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Placement of Particles.

  1. Aug 26, 2006 #1
    Hey everyone,

    I know the rules are post your own work first but I have no clue. I don't want answers though, I simply want to know what to do to get the answer.

    Anyways, if it's ok, here's the question...


    Book: Cutnell & Johnson Physics 7th Edition
    pp. 569 #40


    A particle of charge +12 uC and mass 3.8 x 10^-5 kg is released from rest in a region where there is a constant electric field of +480 N/C. What is the displacement of the particle after a time of 1.6 x 10^-2 s?

    Here's what I can gather to be given...

    Particle Charge: +12 x 10^-6 C
    Particle Mass: 3.8 x 10^-5 kg
    Initial Velocity: ???
    Electric Field: +480 N/C

    I have the equations for Kinematics, Newton's Second Law, and The Electric Field... but I don't know how to implement my data. Am I missing something?

    Thanks for you help everyone!
     
  2. jcsd
  3. Aug 26, 2006 #2

    quasar987

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    The so-called "equations of kinematics" keep track of the position of a particle under the action of a constant force, right? For a constant force F in the x-direction, the equation of kinematics is

    [tex]x(t)=x_0+v_0t+\frac{F}{2m}t^2[/tex]

    Now... what you have is a particle in a constant electric field. How is the electric force F on a particle of charge Q related to the field E (at the location of Q)? Is that a constant force? If so, you can use the above equation to find the motion and solve for the appropriate time.
     
  4. Aug 27, 2006 #3
    I'm sorry... I don't understand :frown:
     
  5. Aug 27, 2006 #4
    It is clear that the initial velocity is ZERO. It says so right in the problem!

    I have helped you in another thread regarding E&M. I think it is in your best interest to read the assigned readings in your book a few times. You seem to be having trouble with the meaning of electric field, acceleration, and whatnot. You need to learn the basics before diving into problem solving.

    You know the E-field, and you know the charge of the particle in the e-field. If you knew what the meaning of the e-field was, it would be trivial for you to determine the force on the particle.

    If you knew the force on the particle you can easily determine the acceleration of th particle using Newton's second law.

    If you knew the acceleration of the particle you can find the displacement of the particle after any given amount of time. You can use the following kinematic equation for when the initial velocity and initial displacement are both zero: displacement = 0.5At^2

    Go to the library and read your book for a few hours. You can't expect to learn this stuff without attempting to look in your book.
     
  6. Aug 27, 2006 #5
    Oh ok thanks, that clears up alot. I don't want you to think I haven't been trying this stuff on my own. I spent close to 3 hours reading the chapter and taking notes before posting yesterday. I'm really trying to understand this stuff... :redface: :redface:

    I appreciate your help... :smile:
     
  7. Aug 27, 2006 #6
    Alright, here's what I've come up with.


    Given:

    q = 12 x 10^-6
    m = 3.8 x 10^-5
    N/C = +480
    t = 1.6 x 10^-2



    Work:

    F = (12 x 10^-6) x (480) = 5.76 x 10^-.03 (I wasn't sure about this. My calculator just says ^-03 so I assume it

    means .03)

    a = (5.76 x 10^-.03) / (3.8 x 10^-5) = 141,461.7

    d = .5(141,461.7) x (1.6x10^-2)^2 = 18.11


    Am I doing this right? I hope so because I'm starting to understand it all now.... :smile:
     
  8. Aug 27, 2006 #7
    No, your calculator means exactly what is says! When it displays 5.76x10^-03 it means exactly that! Why would you think it means 5.76x10^-.03?????

    Otherwise, looks like you did it right. Don't forget your units.
     
  9. Aug 27, 2006 #8
    Oh ok. Why does it say 03? is that just 3?


    Thanks again. Sorry I'm so slow with this stuff... I've never had physics in my life =/

    I got 1.21 m
     
    Last edited: Aug 27, 2006
  10. Aug 27, 2006 #9

    quasar987

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    Yes, 03 is just 3. why it writes 03 instead of 3 is probably because it was programmed to write the units on the right and the "tens" on the left. So 03 means 0 tens, 3 units. Is "tens" the right word ?!
     
  11. Aug 28, 2006 #10

    HaHa! That sounds like a probable reasoning...

    Does anyone know if I have the right answer?

    edit: on the given (1st post) I have Particle Charge: +12 x 10^-6 C ... It's actually +12 uC. Did I convert that right? I'm seeing alot of people putting 1 / +12 x 10^-6
     
  12. Aug 28, 2006 #11

    quasar987

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    leright said in his last post,

    "Otherwise, looks like you did it right. Don't forget your units."


    Tell those people they are mistaken.

    [tex]1 \mu C=10^{-6}C[/tex]
     
  13. Aug 29, 2006 #12

    Great! So 1.12m is right!

    I thought so. I could understand say 1/10^6 in order to keep from having a negative in the powers place.
     
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