# Parallel Plates

1. Sep 13, 2007

### Etopn23

1. The problem statement, all variables and given/known data
An electron starts from one plate of a charged closely spaced (vertical) parallel plate arrangement with a velocity of 1.63 x 10^4 m/s to the right. Its speed on reaching the other plate, 2.10 cm away is 4.15x10^4 m/s

I got parts a and b - part C is what I'm having problems with.

C) If the plates are square with an edge length of 25.4 cm, determine the charge on each.

2. Relevant equations
E=(4πkQ)/A

3. The attempt at a solution

I substituted for all the unknowns except for E and Q - but that leaves me with two unknowns; I'm completely flabbergasted with what to do next.

2. Sep 13, 2007

### learningphysics

Can you post what parts a and b said, and what your answers were?

3. Sep 13, 2007

### Etopn23

A) What type of charge is on each plate?
Left Plate Negative, Right Plate positive

B) What is the direction of the electric field between the plates?
From right to left

4. Sep 13, 2007

### learningphysics

For a capacitor Q = CV.

Try to calculate C and V... then you can get Q.

5. Sep 13, 2007

### Etopn23

Thanks alot - i'll go do that. I didn't have that equation unfortunately xD.

6. Sep 13, 2007

### Etopn23

Ok I plugged in

C=((8.85x10^-12)(9x10^9)(.064516))/(2.10x10^-2)=.2447

Then Q=(.2447)(1.63X10^4)= 3988.61

Q=(.2447)(4.15x10^4)= 10155

I plugged each Q into E=(4pikQ)/A and I got two answers that were a bit off =\ No idea what I did wrong.

I know the final answer is 1.13 x 10^-13 but i'm not getting anywhere near that.

7. Sep 14, 2007

### learningphysics

The V in Q = CV is voltage not speed. You need to find the voltage between the two plates of the capacitor... use energy to find voltage...

8. Sep 14, 2007

### Etopn23

Aaah sorry; just assumed what the variables were since I hadn't encountered this equation before - let me fix this now xD.

9. Sep 14, 2007

### Etopn23

Ok so V=Ue/Qe

Ue= qeEd but E is unknown =\.

I'm reading up on this equation (surprisingly in my book it isn't in the chapter that the problem was assigned from - it's in the next chapter) so I haven't had exposure to it.

Am I doing something wrong? =\

10. Sep 14, 2007

### learningphysics

But you know that V = Ed...

Ue = qe*V

11. Sep 14, 2007

### Etopn23

Unless i'm missing something (which I problem am due to my own ignorance ;_;) I still have multiple unknowns.

UGH, it's irritating having an assignment which seems easy until you get to the last problem.

Any other clues you can prompt me to? =\

12. Sep 14, 2007

### learningphysics

What is the change in kinetic energy of the electron? electric potential energy is being converted to kinetic energy...

13. Sep 14, 2007

### Etopn23

If I calculated it correctly I got 1.147x10^-26 - plugged that in for delta Ue; plugged Ue into the V equation - plugged V into the Q equation - which I plugged in for E and I got the wrong answer =s.

14. Sep 14, 2007

### learningphysics

That's not what I'm getting for change in kinetic energy... can you show your calculations?

15. Sep 14, 2007

### Etopn23

KEo= (1/2)(9.10938188 × 10-31)(1.63x10^4)
KEf= (1/2)(9.10938188 × 10-31)(4.15x10^4)

KEf-KEo=Delta KE which is where I got 1.148E-26

I assumed the change in kinetic energy would be equal to the change in potential energy. So I took the magnitude of it and plugged in all the way through - lead me to a V of 7.17334E-8
which then lead me to a Q of 1.7E-8 which I proceeded to plug into the initial equation; I got a number that I knew was off immediately so I didn't bother writing it down =\.

16. Sep 14, 2007

### learningphysics

you didn't square the velocties... (1/2) mv^2...

17. Sep 14, 2007

### Etopn23

Ouch >.<. K - correcting my mistake.

18. Sep 14, 2007

### learningphysics

also here... Capacitance = epsilon*A/d

I think the k is a mistake.

19. Sep 14, 2007

### Etopn23

Ok corrected; that got my Ue to 6.55452E18 - which when plugged into my V got me 4.09658E37. That lead to Q being 1.00243E37 and E being some huge number =\.

20. Sep 14, 2007

### Etopn23

aaah ok i'll check on that - thanks; I knew something had to be wrong