3.7.4.2 - Parallel Plate Capacitor

Plate geometry and electric field

A parallel-plate capacitor has two flat conducting plates with overlapping area $A$, separated by an insulating gap of width $d$. One plate carries $+Q$ and the other $-Q$. Here $A$ is the overlapping area of one plate, not the sum of both plate areas.

[DIAGRAM: asset_name: 7.4.2 - Parallel Plate Capacitor - Diagram 1; asset_slug: 7.4.2 - Parallel Plate Capacitor - Diagram 1; caption: Parallel conducting plates of overlapping area A and separation d. The field points from the positive plate to the negative plate, the cell polarity matches the plate charges.; description: Parallel conducting plates of overlapping area A and separation d. The field points from the positive plate to the negative plate, the cell polarity matches the plate charges.]

When the separation is small compared with the plate dimensions, the field away from the edges is approximately uniform. Then $E=V/d$. Fringing at the edges makes the ideal model less accurate for widely separated or poorly aligned plates. Plate charge flows through the connecting wires; an intact dielectric does not conduct electrons from plate to plate.

How area, spacing and material set capacitance

Increasing the overlapping plate area allows more charge at a given pd; reducing the separation also increases capacitance. For parallel plates completely filled by a uniform dielectric,

The permittivity of free space is $\varepsilon_0=8.85\times10^{-12}\,\mathrm{F}\,\mathrm{m}^{-1}$. Relative permittivity $\varepsilon_r$ is dimensionless; it is 1 in vacuum and approximately 1 in air. The equation assumes negligible edge effects and no air gaps in the dielectric.

For vacuum, the uniform-field relation $E=Q/(\varepsilon_0 A)$, together with $V=Ed$, gives $C=Q/V=\varepsilon_0 A/d$. The factor $\varepsilon_r$ describes the increase produced by the dielectric. Thus doubling $A$ doubles $C$, whereas doubling $d$ halves it. Doubling a square plate's side length quadruples its area. Thin dielectric layers and large folded or rolled electrode areas allow large capacitance in a small volume.

For square plates of side $0.25\,\mathrm{m}$, separation $1.5\,\mathrm{mm}$ and $\varepsilon_r=2.3$, use $A=0.0625\,\mathrm{m^2}$ and $d=1.5\times10^{-3}\,\mathrm{m}$. This gives

Relative permittivity and its measurement

Relative permittivity, also called the dielectric constant in this context, is the ratio of a material's permittivity to that of free space. For the same plate geometry,

Here $C$ is the fully dielectric-filled capacitance and $C_0$ is the vacuum capacitance (approximated by air). A dielectric constant of 7.0 means seven times the capacitance for the same geometry. It also means seven times the plate charge at the same pd, since $Q/Q_0=C/C_0$. The charge ratio is invalid if the pd changes.

To determine $\varepsilon_r$, measure capacitance using a capacitance meter with air between two plates, then insert a sheet filling the same gap. Keep overlapping area and separation fixed; measure sheet thickness or use spacers, because inserting a thicker sheet would change two variables. Subtract a measured lead/stray capacitance if significant, then calculate $C/C_0$. For corrected readings $C_0=82\,\mathrm{pF}$ and $C=205\,\mathrm{pF}$, $\varepsilon_r=2.50$. Air pockets lower the measured increase.

The same apparatus can investigate $C\propto A$ at fixed $d$, or $C\propto1/d$ at fixed $A$. Measure several spacings; plot $C$ against $1/d$, keeping leads in the same position and plates parallel. A straight line through the origin after stray-capacitance correction supports the model. Its gradient is $\varepsilon_0\varepsilon_r A$; a nonzero intercept can indicate uncorrected stray capacitance.