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Force exerted by an external electric field

  1. Feb 4, 2009 #1
    1. The problem statement, all variables and given/known data
    A particle of mass 1.20 kg with a charge of +2.30 mC is shot into a region with an electric field of strength 18.0 N/C, having a direction opposite the particle's velocity. If the particle has an initial speed of 2.40 m/s, how long does it take to come to a complete (though momentary) standstill? Assume there are no other forces acting on the particle.


    2. Relevant equations
    F = qE
    F= ma


    3. The attempt at a solution
    I'm not sure what equation to start with because of all the information already given. I couldn't find any equations dealing with time in the chapter this problem was in. It sounds like you may have to use kinematics but I'm not sure if that is the correct way to do it.
     
  2. jcsd
  3. Feb 4, 2009 #2

    Hootenanny

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    Kinematics (well dynamics really) sounds good to me. What do you think the first step would be?
     
  4. Feb 4, 2009 #3
    Vf = Vi +at ??
     
  5. Feb 4, 2009 #4

    Hootenanny

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    Looks good to me.
     
  6. Feb 4, 2009 #5
    0 = 2.40 + 9.8 (t)
    t = 9.8 / 2.4 = 4.08333
     
  7. Feb 4, 2009 #6

    Hootenanny

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    Why have you used an acceleration of 9.8 m.s-2?
     
  8. Feb 4, 2009 #7
    I forgot to find the force using F=qe then you can plug it into f=ma

    A = .000035

    .000035t = 2.40 - 0
    t = .000014

    would the acceleration be negative also since the field of strength is opposite the particle
     
  9. Feb 4, 2009 #8
    the problem states millicoulombs not micro - t = .014375
     
  10. Feb 4, 2009 #9

    Hootenanny

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    You're good up until this point. However, your final answer is wrong.
    You may want to check your arithmetric
    Yes, but you took care of that (be it intentionally or otherwise) in your second line above, by multiplying through by -1.
     
  11. Feb 4, 2009 #10

    Hootenanny

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    Your final answer is still wrong. Recheck your basic arithmetic. Notice that you are dividing a number greater than one by a number smaller than one. How can the result be less than one?
     
  12. Feb 4, 2009 #11
    2.4 / .0345 = 69.5652
     
  13. Feb 4, 2009 #12

    Hootenanny

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    You're getting closer, but your acceleration is two orders of magnitude out.
     
  14. Feb 4, 2009 #13
    GOT IT thanks for the help!
     
  15. Feb 4, 2009 #14

    Hootenanny

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    Not a problem :smile:
     
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