Physics · Electrostatics

Capacitance of a parallel-plate capacitor

For large parallel plates fully filled with a uniform linear dielectric, capacitance is proportional to overlap area and permittivity and inversely proportional to perpendicular separation.

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Formula

Calculate

C = ε_{0} ε_{r} \frac{A}{d}

C in F; ε0 in F/m; A in m²; d in m; εr dimensionless.

Variables & units

Variables & units
Symbol	Meaning	Unit
$C$	Capacitance	F
$ε0$	Vacuum permittivity	F/m
$εr$	Relative permittivity	dimensionless
$A$	Overlapping plate area	m²
$d$	Plate separation	m

LaTeX

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

QUANTA STUDYquanta-study.de

Inverse curve with normalized points: half gap double capacitance, unit gap unit capacitance, double gap half capacitance. — At fixed plate area and permittivity, capacitance is inversely proportional to gap. Half the gap doubles capacitance; twice the gap halves it.

Parallel plate capacitance calculator

Find the capacitance, plate area, gap or relative permittivity of a parallel plate capacitor, also with two materials; edge effects are ignored.

Calculation model

Find

Plate area A

Area where the plates face each other, counted once, e.g. 200 cm².

Perpendicular plate separation d

Gap between the plates, small compared with their width, e.g. 1 mm.

Relative permittivity εr

Unitless number for the material filling the gap: 1 for air, 2 to 4 for plastics, 5 to 10 for glass.

Filled with example values. Type your own and the result updates straight away.

Result: Capacitance C

C = 885.4 pF

With A = 200 cm², d = 1 mm and εr = 5, C = 885.4 pF.

Working · rearrange for C, substitute, evaluate

1. Rearrange for C
$C=\varepsilon_0\varepsilon_r\frac Ad$
2. Substitute values (in SI units)
- A = 200 cm² = 0.02 m²
- d = 1 mm = 0.001 m
$\frac{8.85419\times10^{-12}\cdot 5\cdot 0.02}{0.001}=8.85419\times10^{-10}$
3. Result
C = 885.4 pF

ε₀≈8.8541878188·10⁻¹² F/m: rounded measured CODATA2022 value, not an exact SI definition.

Quick tips

Twice the area, twice the capacitance; twice the gap, half the capacitance. A material with εr = 5 filling the whole gap makes it five times larger.
Convert to m² and m first: 200 cm² = 0.02 m², 1 mm = 0.001 m. With air, C ≈ 8.85 pF/m · 20 m ≈ 177 pF.

Assumptions and limits

Large parallel plates; each lateral dimension large compared with the gap, fringe fields neglected. A uniform linear isotropic dielectric fills the whole gap, with εr>0 and d>0. Count A once. A=0 is only a formal zero-area limit, not a real device model. Geometry/material remain fixed in the linear regime.

Derivation & background

This approximation neglects edge effects and assumes d much smaller than plate dimensions and a uniform dielectric. In vacuum $\varepsilon_r=1$ . It complements $C=\frac QU$ with geometric dependence.

Exam blueprint

Validity range

Large parallel plates; d small compared with both lateral dimensions, neglected fringe fields and uniform linear isotropic dielectric throughout the gap. A is one overlap area, not both plate areas added.

Derivation steps

Free surface-charge density determines the field; the uniform voltage integral then gives geometric capacitance.

1Assume large parallel plates with overlap area A and perpendicular separation d.
2The gap is completely filled by a uniform linear isotropic dielectric; neglect fringe fields.
3Free plate-charge magnitude gives $\sigma_f=\frac QA$ .
4Gauss for D and the constitutive law give $E=\frac{Q}{\varepsilon_0\varepsilon_r A}$ .
5In the ideal uniform gap, voltage magnitude is $U=Ed$ .
6Using definition $C=\frac QU$ , Q cancels: $C=\varepsilon_0\varepsilon_r\frac Ad$ .
7Check larger area, smaller separation and complete material filling separately; layers require their own balance.

Rearrangements

Plate separation

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

Neglect edge effects.

Task variant

Large parallel plates have overlap $A=0.010\,\mathrm{m^2}$ and separation $d=1.0\,\mathrm{mm}$ . Approximate air by $\varepsilon_r=1$ and neglect fringe fields. Find capacitance.

With $A=0.010\,\mathrm{m^2}$ , $d=1.0\,\mathrm{mm}$ and approximate air $\varepsilon_r=1$ , $C=\varepsilon_0\frac{0.010\,\mathrm{m^2}}{0.0010\,\mathrm m}\approx88.5\,\mathrm{pF}$ .

Common mistakes

Adding both plate areas or ignoring the air remaining in a partly filled gap.

Use one overlap area. Layers along the field add voltage drops; side-by-side area regions share voltage.

Exam context

Estimate component capacitance, geometric effects and energy in electric fields.

Formula cluster

Electrostatics

Capacitance relates electrode charge and voltage. Geometry determines its value; energy and RC timing require additional model information.

C = Q/U

Capacitance of a Capacitor

E = ½CU²

Energy Stored in a Capacitor

τ = RC; q(t)=Q(1−e^(−t/τ))

RC time constant

Worked example

With $A=0.010\,\mathrm{m^2}$ , $d=1.0\,\mathrm{mm}$ and approximate air $\varepsilon_r=1$ , $C=\varepsilon_0\frac{0.010\,\mathrm{m^2}}{0.0010\,\mathrm m}\approx88.5\,\mathrm{pF}$ .

