How Do You Calculate the Displacement in Thompson's Cathode Ray Tube Experiment?

[David112234]

Homework Statement

What is the distance Δy between the two points that you observe? Assume that the plates have length d, and use e and m for the charge and the mass of the electrons, respectively.
Express your answer in terms of e, m, d, v0, L, and E0.

Homework Equations

kinematics equations
F=qE=ma

The Attempt at a Solution

Split the problem into 2 sections, the first being with acceleration due to the electric field,
t₁ = when particle lease electric field
t₂ = when particle hits screen
so at t₁ a distance d

d = v₀t
Δy₁ = ½at²

t values are unknown so I eliminate t
t=d/v₀
and
qE₀=ma
q=e
(eE₀)/m = a

Δy₁ = ½((eE₀)/m) (d/v₀)²

now for the x and y components of velocity at this point
using the equation
v=v₀+at

v_y = 0 + ((eE₀)/m)(d/v₀
v_x has not changed so = v₀

Now is have the height at d and both components of velocity, I can use kinematics to find the particles motion from d to L using these values (at d) as the initial points.

Δy₂ = Δy₁ + v_y(t)

this t value will be

L= v₀ t
t= L/v₀

plugging it all in

= ½((eE₀)/m) (d/v₀)² +
(eE₀/m)d/v₀*L/v₀

This is not the correct answer for Δy₂
once I know Δy₂ I can find Δy = Δy₂ + Δy₁
But I can not find my mistake in finding Δy₂

 

[TSny]

Δy₁ = ½at²

Δy₁ = ½((eE₀)/m) d/v₀

Did you forget to square the time?

Δy₂ = Δy₁ + v_y(t)

Why did you include Δy₁ as part of Δy₂?

 

[David112234]

Did you forget to square the time?

While typing up my question here I did, but I did not in my work and the answer I tried entering.

Δy2 = Δy1 + vy(t)
Why did you include Δy₁ as part of Δy₂?

I'm not sure what you mean, I wrote Δy2 as a function of Δy2 because of the kinematics equation x=x0 + v(t) when a=0 and Δy1 and Δy2 represent heights.

 

[TSny]

In your diagram, the distance Δy₂ does not include the distance Δy₁.

But your equation Δy₂ = Δy₁ + vy(t) contains Δy₁ on the right hand side.

 

[David112234]

In your diagram, the distance Δy₂ does not include the distance Δy₁.

But your equation Δy₂ = Δy₁ + vy(t) contains Δy₁ on the right hand side.

I still do not see your point, Is that kinematic equation not applicable? Why?
Lets consider this situation. I am in a car, on the x-axis at x=5. I go east at 1 mile/hour. To find the position after 3 hours I would use that equation.
x(t) = 1(t) + 5
x(3) = 1(3) + 5 = 8

I see no difference between that situation and this particle, at d its initial position is Δy₁ and it is moving up at a rate of the velocity v. After a certain amount of time t it will be at Δy₂.

 

[gneill]

I still do not see your point, Is that kinematic equation not applicable? Why?

@TSny is making the observation that your use of variables has strayed from the definitions in the figure. In the figure (and the problem statement) the total displacement is given by $Δy = Δy_1 + Δy_2$. You have essentially redefined $Δy_2$ to be the total displacement.

 

[David112234]

Ah now I see what you mean, Δy₂ is not measured with re
respect to the zero line (whereΔy_1 was), my equation than should be
$Δy =$Δy_1 +v(t)

So how would I find Δy_2 before finding ΔY?

 

[gneill]

So how would I find Δy_2 before finding ΔY?

Consider the "launch" conditions from the Δy₁ location and use geometry (and similar triangles in particular).

You should have both the x and y velocities of the electron at the Δy₁ location.

 

[David112234]

Consider the "launch" conditions from the Δy₁ location and use geometry (and similar triangles in particular).

You should have both the x and y velocities of the electron at the Δy₁ location.

since
Δy = Δy₁ + Δy₂
and
Δy= Δy₁ + v(t)
Δy₁+Δy₂ = Δy₁ + v(t)

Is that the correct approach?

 

[gneill]

since
Δy = Δy₁ + Δy₂
and
Δy= Δy₁ + v(t)
Δy₁+Δy₂ = Δy₁ + v(t)

Is that the correct approach?

The velocity is 2D, that is, it has both x and y components. So unless the v in your equations is a vector, the equations are not correct.

Start by finding the x an y components of the velocity at the "launch" point.

EDIT: Perhaps I spoke too soon. If your v is the y-velocity component and t is the transit time from the launch point to the impact point, then your equations would work. You'll have to show how you get both v and t.

 

[David112234]

The velocity is 2D, that is, it has both x and y components. So unless the v in your equations is a vector, the equations are not correct.

Start by finding the x an y components of the velocity at the "launch" point.

I have found those, time is written in terms of x velocity.

 

[TSny]

since
Δy = Δy₁ + Δy₂
and
Δy= Δy₁ + v(t)
Δy₁+Δy₂ = Δy₁ + v(t)

Is that the correct approach?

This would be correct if you replace the v(t) expression by the expression you wrote in your first post v_y(t). I assume that by v_y(t) you meant v_y⋅t.

 

[David112234]

This would be correct if you replace the v(t) expression by the expression you wrote in your first post v_y(t). I assume that by v_y(t) you meant v_y⋅t.

Yes, because the only effect the x velocity is making it hit the plate before it can reach any higher.

 

[TSny]

Yes, because the only effect the x velocity is making it hit the plate before it can reach any higher.

Yes. In your first post you have the correct expressions for v_y and t to use for Δy₂ = v_y⋅t.