more KE will the proton have when it reaches the other plate?
     
  6. Sep 14, 2007 #5
    the potential energy = Vq so if the voltage increases by a factor of two then the potential energy will increase by a factor of two. Which would then make sense to me that the kenetic energy would have to double to keep conservation of energy?
     
  7. Sep 14, 2007 #6

    Astronuc

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    Correct. The proton (initially at rest, so KE = 0) is accelerated across the potential difference. The electric potential energy is converted to the proton's KE.
     
  8. Sep 14, 2007 #7
    So if initially the proton had a final velocity of 50,000m/s, once the charge on the plates was doubled and the potential energy was doubled and the kinetic was therefor doulbed would the resulting final velocity be double that of the original , 100,000m/s? if Ke= 1/2mv^2. the mass is the same but since the kinetic energy doubled the velocity would as well? Correct line of thought?
     
  9. Sep 14, 2007 #8

    Astronuc

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    Careful. What is the relationship between magnitude of velocity and kinetic energy.

    Take V - and the kinetic energy, KE = 1/2 mV2,

    now let V become 2V and KE = 1/2 m (2V)2.


    Find the expression for V in terms of KE from the first expression.
     
  10. Sep 14, 2007 #9
    If we solve for v in terms of KE from the first expresion

    v= [tex]\sqrt{(2KE)/m}[/tex]

    now that is without using 2v

    using KE= 1/2m2v^2


    v= [tex]\sqrt{KE/m}[/tex]

    or since you have it as KE = 1/2 m (2V)^2 when you solve for V in terms of KE would

    v= 1/2 [tex]\sqrt{(2KE)/m}[/tex]

    my work:

    KE = 1/2 m (2V)^2

    2KE=m(2V)^2

    (2KE)/m = (2V)^2

    squareroot((2KE)/m)= 2V

    1/2 squareroot((2KE)/m) = V
     
    Last edited: Sep 15, 2007
  11. Sep 15, 2007 #10

    learningphysics

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    So if the energy doubles... how does the speed change?
     
    Last edited: Sep 15, 2007
  12. Sep 15, 2007 #11
    rather than have (2V) wouldn't i have (2KE) since we have determined that we have double the PE which become double the KE? so the expression would then be

    2KE = 1/2 mv^2

    which would mean that V in terms of KE would equal

    v = [tex]\sqrt{(4KE)/m}[/tex]

    which simplified would be v = 2 [tex]\sqrt{KE/m}[/tex]?

    would make sense to me that the increase in v due to the increase in KE would be a factor of 2? or should it only be an increase of [tex]\sqrt{2}[/tex]?

    without the increase v= [tex]\sqrt{(2KE)/m}[/tex]
    after v = [tex]\sqrt{(4KE)/m}[/tex] since the only difference is from 2 to 4....... the increase would be by a factor of [tex]\sqrt{2}[/tex]? which would then make the velocity after doubling the charge on the plates v= 70,710.7m/s.
     
    Last edited: Sep 15, 2007
  13. Sep 15, 2007 #12

    learningphysics

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    Yes, that's the answer... if you are unsure, just take the ratio of the new velocity to the old:

    [tex]\sqrt{(2KE)/m}[/tex]

    This is correct for v... so v1 = [tex]\sqrt{(2KE_1)/m}[/tex]

    v2 = [tex]\sqrt{(2KE_2)/m}[/tex]

    What does v2/v1 come out to?
     
  14. Sep 15, 2007 #13

    learningphysics

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    yes.
     
  15. Sep 15, 2007 #14
    well the ratio of the new velocity to the old is going to be [tex]\sqrt{2}[/tex]

    (70,710.7m/s)/(50,000m/s) = [tex]\sqrt{2}[/tex]

    which doesn't really confirm anything for me because i came up with Vfinal by multiplying
    50,000m/s by [tex]\sqrt{2}[/tex] so it only make since that working backwards would give me that? Maybe i missed the point of your question. I'm sorry if this is frustrating. I really appreciate the time and effort from all of you.
     
  16. Sep 15, 2007 #15

    learningphysics

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    Oops... sorry. I understand now... I didn't mean take the ratio of the numbers... I gave expressions for v2 and v1 in terms of E2 and E1... take the ratio of v2/v1 using those two expressions...
     
    Last edited: Sep 15, 2007
  17. Sep 15, 2007 #16
    oh ok now i see that by simplifying

    [tex]\sqrt{(2(2KE))/m}[/tex] / [tex]\sqrt{(2KE)/m}[/tex] = [tex]\sqrt{2}[/tex]

    which confirms that it would increase by a factor of [tex]\sqrt{2}[/tex]. Thank you guys.
     
    Last edited: Sep 15, 2007
  18. Sep 15, 2007 #17

    learningphysics

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    yup. :smile: you're welcome.
     
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