# Electron shot between two charged plates, does it hit?

1. May 16, 2010

### mhen333

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

A uniform, upward electric field E of magnitude 2.00x10^3 N/C has been set up between two horizontal plates by charging the lower plate positively and the upper plate negatively. The plates have length L = 10.0 cm and separation d = 2.00 cm. An electron is then shot between the plates from the left edge of the lower plate. The initial velocity Vi of the electron makes an angle theta = 45.0 degrees with the lower plate and has a magnitude of 6.00x10^6 m/s. (A) Will the electron strike one of the plates? (B) If so, which plate and how far horizontally from the left edge will the electron strike?

2. Relevant equations

$$F = ma_{y}$$

$$a_{y} = \frac{F}{m} = \frac{eE}{m}$$

3. The attempt at a solution

Okay, I think I have this. The components (X and Y) of the velocity are

$$V_{x_{0}} = (6.00 * 10^6 \frac{m}{s})(\cos {\frac{\pi}{4}}) = 4.24 * 10^6 \frac{m}{s} \hat{i}$$

and

$$V_{y_{0}} = (6.00 * 10^6 \frac{m}{s})(\sin {\frac{\pi}{4}}) = 4.24 * 10^6 \frac{m}{s} \hat{j}$$

$$a_{y} = \frac{F}{m} = \frac{eE}{m}$$

$$a_{y} = \frac{(-1.6*10^-19 C)*(2 * 10^3 N/C}{9.109 * 10^-31 kg} a_{y} = -3.5 * 10^14 \frac{m}{s^2}$$

Using substitution and solving for t:
$$t = \frac{-2 * V_{x}}{a} = \frac{-2(4.24 * 10^6 m/s}{-3.5 * 10^14 m/s^2} = 2.4 * 10^{-8} s$$

So in that length of time, the electron will travel distance:

$$\Delta X = V_{x_{o}} * t = (4.24 * 10^{6} m/s) * (2.4 * 10^{-8} s) = .10176 m$$

Since 10.176 cm is greater than the 10cm length of the plates, am I done at this point?

2. May 17, 2010

### kuruman

You are not done. You have not shown that the electron will not hit the top plate before it clears the end of the bottom plate. Calculate how high vertically the electron will rise and compare that with the plate separation.

3. May 17, 2010

### mhen333

Oh, thanks!

So then,

$$\Delta y = v_{0} * t - \frac{a * (t/2)^{2}}{2}$$

$$\Delta y = 4.24*10^6 \frac{m}{s} * t - \frac{3.5 * 10^{14} \frac{m}{s^{2}} * t}{2}$$

Setting vertical displacement equal to 0.02m, and solving for t, I get

$$t = 6.416 * 10^{-9} s$$

for the time when the electron hits.

At that point, finding the X displacement gives me

$$\Delta x = .0272m$$

So it DOES hit, and very early on. Which makes sense, since it's going ridiculously fast to begin with, and angled at 45 degrees.
Thanks for your help, much appreciated.