Charged capacitor, electric field

AI Thread Summary
The discussion revolves around calculating the charge (Q) needed for a spark to appear between two parallel metal disks of a capacitor. The electric field strength at which a spark occurs is approximately 3x10^6 N/C, and the user attempts to derive Q using the formula for the electric field. There is confusion regarding the application of the equations and the constants, particularly the permittivity of free space. Suggestions are made to revisit Gauss' law for a more accurate model of the electric field produced by a large plate. The conversation emphasizes the importance of precise calculations and correct constants in solving the problem.
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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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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?
 
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.
 
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.
 
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