Applications

Estimate component capacitance, geometric effects and energy in electric fields.

How do you calculate with Capacitance of a parallel-plate capacitor?

Here is how to work through a typical Capacitance of a parallel-plate capacitor (C = ε₀ εr A/d) task step by step:

1
Task
Large parallel plates have overlap $A=0.010\,\mathrm{m^2}$ and separation $d=1.0\,\mathrm{mm}$ . Approximate air by $\varepsilon_r=1$ and neglect fringe fields. Find capacitance.
Solution path
With $A=0.010\,\mathrm{m^2}$ , $d=1.0\,\mathrm{mm}$ and approximate air $\varepsilon_r=1$ , $C=\varepsilon_0\frac{0.010\,\mathrm{m^2}}{0.0010\,\mathrm m}\approx88.5\,\mathrm{pF}$ .

Understand and apply the equation

An ideal parallel-plate capacitor’s capacitance depends on geometry and material: greater overlap and permittivity increase it, while greater separation decreases it. The approximation requires an approximately uniform field in the relevant gap. A fully filled gap and two layers show why material arrangement matters as much as an $\varepsilon_r$ value.

Material arrangement matters too

Large overlap and a small gap permit more charge separation at the same voltage. The plate formula summarises this in a geometric approximation. “Large” means large relative to separation: even a small sensor may qualify if both lateral dimensions greatly exceed its gap.[1]

Before substitution, check whether material fills the entire gap uniformly. Layers along the field add voltage drops. Regions side by side between the same plates share a voltage. The two arrangements can therefore have different capacitances despite identical material fractions.[5][4]

How does plate capacitance depend on overlap area, separation and dielectric?

C=\varepsilon_0\varepsilon_r\frac Ad

A is overlap area of one plate, not the sum of both plate areas; d is normal separation.[1][2]

The plates carry equal and opposite free charges. In $C=\frac QU$ , Q denotes their positive magnitude and U voltage magnitude. The model contains a uniform linear isotropic dielectric filling the whole gap, or vacuum with $\varepsilon_r=1$ . In the linear regime Q is proportional to U while geometry and material remain fixed. Doubling charging voltage therefore doubles Q rather than C.[1][2]

Connect field, voltage and free plate charge

\begin{aligned}D_n&=\sigma_f=\frac QA\\E&=\frac{D_n}{\varepsilon_0\varepsilon_r}\\U&=Ed\end{aligned}

In the ideal uniform gap, free surface charge gives normal displacement field; the constitutive relation gives E.[5]

The opposite plates produce approximately uniform field between them. Outside, ideal infinite-plate contributions cancel. With $U=\frac{Qd}{\varepsilon_0\varepsilon_r A}$ , ratio $\frac QU$ becomes $\varepsilon_0\varepsilon_r\frac Ad$ . In vacuum the same form follows directly from Gauss’s electric-field law. For a dielectric, the D description distinguishes free plate charge from bound polarization charge.[5]

\begin{aligned}d&=\frac{\varepsilon_0\varepsilon_r A}{C}\\A&=\frac{Cd}{\varepsilon_0\varepsilon_r}\end{aligned}

Assume positive C, d, A and suitable material values. A calculated d must still be small relative to plate dimensions.

What “large plates” means

Large refers to lateral dimensions relative to separation rather than an absolute minimum size. Both sides of rectangular plates should greatly exceed d; area alone can mislead for a very narrow strip. Fringe fields contribute real additional field energy and affect C. This approximation is not an exact statement for every finite metal surface.[1]

The entire overlap must be uniformly filled. Material layers along the field direction act as a series arrangement; fully filled regions side by side approximately act in parallel. A simple arithmetic average permittivity is therefore not independent of spatial arrangement. Leakage conduction, frequency dependence or nonlinear polarization require further material information.[5][4]

Area and separation in consistent SI units

\begin{aligned}[C]&=\frac{\mathrm F}{\mathrm m}\frac{\mathrm{m^2}}{\mathrm m}=\mathrm F\\1\,\mathrm{cm^2}&=10^{-4}\,\mathrm{m^2}\\1\,\mathrm{pF}&=10^{-12}\,\mathrm F\end{aligned}

Area conversion is quadratic. A millimetre separation is 10⁻³ m with no extra square factor.

We use $\varepsilon_0\approx8.8541878188\cdot10^{-12}\,\mathrm{F/m}$ with published CODATA 2022 uncertainty. Example $\varepsilon_r$ values are linear model inputs, not invented data for named materials. For $A=0.010\,\mathrm{m^2}$ and $d=1.0\,\mathrm{mm}$ in approximate air, $88.5\,\mathrm{pF}$ is suitable rounding. Accurate material modelling would need to check the $\varepsilon_r\approx1$ approximation.[7]

Fully filled gap and two layers

A linear model capacitor

Given: Square parallel plates have $A=0.0180\,\mathrm{m^2}$ and $d=0.900\,\mathrm{mm}$ . A uniform linear dielectric with specified $\varepsilon_r=2.80$ fills the entire gap. $U=300\,\mathrm V$ ; fringe fields and leakage neglected.
Find: Find C, Q and field magnitude.

