The potential difference across the parallel-plate capacitor is given by V = Ed, where d is the separation between the plates. The electric field strength in dry air will break down at 3.0 × 10⁶ V/m. Therefore, the breakdown voltage is given by;Vbreakdown = Ed3.0 × 10⁶ = Ed ∴ d = Vbreakdown / 3.0 × 10⁶Let the amount of charge that can be placed on each plate of the capacitor be Q.The capacitance of the parallel-plate capacitor is given by;C = ε0A/dwhere A is the area of each plate and ε0 is the permittivity of free space.Substituting the value of d and solving for Q we have;Q = CVbreakdownQ = (ε0A/d) Vbreakdown = (ε0A/ Vbreakdown / 3.0 × 10⁶) Vbreakdown = (ε0AVbreakdown/ 3.0 × 10⁶)Since A = 83cm² = 8.3 × 10⁻⁴m², and ε0 = 8.85 × 10⁻¹² C²/(N∙m²);Q = (8.85 × 10⁻¹²C²/(N∙m²)× 8.3 × 10⁻⁴m² × 3.0 × 10⁶V/m) / 3.0 × 10⁶V/mQ = 2.45 × 10⁻⁸ CAnswer:Therefore, the amount of charge that can be placed on the capacitor is 2.45 × 10⁻⁸ C.