# Homework Help: Calculating distance of an electron moving between 2 plates

1. Jan 15, 2012

### Combine

1. The problem statement, all variables and given/known data
Electrons are accelerated from REST by a potential difference of 1.0 x 10^3 volts
They pass midway between two flat parallel plates 0.15m long and 0.030m apart. The electric field between the plates has a magnitude of 2.0 x 10^3 N/C. Determine how far an electron will be deflected (x) from it's original horizontal path by the time that it reaches the other end of the plates

I can't use most of the formulas because there is constant acceleration involved after it reaches the 0.35 m mark, thus I can't get a x nor a y and the acceleration is just in the X axis,
I'm really stuck!

2. Relevant equations
e = 1.602 x 10^-19
m = 9.11 x 10^-11
k = 9.0 x 10^9 N M^2/C^2

ε = V/r
ε = Fe/q
d = v2t - 1/2at^2
d = v1t - 1/2at^2

3. The attempt at a solution
First find the acceleration
F = ma
qε = ma
(1.602x10^-19)(2000) = 9.11 x 10^-31a
a = 3.52 x 10^14

next find the distance between starting position and the end of the plates
ε = V/r
2000 = 1000/r
r = 0.50

after that find the distance between starting position and the start of the plates
0.50 - 0.15 = 0.35m

next you find the final velocity (the second before the electron reach's the plates)
Vf^2 = Vi^2 + 2(3.52x10^24)(0.35)
vf = 1.6 x 10^7

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Last edited: Jan 15, 2012
2. Jan 15, 2012

### technician

The first thing to do is calculate the velocity of electrons accelerated through 1000V
use eV = 0.5mv^2
These electrons then travel into the space betwen the parallel plates where they experience a vertical force.
Hope this helps

3. Jan 15, 2012

### Combine

Well it only kinda helped, I am still unsure if that is the answer.

eV = 1/2mv^2
1.602 x 10^-16 = 4.555 x x10^-31V^2

v = 1.9 x 10^7

t = 0.15/1.9 x 10^7
t = 7.0 x 10^-9

d = V1t + 1/2at^2
d = 0 + 1/2(3.52 x 10^14)(7.0 x 10^-9)
d = 0.008624

The answer makes logical sense, because we know that the space between the plates is 0.030m

and it's only deflected a bit so it would make sense that the vertical deflection would be 0.008624m

can anyone confirm this?