Approach: Complete uniform filling of the overlap permits geometric capacitance; the same voltage gives charge and normal field magnitude.
$\begin{aligned}C&=\varepsilon_0\varepsilon_r\frac Ad\\Q&=CU\\E&=\frac Ud\end{aligned}$
Insert separation in metres and calculate capacitance.
$C=\varepsilon_0(2.80)\frac{0.0180}{0.000900}\approx4.95834518\times10^{-10}\,\mathrm F$
Apply $Q=CU$ and $E=\frac{U}{d}$ to the same model.
$\begin{aligned}Q&\approx1.48750355\times10^{-7}\,\mathrm C\\E&=3.33333333\times10^5\,\mathrm{V/m}\end{aligned}$

Result: $C\approx495.835\,\mathrm{pF}$ and $Q\approx148.750\,\mathrm{nC}$ ; field magnitude ≈ 333.333 kV/m. Capacitance does not depend on the chosen 300 V value.

Check: From $\frac QA$ and $E=\frac{Q}{\varepsilon_0\varepsilon_rA}$ , field magnitude agrees independently. Side length $\sqrt A\approx0.134\,\mathrm m$ is about 149 times d, consistent with the approximation. The energy check gives $\frac12CU^2\approx22.3126\,\mathrm{\mu J}$ .

Two dielectric layers along the field direction

Given: $A=0.0100\,\mathrm{m^2}$ . Layer 1 has $d_1=0.400\,\mathrm{mm}$ and $\varepsilon_{r1}=2.00$ ; layer 2 has $d_2=0.600\,\mathrm{mm}$ and $\varepsilon_{r2}=5.00$ . Both are uniform linear models; no free charge at the internal interface. Total voltage 200 V.
Find: Calculate C and layer voltages.

Approach: Without free interfacial charge, normal displacement is equal in both layers. Their voltage drops add as for series capacitances.
$\begin{aligned}C&=\frac{\varepsilon_0A}{\frac{d_1}{\varepsilon_{r1}}+\frac{d_2}{\varepsilon_{r2}}}\\Q&=CU\\U_i&=\frac{Qd_i}{\varepsilon_0\varepsilon_{ri}A}\end{aligned}$
With the same D, voltage drops add; partial capacitances are effectively in series.
$\begin{aligned}C&=\frac{\varepsilon_0A}{\frac{d_1}{\varepsilon_{r1}}+\frac{d_2}{\varepsilon_{r2}}}\\&\approx276.693369\,\mathrm{pF}\end{aligned}$
Determine free charge from total voltage.
$\begin{aligned}Q&=CU\approx55.338674\,\mathrm{nC}\\D&=\frac QA\end{aligned}$
Determine each field from its own permittivity.
$\begin{aligned}E_1&=312500\,\mathrm{V/m}\\E_2&=125000\,\mathrm{V/m}\\U_1&=125\,\mathrm V\\U_2&=75\,\mathrm V\end{aligned}$

Result: $C\approx276.693\,\mathrm{pF}$ with voltage drops 125 V and 75 V. Smaller $\varepsilon_r$ gives the larger electric field at the same D.

Check: $125\,\mathrm V+75\,\mathrm V=200\,\mathrm V$ . Effective relative permittivity is $\begin{aligned}\varepsilon_{r,\mathrm{eff}}&=\frac{1.00\,\mathrm{mm}}{\frac{0.400\,\mathrm{mm}}2+\frac{0.600\,\mathrm{mm}}5}\\{}&=3.125\text{.}\end{aligned}$ A simple volume average of 3.8 would give incorrect capacitance; field direction through the layers determines series structure.

What changes at fixed charge rather than fixed voltage?

Increasing d while connected to an ideal voltage source keeps U fixed while C and Q fall. In an isolated charged capacitor Q instead stays approximately fixed and $U=\frac QC$ rises. Geometry formula is the same in both cases, but energy accounting differs. Plate movement requires including mechanical work and possible exchange with the source. The capacitance of a capacitor relates plate charge and voltage difference in both cases; geometry fixes their linear ratio.[3][2]

The same applies to dielectric insertion: at fixed Q, voltage and stored energy decrease; at fixed U, charge and stored energy increase, with extra energy from the source. A statement that greater C always means greater energy therefore omits the boundary condition. Energy in the capacitor must therefore be accounted for using the quantity actually held fixed.[2][3]

View diagram: boundary conditions and field energy

Same gap, different energy balances

QUANTA STUDYquanta-study.de

With

x=\frac{d}{d_0}

, an isolated capacitor gives

\frac{W}{W_0}=x

, while an ideal voltage source gives

\frac{W}{W_0}=\frac1x

. The plot shows field energy only; the connected case also requires source energy and mechanical work in the full balance.[3][1]

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Example: same materials, different arrangement

Area fractions instead of layers

Given: Square plates have $A=0.0100\,\mathrm{m^2}$ , $d=1.00\,\mathrm{mm}$ . Linear materials $\varepsilon_{r1}=2$ , $\varepsilon_{r2}=5$ each fill half the area and the full gap. $U=100\,\mathrm V$ . Neglect edge and interfacial corrections.
Find: Find capacitance and charge; compare with equally thick layers.

