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An electron traveling at a speed of 5.50*10^4 m/s enter a region midway between two parallel...

An electron traveling at a speed of 5.50*10^4 m/s enter a region midway between two parallel charged plates. The magnitude of the E-field between the plates is 75.0 N/C. The distance between the plates is d=3.00mm.

A.Find the magnitude of the acceleration of the electron while it is between the plates

B.What is the speed of the electron as it hits one of the plates?
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Answer #1

F = qE = 1.6 x 10^-19 x 75 = 1.2 x 10^-17 N

acceleration = F/m = 1.318 x 10^13 m/s^2

time to reach the other plate =

S = .5 x a x t^2

3 x 10^-3 = .5 x 1.318 x 10^13 x t^2

t = 2.13 x 10^-8 s

velocity perpendicular to the plate = at = 1.318 x 10^13 x 2.13 x 10^-8

v = 2.8 x 10^5 m/s

total velocity = root (2.8 x 10^5^2 + 5.5 x 10^4 ^2)

= 2.8 x 10^6 m/s

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