 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: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 $2ES=\sigma S/{ϵ}_0$, so each sheet contributes $\sigma /(2{ϵ}_0)$ 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 $E=\sigma /{ϵ}_0$. The electric potential difference and capacitance of this parallel-plate capacitor are consequentlyThe approximation requires $d$ much smaller than both transverse sheet dimensions. A uniform filling with permittivity $ϵ$ replaces ${ϵ}_0$ by $ϵ$.