- #1
YamiBustamante
- 17
- 0
Homework Statement
[/B]
An electron is launched at a 45∘ angle and a speed of 5.0×10^6 m/s from the positive plate of the parallel-plate capacitor shown in the figure (Figure 1) . The electron lands 4.0 cm away.
a)
What is the electric field strength inside the capacitor?
b)
What is the smallest possible spacing between the plates?
Homework Equations
F = qE
Equations used for Projectile Motion
The Attempt at a Solution
For Part A[/B]
Okay so first I thought that I may need the acceleration in the y-direction to use F = qE, so I separated the velocity into separate components.
Vx = 5.0×10^6 m/s * cos(45)
Vy = 5.0×10^6 m/s * sin(45)
Then I found the time.
0.04m = (5.0×10^6 m/s * cos(45))*t
t = 1.131*10^-8 seconds
Then I proceeded to finding the acceleration with the time I found by using the highest point reached (0 m/s)
0 m/s = 5.0×10^6 m/s * sin(45) + a (1.131*10^-8 sec)
a = -2.69 * 10^4 m/s^2
After that I found the force on the electron using its mass, 9.10938356 × 10^-31 kilograms
F = (9.10938356 × 10^-31 kg)*(-2.69 * 10^4 m/s^2)
F= -2.459*10^-16 N
Then used that to find E in F= qE using the charge of the electron 1.6*10^-19 C
2.459*10^-16 N = (1.6*10^-19 C)*(E)
E = -1536.57 N/C
But it's wrong. Can someone tell me what I did wrong?For Part B.
I tried calculating the amount of distance the electron traveled in the vertical direction.
0^2 = (5.0×10^6 m/s * sin(45))^2+2(-2.69 * 10^4 m/s^2)*y
y= 0.023 m
But it's wrong. I think I might have misunderstood the question. How do I find the answer?Any help is much appreciated!