How Many Electrons Cause a Spark in a Parallel-Plate Capacitor?

[KillerZ]

Homework Statement

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?

Homework Equations

Q = $$\epsilon$$₀AE

N = Q/e

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

 

[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!

 

[LowlyPion]

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!

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

 

[rebeccc]

The problem is already posted above.

 

[LowlyPion]

OK.

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

 

[rebeccc]

It's not correct, is it?

 

[LowlyPion]

What step do you think is incorrect?

 

[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.