Proton between two plates

  • Thread starter exitwound
  • Start date
  • #1
292
1

Homework Statement



problem.jpg


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
 

Answers and Replies

  • #2
kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
10,824
3,837
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
292
1
E=-dV/ds isn't it?
 
  • #5
292
1
I don't understand how to use that. I have a 20V potential difference. How does that relationship help me?
 
  • #6
kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
10,824
3,837
Integrate the expression.

[tex]\Delta V=-\int Eds[/tex]

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
292
1
[tex]
\Delta V=-\int Eds
[/tex]

[tex]
20=-E\int ds
[/tex]

[tex]
20=-E (2.3E-6 km)
[/tex]

E=-20/2.3E-6 = -8.695E6 V/km
 
  • #8
kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
10,824
3,837
The plate separation is 2.3 mm, not 6 km.
 
  • #9
292
1
That should read 2.3x10^-6 km since the problem will ultimately ask for KM/s velocities.
 
  • #10
kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
10,824
3,837
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
292
1
It's all confusing no matter what units are used.

E=-20/2.3E-3 = -8.695E3 V/m then
 
  • #12
2,483
100
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
292
1
So how would i set that up? Something like this?

[itex]U_i + K_i = U_f + K_f[/itex]

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

[itex]qV_1 + 0 = qV_2 + 1/2mv^2[/itex]

Something like this? Solve for v?
 
  • #14
2,483
100
K1 is not zero.
 
  • #15
292
1
Oh. Would it also be 1/2mv^2 (v being initial velocity?)
 
  • #16
2,483
100
Yes.Good luck with it.
 
  • #17
292
1
[itex]U_i + K_i = U_f + K_f[/itex]

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

[itex]qV_1 + 1/2 mv_i^2 = qV_2 + 1/2mv^2[/itex]

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.

[itex]qV_1 + 1/2 mv_i^2 - qV_2 = 1/2mv^2[/itex]

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

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

[itex]-3.20x10^{-18} = 1/2mv^2[/itex]

[itex]-6.40x10^{-18} = mv^2[/itex]

[tex]\frac{(-6.40x10^{-18})}{(9.1x10^{-31})} = v^2[/tex]

[itex]-7.033x10^{12} = v^2[/itex]

First of all I have this errant negative sign that's killing me. I honestly have no idea whats going on.
 
  • #18
kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
10,824
3,837
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
292
1
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
kuruman
Science Advisor
Homework Helper
Insights Author
Gold Member
10,824
3,837
Yu say in part (a) that the particle is a proton. What is the mass of a proton?
 
  • #21
2,483
100
Gain of PE =70q-50q=20q=loss of KE
 
  • #22
292
1
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:

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

[tex]V=7.139x10^4 m/s[/tex] 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:

Related Threads on Proton between two plates

Replies
5
Views
831
Replies
3
Views
2K
  • Last Post
Replies
1
Views
13K
  • Last Post
Replies
1
Views
734
  • Last Post
Replies
5
Views
9K
  • Last Post
Replies
12
Views
6K
Replies
4
Views
985
Replies
2
Views
1K
  • Last Post
Replies
6
Views
449
Top