Vertical displacement of electron

[willson.slp]

Homework Statement

An electron with a speed of 1.9 × 107 m/s moves horizontally into a region where a constant vertical force of 4.3 × 10-16 N acts on it. The mass of the electron is 9.11 × 10-31 kg. Determine the vertical distance the electron is deflected during the time it has moved 34 mm horizontally.
(btw the answer is in meters)

Homework Equations

F=ma
a=F/m

x-x.o=v.ot+(1/2)a*t^2
y-y.o=v.o yt-(1/2)at^t

The Attempt at a Solution

I've gotton completely lost on this. First I solved for acceleration and got 4.72E14 m/s^2. Next I used that to find time using the horizontal displacement formula and got t=1.20027E-8. When i plugged all that into find the vertical displacement I got .034meters which was incorrect. I ve tried a few other ways but to no avail. Any help would be great.

Thanks

 

[ideasrule]

Your method is correct, but t is not equal to 1.20027E-8 s. What value did you use for "a" in the horizontal displacement formula? It should be 0 because there's no horizontal force.

 

[willson.slp]

When I used a=F/m I got (4.3E-16)/9.11E-31) which equals 4.72E14

when I try a=0 in the horzizontal displacement formula, I got t= .000002

Do I use that in the vertical diplacement formula for t and then 4.72E14 for a?

 

[Redbelly98]

When I used a=F/m I got (4.3E-16)/9.11E-31) which equals 4.72E14

when I try a=0 in the horzizontal displacement formula, I got t= .000002

There are two problems with that answer:
1. Did you use 34 m or 34 mm for the horizontal distance?
2. Your answer has just 1 significant figure, but it should have 2 significant figures.

Do I use that in the vertical diplacement formula for t and then 4.72E14 for a?

Yes, once you get the correct value of t, use it and 4.72e14 m/s². Important: what is the vertical component of the initial velocity?