An electron that has a horizontal initial speed of 4.50 106 m/s enters a region that has an electric field of 120 V/m pointing vertically upward.
(a)How long does the electron take to travel 9 cm in the horizontal direction?
s
(b) In this same time, what is the vertical displacement of the electron? Treat up as positive.
(a) Divide 0.09 meters by the horizontal speed component, which remains constant.
t = 0.09 m/4.5*10^6 m/s = ____ s
(b) First compute the acceleration in the vertical direction,
a = e*E/m
e and m are the electron charge and mass. Look them up.
The electron will accelerate down due to its negative charge in a positive E field.
The distance travelled in time t is
Y = (1/2) a t^2