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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.

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Im really confused with questions asking for vertical deflection...im 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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