# Electric Field (velocity of a particle)

1. Mar 12, 2007

### suspenc3

1. The problem statement, all variables and given/known data
At some instant in the velocity components of an electron moving between two charged parallel plates are $$v_x=1.5x10^5m/s$$ and $$v_y=3.0x10^3m/s$$. Suppose that the electric field between the plates is given by $$\vec{E}=(120N/C)j$$.

a)what is the acceleration of the electron?

b)what will be the velocity of the electron after its x coordinate has changed by 2.0cm?

2. Relevant equations
$$F=ma$$
$$\frac{F}{m}=a_y$$
$$a_y= \frac{q\vec{E}}{m}$$?

3. The attempt at a solution
Do I just have to sub in the values for part a?

Last edited: Mar 12, 2007
2. Mar 12, 2007

### Dick

Yes. I think you have everything you need. Now get started!

3. Mar 12, 2007

### suspenc3

so $$a_y = \frac{(1.6x10^{-19}C)((120N/C)j)}{9.1x10^{-31}Kg}$$
$$a_y=(2.108x10^13m/s^2)j$$?

4. Mar 12, 2007

### Dick

Careful. What sign is the charge on an electron?

5. Mar 12, 2007

### suspenc3

righht, negative, so does that mean that it is accelerating downwards?

6. Mar 12, 2007

### Dick

Don't you believe your equations?

7. Mar 12, 2007

### suspenc3

I suppose...is there a horizontal component of the acceleration?

8. Mar 12, 2007

### Dick

Is there a horizontal component of the field?

9. Mar 12, 2007

### suspenc3

thats what I figured, just making sure.

10. Mar 12, 2007

### suspenc3

and for part b, would I use :$$\vec{E} = k Q / r2$$?

11. Mar 12, 2007

### Dick

No, now that you have the accelerations just use kinematics. How long does it take the electron to go 2cm horizontally? Change in velocity=acceleration*time, etc, etc.

12. Mar 12, 2007

### suspenc3

pardon my stupidness, but Ive always been bad at this kinematic stuff.
Once I find the change in velocity what do I do?

13. Mar 12, 2007

### Dick

This is gonna make you really feel dumb, but you asked for it. Add the change in the velocity to the initial velocity to get the final velocity????

14. Mar 12, 2007

### suspenc3

hahaha, its just one of those days....

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