RF Toolbox

Parallel-Plate Capacitor

Computes the capacitance of a parallel-plate capacitor from the overlapping plate area, the gap between plates, and the dielectric constant of the material between them. A multi-plate stack of N plates acts like N−1 capacitors in parallel. Enter an applied voltage to also see the electric field and the stored energy. (This is the ideal formula; real capacitors add fringing at the plate edges.)

Equations & Parameters ▸
\(C = \dfrac{\varepsilon_0\,\varepsilon_r\,A\,(N-1)}{d} \qquad E = \dfrac{V}{d} \qquad U = \tfrac{1}{2}C V^2\)
AOverlapping plate area (mm²).
dSeparation between adjacent plates (mm).
εrRelative permittivity: air ≈ 1, FR-4 ≈ 4.4, mica ≈ 6, alumina ≈ 9.8, class-I ceramic 15–100.
NNumber of plates (2 for a simple capacitor). A stack of N plates gives N−1 gaps in parallel.
ε0Permittivity of free space = 8.854×10⁻¹² F/m.
References: D. K. Cheng, Field and Wave Electromagnetics, 2nd ed., Addison-Wesley, 1989. · D. M. Pozar, Microwave Engineering, 4th ed., Wiley, 2012.
Inputs
mm²
Overlap area
mm
Plate gap
1 = air
Default 2
V
For field/energy
Results

Capacitance

Capacitance C
C per gap

At applied voltage

Electric field
Stored energy
Charge
Diagram