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

In summary, the problem involves a proton entering a parallel-plate capacitor with a uniform electric field, traveling at a given speed and distance between the plates. The goal is to find the horizontal distance the proton will traverse before hitting the bottom plate. Using the formula for force, acceleration, and projectile motion, the horizontal displacement can be calculated by finding the time taken for the proton to fall and multiplying it by the horizontal velocity. This can be done by using the formula for displacement and time in uniformly accelerating motion.
  • #1
tuggler
45
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!
 
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  • #2
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:
  • #3
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.
 
  • #4
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?
 
  • #5
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?
 
  • #6
I meant "bottom plate". The velocity will depend on the electric field since they both are perpendicular to each other.
 
  • #7
Horizontal or Vertical?
 
  • #8
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
 
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  • #9
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
[tex] v_f^2 = v_i^2 + 2ad?[/tex]
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
 

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

What is electric field distance?

Electric field distance refers to the distance between a point in space and an electric charge. It is a measure of how far the electric field extends from the charge.

How is electric field distance calculated?

Electric field distance is calculated using the equation E = kQ/r^2, where E is the electric field, k is the Coulomb's constant, Q is the electric charge, and r is the distance from the charge.

What is the unit for electric field distance?

The unit for electric field distance is meters (m).

How does electric field distance affect the strength of an electric field?

The farther away you are from an electric charge, the weaker the electric field will be. This is because the electric field strength decreases with distance from the charge.

Can electric field distance be negative?

No, electric field distance cannot be negative. It is always a positive value representing the distance from the charge.

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