# Charged capacitor, electric field

## Homework Statement

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?

## Homework 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)]

## 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?

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## Answers and Replies

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

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?

collinsmark
Homework Helper
Gold Member
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.

collinsmark
Homework Helper
Gold Member
Okay, you have approximated the electric field of a single plate as
$$E = \frac{Q}{\frac{A}{2\epsilon _0}}$$, and that's where i think you went wrong.

Try using Gauss' law again to model the electric field caused by a large plate, or find the correct equation in your textbook.