An electron has an initial velocity of 9.00 106 m/s in the x direction. It enter

Question

An electron has an initial velocity of 9.00 106 m/s in the x direction. It enters a uniform electric field = (400 N/C) , which is in the y direction.
(a) Find the acceleration of the electron. answer in (m/s2)

(b) How long does it take for the electron to travel 10 cm in the x direction in the field? Answer in (µs)

(c) By how much and in what direction is the electron deflected after traveling 10 cm in the x direction in the field? (To indicate direction, use negative for down and positive for upward.) answer in (cm)

Explanation / Answer

a) F=ma = Eq (m=mass, a=acceleration, E=electric Field, q=charge of electron) ma = Eq (9.1 x 10^-31) a = (400)(1.602 x 10^-19) a = 7.04 x 10^13 m/s b) Its speed in the x direction is 9.00 x 10^6 m/s. There is no acceleration on the x axis, so velocity = distance/time time = distance / velocity t = .10 m / 9.00 x 10^6 = 1.11 x 10^-8 seconds, or 11.1 µs c) Now plug in the acceleration into a kinematic equation to get total distance traveled along y axis x=xi + vt + .5at^2 (x=final position, xi = initial position, v=initial velocity, t=time, a=acceleration) x = 0 + 0 + (.5)(7.04 x 10^13)(1.11 x 10^-8)^2 = .004 Now use trig to direction tan x = .004 / .01 x = 23.4 degrees Use pythagorean theorem to get total displacement (.004)^2 + (.01)^2 = x^2 x = .0109 meters, or 1.1 cm

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