# Homework Help: Find distance traveled given electric field + vf and vi

1. Dec 2, 2015

### Shaif

1. The problem statement, all variables and given/known data
Electric field= 425 N/C
Vf = 3.25x10^5 m/S
VI=0
The charge is an electron

2. Relevant equations
An electron gun had a uniform electric field. The electron starts from rest and is fired from the gun, reaching a speed of (vf given above). How far does the electron move relative to the electron field?

3. The attempt at a solution
Ek = .5 mv^2
Electric pot. Energy = -qed
Find intial energy at the end of the motion using .5mvf^2 then use it to solve for d by plugging it into Ee= -qed where D is the distance travelled?

2. Dec 2, 2015

### Orodruin

Staff Emeritus
Not exactly, but close. What is your reasoning in putting Ekin = Epot?

3. Dec 2, 2015

### Shaif

I'm assuming that the vf equals the electrons energy at the end is a complete energy transfer.

But I've talked to another teacher and what I should be doing is apparently this :

Fe= we
Fnet=ma
Fe=Fnet
Isolate for a
a=qe/m
Solve for D.
I believe this is righr

4. Dec 2, 2015

### Orodruin

Staff Emeritus
I suggest you write down exactly what you would get and exactly what your motivation is for it to hold in terms of mathematical expressions. You can use energy arguments, but the way you have presented yours makes me suspect that you have a basic flaw in your reasoning. What do you get when you try to solve the problem?

5. Dec 2, 2015

### Shaif

If the flaw is that Eeinitial =/= Ekfinal but rather Eeinitial =/= Ekfinal + Eefinal but I'm assuming Ee is pretty much 0 but I can see how there's a flaw in that reasoning. That's why I guess the second approach where I use forces to calculate the acceleration and use vf^2=vi^2 +2ad would be a lot better

6. Dec 2, 2015

### Orodruin

Staff Emeritus
The total energy is conserved, but what does that tell you about the relation between the final kinetic energy and the final potential energy? The energy argument is much cleaner and simpler than the force argument.