The voltage value for the electric breakdown is given by
[tex]V_{BD}=E_{DS} d[/tex] (1)
where [tex]E_{DS}=3.0 \cdot 10^6 V/m[/tex] 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
[tex]C= \frac{Q}{V}= \frac{\epsilon A}{d} [/tex] (2)
where Q is the charge on the capacitor, V the voltage applied, [tex]\epsilon=1[/tex] is the dielectric constant in air and [tex]A=6.0 cm^2 = 6.0 \cdot 10^{-4} m^2[/tex] 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:
[tex] \frac{Q}{E_{DS} d}= \frac{\epsilon A}{d} [/tex]
and from this, we can find the maximum charge allowed on the capacitor before the break down:
[tex]Q=\epsilon A E_{DS}= (1)(6\cdot 10^{-4}m^2)(3 \cdot 10^6 V/m)=1800 C[/tex]
