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Deflection of electron

  1. Sep 11, 2006 #1
    An electron with a kinetic energy 5000.0 eV (1 eV = 1.602×10-19 J) is fired horizontally over a charged plate with surface charge density +2.0 μC m-2. Taking the positive direction to be upwards (away from plate), what is the vertical deflection of the electron after it has traveled a horizontal distance 2.0 cm?

    Here is my work:

    Find V of electron - KE = .5mV ---> 41934557 = Velocity

    time = s/V ----> .02m/41934557 --------> 4.769e^-10 = time

    electric field E = sigma/2*Eo -----> 2.0e^-6/(2*(8.85*10^-12) ---->112994 = E

    Force = q*E ------>(1.602*10^-19)(112994) ------->1.810e^-14 = F

    acceleration = F/m -------> 1.810e^-14/(9.11*10^-31) = 1.98e^16. 9.11*10^-31=mass of one proton.

    vertical deflection = d=.5at^2 --------> .00226=.5(1.98e^16)(4.769e^-10)^2

    I come up with .00226 meters, but "the man" still says I'm wrong

    Can you see anything wrong with my set up??
  2. jcsd
  3. Sep 11, 2006 #2
    Can't see anything wrong except the one that has been highlighted.
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