# Proton between two plates

1. Sep 13, 2009

### exitwound

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

2. Relevant equations

Ki + Ui = Kf + Uf

3. The attempt at a solution

I need to find the velocity of the proton when it's 2.3mm's to the right of where it is now. I need to find the acceleration of the proton. I believe that I should use F=ma to do this. The only force acting on the proton is -Eq (to the left, hence the minus sign). I don't know the Electric field though.

Any help?

Edit: or, can I do this:

Ki= 0
Ui=qVi
Kf=1/2mv^2
Uf=qVf

2. Sep 13, 2009

### kuruman

You don't know the electric field but you know that it is uniform between the plates and you know the potential difference. How is E related to the potential difference in this case?

3. Sep 13, 2009

### exitwound

E=-dV/ds isn't it?

4. Sep 13, 2009

### kuruman

That's it.

5. Sep 13, 2009

### exitwound

I don't understand how to use that. I have a 20V potential difference. How does that relationship help me?

6. Sep 13, 2009

### kuruman

Integrate the expression.

$$\Delta V=-\int Eds$$

You know ΔV = 20 Volts, you know that E is constant and you know the plate separation. Can you do the integral and solve for E?

7. Sep 13, 2009

### exitwound

$$\Delta V=-\int Eds$$

$$20=-E\int ds$$

$$20=-E (2.3E-6 km)$$

E=-20/2.3E-6 = -8.695E6 V/km

8. Sep 13, 2009

### kuruman

The plate separation is 2.3 mm, not 6 km.

9. Sep 13, 2009

### exitwound

That should read 2.3x10^-6 km since the problem will ultimately ask for KM/s velocities.

10. Sep 13, 2009

### kuruman

You can convert later. For the time being, calculate the field in Volts/meter, which the conventional unit for electric field. It is less confusing this way.

11. Sep 13, 2009

### exitwound

It's all confusing no matter what units are used.

E=-20/2.3E-3 = -8.695E3 V/m then

12. Sep 13, 2009

The proton moves through a potential difference of 20V so therefore gain of PE=20e(e=charge of proton/electron).Now you can use PE gained =KE lost.This method may be a bit quicker.

13. Sep 13, 2009

### exitwound

So how would i set that up? Something like this?

$U_i + K_i = U_f + K_f$

$U_i = qV_1[/tex] [itex]K_i = 0$
$U_f = qV_2$
$K_f = 1/2mv^2$

$qV_1 + 0 = qV_2 + 1/2mv^2$

Something like this? Solve for v?

14. Sep 13, 2009

K1 is not zero.

15. Sep 13, 2009

### exitwound

Oh. Would it also be 1/2mv^2 (v being initial velocity?)

16. Sep 13, 2009

Yes.Good luck with it.

17. Sep 13, 2009

### exitwound

$U_i + K_i = U_f + K_f$

$U_i = qV_1[/tex] [itex]K_i = 1/2 mv_i^2$
$U_f = qV_2$
$K_f = 1/2mv_f^2$

$qV_1 + 1/2 mv_i^2 = qV_2 + 1/2mv^2$

The answer still doesn't make any sense. When you plug all the numbers in, the final velocity comes out to be WAY too fast.

$qV_1 + 1/2 mv_i^2 - qV_2 = 1/2mv^2$

$$(1.6x10^{-19})(-70) + (.5)(9.1x10^{-31})(94.5^2) - (1.6x10^{-19})(-50) = 1/2mv^2$$

$(-1.12x10^{-17} + (4.063x10^{-27}) - (-8x10^{-18}) = 1/2mv^2$

$-3.20x10^{-18} = 1/2mv^2$

$-6.40x10^{-18} = mv^2$

$$\frac{(-6.40x10^{-18})}{(9.1x10^{-31})} = v^2$$

$-7.033x10^{12} = v^2$

First of all I have this errant negative sign that's killing me. I honestly have no idea whats going on.

18. Sep 13, 2009

### kuruman

Your potential energy is in Joules but your kinetic energy is not. You need to express the velocity in meters per second not kilometers per second then put it in the equation.

19. Sep 13, 2009

### exitwound

yeah I did that too, but it still comes out way too high of a value and still attempting to square root a negative.

20. Sep 13, 2009

### kuruman

Yu say in part (a) that the particle is a proton. What is the mass of a proton?