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Homework Help: Electric Fields , Finding final velocity. NEED HELP

  1. Jun 5, 2012 #1
    1. The problem statement, all variables and given/known data
    Mass= 15g
    Q1= 8.0uC
    Q2= 25 uC (stationary)

    Q1 is released from a point 12cm from Q2. How fast will Q1 be moving when the separation is 20cm.


    2. Relevant equations

    F=ma
    F=qe
    F=kQq/r2
    E=kQ/r2
    Ke=mv2/2
    W=FD
    vf2=vi2+2ad

    3. The attempt at a solution
    F=(9x109)(25x10^-6)(8x10^-6)/(.12^2)
    F=125N

    F=MA
    125/0.015= A
    A= 8333m/s2

    vf2=vi2+2ad
    vf2=0+2(8333)(.20-.12)
    vf=36.5m/s

    ACTUAL ANSWER IS : 28.3 according to the answer key

    Thanks for help!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 5, 2012 #2
    Hi skye204!! Welcome to PF :smile:

    The formula that you've used i.e "vf2=vi2+2ad" is applicable only when the acceleration is constant. In this case acceleration changes, so this is not applicable. I would suggest you apply the energy conservation principle instead :wink:
     
  4. Jun 5, 2012 #3

    pcm

    User Avatar

    vf2=vi2+2ad
    vf2=0+2(8333)(.20-.12)
    vf=36.5m/s



    you can not use the above formula as the acceleration is not constant.
    you should try using energy method.
     
  5. Jun 5, 2012 #4
    I get it

    W=FD
    W=125N x.12
    W= 15

    F=(9x109)(25x10^-6)(8x10^-6)/(.2^2)
    F=45N


    W=45 x .2
    W= 9

    pe=pe' + ke'
    15= 9 + 0.015(v^2)/2
    v= 28.28

    Thanks!
     
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