Finding the Electric Field of a Parallel Plate Capacitor

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1. Jul 26, 2016

arileah

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

F = ma = qE

v = d/t

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

x = 0.04 m

3. The attempt at a solution

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!

2. Jul 26, 2016

BvU

Seems strange. Doesn't have the same dimensions left and right ....

3. Jul 26, 2016

I think your error is the equation t=x/cos(45). It needs a $v_o$ in it.

4. Jul 26, 2016

arileah

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)