1. Jul 21, 2009

### DPavel

1. The problem statement, all variables and given/known data
An electron with a speed of 5 E6 meter/second moves into a uniform electric field with a value of 1,000 volt/meter. The field is parallel to the electron's motion. Calculate the distance it travels before being brought to rest.

2. Relevant equations

3. The attempt at a solution

2. Jul 21, 2009

### negitron

You've provided the problem, but no attempt at a solution or even an attempt at stating relevant equations. The harder you try, the more we'll help you. Zero effort gets zero help.

3. Jul 21, 2009

### DPavel

wow that is very rude! the whole reason i posted it is because i am not sure where to begin. i am a PT major just trying to make it through this class. sheesh. next time, keep your mean comments to yourself!

4. Jul 21, 2009

### ralilu

multiply the charge of an electron by the electric field, this should give you a resultant force. divide the force by the mass of an electron to get the decelleration. so now u got the initial velocity, the decelleration and the final velocity (zero) which is enough to calculate the distance using: final velocity squared = initial velocity squared + 2 x acelleration x distance

5. Jul 21, 2009

### ralilu

negitron has a point. u must atleast attempt problems on your own if you want any hope of surviving the course. not just post it on the web coz u dnt feel like trying

6. Jul 21, 2009

### DPavel

but then the answer would be zero...would it not? because to find the distance you would have to divide that answer by zero??

7. Jul 21, 2009

### Pengwuino

You should not be surprised by the response if you read the forum rules. We do not answer your questions without any work, we do not attempt to guess at what you're thinking or asking, we provide support to students who has attempted problems and are having trouble. The template you used should have hinted at what you needed to provide.

For this question, do you know the force acted on a charge in a constant electric field?

8. Jul 21, 2009

### DPavel

i believe you have it all wrong. i am just trying to study for my final, and this is a chapter i have trouble with. and ive been searching for an answer for a few hours on this. anyway, i appreciate the help that you have given me. and really, to say that i am just posting it on the web because i do not feel like trying is...well, wrong. it was kind of a last resort. summer class for physics II is trying to teach a 15 chapters in 6 weeks with a book that I could not afford. So yes, I do use the internet to search for equations and this is the one problem that I cannot locate.

again, thanks for the help.

9. Jul 21, 2009

### DPavel

the K constant? 9 E 9 i believe..

10. Jul 21, 2009

### Dick

You don't need that. Use conservation of energy. How much initial kinetic energy does the electron have? How much energy does it lose every meter it moves through the field? Do you have a formula for any of these things?

11. Jul 21, 2009

### Pengwuino

Actually in this question, you do not need the coulomb constant. If you know the electron is in a constant electric field, E, it experiences a force of F = qE where q is the charge or in this case, -1.602e-19 C. We also know that F = ma. Equating the two, you see that f = qE/m. At this point, we can use our kinematics. Remember, the force should be negative since it's an electron with negative charge and a positive electric field.

$$\begin{array}{l} x_f = x_i + v_i t + \frac{{at^2 }}{2} \\ v_f = v_i + at \\ \end{array}$$

So what we're looking for is the final position with this constant acceleration, a. We know that the initial position is 0 and our initial velocity is 5e6 m/s while the final velocity will be 0. Using the initial and final velocity, we can solve for 't' in the second equation. Finally, plug this value of 't' into the first equation and you can solve for $$x_f$$.

12. Jul 21, 2009

### Pengwuino

Oh yah, this is actually by far an easier method of solving it.

13. Jul 21, 2009

### Phrak

Not from what I've seen. DPavel knows how to work a crowd, hey?

14. Jul 21, 2009

### Phrak

Not from what I've seen. DPavel knows how to work a crowd. The squeaky-wheel approach to higher learning is one of the infallible methodologies. We can't fault him for not learning his lessons.

15. Jul 21, 2009

### ralilu

no the answer wouldnt be zero, coz u will subtract initial velocity squared from zero then divide it by 2 x acelleration

16. Jul 21, 2009

### ralilu

let me knw what the answer is, i am interested.

17. Jul 21, 2009

### Dick

DPavel, that's not a mean comment. It's a standard response to a post without effort. Note the "harder you try, the more we'll help you" part. Considering it's your first post, I think other people could be easier on you. Now, can you help us to help you?

Last edited: Jul 22, 2009
18. Jul 21, 2009

### Dick

"negitron has a point. u must atleast attempt problems on your own if you want any hope of surviving the course. not just post it on the web coz u dnt feel like trying"

Why don't you try? It's NOT even HARD.

19. Jul 22, 2009

### ralilu

lol. i am content with figuring out the problem. i dnt really need a numerical value. and yes it is a really simple problem.

20. Jul 22, 2009

### DPavel

you physics guys are funny. Dick, you may not think it is a hard problem, but for a girl who has only had one other class in physics and typically studies bones and muscles, it really is. Maybe you can think of a subject that you just couldnt grasp? idk. Or maybe you are just perfect in everything. Maybe you have forgotten how it was when you first learned this stuff? I feel like alot of teachers forget this. And ralilu, thanks for the help, i am going to get back down to this problem right now and i will post the answer i come up with.

21. Jul 22, 2009

### DPavel

my final answer is 0.071 meters. Thanks all for your help, it really makes some sense now! Final on Thursday.

22. Jul 22, 2009

### ralilu

cool, that sounds correct. my advice for your final is try and understand the energy and work concepts the most and you will find that you can solve almost all of the problems in your course using a simple energy equation.

e.g in this problem: work done = change in kinetic energy
work = force times distance
change in kinetic energy = (1/2)(mass)(change in velocity)
gudluck! and i struggled with muscles and bones...

23. Jul 22, 2009

### Dick

I wasn't talking to you when I said it's not a hard problem. I was talking to ralilu. Every problem is hard for somebody. Most problems are easy for somebody else. But if somebody else posts advice on a problem, I would expect them to be able to solve it, not post back saying send me the answer when you get it. Sheesh.

Last edited: Jul 22, 2009
24. Jul 23, 2009

### ralilu

if u go back a few posts you will find that i did solve the problem i just didnt plug in the numbers...
i asked her to let me knw the answer coz i wanted to knw wether or not she solved it.

25. Jul 23, 2009

### Dick

Ok. That's a good reason.