# Thompson's cathode experiment

1. Nov 27, 2016

### David112234

1. The problem statement, all variables and given/known data

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.

2. Relevant equations
kinematics equations
F=qE=ma

3. 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

vy = 0 + ((eE0)/m)(d/v0
vx has not changed so = v[SUB]0[/SUB]

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 + vy(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

Last edited: Nov 27, 2016
2. Nov 27, 2016

### TSny

Did you forget to square the time?

Why did you include Δy1 as part of Δy2?

3. Nov 27, 2016

### David112234

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

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.

4. Nov 27, 2016

### 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.

5. Nov 28, 2016

### David112234

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.

6. Nov 28, 2016

### Staff: Mentor

@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.

7. Nov 28, 2016

### 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?

8. Nov 28, 2016

### Staff: Mentor

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.

9. Nov 28, 2016

### David112234

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

Is that the correct approach?

10. Nov 28, 2016

### Staff: Mentor

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.

11. Nov 28, 2016

### David112234

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

12. Nov 28, 2016

### TSny

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

13. Nov 28, 2016

### David112234

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

14. Nov 28, 2016

### TSny

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