# Homework Help: Parallel Plates Question

1. Jan 7, 2013

### skg94

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

An electron traveling 1.3*10^2m/s enters the region between two parallel plates at the distances shown. When it collides with the plates, it has a speed of 1.8*10^3 m/s. What is the electric field between the plates? What is the electrostatic force on the electron? And how far along the plate does the electron hit?

------- +
|3.0cm (3cm between electron and top plate (positive)
|
electron
|
|
|10cm distance between electron and negative (bottom plate)
------------- negative
2. Relevant equations

3. The attempt at a solution

I tried using V= ΔE/q but than realized that i probably had to use E = ΔV/ΔD

I really dont know how to go about this. You dont have to give me all the answers, just some steps please

2. Jan 7, 2013

### Staff: Mentor

Why don't you start with the change in kinetic energy? You're given the initial and final speeds. How does the change in KE relate to the change in electric potential over the given distance?

3. Jan 7, 2013

### skg94

I tried change in energy divided by charge it didnt work for potential difference

4. Jan 7, 2013

### Staff: Mentor

Sure it will. But keep in mind it's the potential difference through which the electron falls; the change in potential between the initial position of the electron (distance 3.0 cm from the + plate) to the final position of the electron (at the + plate); it's not the full potential difference between the two plates.

5. Jan 7, 2013

### skg94

ok so i did V= .5 (9.11*10^-31) * ((1800^2) - (130^2)) / 1.6*10^-19

= 9.175762813*10^-6 - its on the sheet but is that right? i must have done something wrong before

Than for electric field i would do 9.17...*10^-6 / (10-3cm)?

Fe= Eq

than how about how far along the plate how do i find that?

6. Jan 7, 2013

### Staff: Mentor

Looks okay.
Make that 3x10-2 m, not cm. The field is the change in potential divided by the distance, and here the distance is 3 cm but to make the units match write 3x10-2 m
It's a projectile problem. Use the force and known mass to find the acceleration in the y-direction.

Last edited: Jan 8, 2013
7. Jan 8, 2013

### skg94

so (10*10^-2 - 3*10^-2) ?

and i forget how to do projectile can you guide me in the right direction?

8. Jan 8, 2013

### Staff: Mentor

No, just 3x10-2. The electron is only falling through 3 cm. It just so happens that the field between the plates is (essentially) uniform, so knowing the field strength anywhere will give it to you for the whole region.
Start with the separate equations of motion for the x and y components of the trajectory. Find out how long (time) it takes for the electron to strike the plate (y-direction motion).

9. Jan 8, 2013

### skg94

Is it not falling through 10cm? because it starts at 3cm below the top plate, than drops 10cm to the bottom plate?

And like kinematics equations? d=1/2at-vit^2? and such?

10. Jan 8, 2013

### Staff: Mentor

Which direction is the electric force acting on the electron? (Which plate will attract the negatively charged electron? Which repels it?)
Yup.

11. Jan 8, 2013

### skg94

Isnt it positive to negative, so down?

Ah alright

12. Jan 8, 2013

### Staff: Mentor

Can you clarify that? Isn't what positive to negative?

According to your diagram the top plate is positive, the bottom plate is negative. There are positive charges on the top plate, negative charges on the bottom plate. Will the negative electron be attracted to the negative charges on the bottom, or to the positive charges on the top?

13. Jan 8, 2013

### skg94

OH positive to negative i think i was thinking of magnetism lol even than i dont even remember if thats correct. It will be attracted to the top plate since electron is negative i got it.

14. Jan 9, 2013

### skg94

I dont think i got the potential difference right, because i tried all 9.2*10^-6 / 3*10^-2 and 10^-2 and 7^-2 i tried all the denominators and i never got an answer on the sheet

15. Jan 10, 2013

### CAF123

Do they give you the distance between the plates in the problem statement? I don't see how you can tell what distance the electron has travelled without knowing this distance. If I understand the problem statement correctly, they give you the relative distances of the electron to the plates at either end but not the separation. Or maybe it's the case that it enters at 3cm below top plate in which case you would know the distance.

16. Jan 10, 2013

### Staff: Mentor

Well, 3.0 cm is 0.03 m, or 3 x 10-2 m.

Why, what are the choices on the sheet? Is the problem statement exactly as you've written it?

17. Jan 10, 2013

### skg94

Yes it is the distance between plates is given the length of plates is not.

The distance is 3cm from the top plate to the electron ( just coming in) an 10cm from the electron to the bottom plate.

Would itbe bettet if i took a picture of it?

18. Jan 10, 2013

### Staff: Mentor

This is my understanding of the setup:

#### Attached Files:

• ###### Fig1.gif
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19. Jan 10, 2013

### skg94

Yes thats right except the curvature isnt drawn on mine but that diagram is accurate

The 4ans given are
7.1 x10^-5
6.1x10^-5
1.2x10^-6
1x10^-6
All N/C

20. Jan 10, 2013

### Staff: Mentor

It's possible that the given answers are incorrect (they may correspond to a previous revision of the problem where the given values were different).

What value did you calculate for E?

21. Jan 10, 2013

### skg94

I highly douby its wrong its a pracice diploma but i calculated using .03 as d

3.06666...x10^-4

Using the 9.2x10^-6V

22. Jan 10, 2013

### Staff: Mentor

Well, that looks okay to me.

You can confirm by determining the work done on the electron by that field in moving through the 3.0 cm, and from that find the final velocity in the y-direction (work-energy theorem). The net velocity (add the components) should equal the final velocity as given.

23. Jan 10, 2013

### skg94

Wait isnt the energy work E=hf-w right? Im confused as to hat you just said

24. Jan 10, 2013

### Staff: Mentor

The work done in moving the electron through distance h with force f is fh. That will also be the KE imparted. So $\frac{1}{2}m_e v_y^2 = f h$. Solve for the $v_y$ and see if it, combined with the initial x-direction velocity, produces the given final velocity.

25. Jan 10, 2013

### CAF123

To find the distance along the plates, use the kinematic relations. You know the resultant velocity vector is of magnitude 1800m/s with an x component 130m/s. The x component of velocity stays a constant 130m/s since the only force acting on it (in addition to gravity) is the electric force which is perpendicular to the x direction. From this information, find $v_y$. Alternatively, gneill's method works too.

You should also be able to find the electron's acceleration and then use kinematics.