Parallel-Plate Capacitor

In one sentence: The parallel-plate formula says a capacitor holds more charge per volt when its plates are bigger, closer together, or separated by a material with a higher permittivity.

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

Read it out loud: “C equals epsilon sub zero times epsilon sub r times A, divided by d. Read C as the capacitance, in farads. Epsilon sub zero, said EP-sih-lon zero, is the permittivity of free space, a fixed constant. Epsilon sub r is the relative permittivity of the material between the plates, a plain number. A is the area of one plate, in square meters. d is the gap between the plates, in meters.”

Before you start

The ideas and signs this page uses, explained from zero. For powers of ten, Greek letters and other notation in general, see Math basics.

Powers of ten, like $10^{-12}$ and $8.85 \times 10^{-12}$: Physical constants are often written with powers of ten. $10^{-12}$ means divide by 10 twelve times, which is 0.000000000001, a trillionth. In $8.85 \times 10^{-12}$ the number 8.85 is multiplied by that tiny value. Multiplying powers adds their exponents, and dividing subtracts them. The exponent tells you where the decimal point goes.
Permittivity and the Greek letter epsilon ( $\varepsilon$ ): The Greek letter $\varepsilon$ , said EP-sih-lon, is the usual symbol for permittivity. Permittivity measures how a material responds to an electric field, by lining up tiny charges inside it. A material with a high permittivity lets a capacitor store more charge per volt. The value for empty space is written $\varepsilon_0$ , the permittivity of free space.
Relative permittivity ( $\varepsilon_r$ ) and dielectrics: The relative permittivity $\varepsilon_r$ is a plain number that compares a material with empty space. It is exactly 1 for a vacuum and about 1.0006 for air. Glass and many plastics have values several times bigger. A dielectric is an insulating material placed between the plates, and the material’s permittivity is $\varepsilon_0 \varepsilon_r$ .
Area, the square meter and the square centimeter: Area is the flat space a shape covers. A square meter, written $\text{m}^2$ , is a square one meter long on each side. A square centimeter, written $\text{cm}^2$ , is a square one centimeter on each side. Since one centimeter is 0.01 m, one square centimeter is $10^{-4}$ m $^2$ , so 100 cm $^2$ equals 0.01 m $^2$ .
The prefix milli (m) and millimeters: A millimeter, written mm, is one thousandth of a meter, so 1 mm = 0.001 m. The prefix milli always means divide by 1000. A gap given in millimeters must be converted to meters before it goes into the formula, because the formula needs every length in meters.
The farad (F) and the picofarad (pF): The farad, written F, is the unit of capacitance: one coulomb of charge per volt. Most capacitors are much smaller than a farad. The prefix pico, written p, means one trillionth, $10^{-12}$ , so 1 pF = $10^{-12}$ F. Small capacitors in circuits are often measured in picofarads.

What each symbol means

Symbol	Meaning
$C$	The capacitance of the capacitor, measured in farads (F). One farad is one coulomb of charge per volt. The example gives about 354 picofarads, which is a small capacitor.
$\varepsilon_0$	The permittivity of free space, a fixed constant measured in farads per meter (F/m). Its value is $8.8541878128 \times 10^{-12}$ F/m. It sets the scale of the formula, so that the capacitance comes out in farads when the sizes are given in meters. The name refers to empty space.
$\varepsilon_r$	The relative permittivity of the material between the plates. It is a plain number with no unit. It is exactly 1 for a vacuum and about 1.0006 for air. Larger values, such as 4 in the example, raise the capacitance by that factor.
$A$	The area of one plate, measured in square meters ( $\text{m}^2$ ). It is the flat space of the plate that holds the charge. The example uses 0.01 m $^2$ , which is 100 cm $^2$ , a square about 10 cm on each side.
$d$	The separation, the gap between the plates, measured in meters (m). It is the distance between the two plates at their closest point. The example uses 1 mm, which is 0.001 m. A smaller gap gives a bigger capacitance.

