# Proton in a uniform electric field

1. Aug 28, 2014

### Feodalherren

1. The problem statement, all variables and given/known data
A proton is projected in the positive x direction into a region of uniform electric field E = (-5.80 ✕ 10^5) N/C at t = 0. The proton travels 6.50 cm as it comes to rest.

Find the initial velocity and the time it takes for the proton to stop.

2. Relevant equations

3. The attempt at a solution

The charge of the proton multiplied by the electric field must equal the mass of the proton times the acceleration

qE=ma

hence

a= (-5.556x10^13)

Constant accelecration, integrate with respect to time to get velocity

V= Vi - 5.556x10^13 t

X= 0 + Vi(t) - 2.2778 t^2

Since initial position = 0.

Using the position function to find the time it takes to stop:

.065m = (5.556x10^13)t^2 - (2.2778x10^13)t^2

solving for t, t= 4.8x10^-8 s

Plugging this back to the velocity equation where at t= 4.8x10^-8 s the final velocity = 0 therefore:

Vi=(5.556x10^13 (m/s^2))(4.8x10^-8 s)

Clearly not right, it's faster than the SOL.
Where am I going wrong?

ps. I don't like to memorize stuff so please don't refer me to the kinematic equations. I prefer to derive them by myself.

2. Aug 28, 2014

### Zondrina

You need to solve $v dv = a dx$ assuming that $v = v_i$ at $x = x_0 = 0$.

Or just use $v_f^2 = v_i^2 + 2a \Delta x$ to save some time.

Also that acceleration should be positive I think.

Last edited: Aug 28, 2014
3. Aug 28, 2014

### Feodalherren

That's exactly what I did? It doesn't seem to be working. I missed something along the way.

Acceleration can't be positive since it's slowing down and moving in the positive x-direction.

4. Aug 28, 2014

### Zondrina

Ahh so the electric field is in the $(- \hat i)$ direction.

You seem to be solving $a = \frac{dv}{dt}$.

5. Aug 28, 2014

### Feodalherren

Well yes, but only as an initial step. Once I have that I integrate once more to solve for the position function. I can't see anything wrong with it :/.

6. Aug 28, 2014

### Zondrina

You can find a relationship between velocity and position because you have the constant acceleration.

$$v dv = a dx$$
$$\int_{v_i}^{0} v dv = \int_{0}^{0.065} a dx$$

This lets you solve for the initial velocity. The time taken can now be found from $a = \frac{dv}{dt}$, or simply use $v_f = v_i + at$.

7. Aug 28, 2014

### Feodalherren

But that's exactly what I did....

8. Aug 28, 2014

### Zondrina

Look at the integral limits carefully and ask yourself where you went wrong.

9. Aug 28, 2014

### Feodalherren

I honestly have no clue. This is how I did all of my mechanics in Physics I. I never bothered to remember the kinematic equations as just integrating / taking derivatives worked for everything. I'm missing something fundamental.

Why can't I just integrate from acceleration to velocity to position?

10. Aug 28, 2014

### Zondrina

Everything looks correct to me too, I'm not sure then.

11. Aug 28, 2014

### ehild

What numerical value do you get for Vi? And what is the speed of light?

ehild

Vi≈3*106

12. Aug 29, 2014

### Feodalherren

Vi=(5.556x10^13 (m/s^2))(4.8x10^-8 s)= 2,666,880 m/s

...... my bad I thought the SOL was about 3 x 10^6 m/s but that's km/s.

Thanks ehild.

13. Aug 29, 2014

### ehild

No. So what is the SOL? You need to know the magnitude...
3x10^5 km/s = 3x10^8 m/s

ehild

Last edited: Aug 29, 2014
14. Aug 29, 2014

### Feodalherren

Don't worry about it, I solved it. Thanks again :).