Parallel Plate Capacitor Calculator - C from Area, Gap, εr

Parallel plate capacitance, before and after you pick a dielectric

Two plates of 10 cm² half a millimeter apart hold 17.719 pF with air between them and 478.13 pF with tantalum pentoxide, the same geometry and 27 times the capacitance. Enter the plate area, the gap and the dielectric, optionally the number of stacked plates and a voltage, and the calculator returns the capacitance from C = ε₀ × εr × A ÷ d, the charge and energy at that voltage, and a table of what every other listed dielectric would give with the same plates.

Estimating by eye

Area in square inches and gap in mils mixed with metric constants
The plate count read as the number of layers, one too many
Permittivity taken as "about the same" for every plastic
No idea whether the gap is small enough for the formula to hold

With the calculator

in², cm², mm², m² and mm, mil, µm, in converted for you
(n − 1) active layers counted from the plates you stack
Eleven dielectrics side by side for your exact plates
A warning once the gap exceeds 10% of the plate width

Six entries on the form

Plate area and its unit - the area of one plate where it overlaps the other, in in², cm², mm² or m². A rectangle is length times width; a disc is π × r².
Gap and its unit - the dielectric thickness in mm, mil (a thousandth of an inch), µm or inches. One mil is 25.4 µm.
Dielectric - pick one of eleven materials, or "Other" to type the relative permittivity from a datasheet (at least 1).
Number of plates - optional, 2 if left empty. Interleaved plates give n − 1 active layers.
Voltage - optional. With it the result adds charge, stored energy and the field in the gap.
Read the result - the capacitance with its prefix (fF, pF, nF, µF), the inputs as used, and the comparison table.

How the capacitance is computed

For two parallel plates the capacitance is C = ε₀ × εr × A ÷ d. ε₀ is the vacuum permittivity, 8.8541878188 × 10⁻¹² F/m in the current CODATA value; εr is how much the material between the plates multiplies it; A is the overlapping area and d the gap. The model holds when the gap is much smaller than the plates, because it ignores the field that bulges out at the edges. Stacking n interleaved plates puts n − 1 such capacitors in parallel, so the result is multiplied by n − 1.

C = ε₀ × εr × A × (n − 1) ÷ d
Q = C × V    W = C × V² ÷ 2    E = V ÷ d

Vacuum, 10 cm², 0.5 mm, 100 V. 8.8542 × 10⁻¹² × 0.001 m² ÷ 0.0005 m = 17.708 pF. At 100 V the plates hold 1.7708 nC, store 88.542 nJ, and the field in the gap is 200 kV/m.

Polyimide film, 1 in², 1 mil, 12 V. 1 in² is 6.4516 cm² and 1 mil is 25.4 µm, so C = 8.8542 × 10⁻¹² × 3.4 × 0.00064516 ÷ 0.0000254 = 764.65 pF. At 12 V: 9.1758 nC and 55.055 nJ.

Strontium titanate, 1 cm², 0.1 mm, 11 plates, 50 V. Ten active layers with εr 310: 27.448 nF, holding 1.3724 µC and 34.31 µJ at 50 V, with 500 kV/m across each layer.

What each change gains you

Every row starts from the same reference, 10 cm² of plate, a 0.5 mm air gap and 17.719 pF, and changes one thing. The gain column is the new capacitance divided by the old.

Change	Before	After	Gain
Air replaced by PTFE	17.719 pF	37.188 pF	× 2.099
Gap cut from 0.5 mm to 0.05 mm	17.719 pF	177.19 pF	× 10
2 plates replaced by 11 interleaved plates	17.719 pF	177.19 pF	× 10
Plate shrunk from 10 cm² to 1 in²	17.719 pF	11.431 pF	× 0.645
Air replaced by tantalum pentoxide	17.719 pF	478.13 pF	× 26.98
Largest single change in this table	17.719 pF	478.13 pF	× 26.98

Two routes give exactly ten times: a gap ten times thinner or ten active layers instead of one. Multilayer ceramic and stacked film capacitors take the second route, since a thinner dielectric also withstands less voltage. The dielectric swap is the largest lever of the three.

