# Electric force between 2 parallel plates

1. ### Oblivion77

113
1. The problem statement, all variables and given/known data

An electron is projected with an initial velocity of 1.6x10$$^{6}$$ m/s. If the electron just misses the upper plate as it emerges from the field, find the speed of the electron as it emerges from the field?

2. Relevant equations

Electric force equation

3. The attempt at a solution

I am stuck trying to figure out the magnitude of the electric field, once I can figure this out I know how to solve the problem. Any pointers on how to find the magnitude of the electric field?

2. ### LowlyPion

5,332
Use ordinary kinematic means to determine the velocity.

You know the speed, hence how long for it to emerge.

In that time you also know the deflection so you can determine the acceleration.

That acceleration then yields the additional sideways component of velocity to calculate it's speed at that point right?

### Staff: Mentor

You don't need to know the magnitude of the electric field.

4. ### Oblivion77

113
I was planning to use Vf^2 = Vo^2 +2ad to find the final velocity, but I am missing the acceleration component. To find the acceleration I wanted to use F = ma (knowing the mass of the electron). But I would need the magnitude of the electric field to find the force from E = F / q.

5. ### Oblivion77

113
So how would I do this without using the electric field? I don't know what you mean about using the "deflection" to find the acceleration. How would I calculate that?

Last edited: Jan 24, 2009
6. ### LowlyPion

5,332
The horizontal velocity to the end of the plate gives you time.

Use the distance, acceleration, time relationship to determine acceleration.

Then you can use your V2, acceleration and distance.

7. ### Oblivion77

113
Thanks for the advice, but this does not work. I get the answer wrong, the magnitude of the electric field is 364N/C, I am just not sure how to find it. When I use this electric field with the method I stated above I get the right answer. I just can't figure out how they got 364N/C :(

Last edited: Jan 24, 2009
8. ### LowlyPion

5,332
Once you determine the acceleration from the trajectory, then you can use f = ma to determine field intensity.

They ask really though for just the speed.

That equals (Vx2 +Vy2)1/2

9. ### Oblivion77

113
Yes! Thank you so much, I figured it out now =)

10. ### LowlyPion

5,332
I wondered if that wasn't it.

Glad you got it.

Good luck.

11. ### NamrataJ

4
Please will someone explain how I am meant to calculate the time taken for the electron to reach the end of the plates. I'm just not getting it.

12. ### songoku

803
you have the horizontal distance and the velocity in x-direction is constant, so t = v / s

13. ### NamrataJ

4
oh dear. that was rather dense of me. thank you! :)

14. ### songoku

803
You're welcome ^^

15. ### NamrataJ

4
Sorry, i'm not getting the right answer..not sure where i'm going wrong..should it not be t=s/v because t=v/s yields an answer of 80,000,000

### Staff: Mentor

To check if an equation makes sense, look at the units. t = v/s → (m/s)/s = m/s^2; these are units of acceleration, not time, so this equation makes no sense.

Since v = s/t, t = s/v is correct. The units would be m/(m/s) = m(s/m) = s. Makes sense.

17. ### songoku

803
oh sorry, it's my mistake
it should be t = s/v

4
thank you :)

19. ### songoku

803
You're welcome
sorry for the mistake earlier ^^

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