Approach: Both area regions share the same voltage. Their charges and capacitances add.
$\begin{aligned}C_{\parallel}&=\frac{\varepsilon_0 A}{2d}(\varepsilon_{r1}+\varepsilon_{r2})\\Q&=C_{\parallel}U\\C_{\mathrm{layers}}&=\frac{\varepsilon_0 A/d}{\frac1{2\varepsilon_{r1}}+\frac1{2\varepsilon_{r2}}}\end{aligned}$
Substitution: Vacuum capacitance supplies the common geometric scale.
$\begin{aligned}C_0&\approx88.5418782\,\mathrm{pF}\\C_{\parallel}&=3.5C_0\approx309.896574\,\mathrm{pF}\\Q&\approx30.9896574\,\mathrm{nC}\\C_{\mathrm{layers}}&=\frac{20}{7}C_0\approx252.976795\,\mathrm{pF}\end{aligned}$

Result: Side-by-side regions give about $309.897\,\mathrm{pF}$ ; equal-thickness layers give about $252.977\,\mathrm{pF}$ . Material fractions alone do not specify capacitance.

Check: Both regions have $E=\frac{U}{d}=100\,\mathrm{kV/m}$ , but different displacement fields. Charges $8.85419\,\mathrm{nC}$ and $22.13547\,\mathrm{nC}$ sum to total charge. Lowering voltage does not remove the capacitance difference in the linear model.

Five geometry checks

Calculate · Foundation

Vacuum plates with $A=0.0200\,\mathrm{m^2}$ and $d=2.00\,\mathrm{mm}$ : find C.

Show hint

Vacuum has $\varepsilon_r=1$ .

Learning goal: Calculate vacuum capacitance from geometry.

Show solution and calculation

Approach: $C=\varepsilon_0\frac Ad$ .
$C\approx88.5419\,\mathrm{pF}$ .

Check: $\frac Ad=10.0\,\mathrm m$ ; $\varepsilon_0$ times 10.0 m gives farads.

C\approx88.5419\,\mathrm{pF}

Transfer · Application

At unchanged material, overlap doubles and separation halves. How does C change?

Show hint

Multiply both geometry factors.

Learning goal: Understand combined geometry scaling.

Show solution and calculation

Approach: $\frac{C_2}{C_1}=\frac{A_2}{A_1}\frac{d_1}{d_2}$ .
$\frac{C_{\mathrm{new}}}{C_{\mathrm{old}}}=\frac2{\frac12}=4$ .
Voltage or charge additionally needs the boundary condition and is not determined by this ratio alone.

\frac{C_2}{C_1}=4

Rearrange · Application

A fully filled linear plate capacitor must have $C=1.00\,\mathrm{nF}$ with $A=0.0150\,\mathrm{m^2}$ and $\varepsilon_r=4.00$ . Find d.

Show hint

Insert C in farads first.

Learning goal: Find gap for a fully filled dielectric.

Show solution and calculation

Approach: $d=\frac{\varepsilon_0\varepsilon_r A}{C}$ .
$\begin{aligned}d&=\frac{\varepsilon_0\cdot4.00\cdot0.0150\,\mathrm{m^2}}{10^{-9}\,\mathrm F}\approx0.000531251\,\mathrm m\\{}&=0.531251\,\mathrm{mm}\text{.}\end{aligned}$ Lateral plate dimensions must still greatly exceed this value.

d\approx0.531251\,\mathrm{mm}

Interpret · Application

Two identical ideal vacuum capacitors initially have capacitance C0 and voltage U0. Both gaps are doubled quasistatically; one remains on an ideal voltage source and the other is isolated. Compare C, Q and U with initial values; area stays fixed and charge does not leak.

Show hint

Geometry sets C; source or isolation fixes U or Q.

Learning goal: Separate geometry change from electrical boundary condition.

Show solution and calculation

Approach: $C_2=\frac{C_0}{2}$ .
Inverse gap dependence halves both capacitances.
With the source connected, voltage stays fixed and charge halves.
In isolation, charge stays fixed and voltage doubles.
Both satisfy charge equal to capacitance times voltage.
Different voltage responses do not contradict identical geometry changes.

\begin{aligned}C_2&=\frac{C_0}{2}\\{}U_{\mathrm{connected}}&=U_0,&Q_{\mathrm{connected}}&=\frac{Q_0}{2}\\{}Q_{\mathrm{isolated}}&=Q_0,&U_{\mathrm{isolated}}&=2U_0\end{aligned}

[1][3]

Error analysis · Challenge

Half the gap d between large plates is filled with a linear dielectric $\varepsilon_r=3$ and half with vacuum. Area A is shared and fringing neglected. Someone uses factor 3 across the entire d. Find the correct factor relative to the empty capacitor.

Show hint

Layers lie along the field path; their voltage drops add.

Learning goal: Account for an air gap in layered dielectric.

