Physics · Electricity

Capacitance of a capacitor

Capacitance relates the charge magnitude on one capacitor plate to the voltage between plates. For linear material and unchanged geometry, C is constant.

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Formula

Calculate

C = \frac{Q}{U}

C in farads [F] = [C/V] · Q in coulombs [C] · U in volts [V]

Variables & units

Variables & units
Symbol	Meaning	Unit
$C$	Capacitance in the chosen linear model	F
$Q$	Charge magnitude on one plate, not total net charge of both plates	C
$U$	Voltage magnitude between plates	V

LaTeX

C = \frac{Q}{U}

QUANTA STUDYquanta-study.de

Charge-voltage plot with two straight lines through the origin. The steeper line represents 100 µF, the shallower 50 µF. Endpoints at 12 V mark 1,200 and 600 µC. — At constant C, Q is proportional to U. At 12 V, one plate carries 600 µC for 50 µF or 1,200 µC for 100 µF. The other plate carries the opposite charge.

Capacitance calculator

Find the capacitance, charge or voltage of a capacitor: enter two values and the third follows from Q = C · U.

Find

Charge magnitude |Q|

Charge on one plate as a magnitude, not the sum of both plates, e.g. 600 µC.

Voltage magnitude |U|

Voltage between the plates as a magnitude, e.g. 12 V.

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

Working · rearrange for C, substitute, evaluate

1. Rearrange for C
$C=\frac{|Q|}{|U|}$
2. Substitute values (in SI units)
- |Q| = 600 µC = 0.0006 C
$C=\frac{0.0006}{12}=0.00005$
3. Result
C = 50 µF

Result: Capacitance C

C = 50 µF

With |Q| = 600 µC and |U| = 12 V, C = 50 µF.

Quick tips

Formula triangle: Q on top, C and U below. Cover the quantity you need and the rest is the calculation, e.g. C = Q ÷ U.
Unit check: 1 F = 1 C/V. One farad is huge; typical capacitors range from pF to mF, that is 10⁻¹² to 10⁻³ F.

Assumptions and limits

Linear capacitor with C > 0: |Q| = C|U|. Charge and voltage are magnitudes; plate polarity is not determined. At Q = U = 0, capacitance cannot be inferred.

Derivation & background

In the linear electrostatic model, Q and U are proportional. C depends on geometry and material. For large parallel plates with a uniform linear dielectric and neglected fringe fields, $C=\varepsilon_0\varepsilon_r\frac A d$ : increasing area or reducing separation raises C. A real component's material response may depend on operating conditions. Typical values lie in the micro-, nano- and picofarad range.

Exam blueprint

Validity range

Two-terminal capacitor in an electrostatic state; Q is the magnitude on one plate, U the voltage magnitude between plates. Constant C requires linear material and unchanged geometry within the chosen range. Nonlinear characteristics also need differential capacitance.

Derivation steps

In the linear electrostatic model, plate charge, field and voltage scale together; their ratio defines capacitance.

1Consider two conductors with equal opposite charge and fixed geometry. Q is the magnitude on one plate; the net charge of both plates can be zero.
2For linear material, doubling charge doubles the electric field, provided geometry and boundary conditions remain unchanged.
3Voltage is the field integral between conductors and then scales with Q as well.
4 $C=\frac{Q}{U}$ defines the constant proportionality factor for positive U; $Q=CU$ also describes the uncharged state.
5 $1\,\mathrm F=\frac{1\,\mathrm C}{1\,\mathrm V}$ . Material and geometry determine C in this model; a nonlinear characteristic needs separate specification.

Rearrangements

Charge from capacitance and voltage

Q = C \cdot U

The charge grows linearly with the applied voltage.

Voltage from charge and capacitance

U = \frac{Q}{C}

Small capacitance means little charge already creates a high voltage.

Parallel-plate capacitor

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

Large area, small separation and a dielectric increase C.

Task variant

A 100 µF capacitor is at $U=9\,\mathrm V$ . How much charge does it store?

$Q=CU=(100\times10^{-6}\,\mathrm F)(9\,\mathrm V)=0.9\,\mathrm{mC}$

$C=150\,\mathrm{\mu F}$ carries $Q=3\times10^{-3}\,\mathrm C$ . Find the voltage.

$U=\frac{Q}{C}=\frac{3\times10^{-3}\,\mathrm C}{150\times10^{-6}\,\mathrm F}=20\,\mathrm V$

Common mistakes

Not converting µF, nF and pF to farads.

$\mu{} = 10^{-6}$ , $n = 10^{-9}$ , $p = 10^{-12}$ ; convert before substituting.

Confusing capacitance with charge.

C is the storage capability per volt, Q the actually stored charge.

Expecting double capacitance when doubling the plate separation.

$C\propto\frac1d$ ; doubling the separation halves the capacitance.

Exam context

Typical: parallel-plate capacitor with dielectric, charge/voltage after switching, series and parallel combinations of capacitors.

Formula cluster

The capacitor

Capacitance defines the component, the energy formula its use.

E = ½CU²

Energy Stored in a Capacitor

F = ke·|q₁q₂|/r²

Coulomb's Law

P = U·I

Electric Power

E⃗ = F⃗el/q; Ehom = |U|/d

Electric Field Strength

Worked example

One plate carries $Q=6\times10^{-4}\,\mathrm C$ at $U=12\,\mathrm V$ . Then $C=\frac Q U=\frac{6\times10^{-4}\,\mathrm C}{12\,\mathrm V}=50\,\mu\mathrm F$ . Check: $CU=600\,\mu\mathrm C$ .

