# Electric Fields and Electric Potential

1. Jul 20, 2017

### gbajwa

1. The problem statement, all variables and given/known data
An Electron enters a parallel plate apparatus that is 8cm long and 4cm wide. The electron has a horizontal speed of 6*10^7 m/s. The potential difference between the plates is 6*10^2 V. Calculate the electrons velocity as it leaves the plates. (ans. 6*10^7 m/s [E 3.3° N])

2. Relevant equations
ΔE=qΔV
V=kq/r
E=kq/r2
θ = Tan-1 = (Vfy/Vfx)
ay=qE/m
ΔT=L/vi
Vf = (Vfy2 + Vfx2)1/2

3. The attempt at a solution
I FOUND OUT HOW TO DO

Electric Field Magnitude --> E

E = -
ΔV/ΔD
E = -600V/0.04m
E = -1500 V/m

Vertical Acceleration:
ay = (qE)/m
ay = ((-1.6022*10-19C)(-1500V/m))/0.08m
ay=2.64098901*1014m/s2

Time to reach end of apparatus:
vi = L/Δt
Δt = L/vi
Δt = 0.08m/6*107m/s
Δt = 4/3*10-9s

Final Vertical Velocity:
Vfy = AyΔt
Vfy = (2.64098901*1014 m/s2)(4/3*10-9s)
Vfy = 352131.868m/s

Vf = (V2fx+V2fy)0.5
Vf = 60001033.2982m/s

Direction of speed:
θ = Tan-1(Vfy/Vfx)
θ = 3.34°

Therefore, final velocity of electron as it exits apparatus is 60001033.2982m/s [E 3.34° N]

Last edited: Jul 20, 2017
2. Jul 20, 2017

### NFuller

This is a step in the right direction. With this, what is the force on the electron? Once you have the force, what is the acceleration?

3. Jul 22, 2017

### rude man

Since you're computing to 9-12 significant digits you'll have to correct for relativistic kinematics!