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Particles in a Field

Discussion in 'Electronics Homework Help' started by chopnhack, Jul 8, 2017.

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  1. chopnhack

    chopnhack

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    Apr 28, 2014
    Hi all!

    I am taking a class so no spoilers please!

    Question:
    A proton moving to the right at 6.2 × 10^5 ms-1 enters a region where there is an electric field of 62k NC-1 directed to the left. Describe qualitatively the motion of the proton in this filed. What is the time taken by the proton to come back to the point where it entered the field?

    So far I have surmised that the proton should come to a stop shortly after entering the field. It should then accelerate towards the left, retracing where it came in from.

    The next part I wanted to solve using kinematics. I thought that I could determine force by dividing the field by the charge of the proton, thus producing the net force of the field on the proton, so 62k NC-1 / 1.602 x 10^-19 C = 3.87x10^23 N
    a=F*m which is 3.87x10^23 N / 1.673x10^-27 kg a = 2.31x10^52 m.s-2

    That seems ludicrously fast....

    Now, looking for a way to use kinematics, I am stymied. I know that we have an initial velocity, but I don't know time or x position. Am I approaching this the correct way
     
  2. Ratch

    Ratch

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    Mar 10, 2013
    What's the hangup? The force on the proton can be calculated by its charge and the electric field strength. You know the mass of the proton and the force on it so the deceleration can be calculated. From the deceleration you can calculate the time when the velocity will be zero. The same time is needed to return the proton to where it started.

    If you have to express velocity units, don't use ms-1. Nobody will know what you mean. Use meters/sec or m/s. Same with other units.

    Ratch
     
    chopnhack likes this.
  3. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    or use ms-1 which pretty much everyone will understand.
     
    chopnhack likes this.
  4. davenn

    davenn Moderator

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    rethink that bit ... it's incorrect and it's why you got a huge and incorrect number ;)
     
    chopnhack likes this.
  5. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Also note the strength of the field.

    What are the units, and what does that represent?

    Sorry, no spoiler from me
     
  6. chopnhack

    chopnhack

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    Apr 28, 2014
    Thanks all, I think I got it.

    Thank you dave - thats actually a typo. The error I made was in thinking I could divide the field by the charge - the field is force divided by coulomb and so it has to be multiplied. It still gave me pause because it yielded a subatomic particle moving faster than light, by magnitudes!!! Which I didn't think was possible.

    So the correct start of the solution is F = QE or 9.93x10^-15 N
    Applying F = ma
    a = 5.95 x 10^12 m.s-2 --------- (sorry @Ratch , this is the convention the professor wants us to use)

    The initial velocity will carry the particle x meters into the field before it stops.
    Vf=Vi+at
    t1 = 1.04x10^-7 sec to stop particle.

    using the same equation we can calculate Vf = 6.18x10^5 m.s-1 which we later use to calculate the time taken for the last leg of the trip, the time from when the velocity negatively accelerated to a stop and then reversed direction and exited the field.

    d = 1/2at^2
    2.975x10^12 m.s.-2 * 1.0816x10^-14 sec2
    d = 0.032m

    d = [(Vi+Vf)/2]*t
    0.032 = 309,000 * t
    t2 = 1.03x10^-7 sec - this is the time for exiting the field from a stop inside it

    total time = t1+t2 t = 2.07x10^-7 sec

    Have I followed this logically to completion?
    @(*steve*) - sorry if you have posted something similar in detail, but I haven't looked yet, I want to make sure I got this correct!
     
  7. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    They can't, but using those equations they can. If you start getting into higher velocities you need to use more complex equations. But equally, you're not going to be given problems that get you into that sort of problem.
     
  8. chopnhack

    chopnhack

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    That's a relief!
    Did I work this problem out correctly?
     
  9. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Correct. The field strength is in Newtons per coulomb. But what do you need to multiply this by to get the force?

    Then what is the acceleration?

    Then I would solve for time the simple Newtonian equation involving initial and final velocity, acceleration, and time. Where is the particle now?

    Also consider what the maximum speed of the particle can be. Is the return journey symmetrical to the first part of the journey into the field?

    nup, still no spoiler :(
     
  10. chopnhack

    chopnhack

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    Apr 28, 2014
    Interesting your version sounds like what I did in post #6. Is my version wrong?

    To answer your question, the field strength in N/C needs to be multiplied by the charge of the particle to yield the force in newtons on the particle. In this case the charge of the proton is e - 1.602x10^-19 C multiplied by the field which is 62,000 N/C. This gives 9.93x10^-15 N .
     
  11. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Yep, then what is the acceleration? (I want to see working)
     
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  12. davenn

    davenn Moderator

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    it isn't, so yes, you have still made a mistake

    As you didn't show a correction, did you understand the mistake you made in the bit I pointed out that I asked you to rethink ?
     
  13. Ratch

    Ratch

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    chopnhack,

    What is all this scratching around for? The initial starting velocity was given, the final arrival velocity is zero. The time to return is the same as the time to get there, so just multiply your arrival time by two to get the total time.

    Ratch
     
  14. chopnhack

    chopnhack

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    Apr 28, 2014

    F = ma
    rewrite to solve for a, a = F/m

    F = 9.93x10^-15 N
    mass of proton = 1.6726219 × 10-27 kilograms

    in dividing the two we get 5.94 x 10^12 m.s-2 as acceleration --> this is magnitudes larger than speed of light. Have I miscalculated this part?:confused:
     
    Last edited: Jul 9, 2017
    davenn likes this.
  15. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Yes. (I read that as misinterpreted) An acceleration is not a velocity.

    So what is an equation with 2 velocities, acceleration, and time?
     
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  16. Ratch

    Ratch

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    Are you comparing acceleration to velocity? What is the acceleration of light?

    Ratch
     
  17. chopnhack

    chopnhack

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    But a newton is kg/ms^-2 - if we divide it by kg, do we not end up with ms^-2, which is acceleration?

    Vf=Vi+at
     
  18. chopnhack

    chopnhack

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    In a manner, yes. I was looking at the fact that nothing is faster than light, so that would mean that nothing can accelerate faster than the speed of light.
     
  19. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Yes. But acceleration is not velocity.

    I could be continuously subject to that acceleration but never reach a significant velocity (I would be pulled apart into my constituent particles, but none of them might move very fast).

    It's the time that this acceleration applies which is the issue. If it we're alternating very fast, or I was orbiting the source of this acceleration, my speed may be quite small compared to C.
     
  20. Ratch

    Ratch

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    Mar 10, 2013
    No, a newton is kg-meter/sec^2

    Ratch
     
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