The formula piece by piece

$A$: Start with the plate area, the flat space that holds the charge. A bigger area gives more room for charge on the plates, so the capacitance rises in step with the area.
$d$: The gap between the plates. It sits in the bottom of the formula, because a bigger gap weakens the field between the plates for the same voltage. A smaller gap lets the same voltage hold more charge, so the capacitance falls as the gap grows.
$\varepsilon_0 \varepsilon_r$: The permittivity of the material between the plates. $\varepsilon_0$ is the fixed value for empty space, and $\varepsilon_r$ multiplies it for the material in the gap. Together they give the permittivity that the field meets as it crosses the gap.
$\frac{\varepsilon_0 \varepsilon_r A}{d}$: The permittivity times the plate area, divided by the gap. This fraction gives the capacitance in farads when every size is in meters.
$C = \frac{\varepsilon_0 \varepsilon_r A}{d}$: The capacitance of two flat, parallel plates equals the permittivity of free space times the material’s relative permittivity times the plate area, divided by the gap between the plates.

What the formula is doing, and why

A parallel-plate capacitor is two flat conducting plates facing each other, with a thin gap between them. When a voltage is applied, charge builds up on the plates. The electric field across the gap is what holds that charge. The formula tells you how much charge the capacitor stores per volt, for a given plate size, gap and material.

Why is the area on top? The area $A$ is in the top of the formula, so a bigger plate holds more charge for the same voltage. A larger plate has more room for the charge to spread across. If you double the area, you double the capacitance.

Why is the gap on the bottom? The gap $d$ is in the bottom, so moving the plates apart lowers the capacitance. For a given voltage, the field between the plates is the voltage divided by the gap. A smaller gap makes the field stronger, so more charge sits on the plates. Halving the gap doubles the capacitance.

Why does the material multiply? The constant $\varepsilon_0$ sets the scale in SI units. A dielectric, which is an insulating material between the plates, has $\varepsilon_r$ greater than 1. Its atoms line up with the field, which weakens the field inside the material, so more charge can sit on the plates for the same voltage. The capacitance is multiplied by $\varepsilon_r$ . A material with $\varepsilon_r = 4$ makes the capacitor four times bigger than it would be with empty space.

When the formula does not apply. It is an approximation. It works well when the plates are large compared with the gap, so that the field between them is nearly even. Near the edges the field spreads out, which adds a small extra capacitance that the formula leaves out. Curved plates, or plates far apart, need a different calculation.

An everyday picture

Think of two large trays facing each other with a small gap. Wider trays can hold more charge for the same push. Closer trays pull on each other’s charge more strongly, so each volt can push more charge onto them. Slide them apart and the same voltage holds less charge. Putting an insulating material between them is like filling the gap with something that lets the trays hold even more charge for the same push.

Worked example: A small capacitor from two plates

Two flat plates each have area $A = 0.01\ \text{m}^2$ , which is 100 cm $^2$ , a square about 10 cm on each side. They are separated by a gap $d = 1\ \text{mm} = 0.001\ \text{m}$ filled with a material of relative permittivity $\varepsilon_r = 4$ . Find the capacitance in farads and in picofarads. The plates are an illustration, not a specific product.

