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October 24, 2014

October 24, 2014

Posted by **Lauren** on Friday, February 26, 2010 at 3:16pm.

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 -
**drwls**, Friday, February 26, 2010 at 4:25pmThe 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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