Electric Field (velocity of a particle)

[suspenc3]

Homework Statement

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?

Homework Equations

$$F=ma$$
$$\frac{F}{m}=a_y$$
$$a_y= \frac{q\vec{E}}{m}$$?

The Attempt at a Solution

Do I just have to sub in the values for part a?

 

[Dick]

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

 

[suspenc3]

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

 

[Dick]

Careful. What sign is the charge on an electron?

 

[suspenc3]

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

 

[Dick]

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

Don't you believe your equations?

 

[suspenc3]

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

 

[Dick]

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

Is there a horizontal component of the field?

 

[suspenc3]

thats what I figured, just making sure.

 

[suspenc3]

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

 

[Dick]

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

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.

 

[suspenc3]

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

 

[Dick]

This is going to 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?

 

[suspenc3]

hahaha, its just one of those days...