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Maximum initial speed the electron can have without hitting the negative plate

  1. Jan 31, 2012 #1
    Two parallel plates are 2.0 cm apart and the electric field strength between
    them is 1.0 × 10^4 N/C. An electron is launched at a 45 degree angle from the
    positive plate. What is the maximum initial speed v0 the electron can have
    without hitting the negative plate?

    i think the answer is 1.3x10^7 m/s. can someone confirm this?
     
  2. jcsd
  3. Jan 31, 2012 #2

    berkeman

    User Avatar

    Staff: Mentor

    It would be much easier for us to confirm it if you posted your work for us to see. Remember, that's part of the PF rules -- posting your attempt at the solution.
     
  4. Jan 31, 2012 #3
    The acceleration = F/m = E*q/m

    So at y max vy = 0 so

    vy^2 = vy0^2 - 2*E*q/m*ymax

    So vy0 = sqrt(2*E*q/m*ymax) = sqrt(2*1.20x10^4*1.60x10^-19/9.11x10^-31… = 9.18x10^6m/s

    So v0 = vy0/sin(45) = 9.18x10^6m/s/sin(45) = 1.30x10^7m/s
     
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