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Electron released into an electric field from rest

  1. Sep 4, 2016 #1
    1. The problem statement, all variables and given/known data
    An electron is released from rest in a weak electric field given by boldE.gif = -2.70 multiply.gif 10-10 N/C [PLAIN]http://www.webassign.net/images/lowercase/jhatbold.svg. [Broken] After the electron has traveled a vertical distance of 1.4µm, what is its speed? (Do not neglect the gravitational force on the electron.)

    2. Relevant equations
    I know that F_E = qE, and F_q = mg, as well as W_E= qEh, and W_G = mgh
    also, v2=v02+2aΔy
    but I'm not sure how im supposed to actually combine all of these equations to get the answer

    3. The attempt at a solution
    If I find the accelerations from gravity and the electrical field I get:
    FE=qE=1.602*10-19*-2.4*10-10
    aG=9.81
    a=FE/m+9.81 = -52.01
    v2=v02+2aΔy, so v2=2(-52.01)(1.4*10-6)
    v = 0.01405, which isn't correct. I didn't use the work equations because we haven't talked about work at all in class, so I figured we wouldn't need those equations.
     
    Last edited by a moderator: May 8, 2017
  2. jcsd
  3. Sep 5, 2016 #2

    gneill

    User Avatar

    Staff: Mentor

    Hi Adam21197, Welcome to Physics Forums!

    You don't state what the orientation of the electric field is. We might guess that it is oriented vertically and that it is directed downwards, but it would be nice to have confirmation. Perhaps there was a figure accompanying the problem? I see a "broken" image link in your problem statement. It will be quite a different result if, for example, E is directed parallel to the x-axis.

    Your problem statement says that ##E = -2.70 \times 10^{-10}~N/C##, but you've used ##-2.40 \times 10^{-10}~N/C## in your calculation. Which is correct?
     
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