# Homework Help: Potential difference to stop an electron question

1. Feb 25, 2009

### cwesto

1. The problem statement, all variables and given/known data

An electron with an initial speed of 4.70×105m/s is brought to rest by an electric field.
What was the potential difference that stopped the electron?

Unknown:
potential difference or $$\Delta$$V

Known:
vo=initial speed
m=mass of electron
q=charge of electron

2. Relevant equations

$$\Delta$$Vq+$$\frac{1}{2}$$*mv2=$$\Delta$$Vq

3. The attempt at a solution

I figure basically plug and chug but my equation doesn't seem right. I sure it's not. Basically I don't know where to go from here.

Last edited: Feb 25, 2009
2. Feb 25, 2009

### Kruum

Welcome to PF!

Two questions. Are you sure this is the whole question? With these values, we could use almost any kind of electric field to stop the electron. Where'd you get that equation?

3. Feb 25, 2009

### Sonolum

Well, I'll take a stab at helping out...

You need to stop a moving electron, right? Does that mean you will have an acceleration (or deceleration, same thing)?

If you have an acceleration, there's a force. And the force in this case is related to the electric field, right?

So you need an electric field such that it creates a force capable of stopping your electron.

If you can find the electric field, you *should* be able to relate the electric field to the potential energy, relate that to the electric potential, and then find a potential difference...

But then, I haven't worked a problem like this in awhile and I may not be solving it with the tools that you already have... Have you done line integrals and all that stuff yet?

4. Feb 25, 2009

### Kruum

Sonolum, if the given problem is the whole problem, even the smallest force is enough to stop the electron eventually. Why go for the trouble of using line integrals?

5. Feb 25, 2009

### Sonolum

I don't really see them so much as trouble, myself... I was mostly trying to get a feel for the level of this problem and help out.

6. Feb 25, 2009

### cwesto

I also told the electron will move to a region of lower potential. I got the equation from a previous question that is simliar posted on PF last year. That's all I have. The link to the previous question is below.

7. Feb 25, 2009

### Sonolum

Ah, see, in that problem we know how much distance we have to stop the electron. Here, we don't know that - do we have to stop the electron in a certain time? In a certain amount of distance?

Kruum's 100% right, I didn't realize it at first, but yeah, any electric field that opposes the electron's motion will *eventually* stop it. It kinda depends on where/when you want to stop it to get anything specific.

8. Feb 25, 2009

### LowlyPion

I think you need to understand that the kinetic energy is to be absorbed by work from some source. In this case measuring the work can best be done by observing that changing the voltage of a charge can be determined by:

W = q*ΔV and that would equal ½mv² in order to stop it.

Hence won't the needed ΔV = ½mv²/q ?

9. Feb 25, 2009

### cwesto

So in other words I need to know something else?

10. Feb 25, 2009

### LowlyPion

Like what?

11. Feb 25, 2009

### Sonolum

I dunno, Pion's method seems a heckuva lot more eloquent and to the point... But I pretty sure it's (Delta)V = - W/q...

12. Feb 25, 2009

### Kruum

Oh crap! I should read the question a bit more carefully! Sorry for the confusion cwesto!

13. Feb 25, 2009

### cwesto

The distance in which they want the electron to stop.

14. Feb 25, 2009

### LowlyPion

Happily q is (-) electron which makes the ΔV (-).

15. Feb 25, 2009

### LowlyPion

No distance required.

If the E-Field is 1000 V/m or 10 V/m the distance would change of course but not the ΔV required.

16. Feb 25, 2009

### cwesto

Wouldn't "ΔV = ½mv²/q" give me ΔV while the speed is 4.70×105m/s?
I want to know the ΔV when the speed is 0.

17. Feb 25, 2009

### cwesto

I'll ask my professor then post. I have class now. Thanks though.

18. Feb 25, 2009

### Sonolum

Ah, but see, that's why there was mention of changing the kinetic energy into work, right? The energy had to go*somewhere*, and the energies should be equivalent before and after, right?

19. Feb 25, 2009

Oh yeah!
ΔV = ½mv²/q