In the figure below, a uniform, upward electric field E of magnitude 2.00 times
ID: 2128008 • Letter: I
Question
In the figure below, a uniform, upward electric field E of magnitude 2.00 times 103 N/C has been set up between two horizontal plates by charging the lower plate positively and the upper plate negatively. The plates have length L = 11 cm and separation d = 1.9 cm. An electron is then shot between the plates from the left edge of the lower plate. The initial velocity v of the electron makes an angle ? = 59 degree with the lower plate and has a magnitude of 6.10 times 106 m/s. Which, if any plate will the electron hit? How far horizontally from the left edge will the electron strike?Explanation / Answer
Force on the electron = E*q = 2*10^3 * 1.6*106-19 = 3.2*10^-16 N
Acceleration of electron downwards = F/m = 3.518*10^14 m/s^2
Now this is similar to a projectile, with g = 3.518*10^14 m/s^2
1. Maximum height = u^2 sin^2x /2g
= 0.0388 m = 3.88 cm...
So, it hits the upper plate...
2. Lets find the time taken to hit the upper plate...
h = usinx *t-1/2gt^2
0.019 = 6.1*10^6 *sin59 *t - 1/2 * 3.518*10^14 * t^2
1.759*10^14 t^2 -5.22*10^6 t + 0.019 = 0
t = 4.248*10^-9 s
Horizontal distance traveled = ucos59 * t
= 0.0133 m = 1.33 cm