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An electron enters a parallel plate apparatus 10.0cm long and 2.0cm wide, moving horizontally
at 8.0 × 10^7 m/s. (top plate is positive, electron is 1.0cm from each plate..) The potential difference between the plates is 6.0 × 10^2 V. Calculate...

(a) the vertical deflection of the electron from its original path.
(b) the velocity with which the electron leaves the parallel plate apparatus.


Im really confused with questions asking for vertical not sure what formula to use..should i be using the formula for energy where U=mgh and solve for h? this is the only thing i can think of..any help would be great

  • Physics -

    The vertical motion will be towards the positively charged plate due to the E field in that direction. The acceleration rate will be
    a = F/m = e E/m = e V/(d m)
    where d is the plate separation (0.2 m), V is the voltage, and m is the electron mass.

    Once you know a, calculate how far it deflects vertically using
    y = (1/2) a t^2

    t is the time of flight going through the plates, t = 0.10 m/8*10^7 m/s

    The horizontal velcoity component will not change. The vertical velocity acquired is a * t

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