Show solution and calculation

Approach: $C=\frac{\varepsilon_0 A}{\frac{d}{6}+\frac d2}$ .
The two equally thick layers act in series.
Inverse capacitance adds the vacuum path and the dielectric path reduced electrically by a factor of three.
The resulting overall factor is 1.5, not 3.
Applying relative permittivity three across the full gap would omit the remaining vacuum voltage drop.

\begin{aligned}\frac1C&=\frac{d}{2\varepsilon_0A}+\frac{d}{6\varepsilon_0A}\\{}C&=\frac32\frac{\varepsilon_0A}{d}=1.5C_0\end{aligned}

[4][2]

What real components additionally require

A practical capacitor may have many wound layers, contact areas and temperature- or frequency-dependent materials. The geometric formula is a local idealisation rather than a complete data sheet. Breakdown, leakage, series resistance and mechanical deformation cannot be determined from C alone. Even large static capacitance does not guarantee arbitrary operating voltage or lossless rapid energy delivery.[6][5]

In brief

Formula $C=\varepsilon_0\varepsilon_r\frac Ad$ belongs to a specific field geometry. Use one plate’s overlap and normal separation, checking fringe fields and material arrangement. Geometry determines $C$ ; source connection or isolation then determines charge, voltage and energy response.[1][3]

Sources for this explanation

Quanta · Updated October 2, 2026

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October 2, 2026Refined question headings and relevant paragraph links; preserved existing scientific explanations, worked examples and exercises.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Clarified introduction, model connections and solution steps; added an original third example, quantitative diagram and concise summary.

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Quanta. Capacitance of a parallel-plate capacitor. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/plattenkondensator-kapazitaet-geometrie

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Frequently asked questions about Capacitance of a parallel-plate capacitor

What does the capacitance of a parallel-plate capacitor describe physically?+

The geometric relation gives capacitance for two large parallel plates with a uniformly filled gap. Overlapping area increases charge stored per volt, while larger separation decreases it. Linear relative permittivity accounts for the specified dielectric. Use the area of one overlap, not the sum of both plate areas. At 0.010 m² and 1.0 mm, approximate air modelling gives about 88.5 pF. This value characterises the idealised configuration. An applied voltage then determines charge without itself changing constant linear capacitance. Real materials may depart from that assumption, requiring voltage-dependent or other measured characteristics.

How can I check a calculation involving the capacitance of a parallel-plate capacitor?+

Determine the actual overlapping area and perpendicular plate separation. Convert both to SI units and insert the permittivity of the material completely filling the gap. The worked example with 0.0180 m², 0.900 mm and model relative permittivity 2.80 gives about 495.835 pF. At 300 V this implies approximately 148.750 nC charge. Independently check the field through voltage and separation, then through free surface-charge density and permittivity. Both routes must agree. Also compare both lateral dimensions with separation; a large area alone does not guarantee small fringe effects for a very narrow strip.

What assumptions and limits apply to the capacitance of a parallel-plate capacitor?+

Plates should be parallel, with separation small compared with both lateral dimensions. Fringe fields are neglected. A uniform linear isotropic dielectric must fill the complete overlap and entire gap. Layers arranged along field direction cannot simply be replaced by an arithmetic average permittivity: different fields arise, and their voltage drops add. Side-by-side regions can instead approximately behave as parallel subcapacitors. Leakage, breakdown, frequency dependence and mechanical deformation are not fully described by the geometric formula. Practical components therefore also require material characteristics and actual construction data, particularly when ideal assumptions no longer match their operation.

How do I check units and scale for the capacitance of a parallel-plate capacitor?+

Permittivity in farads per metre, multiplied by square metres and divided by metres, gives farads. Converting square centimetres introduces a factor of 10⁻⁴, whereas millimetre separation introduces 10⁻³. Also distinguish picofarads from microfarads. For otherwise identical geometry, doubling area must double capacitance, and halving separation must also double it. Together those changes give a factor of four. This scaling provides an independent plausibility check on numerical substitution. Relative permittivity is dimensionless; it adds no unit factor and must correspond to the stated material model rather than an unrelated tabulated value.

When should I combine the capacitance of a parallel-plate capacitor with another relation?+

Combine capacitance and voltage with the charge relation to obtain charge on one plate. Add the linear capacitor-energy expression for stored energy, and the uniform voltage relation for ideal field magnitude. Before changing geometry, establish the electrical boundary condition. An isolated capacitor approximately retains charge; one connected to an ideal voltage source retains voltage. Charge, voltage and energy therefore change differently in those cases. Their comparison also requires mechanical work and, where applicable, energy exchanged with the source. The capacitance formula alone determines neither the complete energy flow nor the permissible operating voltage of the actual configuration.

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, University Physics Vol. 2, §8.1: Capacitors and Capacitance

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Capacitance of a parallel-plate capacitor

For large parallel plates fully filled with a uniform linear dielectric, capacitance is proportional to overlap area and permittivity and inversely proportional to perpendicular separation.

AdvancedExam-relevantUpper secondaryUniversity/collegeBachelor foundations

Free to browse · flashcards or a multiple-choice quiz · import from Essential

Symbol

Meaning

Unit

C

Capacitance

ε0

Vacuum permittivity

F/m

εr

Relative permittivity

dimensionless

A

Overlapping plate area

m²

d

Plate separation

Exam blueprint

Validity range

Derivation steps

Free surface-charge density determines the field; the uniform voltage integral then gives geometric capacitance.

1Assume large parallel plates with overlap area A and perpendicular separation d.
2The gap is completely filled by a uniform linear isotropic dielectric; neglect fringe fields.
3Free plate-charge magnitude gives $\sigma_f=\frac QA$ .
4Gauss for D and the constitutive law give $E=\frac{Q}{\varepsilon_0\varepsilon_r A}$ .
5In the ideal uniform gap, voltage magnitude is $U=Ed$ .
6Using definition $C=\frac QU$ , Q cancels: $C=\varepsilon_0\varepsilon_r\frac Ad$ .
7Check larger area, smaller separation and complete material filling separately; layers require their own balance.

How do you calculate with Capacitance of a parallel-plate capacitor?

