# Proton between two plates

## Homework Equations

Ki + Ui = Kf + Uf

## 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

kuruman
Homework Helper
Gold Member
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?

E=-dV/ds isn't it?

kuruman
Homework Helper
Gold Member
That's it.

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

kuruman
Homework Helper
Gold Member
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?

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

$$20=-E\int ds$$

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

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

kuruman
Homework Helper
Gold Member
The plate separation is 2.3 mm, not 6 km.

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

kuruman
Homework Helper
Gold Member
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.

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

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

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.

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?

K1 is not zero.

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

Yes.Good luck with it.

$U_i + K_i = U_f + K_f$

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

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

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.

kuruman
Homework Helper
Gold Member
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.

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

kuruman
Homework Helper
Gold Member
Yu say in part (a) that the particle is a proton. What is the mass of a proton?

Gain of PE =70q-50q=20q=loss of KE

First of all, true. I was using the mass of an electron like an idiot.

Secondly, I know i can use a relative zero for the voltage difference. I could set the left plate to 0, and the right plate to 20. I dont understand why it comes out to 20V when Vi-Vf= -70-(-50)= -20.

So when I plug in all the numbers:

$$(1.6x10^{-19})(-70) + (.5)(1.67x10^{-27})((94.5x10^3)^2) - (1.6x10^{-19})(-50) = 1/2mv^2$$

$$V=7.139x10^4 m/s$$ or 7.139x10^1 Km/s which is 71.39 Km/s . It's a reasonble number, I guess. am I on the right track yet?

Last edited: