An electron that has a horizontal initial speed of 4.50 106 m/s enters a region

Question

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 20 cm in the horizontal direction?
____7.18E6_________s

(b) In this same time, what is the vertical displacement of the electron? Treat up as positive.
________________cm

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I am having trouble the second portion of this problem. Heres my work so far:

I calculated the force by F= (1.60E-19J)(120 V/m)=1.92E-17N

Then I calculated the acceleration: a= (1.92E-17N)/(9.11E-31kg)= 2.1E13 m/s2 upward

Using the equation: Y=Voyt + (1/2)ayt^2

Y= (0) + .5(2.1E13 m/s^2)(4.44E-8)^2

Answer: 2.6 cm is wrong

Explanation / Answer

Along the horizontal direction we have x = vx * t ..........1 Along the vertical direction we have y = vy * t + 0.5 ay t^2 = 0 + 0.5 ay t^2 = 0.5 (qE/m) t^2 .......2 From 1 and 2 we have y = 0.5 (qE/m) * (x / vx)^2 = 0.5 * ( 1.6 x 10-19 * 120 / 9.1 x 10-31) * ( 0.2 / 4.5 x 10^6) ^2 = 0.02083 m = 2.083 cm

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