Last updated: 2026-06-15

Parallel-Plate Capacitor Capacitance

A parallel-plate capacitor consists of two flat conducting plates separated by an insulating material called a dielectric. When a voltage is applied across the plates, opposite charges accumulate on each plate, creating a nearly uniform electric field in the gap between them. The capacitance — the amount of charge stored per unit voltage — depends on the geometry and the material:

$C = \varepsilon_0 \varepsilon_r \frac{A}{d}$

where $\varepsilon_0 = 8.8542 \times 10^{-12}\ \text{F/m}$ is the permittivity of free space, $\varepsilon_r$ is the relative permittivity (dielectric constant) of the material between the plates, $A$ is the plate area in square metres, and $d$ is the plate separation in metres.

The Role of the Dielectric

The dielectric constant $\varepsilon_r$ quantifies how much the insulating material between the plates amplifies the capacitance relative to vacuum ($\varepsilon_r = 1$). When a dielectric is inserted, the polar molecules in the material align with the electric field, reducing the net field in the gap and allowing more charge to be stored for the same voltage. Representative values:

A dielectric with $\varepsilon_r = 10$ gives ten times the capacitance of the same plate geometry in air.

Formula

Capacitance increases with plate area and decreases with plate separation. Halving the gap doubles the capacitance; halving the area halves it.

Worked Example

Two square aluminium plates, each 10 cm × 10 cm (area $A = 100\ \text{cm}^2 = 0.01\ \text{m}^2$), are separated by a 1 mm (0.001 m) air gap. Find the capacitance.

Entering 100 cm², 1 mm, and $\varepsilon_r = 1$ in the calculator gives the same result. Now suppose the air gap is replaced with a glass dielectric ($\varepsilon_r = 5$). The capacitance becomes $5 \times 88.5 \approx 443\ \text{pF}$, a fivefold increase with no change in geometry.

Increasing Capacitance in Practice

Real capacitors achieve high capacitance in small packages through three strategies used simultaneously. First, the plate area is maximised by rolling thin foil sheets into a cylinder or stacking many layers. Second, the plate separation is minimised by using very thin dielectric films, sometimes just a few micrometres. Third, high-$\varepsilon_r$ materials such as barium titanate ceramics are used. A multilayer ceramic capacitor can achieve several hundred microfarads in a package a few millimetres across — roughly $10^{12}$ times the capacitance of the simple 10 cm plate example above.

Limitations of the Model

The formula $C = \varepsilon_0 \varepsilon_r A / d$ assumes the plates are infinite, so that the electric field between them is perfectly uniform and fringing fields at the edges are negligible. In practice the formula is accurate to within about 1 % when the plate dimensions are at least ten times the separation. For small plates or large gaps the actual capacitance is higher than the formula predicts, because the fringing fields add additional stored energy. More precise results require numerical field solvers or correction factors.

The formula also assumes the dielectric is linear and uniform. In practice $\varepsilon_r$ for many ceramics varies with voltage, temperature, and frequency, so manufacturers specify values at particular test conditions.