Here is how to work through a typical Capacitance of a parallel-plate capacitor (C = ε₀ εr A/d) task step by step:

Task

Large parallel plates have overlap $A=0.010\,\mathrm{m^2}$ and separation $d=1.0\,\mathrm{mm}$ . Approximate air by $\varepsilon_r=1$ and neglect fringe fields. Find capacitance.

Solution path

With $A=0.010\,\mathrm{m^2}$ , $d=1.0\,\mathrm{mm}$ and approximate air $\varepsilon_r=1$ , $C=\varepsilon_0\frac{0.010\,\mathrm{m^2}}{0.0010\,\mathrm m}\approx88.5\,\mathrm{pF}$ .

Understand and apply the equation

\varepsilon_r

value.

Material arrangement matters too

How does plate capacitance depend on overlap area, separation and dielectric?

C=\varepsilon_0\varepsilon_r\frac Ad

A is overlap area of one plate, not the sum of both plate areas; d is normal separation.[1][2]

Connect field, voltage and free plate charge

\begin{aligned}D_n&=\sigma_f=\frac QA\\E&=\frac{D_n}{\varepsilon_0\varepsilon_r}\\U&=Ed\end{aligned}

In the ideal uniform gap, free surface charge gives normal displacement field; the constitutive relation gives E.[5]

\begin{aligned}d&=\frac{\varepsilon_0\varepsilon_r A}{C}\\A&=\frac{Cd}{\varepsilon_0\varepsilon_r}\end{aligned}

Assume positive C, d, A and suitable material values. A calculated d must still be small relative to plate dimensions.

What “large plates” means

Area and separation in consistent SI units

\begin{aligned}[C]&=\frac{\mathrm F}{\mathrm m}\frac{\mathrm{m^2}}{\mathrm m}=\mathrm F\\1\,\mathrm{cm^2}&=10^{-4}\,\mathrm{m^2}\\1\,\mathrm{pF}&=10^{-12}\,\mathrm F\end{aligned}

Area conversion is quadratic. A millimetre separation is 10⁻³ m with no extra square factor.

Fully filled gap and two layers

A linear model capacitor

Given: Square parallel plates have $A=0.0180\,\mathrm{m^2}$ and $d=0.900\,\mathrm{mm}$ . A uniform linear dielectric with specified $\varepsilon_r=2.80$ fills the entire gap. $U=300\,\mathrm V$ ; fringe fields and leakage neglected.
Find: Find C, Q and field magnitude.

Approach: Complete uniform filling of the overlap permits geometric capacitance; the same voltage gives charge and normal field magnitude.
$\begin{aligned}C&=\varepsilon_0\varepsilon_r\frac Ad\\Q&=CU\\E&=\frac Ud\end{aligned}$
Insert separation in metres and calculate capacitance.
$C=\varepsilon_0(2.80)\frac{0.0180}{0.000900}\approx4.95834518\times10^{-10}\,\mathrm F$
Apply $Q=CU$ and $E=\frac{U}{d}$ to the same model.
$\begin{aligned}Q&\approx1.48750355\times10^{-7}\,\mathrm C\\E&=3.33333333\times10^5\,\mathrm{V/m}\end{aligned}$

Result: $C\approx495.835\,\mathrm{pF}$ and $Q\approx148.750\,\mathrm{nC}$ ; field magnitude ≈ 333.333 kV/m. Capacitance does not depend on the chosen 300 V value.

Two dielectric layers along the field direction

Given: $A=0.0100\,\mathrm{m^2}$ . Layer 1 has $d_1=0.400\,\mathrm{mm}$ and $\varepsilon_{r1}=2.00$ ; layer 2 has $d_2=0.600\,\mathrm{mm}$ and $\varepsilon_{r2}=5.00$ . Both are uniform linear models; no free charge at the internal interface. Total voltage 200 V.
Find: Calculate C and layer voltages.

Approach: Without free interfacial charge, normal displacement is equal in both layers. Their voltage drops add as for series capacitances.
$\begin{aligned}C&=\frac{\varepsilon_0A}{\frac{d_1}{\varepsilon_{r1}}+\frac{d_2}{\varepsilon_{r2}}}\\Q&=CU\\U_i&=\frac{Qd_i}{\varepsilon_0\varepsilon_{ri}A}\end{aligned}$
With the same D, voltage drops add; partial capacitances are effectively in series.
$\begin{aligned}C&=\frac{\varepsilon_0A}{\frac{d_1}{\varepsilon_{r1}}+\frac{d_2}{\varepsilon_{r2}}}\\&\approx276.693369\,\mathrm{pF}\end{aligned}$
Determine free charge from total voltage.
$\begin{aligned}Q&=CU\approx55.338674\,\mathrm{nC}\\D&=\frac QA\end{aligned}$
Determine each field from its own permittivity.
$\begin{aligned}E_1&=312500\,\mathrm{V/m}\\E_2&=125000\,\mathrm{V/m}\\U_1&=125\,\mathrm V\\U_2&=75\,\mathrm V\end{aligned}$

Result: $C\approx276.693\,\mathrm{pF}$ with voltage drops 125 V and 75 V. Smaller $\varepsilon_r$ gives the larger electric field at the same D.

What changes at fixed charge rather than fixed voltage?