Write the formula and list the values in meters. The formula is $C = \frac{\varepsilon_0 \varepsilon_r A}{d}$ . The area is already in square meters, so $A = 0.01\ \text{m}^2$ . The gap is 1 mm, and a millimeter is one thousandth of a meter, so $d = 0.001\ \text{m}$ . The relative permittivity is a plain number with no unit, so $\varepsilon_r = 4$ .
Multiply the two permittivities. Multiply $\varepsilon_0$ by $\varepsilon_r$ : $8.8541878128 \times 10^{-12}\ \text{F/m} \times 4 = 35.4167512 \times 10^{-12}\ \text{F/m}$ . To write this in standard form, move the decimal point one place to the left and raise the exponent by one: $3.54167512 \times 10^{-11}\ \text{F/m}$ .
Multiply by the area and divide by the gap. First, $\frac{A}{d} = \frac{0.01\ \text{m}^2}{0.001\ \text{m}} = 10\ \text{m}$ . Dividing the numbers gives $0.01 \div 0.001 = 10$ , and one square meter divided by one meter leaves one meter. Then $C = 3.54167512 \times 10^{-11}\ \text{F/m} \times 10\ \text{m} = 3.54167512 \times 10^{-10}\ \text{F}$ . The meters cancel, leaving farads.
Convert the answer to picofarads. One picofarad is $10^{-12}$ F, so to count in picofarads we divide by $10^{-12}$ , which is the same as multiplying by $10^{12}$ . The exponent $-10$ rises by 12 to $+2$ , so $3.54167512 \times 10^{-10}\ \text{F} = 354.17\ \text{pF}$ , which is about 354 pF.
Check how the answer scales with the area. The area is on top of the formula, so doubling the plate area doubles the capacitance to about $2 \times 354 = 708$ pF. Halving the gap would also double it, because the gap is on the bottom. Both changes make sense, because the plates then hold twice the charge at the same voltage.
Check the effect of the material. Compare with empty space, where $\varepsilon_r = 1$ . The same plates would then have a capacitance four times smaller, $354.17 / 4 \approx 88.5$ pF. So the material in the gap is responsible for the factor of four in the result.

Answer: The capacitance is about 354 pF, which is $3.54 \times 10^{-10}$ F. Doubling the plate area would double it, and halving the gap would double it too. With empty space between the plates the value would be about 88.5 pF.

The numbers in this example are recomputed by script every time the site is built.

Common mistakes

Using centimeters in the formula. Convert to meters first. $1\ \text{cm}^2 = 10^{-4}\ \text{m}^2$ , so 100 cm $^2$ is 0.01 m $^2$ , not 100 m $^2$ .
Leaving out $\varepsilon_r$ or setting it to 0. If the gap holds only vacuum or air, use $\varepsilon_r = 1$ , which is very close to the value for air. A relative permittivity of 0 would give zero capacitance, which is never right for a gap that holds charge.
Thinking the capacitance grows when the plates move apart. The gap is on the bottom of the formula, so a bigger gap gives a smaller capacitance.
Using the formula for plates that are far apart or curved. Large gaps and curved electrodes need a different calculation, because the field no longer runs evenly between flat plates.

Where you will meet it

A first estimate for the capacitance of thin-film capacitors, which are made from very thin layers of material on circuit boards and chips.
Capacitive touch and proximity sensors, which detect a finger or an object by the change in capacitance it causes.
Capacitive microphones and pressure sensors, where a thin membrane moves and changes the gap between the plates.
Variable capacitors in radio tuners, where moving the plates or changing how much of their area overlaps tunes the circuit.

Questions people ask

What is permittivity?

Permittivity describes how a material responds to an electric field. The material’s tiny charges line up with the field, which weakens the field inside the material, so more charge can sit on the plates for the same voltage. Vacuum is the reference value, $\varepsilon_0$ , and materials multiply it by $\varepsilon_r$ .

Does the formula work for round capacitors?

Not exactly. The formula assumes flat plates close together. Cylindrical and spherical capacitors need their own formulas.

Is a bigger capacitor always better?

Not always. A bigger plate area gives more capacitance, but a larger capacitor takes more space, and its voltage rating and leakage also matter.

Capacitance: Capacitance tells you how much electric charge a capacitor stores for each volt across it.
Capacitors in Series and Parallel: Capacitors connected side by side add their capacitances together, while capacitors connected in a chain give a smaller total that you find with a special rule.
Energy Stored in a Capacitor: A charged capacitor holds energy in the electric field between its two plates, and doubling the voltage across it makes the stored energy four times bigger.