- #1

gbajwa

## Homework Statement

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])

## Homework Equations

ΔE=qΔV

V=kq/r

*E*=kq/r

^{2}

θ = Tan

^{-1}= (Vfy/Vfx)

ay=q

*E*/m

ΔT=L/vi

Vf = (Vfy

^{2}+ Vfx

^{2})

^{1/2}

## The Attempt at a Solution

I FOUND OUT HOW TO DO

Electric Field Magnitude -->

*E*

E = -ΔV/ΔD

E = -

*E*= -600V/0.04m

*E*= -1500 V/m

Vertical Acceleration:

ay = (q

*E)*/m

ay = ((-1.6022*10

^{-19}C)(-1500V/m))/0.08m

ay=2.64098901*10

^{14}m/s

^{2}

Time to reach end of apparatus:

vi = L/Δt

Δt = L/vi

Δt = 0.08m/6*10

^{7}m/s

Δt = 4/3*10

^{-9}s

Final Vertical Velocity:

Vfy = AyΔt

Vfy = (2.64098901*10

^{14}m/s

^{2})(4/3*10

^{-9}s)

Vfy = 352131.868m/s

Velocity Vectors added:

Vf = (V

_{2}fx+V

_{2}fy)

^{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]

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