Answer:

Explanation:
The capacitance of a parallel-plate capacitor is given by

where
is the vacuum permittivity, A is the surface area of the plates, d their separation.
We also know the following relationship

where Q is the charge stored on the capacitor and V the potential difference between the plates.
Combining the two equations,

We also know that for a uniform electric field (such as the one between the plates of a parallel-plate capacitor), we have

where E is the magnitude of the electric field. Substituting into the previous equation and re-arranging it,

For the capacitor in the problem:
is the area of the plates
is the maximum electric field before a spark is produced
Solving for Q, we find the maximum charge that can be added before that occurs:
