Capacitance calculator for parallel plate capacitors. Compute stored charge, energy density, electric field & electrostatic force with automatic SI prefix conversion for any dielectric material.
The Parallel-Plate Capacitance Formula
This tool runs the ideal parallel-plate model from classical electrostatics — the same relationship you get from applying Gauss’s law to two flat, oppositely charged conductors separated by a uniform gap. Capacitance comes down to three physical facts: how much conductor area faces the gap, how far apart the plates sit, and what fills the space between them.
$$C = \varepsilon_r \varepsilon_0 \frac{A}{d}$$
$\varepsilon_0$, the permittivity of free space, is a fixed physical constant. It never changes, no matter what you’re measuring. $\varepsilon_r$, the relative permittivity or dielectric constant, is the only material variable in the equation — air and vacuum both sit at 1, while ceramics, mica, and plastic films push it higher and let the same geometry hold more charge per volt.
| Symbol | Meaning | Value / Unit |
|---|---|---|
| C | Capacitance — the result | farads (F) |
| $\varepsilon_r$ | Relative permittivity of the material between the plates | dimensionless, minimum 1 |
| $\varepsilon_0$ | Permittivity of free space | 8.8541878128 × 10⁻¹² F/m |
| A | Area of plate overlap facing the gap | square meters (converted from mm², cm², or in²) |
| d | Distance separating the plates | meters (converted from µm, mm, cm, or in) |
Geometry does the heavy lifting here, so it helps to see where A and d actually sit on the plates.
A 9-Volt Film Capacitor, Worked Through by Hand
Take a small film capacitor: plates measuring 4 cm² facing each other, separated by 0.1 mm of polyester film ($\varepsilon_r \approx 3$), charged to 9 V. Nothing exotic — the kind of numbers you’d actually find on a component datasheet.
Convert everything to SI units first, because the formula only works in meters and volts, not centimeters and millimeters. 4 cm² becomes $4 \times 10^{-4}$ m². 0.1 mm becomes $1 \times 10^{-4}$ m. Divide the two: $4 \times 10^{-4} / 1 \times 10^{-4} = 4$.
Multiply that ratio by $\varepsilon_0$ to get the vacuum capacitance this geometry would have with nothing but air in the gap: $8.8541878128 \times 10^{-12} \times 4 = 35.42$ pF. Multiply by $\varepsilon_r = 3$ to bring the dielectric in, and the working capacitance comes out to 106.25 pF.
From there, everything else follows directly. Charge is $Q = CV$: 106.25 pF times 9 V gives 956.25 pC, or roughly 5.97 × 10⁹ electrons sitting on the positive plate. Stored energy is $U = \frac{1}{2}CV^2$: half of 106.25 pF times 81 gives 4.30 nJ. The field between the plates is $E = V/d$: 9 V over $1 \times 10^{-4}$ m works out to 90,000 V/m — a reminder that even modest voltages produce strong fields once the gap gets small.
| Quantity | Formula | Result |
|---|---|---|
| Capacitance | $C = \varepsilon_r \varepsilon_0 A/d$ | 106.25 pF |
| Stored charge | $Q = CV$ | 956.25 pC |
| Stored energy | $U = \frac{1}{2}CV^2$ | 4.30 nJ |
| Electric field | $E = V/d$ | 90 kV/m |
| Surface charge density | $\sigma = Q/A$ | 2.39 µC/m² |
| Force between plates | $F = \frac{1}{2}QE$ | 43.03 µN |
| Electrostatic pressure | $P = F/A$ | 0.108 Pa |
One detail worth noticing: the pressure between the plates (0.108 Pa) and the energy density in the gap (0.108 J/m³) come out identical. That’s not a rounding coincidence — pressure and energy density carry the same units in field theory, and for a uniform field they’re numerically the same quantity viewed two ways.
Why Capacitance Ignores Voltage, But Energy Doesn’t
Notice that V never appears in the capacitance formula itself. For an ideal linear dielectric like the one this calculator assumes, capacitance is fixed by geometry and material alone — apply 1 V or 100 V to the same physical capacitor and C stays at 106.25 pF either way.
Charge scales linearly with voltage, and stored energy scales with the square of it, so doubling the applied voltage quadruples the energy even though the capacitance number on the label never moves.
There’s no compliance threshold to check here — this isn’t a wiring calculation with a pass/fail limit. What’s useful instead is context: 106 pF sits squarely in ordinary component territory.
Small ceramic and film capacitors commonly run from a few picofarads up through several nanofarads, while electrolytic types climb into the microfarad-to-farad range for bulk energy storage. A result in the tens or low hundreds of picofarads reads as a small-signal or timing-circuit part, not a power-storage component.
What Actually Moves the Capacitance Number
Separation distance dominates the result, and the reason is arithmetic as much as physics. Because d sits in the denominator and is usually entered in micrometers or millimeters, a small absolute change swings C by a large relative amount. Halve the gap and capacitance doubles; double it and capacitance is cut in half. Area, by contrast, scales in a straight line — twice the plate area, twice the capacitance, no surprises.
The dielectric constant has a hard floor built into the calculation: $\varepsilon_r$ can’t drop below 1, since vacuum is the physical baseline and nothing conducts less capacitance-boosting ability than empty space. From there it only goes up, and picking the wrong dielectric constant for a real material is one of the easier ways to misjudge a result before it’s even calculated.
Unit selection is the other common trap. The calculator accepts area in mm², cm², or in², and separation in µm, mm, cm, or in — pick the wrong one from the dropdown and the output is off by a clean factor of 100, 1,000, or more, not a rounding error you’d catch by eye. Voltage, meanwhile, changes nothing about capacitance directly. It only shows up once you move on to charge and energy.
Where the Ideal-Plate Model Runs Out
This formula assumes a perfectly uniform field between infinite flat plates, so it leaves out fringing fields at the plate edges, the dielectric’s voltage breakdown limit, and any temperature-driven drift in permittivity — a real component’s datasheet will carry tolerances and derating this calculation alone doesn’t capture.
Common Questions About Plate Capacitance
What is the formula for parallel-plate capacitance?
$C = \varepsilon_r \varepsilon_0 (A/d)$, where A is the facing plate area, d is the separation distance, and $\varepsilon_r$ is the dielectric constant of whatever fills the gap. All three terms have to be in consistent SI units before the answer means anything in farads.
Does increasing the voltage increase capacitance?
No, not for an ideal linear dielectric. Capacitance is set entirely by plate geometry and the material between them; voltage only determines how much charge and energy that fixed capacitance ends up holding.
Why does adding a dielectric increase capacitance?
A dielectric’s molecules polarize in response to the field, which partially cancels the field inside the material. That lets the plates hold more charge for the same applied voltage, and $\varepsilon_r$ is exactly the multiplier that describes how much more.
What happens to capacitance if the plate separation doubles?
Capacitance is cut in half, since d sits in the denominator and everything else — area, dielectric constant — stays the same. The relationship is a direct inverse, not an approximation.
Can relative permittivity be less than 1?
No. Vacuum defines the floor at exactly 1, and every real dielectric sits at or above that value. A result below 1 signals a data entry error, not a valid material property.
What’s the difference between relative and absolute permittivity?
Relative permittivity ($\varepsilon_r$) is a dimensionless ratio comparing a material to vacuum. Absolute permittivity ($\varepsilon = \varepsilon_r \varepsilon_0$) carries units of farads per meter and is the actual value plugged into the capacitance equation.
How is stored energy different from the capacitance rating?
Capacitance is a fixed property of the geometry and material, unchanged by how the capacitor is used. Stored energy depends on the voltage applied at any given moment, growing with the square of it, so the same capacitor can hold wildly different amounts of energy depending on charge level.
Why is electric field equal to voltage divided by distance?
In a uniform field between parallel plates, voltage is defined as the field strength multiplied by the distance over which it acts. Rearranging that definition gives $E = V/d$ directly, without needing any additional physical assumption beyond uniformity.