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Electrical kinematics

  1. Jun 26, 2011 #1
    1. The problem statement, all variables and given/known data
    An electron is projected at an angle of 30.8° above the horizontal at a speed of 8.20×105 m/s in a region where the electric field is E = 388j N/C. Neglecting the effects of gravity, calculate the time it takes the electron to return to its initial height.

    Answer: 1.23e-08 s
    2. Relevant equations
    v = v_0+at
    F = qE = MA


    3. The attempt at a solution
    Components of Velocity
    V_x = Vcos30.8
    V_y = Vsin30.8

    F = qE = ma
    a = qE/m
    = 1.6e-19*388/9.11e-31
    = 6.81e13

    V = v_0+at (set V = 0 cuz max height)
    0 = v_y+at
    t = 6.17e-9

    Now i was playing with nums and found that if i multiplied V_y by 2 then divided it by a that is the right answer, why?
     
  2. jcsd
  3. Jun 26, 2011 #2
    I think you might have misinterpreted the question. The electron's path is parabolic. What is its velocity going to be when it returns to its initial height? Hint: non-zero.
     
  4. Jun 26, 2011 #3
    so... if i assume it is a parabola... lets say it is initially at height H.

    It starts going to a max with V initial at V_y. It goes to a max and starts coming down. Once it passes height H again.. will it have the same velocity again?

    If so... does that mean for my equation v = v_0+at
    -V_y = V_y+at
     
  5. Jun 27, 2011 #4
    That's correct! Now solve that equation, and what do you get? :)
     
  6. Jun 27, 2011 #5
    oh thats how the 2 comes about. thanks <3
     
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