Either straight-line kinematics or Work-Energy equations should
give the right answer.
The decelerating force is F = qE = -1.6*10^-19 * 1.18*10^4 =
-1.888*10^-15 N
so the deceleration is a = F/m = -1.888*10^-15 / 9.11*10^-31kg =
-2.0724*10^15 m/s^2
a)
Vf^2=0 = 2ad+Vo^2
so d = -Vo^2 / 2a = -(2.55*10^7)^2 / -2*2.0724*10^15 = 0.1568
mm
d = 1.586*10^-4m
b)
time to stop is t = -Vo/a = -2.55*10^7 / -2.0724*10^15 =
1.230*10^-8 s
Constants Periodic Table PartA n electron with speed 2 55* 10 m,/s straveling paralel to a...
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