View diagram: boundary conditions and field energy

Same gap, different energy balances

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With

x=\frac{d}{d_0}

, an isolated capacitor gives

\frac{W}{W_0}=x

, while an ideal voltage source gives

\frac{W}{W_0}=\frac1x

. The plot shows field energy only; the connected case also requires source energy and mechanical work in the full balance.[3][1]

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Example: same materials, different arrangement

Area fractions instead of layers

Given: Square plates have $A=0.0100\,\mathrm{m^2}$ , $d=1.00\,\mathrm{mm}$ . Linear materials $\varepsilon_{r1}=2$ , $\varepsilon_{r2}=5$ each fill half the area and the full gap. $U=100\,\mathrm V$ . Neglect edge and interfacial corrections.
Find: Find capacitance and charge; compare with equally thick layers.

Approach: Both area regions share the same voltage. Their charges and capacitances add.
$\begin{aligned}C_{\parallel}&=\frac{\varepsilon_0 A}{2d}(\varepsilon_{r1}+\varepsilon_{r2})\\Q&=C_{\parallel}U\\C_{\mathrm{layers}}&=\frac{\varepsilon_0 A/d}{\frac1{2\varepsilon_{r1}}+\frac1{2\varepsilon_{r2}}}\end{aligned}$
Substitution: Vacuum capacitance supplies the common geometric scale.
$\begin{aligned}C_0&\approx88.5418782\,\mathrm{pF}\\C_{\parallel}&=3.5C_0\approx309.896574\,\mathrm{pF}\\Q&\approx30.9896574\,\mathrm{nC}\\C_{\mathrm{layers}}&=\frac{20}{7}C_0\approx252.976795\,\mathrm{pF}\end{aligned}$

Result: Side-by-side regions give about $309.897\,\mathrm{pF}$ ; equal-thickness layers give about $252.977\,\mathrm{pF}$ . Material fractions alone do not specify capacitance.

Five geometry checks

Calculate · Foundation

Vacuum plates with $A=0.0200\,\mathrm{m^2}$ and $d=2.00\,\mathrm{mm}$ : find C.

Show hint

Vacuum has $\varepsilon_r=1$ .

Learning goal: Calculate vacuum capacitance from geometry.

Show solution and calculation

Approach: $C=\varepsilon_0\frac Ad$ .
$C\approx88.5419\,\mathrm{pF}$ .

Check: $\frac Ad=10.0\,\mathrm m$ ; $\varepsilon_0$ times 10.0 m gives farads.

C\approx88.5419\,\mathrm{pF}

Transfer · Application

At unchanged material, overlap doubles and separation halves. How does C change?

Show hint

Multiply both geometry factors.

Learning goal: Understand combined geometry scaling.

Show solution and calculation

Approach: $\frac{C_2}{C_1}=\frac{A_2}{A_1}\frac{d_1}{d_2}$ .
$\frac{C_{\mathrm{new}}}{C_{\mathrm{old}}}=\frac2{\frac12}=4$ .
Voltage or charge additionally needs the boundary condition and is not determined by this ratio alone.

\frac{C_2}{C_1}=4

Rearrange · Application

A fully filled linear plate capacitor must have $C=1.00\,\mathrm{nF}$ with $A=0.0150\,\mathrm{m^2}$ and $\varepsilon_r=4.00$ . Find d.

Show hint

Insert C in farads first.

Learning goal: Find gap for a fully filled dielectric.

Show solution and calculation

Approach: $d=\frac{\varepsilon_0\varepsilon_r A}{C}$ .
$\begin{aligned}d&=\frac{\varepsilon_0\cdot4.00\cdot0.0150\,\mathrm{m^2}}{10^{-9}\,\mathrm F}\approx0.000531251\,\mathrm m\\{}&=0.531251\,\mathrm{mm}\text{.}\end{aligned}$ Lateral plate dimensions must still greatly exceed this value.

d\approx0.531251\,\mathrm{mm}

Interpret · Application

Two identical ideal vacuum capacitors initially have capacitance C0 and voltage U0. Both gaps are doubled quasistatically; one remains on an ideal voltage source and the other is isolated. Compare C, Q and U with initial values; area stays fixed and charge does not leak.

Show hint

Geometry sets C; source or isolation fixes U or Q.

Learning goal: Separate geometry change from electrical boundary condition.

Show solution and calculation

Approach: $C_2=\frac{C_0}{2}$ .
Inverse gap dependence halves both capacitances.
With the source connected, voltage stays fixed and charge halves.
In isolation, charge stays fixed and voltage doubles.
Both satisfy charge equal to capacitance times voltage.
Different voltage responses do not contradict identical geometry changes.

\begin{aligned}C_2&=\frac{C_0}{2}\\{}U_{\mathrm{connected}}&=U_0,&Q_{\mathrm{connected}}&=\frac{Q_0}{2}\\{}Q_{\mathrm{isolated}}&=Q_0,&U_{\mathrm{isolated}}&=2U_0\end{aligned}

[1][3]

Error analysis · Challenge

Half the gap d between large plates is filled with a linear dielectric $\varepsilon_r=3$ and half with vacuum. Area A is shared and fringing neglected. Someone uses factor 3 across the entire d. Find the correct factor relative to the empty capacitor.

Show hint

Layers lie along the field path; their voltage drops add.

Learning goal: Account for an air gap in layered dielectric.

