Calculating Horizontal Distance of a Proton in a Parallel-Plate Capacitor

Click For Summary

Homework Help Overview

The problem involves a proton moving through a parallel-plate capacitor, where it is subjected to a uniform electric field. The proton's initial horizontal speed and the distance between the plates are provided, but the width of the capacitor is not specified. The goal is to determine the horizontal distance the proton travels before hitting the bottom plate.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the forces acting on the proton and its motion, considering the effects of the electric field and the lack of gravitational forces. Questions arise about how to calculate the time it takes for the proton to reach the bottom plate and how to determine the horizontal distance traveled during that time.

Discussion Status

Some participants have provided insights into the relationship between horizontal and vertical motion, likening the situation to projectile motion. There is ongoing exploration of the relevant equations and how to apply them to find the horizontal displacement. Multiple interpretations of the problem are being discussed, with no explicit consensus reached yet.

Contextual Notes

Participants note that the absence of the capacitor's width complicates the problem. The discussion includes clarifications about the direction of acceleration and the nature of the proton's motion within the electric field.

tuggler
Messages
45
Reaction score
0

Homework Statement



A proton enters a parallel-plate capacitor traveling to the right at a speed of 1.276 x 10-5 m/s. The distance between the two plates is 1.59 cm. The proton enters the capacitor halfway between the top plate and the bottom plate; that is, a distance r = 0.795 cm from each plate. The capacitor has a 2.95 x 10-4 N/C uniform electric field between the plates that points downward from the top plate to the bottom plate. Neglecting gravitational forces, what horizontal distance does the proton traverse before the proton hits the bottom plate?

Homework Equations





The Attempt at a Solution



Since the electric field is 2.95 x 10-4 N/C I used that to get my force by multiplying the charge of an electron 1.6E-19 thus I got F = 4.72E-23 from that I used F = ma to calculated the acceleration which I got 28265 m/s^2. But now I am stuck. The question will have been easier if they gave the width of the capacitor but they gave me the distance between. How can I do this problem? Thanks!
 
Physics news on Phys.org
tuggler said:

Homework Statement



A proton enters a parallel-plate capacitor traveling to the right at a speed of 1.276 x 10-5 m/s. The distance between the two plates is 1.59 cm. The proton enters the capacitor halfway between the top plate and the bottom plate; that is, a distance r = 0.795 cm from each plate. The capacitor has a 2.95 x 10-4 N/C uniform electric field between the plates that points downward from the top plate to the bottom plate. Neglecting gravitational forces, what horizontal distance does the proton traverse before the proton hits the bottom plate?

Homework Equations


The Attempt at a Solution



Since the electric field is 2.95 x 10-4 N/C I used that to get my force by multiplying the charge of an electron 1.6E-19 thus I got F = 4.72E-23 from that I used F = ma to calculated the acceleration which I got 28265 m/s^2. But now I am stuck. The question will have been easier if they gave the width of the capacitor but they gave me the distance between. How can I do this problem? Thanks!

What is the direction of the acceleration?
The motion of the proton can be resolved into a horizontal one, parallel with the plates and a vertical one, perpendicular to the plates. How long time does it take for the proton to reach one plate? How far does it travel horizontally during that time?

ehild
 
Last edited:
The direction of the acceleration will be towards the bottom of the plate, kind of looks like a parabola. But other than that I don't know how to finish this problem.
 
The solution will be similar to projectile motion. The question is almost same as a rock thrown from a cliff with a horizontal velocity.
So what does the horizontal velocity of the proton depend upon? Does it change? What about its vertical velocity?
 
tuggler said:
The direction of the acceleration will be towards the bottom of the plate,

You mean the bottom plate or bottom of the plate?
 
I meant "bottom plate". The velocity will depend on the electric field since they both are perpendicular to each other.
 
Horizontal or Vertical?
 
tuggler said:
I meant "bottom plate". The velocity will depend on the electric field since they both are perpendicular to each other.

Only the initial velocity (horizontal) is perpendicular to the vertical electric field. The proton accelerates vertically towards a plate.
You are right saying that it is like horizontal projectile motion. What are the equations for the projectile?

ehild
 
Last edited:
Yup, I understand that part now. The velocity is perpendicular to the field where the acceleration is perpendicular to the velocity vector.

But how can continue to get the desired answer?
 
  • #10
How does a projectile move? Do you recall any equations?

ehild
 
  • #11
v_f^2 = v_i^2 + 2ad?
Will this equation work?
 
  • #12
It might work if you use the correct d, vi and vf, but you need the horizontal displacement.

what horizontal distance does the proton traverse before the proton hits the bottom plate?

ehild
 
  • #13
The horizontal displacement is what I am having trouble figuring out. If I knew the horizontal displacement I will know what to do from there. How can I get it?
 
  • #14
The horizontal displacement is the question in the problem.

As for projectile motion, this link may help.

http://faculty.wwu.edu/vawter/PhysicsNet/Topics/Vectors/ProjectilesMotion.html

ehild
 
Last edited by a moderator:
  • #15
I am still confused
 
  • #16
Well, just calculate the time taken for proton to fall in the plate and multiply it to horizontal velocity. Remember vertical velocity is zero.
 
  • #17
The proton accelerates vertically with a=28265 m/s2. It travels Δy=0.795 cm distance vertically. How long time does it travel till it reaches the bottom plate?
You need the formula between displacement and time for uniformly accelerating motion.
Horizontally, the proton travels with constant speed. How far does it reach (horizontally) during its time of flight?

ehild
 

Similar threads

  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 58 ·
2
Replies
58
Views
6K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 20 ·
Replies
20
Views
4K
Replies
14
Views
3K
  • · Replies 26 ·
Replies
26
Views
3K
  • · Replies 6 ·
Replies
6
Views
14K
  • · Replies 16 ·
Replies
16
Views
2K