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Determining the electron speed in parallel plates

  1. Jan 27, 2013 #1
    1. The problem statement, all variables and given/known data

    Two parallel plates labelled W (negative) and X (positive) are separated by 5.2 cm. The electric potential between the plates is 150 V. An electron starts from rest at time tw and reaches plate X at time tx. The electron continues through an opening in plate X and reaches point P at time tP. Determine the speed at tx and tP

    e = -1.6 x 10-19 C
    me = 9.1 x 10-31

    2. Relevant equations

    V = kq1 / r
    ΔV = ΔE/q
    ε = ΔV / r
    Ek = 0.5mv2

    3. The attempt at a solution

    I'm not sure how to determine v at tw and tx, but I did find the field strength: ε = 2884.6. We also know the electron starts from rest, so we could use the conservation of energy law to generate the following eq'n.

    v2 = 2m-1kq1q2(ra-2 - rb2). But we don't have a "q2" so I'm quite lost. Any help is appreciated!

    God bless and thanks!
     
  2. jcsd
  3. Jan 27, 2013 #2

    gneill

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    Staff: Mentor

    The Volt is a name given to the unit combination Joules/Coulomb :wink:

    You should be able to find (in your notes or text) an expression for the work done on a charge falling through a potential difference. The work done shows up as kinetic energy...
     
  4. Jan 27, 2013 #3
    You can start by finding the acceleration of the electron.
    F = ma = qE
    a = qE/m
    Then it's a kinematics problem
    vf^2 = 2*a*x Thats good for finding Vx
     
  5. Jan 27, 2013 #4
    I used your equation (not on my eq'n sheet), a = qE/m, substituted Vq for E, leaving me with Vq˄2 / m. I found acc. to be 3.376 x 10^-6.

    Plugged that into eq'n v22 = v12 + 2ad,

    which reduces to v2 = sqr(2ad), and I got a velocity (vx) of 5.93 x 10^-4 m/s which seems really slow to me.
     
  6. Jan 27, 2013 #5

    gneill

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    Staff: Mentor

    Why would you substitute Vq for E? E here would be the electric field strength (units of [N]/[C] or [V]/[m]). You calculated the electric field strength previously, calling it ε.

    However, a more slick approach is to use q*ΔV, with q being the charge on the electron and ΔV the change in electric potential (the 150V). The units are then [C][J]/[C] → [J], which is energy. That's the work done on the electron as it falls through the field...
     
  7. Jan 27, 2013 #6
    Ok that makes more sense. When you wrote E, I figured you meant energy, not ε. Tak.
     
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