Electron released into an electric field from rest

[Adam21197]

Homework Statement

An electron is released from rest in a weak electric field given by

= -2.70

10-10 N/C [PLAIN]http://www.webassign.net/images/lowercase/jhatbold.svg. After the electron has traveled a vertical distance of 1.4µm, what is its speed? (Do not neglect the gravitational force on the electron.)

Homework Equations

I know that F_E = qE, and F_q = mg, as well as W_E= qEh, and W_G = mgh
also, v²=v₀²+2aΔy
but I'm not sure how I am supposed to actually combine all of these equations to get the answer

The Attempt at a Solution

If I find the accelerations from gravity and the electrical field I get:
F_E=qE=1.602*10^-19*-2.4*10^-10
a_G=9.81
a=F_E/m+9.81 = -52.01
v²=v₀²+2aΔy, so v²=2(-52.01)(1.4*10^-6)
v = 0.01405, which isn't correct. I didn't use the work equations because we haven't talked about work at all in class, so I figured we wouldn't need those equations.

 

[gneill]

Hi Adam21197, Welcome to Physics Forums!

You don't state what the orientation of the electric field is. We might guess that it is oriented vertically and that it is directed downwards, but it would be nice to have confirmation. Perhaps there was a figure accompanying the problem? I see a "broken" image link in your problem statement. It will be quite a different result if, for example, E is directed parallel to the x-axis.

Your problem statement says that $E = -2.70 \times 10^{-10}~N/C$, but you've used $-2.40 \times 10^{-10}~N/C$ in your calculation. Which is correct?