Four layouts, from a chip to a desk demo

Each of these is one run of the calculator, with the inputs as typed and the figures it prints.

A 1 mm² oxide pad
Silicon dioxide, 100 µm thick, 5 V: 345.31 fF, a charge of 1.7266 pC and 4.3164 pJ. Femtofarads are the normal scale for structures this small.

Two 2 × 2 in plates an inch apart
Air, 4 in², 1 in gap: 900.12 fF. The gap is 50% of the plate width, so the calculator flags the result as a lower estimate.

Three oxides on 1 cm² at 10 µm
Anodic aluminum oxide 850 pF, tantalum pentoxide 2.3906 nF, niobium pentoxide 3.6302 nF. Same plates, only εr changes.

Ten times either way
A 0.05 mm air gap and eleven plates at 0.5 mm both land on 177.19 pF for 10 cm², ten times the two-plate value.

The permittivity figures, and where they come from

Relative permittivity depends on frequency, temperature and the exact grade of a material, so the calculator uses one representative value and the table shows what the sources print. Plastics, glass and air come from the Wikipedia list of relative permittivities (room temperature, mostly at 1 kHz); the three oxides are the amorphous anodic films in electrolytic capacitors, from the comparison table in the Wikipedia article on that type.

Material	Value used	As printed in the source
Vacuum	1	1 by definition
Air	1.00059	1.00058986 at STP, 900 kHz
PTFE (Teflon)	2.1	2.1
Polyethylene	2.25	2.25
Polyimide	3.4	3.4
Silicon dioxide	3.9	3.9
Pyrex glass	4.7	4.7, glasses in general 3.7 to 10
Aluminum oxide, anodic	9.6	9.6 amorphous, 11.6 to 14.2 crystalline
Tantalum pentoxide	27	27 amorphous
Niobium pentoxide	41	41 amorphous
Strontium titanate	310	310

Ceramic capacitors are left to the "Other" option on purpose. Class 1 ceramics span a relative permittivity of 6 to 200 and class 2 barium titanate grades 200 to 14,000, which is too wide for one preset; the datasheet value belongs in the box.

Capacitor questions people type in

How do I calculate the capacitance of two parallel plates?

Multiply ε₀ by the relative permittivity and the plate area, then divide by the gap, all in SI units. 10 cm² of plate and a 0.5 mm vacuum gap give 17.708 pF.

Why does the formula underestimate wide gaps?

It assumes the field stays between the plates. With a large gap part of the field spreads around the edges and adds capacitance. 100 cm² plates 50 mm apart compute to 1.7719 pF, but the gap is 50% of the plate width, so the calculator marks that figure as a lower estimate.

Does doubling the plates double the capacitance?

Not quite. Three interleaved plates make 2 layers, so going from 2 to 3 plates doubles it, while 11 plates give 10 layers and ten times the two-plate value. What grows is the number of layers, n − 1, not the number of plates.

What is a mil in capacitor film thickness?

A thousandth of an inch, 25.4 µm. A 1 in² capacitor with 1 mil of polyimide computes to 764.65 pF.

Why do electrolytic capacitors reach such large values?

Their dielectric is an oxide only nanometers thick grown on an etched, very large surface. With the same plates and a 10 µm oxide, tantalum pentoxide at εr 27 gives 2.3906 nF per cm², against 850 pF for anodic aluminum oxide at 9.6.

Is air really different from vacuum?

Barely. εr of dry air is about 1.00059, so the same plates hold 17.719 pF in air and 17.708 pF in vacuum, a difference under 0.1%.

Why is a thinner dielectric not always the better choice?

Capacitance rises as the gap shrinks, but the field rises too: 50 V across 0.1 mm is already 500 kV/m. Every insulator breaks down above some field strength, so a thinner layer means a lower voltage rating. The calculator prints the field so you can compare it with the dielectric strength in a datasheet.

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