How Do Electrons Behave Between Charged Parallel Plates?

[skg94]

Homework Statement

An electron beam enters the region between two oppositely charged parallel plates near the negatively charged plate, as shown below. The plates are 0.150m long and are 0.040m apart. There is an electrical potential difference of 60.0V across the two plates which creates an electric field in the region.

The path of the electrons just touches the edge of one of the plates as the beam exits the region between the parallel plates.

---------------------- negative plate
|
|.040m 60.0V
|
----------------------- positive plate
.150m

Homework Equations

e=ΔV/ΔD
a=Fe/m

The Attempt at a Solution

1. E=v/d
60v/.040m = 1500v/m

a=fe/m=qe/m=(1.6*10^-19)(1500)/(9.11*10^-31)= 2.634467618*10^14 down

t=$\sqrt{2d/a}$ = $\sqrt{(2*.040)/2.63...*10^14}$
= 1.742603416*10^-8

so using projectile motion i made a chart and listed variables i knew

y
a=2.6...*10^14
V-intial = 0
d=.040m

x
Vi=?
d=.150m

So i found time since time is the same in these types of question if I am correct.

V=d/t
v=.150/1.74*10^-8
=8.61*10^m/s

is this right? i have no answer sheet and I am wondering if i did it right

thanks

 

[ehild]

so using projectile motion i made a chart and listed variables i knew

y
a=2.6...*10^14
V-intial = 0
d=.040m

x
Vi=?
d=.150m

So i found time since time is the same in these types of question if I am correct.

V=d/t
v=.150/1.74*10^-8
=8.61*10^?m/s

is this right? i have no answer sheet and I am wondering if i did it right

thanks

The power of 10 is missing from your result. Otherwise the solution is correct.

 

[skg94]

The power of 10 is missing from your result. Otherwise the solution is correct.

oops sorry, i got to the power of 6, ill recheck my work perphaps i rounded wrong or perphaps you might have

 

[ehild]

Power of 6 is OK. I checked your solution, it is correct.

ehild