A parallel-plate capacitor filled with air is to be constructed to...

Given $Q = 12 \times 10^{-6} \text{ C}$, $V = 1200 \text{ V}$, and E_\max = 3 \times 10^6 \text{ V m}^{-1}.

The capacitance of the capacitor is given by $C = \dfrac{Q}{V}$.

For a parallel-plate capacitor, $C = \dfrac{\varepsilon_0 A}{d}$.

The electric field between the plates is $E = \dfrac{V}{d}$, which gives $d = \dfrac{V}{E}$.

Substituting $d$ in the capacitance formula:

$C = \dfrac{\varepsilon_0 A E}{V}$

Equating the two expressions for $C$:

$\dfrac{Q}{V} = \dfrac{\varepsilon_0 A E}{V}$

$\Rightarrow A = \dfrac{Q}{\varepsilon_0 E}$

For the minimum possible plate area, the electric field must be at its maximum permissible value (dielectric strength), i.e., E = E_\max.

A_\min = \dfrac{Q}{\varepsilon_0 E_\max}

A_\min = \dfrac{12 \times 10^{-6}}{8.85 \times 10^{-12} \times 3 \times 10^6}

A_\min = \dfrac{4}{8.85} \approx 0.45 \text{ m}^2