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Finding the Electric Field of a Parallel Plate Capacitor

  • #1
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Homework Statement



An electron is launched at a 45∘ angle and a speed of Vo = 5.0×10^6 m/s from the positive plate of the parallel-plate capacitor towards the negative plate. The electron lands 4.0 cm away.

Homework Equations



F = ma = qE

v = d/t

s = vt -(1/2)at^2

x = 0.04 m

The Attempt at a Solution


[/B]
We can find t by using trigonometry:

t = x/(cos45)

yf - yi = 0 = Vo * sin(45) * t - (1/2)at^2

Vo*sin(45) = (1/2)at

using F = ma = qE,

a = qE/m

so,

Vo*sin(45) = t * qE/(2m)

and using t =x/cos(45)

Solving for E,

E = (2m * Vo * sin(45) * cos(45))/(qx)

but cos(45)*sin(45) = 1/2

Therefore,

E = (Vo * m)/(q*x)

At this point I plug in my constants:

Vo = 5*10^6 m/s
m = 9.11 * 10^-31
q = 1.6 * 10^-19
x = 0.04

However I am getting a wrong answer. Could anybody point out where it is I've gone wrong? Thanks!
 

Answers and Replies

  • #2
BvU
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t = x/(cos45)
Seems strange. Doesn't have the same dimensions left and right ....:rolleyes:
 
  • #3
Charles Link
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I think your error is the equation t=x/cos(45). It needs a ## v_o ## in it.
 
  • #4
8
2
I think your error is the equation t=x/cos(45). It needs a ## v_o ## in it.
Of course -_- . Thanks! Silly mistake.

For anyone else who wants to use this thread:

My mistake was writing t=x/cos(45) instead of t=x/(cos(45)*Vo)

The end formula would then be:

E = (m* Vo^2)/(q*x)
 
  • Like
Likes Charles Link and BvU

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