# Electric Field (velocity of a particle)

## 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?

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## Answers and Replies

Dick
Science Advisor
Homework Helper
Yes. I think you have everything you need. Now get started!

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

Dick
Science Advisor
Homework Helper
Careful. What sign is the charge on an electron?

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

Dick
Science Advisor
Homework Helper
righht, negative, so does that mean that it is accelerating downwards?

Don't you believe your equations?

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

Dick
Science Advisor
Homework Helper
I suppose...is there a horizontal component of the acceleration?

Is there a horizontal component of the field?

thats what I figured, just making sure.

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

Dick
Science Advisor
Homework Helper
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.

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

Dick
Science Advisor
Homework Helper
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????

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