 Parallel-plate capacitor

A parallel-plate capacitor consists of two parallel electrical conductors with overlapping area $A$ and separation $d$. If the gap is small compared with both plate widths and is filled by a uniform material of positive permittivity $ϵ$, its capacitance is approximatelyThis neglects the fringing field of a capacitor near the plate edges. The interior electric field points from the plate of positive electric charge to the plate of negative electric charge, and is approximately uniform. ^[ref].

For a worked vacuum example, stipulate $A=0.01{\mathrm{m}}^2$, $d=0.001\mathrm{m}$, and vacuum permittivity ${ϵ}_0\approx 8.854\times 10^{-12}F/m$. This rounds the value in Fundamental Physical Constants — Complete Listing by National Institute of Standards and Technology and others. Within the approximation,At terminal electric potential difference $V=10\mathrm{V}$, the electric field magnitude is $E\simeq V/d=10^{4}V/m$, the plate electric charge magnitude is $Q=CV=0.8854\mathrm{nC}$, and the capacitor energy is $U=C{V}^2/2=4.427\mathrm{nJ}$. These numbers use the stipulated rounded vacuum permittivity; finite plates also have corrections from the fringing field of a capacitor.

The formula makes design changes easy to predict: doubling the overlapping area doubles the capacitance, and halving the gap doubles the capacitance, with permittivity fixed and the fringing field of a capacitor neglected. Doubling permittivity also doubles capacitance if the material uniformly fills the gap and remains linear. This lets a compact device store more separated electric charge at the same terminal electric potential difference.

The short derivation uses surface charge density $\sigma =Q/A$. In the interior, Gauss's law gives electric field magnitude $E\simeq \sigma /ϵ=Q/(ϵA)$. The electric potential difference is $V\simeq Ed=Qd/(ϵA)$, so dividing $Q$ by $V$ gives $C\simeq ϵA/d$. A narrow curved capacitor gap has the same leading local approximation when its width is much less than the curvature radius.