- #1
jpierce879
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Homework Statement
In the figure below, an electron is shot at an initial speed of v0 = 2.25 × 106 m/s, at angle θ0 = 20.0o from an x axis. It moves through a uniform electric field = (5.04 N/C) . A screen for detecting electrons is positioned parallel to the y axis, at distance x = 3.23 m. What is (a) the x component and (b) the y component of the velocity of the electron when it hits the screen?
Homework Equations
E = F/q; F = E*q
F = m*a; a = F/m = E*q/m
t = Dx/Vx
Vy = Vo + Ay*t
The Attempt at a Solution
It would probably be easier to look at the attached image to see my work, but I'll try explaining my logic here anyway. The x-component of the velocity is the same as the initial x-component since there is no horizontal acceleration - it turns out to be Vo*cos(20°).
The second part, however, I've had a lot of trouble on. I've tried it multiple times so it doesn't seem to be a calculator error. My logic here is that since fields point from + to - and the magnitude of field strength at any point doesn't depend on the distance from whatever's causing the field, the electron (being negatively charged) accelerates in the -y direction. My work is shown in the attached image. I was just wondering if my logic is correct, and if so then is it just a calculation error on my part?