= Solution
In <vacuum>, <Gauss's law> states that the outward <electric flux> through a closed <surface> is the enclosed <electric charge> divided by the <vacuum permittivity>:
$$
\oint_S\mathbf E\cdot d\mathbf A=\frac{Q_{\rm enc}}{\epsilon_0}.
$$
Take opposite <electric charges> $\pm Q$ on the two <electrical conductors>, with <surface charge densities> $\pm\sigma=\pm Q/A$. A <Gaussian pillbox> crossing one isolated large sheet gives $2ES=\sigma S/\epsilon_0$, so each sheet contributes $\sigma/(2\epsilon_0)$ on either side. By the <superposition principle>, the <electric fields> add between the sheets and cancel outside. Ignoring the <fringing field of a capacitor>, the gap <electric field> is $E=\sigma/\epsilon_0$. The <electric potential difference> and <capacitance> of this <parallel-plate capacitor> are consequently
$$
V=\int_0^d E\,dx=\frac{Qd}{\epsilon_0A},\qquad
\boxed{C=\frac QV=\frac{\epsilon_0A}{d}.}
$$
The approximation requires $d$ much smaller than both transverse sheet dimensions. A uniform filling with <permittivity> $\epsilon$ replaces $\epsilon_0$ by $\epsilon$.
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