# Homework Help: Charged capacitor, electric field

1. Feb 19, 2010

1. The problem statement, all variables and given/known data
A capacitor consists of two large metal disks of radius 2.4 meters placed parallel to each other, a distance of 0.7 millimeters apart. The capacitor is charged up to have an increasing amount of charge +Q on one disk and -Q on the other. At about what value of Q does a spark appear between the disks?

2. Relevant equations
E = [Q/(A/2(epsilon))](1- Z / (R^2+Z^2)^1/2)
R>>Z

E = [Q / (pi(r^2) / 2 * 8.85E-12)]

3. The attempt at a solution
I know that since one disk is positive and one disk is negative, the field points in one direction, so it should be E1 + E2. I tried plugging it into the equation, but then I'd still have 2 variables, E and Q. How do you find the value of Q when a spark appears? Is there a value for this that they did not provide?

Last edited: Feb 20, 2010
2. Feb 19, 2010

### collinsmark

look up "dielectric breakdown." I'm assuming there is an air gap between the plates (as opposed to a vacuum).

3. Feb 20, 2010

Thanks for that. It says that sparks occur when the electric field strength is 3x10^6 N/C.

So if you have that, then:

E = E1 + E2

E = [Q/(A/2e)] + [Q/(A/2e)] (R>>Z)

3x10^6 = 2[Q/(A/2e)]

3x10^6 = 2[Q/(pi(2.4)^2/2(8.58E-12)]

Q = 3E20 N/C.

What am I doing wrong here?

4. Feb 22, 2010

### collinsmark

It's difficult for me to interpret your math the way it is written. Perhaps if you could more carefully writing it out, including the units (when appropriate), it would help. But I think you might be dividing by something when you should be multiplying, or vise versa.

One other minor thing I noticed is that the constant you are using for the permittivity of free space is a little off.

5. Feb 22, 2010

$$E = \frac{Q}{\frac{A}{2\epsilon _0}}$$, and that's where i think you went wrong.