# Parallel-plate capacitor

1. Jan 21, 2009

### KillerZ

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

Air "breaks down" when the electric field strength reaches 3 x 10^6 N/C, causing a spark. A parallel-plate capacitor is made from two 4.0cm-diameter disks. How many electrons must be transferred from one disk to the other to create a spark between the disks?

2. Relevant equations

Q = $$\epsilon$$0AE

N = Q/e

3. The attempt at a solution

Ok, I think this is right but I am not sure:

Q = $$\epsilon$$0AE

A=$$\pi$$r^2

=(8.85x10^-12 C^2/Nm^2)($$\pi$$(0.02m)^2)(3x10^6 N/C)

=3.3x10^-8 C = 33nC

N = Q/e

= (3.3x10^-8 C)/(1.60x10^-19 C/electron)

= 2.1x10^11 electrons

2. Feb 12, 2009

### rebeccc

I'm trying to figure out this same problem only with diff. numbers and I'm stuck.. so if anyone has any ideas, please share!

3. Feb 12, 2009

### LowlyPion

Perhaps if you post your problem, with what you are having trouble reconciling, someone may be able to help clarify things?

4. Feb 12, 2009

### rebeccc

The problem is already posted above.

5. Feb 12, 2009

### LowlyPion

OK.

What don't you understand about the solution he posted?

6. Feb 12, 2009

### rebeccc

It's not correct, is it?

7. Feb 12, 2009

### LowlyPion

What step do you think is incorrect?

8. Sep 23, 2009

### MiNa-mOO

I used the same process! The only thing i did differently was use more significant figures. When done in this manner you get 2.08335*10^11 electrons.

The correct equations were used to solve this problem.