In the figure below, an election is shot at an initial speed v_0 = 2.00 * 10^6 m
ID: 2000528 • Letter: I
Question
In the figure below, an election is shot at an initial speed v_0 = 2.00 * 10^6 m/s, at an angle theta_0 = 30.0 degree from an x axis. It moves through a uniform electric field E^rightarrow = (5.00 N/C)j. A screen for detecting electrons is positioned parallel to the y axis, at a distance of x = 4.20m. In unit-vector notation/ what is the velocity of the electron when it hits the screen? Do you realize that this problem is like the projectile problems of Chapter 4? How is electric field related to electrostatic force? Do you see that this particle is negatively charged? How does that affect the direction of the force? How is force related to acceleration? Do you recall how use the constant acceleration equation of chapter 2?Explanation / Answer
t = x/(Vo cos )
t = 4.2/(2 E6 cos 30°)
t = 2.424 E-6 s
Vx = Vo cos
Vx = (2 E6) cos 30° = 17.3 E5 m/s
a = F/m = q E/m = (-1.6 E-19)(5)/(9.11 E-31) = -8.78 E11 m/s²
Vy = Vo sin + a t
Vy = (2 E6) sin 30° + (-8.78 E11)(2.424 E-6)
Vy = -11.28 E5 m/s