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What is the speed of the electron as it strikes the second plate?

  1. Sep 10, 2010 #1
    1. The problem statement, all variables and given/known data

    A uniform electric field exists in a region between two oppositely charged plates. An electron is released from rest at the surface of the negatively charged plate and strikes the surface of the opposite plate, 2.6 cm away, in a time 3.8 x 10-8 s. (a) What is the speed of the electron as it strikes the second plate? (b) What is the magnitude of the electric field?

    2. Relevant equations

    F = q E
    E = F/ q = q/4pi[tex]\epsilon[/tex] r^2
    V[tex]_{F}[/tex]^2 = V[tex]_{0}[/tex]^2 + 2a( x - x[tex]_{0}[/tex])
    F = ma

    3. The attempt at a solution

    I thought finding the speed was obvious
    i just converted the cm into meters and divided by the time
    it came out to about 6.8e5
    ...which was wrong

    I know how to relate the E field equations to speed and acceleration
    and this was my final simplified eqation without pluging in the numbers

    E = (V^2)m / (2 q x)

    q was the charge of an electron
    x was 0.026 meters
    V was the 6.8e5
    m was the mass of an electron

    is the velocity the only thing holding me back from the right answer?
    or is there something else that's wrong?
  2. jcsd
  3. Sep 10, 2010 #2


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    Homework Helper
    Gold Member

    Yes, it is not correct. What you've done is assumed the electron moved from one plate to the other at a constant velocity.

    But it doesn't do that. It experiences a constant force, meaning it travels from one plate to the other under constant acceleration. Use your kinematics equations for constant acceleration. (You have one already listed in your relevant equations, but you'll need to use at least one other one that incorporates time).
    Yeah, I think your simplified equation is correct for this particular problem. Just find the correct velocity. :smile:
    Last edited: Sep 10, 2010
  4. Sep 10, 2010 #3
    thank you X)
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