# Velocity of Electron in an electric field

1. Nov 8, 2007

### evanist00

1. The problem statement, all variables and given/known data
At some instant the velocity components of an electron moving between two charged parallel plates are vx = 1.6 x 105 m/s and vy = 3.8 x 103 m/s. Suppose the electric field between the plates is given by (120 N/C). In unit-vector notation, what are (a) the electron's acceleration in that field and (b) the electron's velocity when its x coordinate has changed by 1.4 cm?

2. Relevant equations

E = F / q

F = m*a

Vf = Vi + at

3. The attempt at a solution

First, I found the force by multiplying the electric field by the charge of the electron. I divided that force by the mass of the electron to find the acceleration and go it wrong. What should I do?

2. Nov 8, 2007

### Dick

You should show us what you actually did. What numbers did you use and what did you get? It's hard to guess. You do know the force will be in the z direction, yes?

3. Nov 8, 2007

4. Nov 8, 2007

### evanist00

I start with F = E q to find the force. Then I divide the force by the mass because of a=F/m. So the whole thing is as follows:

= (120) (1.602 * 10^-19) / (9.109 * 10^-31) = 2.11044* 10^13

Does this part make sense?

5. Nov 8, 2007

### Dick

That makes a lot of sense, if you mean m/sec^2. Put units on numbers please.

6. Nov 8, 2007

### evanist00

It'd doesn't work though for some reason. I have no idea what I'm doing wrong

7. Nov 8, 2007

### Dick

What do you mean it doesn't work? Your answer is correct to like five decimal places. At least I get the same thing. Did you put in the direction in terms of the direction vector k? The question asks you to.

8. Feb 4, 2008

### mussgo

im having the same trouble with the acceleration i did the same proccess he did and my webassign says its wrong..

9. Feb 4, 2008

### mussgo

my numbers are Vx= 1.2 x 10 ^5 Vy=1.7x 10^3 and E= (105 N/C) j

first i tried to get the acceleration with the equation a = eE/m

1.602x10^-19 * 105 / 9.11x10^-31

and i get 1.844 x10^13 i think i did the math right but still it says its wrong...

anyy ideas what i might of done wrong?

10. Feb 4, 2008

### mussgo

nvm it was just the negative sign i was missing n.n