Parallel Plate Capacitor with Dialectric

[AdkinsJr]

Homework Statement

How much charge can be placed on a capacitor with air between the plates before it breaks down if the area of the plates is 5 cm^2.

Homework Equations

$$C=\frac{Q}{\Delta V}$$

$$C=kC_o$$

$$C=k\frac{\varepsilon _o A}{d}$$

Where "k" is the dialectric constant.

The Attempt at a Solution

I'm not sure where to begin with the equations I know, I also don't know what is meant by "breaks down," how to interpret it mathematically; and I feel like I don't have enough information to solve the problem. Any help appreciated...

 

[TSny]

If the electric field gets too strong, the dielectric material (air in this case) can "break down" and become conductive. See this short video. The minimum electric field that will cause a material to break down is called the "dielectric strength" of the material. To work this problem you will need to know the dielectric strength of air. If you are using a textbook, you can probably look up the value there.

So, essentially, you need to find the charge on the plates of the capacitor that will produce an electric field equal to the dielectric strength of air. (The dielectric constant, k, of air can be taken to be 1 for this problem.)