Show solution and calculation

Approach: $C=\frac{\varepsilon_0 A}{\frac{d}{6}+\frac d2}$ .
The two equally thick layers act in series.
Inverse capacitance adds the vacuum path and the dielectric path reduced electrically by a factor of three.
The resulting overall factor is 1.5, not 3.
Applying relative permittivity three across the full gap would omit the remaining vacuum voltage drop.

\begin{aligned}\frac1C&=\frac{d}{2\varepsilon_0A}+\frac{d}{6\varepsilon_0A}\\{}C&=\frac32\frac{\varepsilon_0A}{d}=1.5C_0\end{aligned}

[4][2]

What real components additionally require

In brief

Sources for this explanation

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Capacitance of a parallel-plate capacitor

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Parallel plate capacitance calculator

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Task variant

Common mistakes

Exam context

Electrostatics

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How do you calculate with Capacitance of a parallel-plate capacitor?

Material arrangement matters too

How does plate capacitance depend on overlap area, separation and dielectric?

Connect field, voltage and free plate charge

What “large plates” means

Area and separation in consistent SI units

Fully filled gap and two layers

A linear model capacitor

Two dielectric layers along the field direction

What changes at fixed charge rather than fixed voltage?

Same gap, different energy balances

Example: same materials, different arrangement

Area fractions instead of layers

Five geometry checks

Vacuum plates with A=0.0200 m2A=0.0200\,\mathrm{m^2}A=0.0200m2 and d=2.00 mmd=2.00\,\mathrm{mm}d=2.00mm: find C.

At unchanged material, overlap doubles and separation halves. How does C change?

A fully filled linear plate capacitor must have C=1.00 nFC=1.00\,\mathrm{nF}C=1.00nF with A=0.0150 m2A=0.0150\,\mathrm{m^2}A=0.0150m2 and εr=4.00\varepsilon_r=4.00εr​=4.00. Find d.

Two identical ideal vacuum capacitors initially have capacitance C0 and voltage U0. Both gaps are doubled quasistatically; one remains on an ideal voltage source and the other is isolated. Compare C, Q and U with initial values; area stays fixed and charge does not leak.

Half the gap d between large plates is filled with a linear dielectric εr=3\varepsilon_r=3εr​=3 and half with vacuum. Area A is shared and fringing neglected. Someone uses factor 3 across the entire d. Find the correct factor relative to the empty capacitor.

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Frequently asked questions about Capacitance of a parallel-plate capacitor

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Retain Capacitance of a parallel-plate capacitor for exams

Capacitance of a parallel-plate capacitor

Variables & units

Parallel plate capacitance calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Electrostatics

Worked example

Applications

How do you calculate with Capacitance of a parallel-plate capacitor?

Material arrangement matters too

How does plate capacitance depend on overlap area, separation and dielectric?

Connect field, voltage and free plate charge

What “large plates” means

Area and separation in consistent SI units

Fully filled gap and two layers

A linear model capacitor

Two dielectric layers along the field direction

What changes at fixed charge rather than fixed voltage?

Same gap, different energy balances

Example: same materials, different arrangement

Area fractions instead of layers

Five geometry checks

Vacuum plates with A=0.0200 m2A=0.0200\,\mathrm{m^2}A=0.0200m2 and d=2.00 mmd=2.00\,\mathrm{mm}d=2.00mm: find C.

At unchanged material, overlap doubles and separation halves. How does C change?

A fully filled linear plate capacitor must have C=1.00 nFC=1.00\,\mathrm{nF}C=1.00nF with A=0.0150 m2A=0.0150\,\mathrm{m^2}A=0.0150m2 and εr=4.00\varepsilon_r=4.00εr​=4.00. Find d.

Two identical ideal vacuum capacitors initially have capacitance C0 and voltage U0. Both gaps are doubled quasistatically; one remains on an ideal voltage source and the other is isolated. Compare C, Q and U with initial values; area stays fixed and charge does not leak.

Half the gap d between large plates is filled with a linear dielectric εr=3\varepsilon_r=3εr​=3 and half with vacuum. Area A is shared and fringing neglected. Someone uses factor 3 across the entire d. Find the correct factor relative to the empty capacitor.

What real components additionally require

In brief

Sources for this explanation

Frequently asked questions about Capacitance of a parallel-plate capacitor

Subject references and further reading

Vacuum plates with $A=0.0200\,\mathrm{m^2}$ and $d=2.00\,\mathrm{mm}$ : find C.

A fully filled linear plate capacitor must have $C=1.00\,\mathrm{nF}$ with $A=0.0150\,\mathrm{m^2}$ and $\varepsilon_r=4.00$ . Find d.

Half the gap d between large plates is filled with a linear dielectric $\varepsilon_r=3$ and half with vacuum. Area A is shared and fringing neglected. Someone uses factor 3 across the entire d. Find the correct factor relative to the empty capacitor.

Vacuum plates with $A=0.0200\,\mathrm{m^2}$ and $d=2.00\,\mathrm{mm}$ : find C.

A fully filled linear plate capacitor must have $C=1.00\,\mathrm{nF}$ with $A=0.0150\,\mathrm{m^2}$ and $\varepsilon_r=4.00$ . Find d.

Half the gap d between large plates is filled with a linear dielectric $\varepsilon_r=3$ and half with vacuum. Area A is shared and fringing neglected. Someone uses factor 3 across the entire d. Find the correct factor relative to the empty capacitor.