"(a) how much charge can be placed on a capacitor with air between the plates before it breaks down if the area of each plate is 6.00 cm2? (assume air has a dielectric strength of 3.00 ✕ 106 v/m and dielectric constant of 1.00.)"

The voltage value for the electric breakdown is given by
(1)
where  is the dielectric strenght of the air while d is the distance between the two plates of the capacitor.

For a parallel plate capacitor, the capacitance is given by
(2)
where Q is the charge on the capacitor, V the voltage applied,  is the dielectric constant in air and  is the area of the plates in our problem. 

If we use the breakdown voltage given by equation (1) and replace V in equation (2) with this value, we find:

and from this, we can find the maximum charge allowed on the capacitor before the break down:

Ibelieve the answer is the dependent variable! hope this !