Applications

Smoothing in power supplies, camera flash, timer circuits (RC element), touchscreens, DRAM memory cells

How do you calculate with Capacitance of a Capacitor?

Here is how to work through a typical Capacitance of a Capacitor (C = Q/U) task step by step:

1
Task
A 100 µF capacitor is at $U=9\,\mathrm V$ . How much charge does it store?
Solution path
$Q=CU=(100\times10^{-6}\,\mathrm F)(9\,\mathrm V)=0.9\,\mathrm{mC}$
2
Task
$C=150\,\mathrm{\mu F}$ carries $Q=3\times10^{-3}\,\mathrm C$ . Find the voltage.
Solution path
$U=\frac{Q}{C}=\frac{3\times10^{-3}\,\mathrm C}{150\times10^{-6}\,\mathrm F}=20\,\mathrm V$

Understand and apply the equation

Capacitance describes the charge separation associated with a given voltage. For a linear capacitor the ratio stays constant as charge and voltage grow together. A charge measurement and a capacitor disconnected from its battery show why voltage, stored charge and component capacitance are three distinct quantities.

Capacitance is not a charge gauge

Two capacitors can have the same capacitance while carrying different charges. Capacitance relates charge separation to voltage; actual charge follows from the operating state. Before calculating, identify the electrode charge, the terminals used to measure voltage and whether material and geometry remain unchanged.[1]

We begin with a linear two-terminal component, using nonnegative magnitudes for one plate’s charge and voltage. Then we distinguish source-connected from isolated operation. Only the optional extension introduces a nonlinear characteristic, where one capacitance value no longer describes everything.[1][5]

Which plate charge and voltage belong in C = Q/U?

\begin{aligned}C&=\frac QU\\Q&=CU\\U&=\frac QC\end{aligned}

Q is the positive charge magnitude on one plate; U is the magnitude of the potential difference between the plates.[1]

An ideal two-plate capacitor can remain electrically neutral overall: its electrodes carry equal and opposite charges. The capacitance formula uses the magnitude on one electrode, rather than their sum. The other quantity is a voltage between two terminals, rather than one potential relative to an arbitrary zero. Signed circuit quantities require matching terminal orientation for charge and voltage.[1]

In the linear range, component capacitance is the slope of the charge-voltage line. A larger value means more charge per volt. A rating of $50\,\mu\mathrm F$ alone states neither the present charge nor the maximum permitted voltage. The voltage limit needs a separate specification.[1][4]

Same voltage, different plate charge

QUANTA STUDYquanta-study.de

U=4\,\mathrm V

, the selected plates carry

8\,\mathrm{\mu C}

and

20\,\mathrm{\mu C}

. Both components can remain neutral overall. Slope shows capacitance, not net charge.[1]

Free with a Quanta account. Your download stays ready after registration.

Why is the ratio constant in the linear model?

Keep geometry and material state fixed. In a linear electrostatic field problem, multiplying electrode charge also multiplies the field and potential difference by the same factor. Their ratio therefore remains unchanged. This justifies constant capacitance within the model, rather than asserting that all real components behave this way.[1][4]

\begin{aligned}Q&\longmapsto kQ\\U&\longmapsto kU\\C&=\frac{kQ}{kU}=\frac{Q}{U}\end{aligned}

A common scaling factor cancels in the ratio; at zero voltage the line slope is used.[1]

\begin{aligned}E&=\frac{Q}{\varepsilon A}\\U&=Ed\\C&=\frac{Q}{U}=\varepsilon\frac{A}{d}\end{aligned}

For large parallel plates the uniform field explains the geometry relation; material and boundary conditions remain part of the model.[1][4]

The relation $C=\varepsilon\frac{A}{d}$ is one particular geometry. It assumes plate separation small relative to lateral dimensions and neglected fringe fields. For other shapes $\frac{Q}{U}$ may remain meaningful although the parallel-plate relation is unsuitable. A dielectric changes the field through its material response, so capacitance does not depend only on external shape. The capacitance of a parallel-plate capacitor explores this area/separation dependence in detail; it is a particular geometry within the general component definition.[1][4]

Calculating capacitance, charge and voltage: farads and coulombs

\begin{aligned}1\,\mathrm F&=\frac{1\,\mathrm C}{1\,\mathrm V}\\1\,\mu\mathrm F&=10^{-6}\,\mathrm F\\1\,\mathrm{nF}&=10^{-9}\,\mathrm F\end{aligned}

The variable C denotes capacitance; the upright unit symbol C instead denotes coulombs.[1]

Convert charge and capacitance to compatible SI units before substitution. Microcoulombs divided by volts give microfarads; explicit powers of ten help when prefixes differ. A voltage calculation must yield volts. Millimetres or square metres are needed only when geometry is also being calculated.[1]

How do you calculate capacitance from charge and voltage?

Find capacitance from plate charge and voltage

Given: A linear capacitor carries $Q=6.0\times10^{-4}\,\mathrm C$ on one plate; the voltage between the plates is $U=12\,\mathrm V$ .
Find: Find capacitance in farads and microfarads.

Use the ratio of one plate’s charge to terminal voltage.
$C=\frac{Q}{U}$
Substitute SI values and check the unit.
$C=\frac{6{,}0\times10^{-4}\,\mathrm C}{12\,\mathrm V}=5{,}0\times10^{-5}\,\mathrm F$
Convert farads to microfarads.
$C=50\,\mu\mathrm F$

Result: The capacitance is $50\,\mu\mathrm F$ .

Check: Back-substitution: $CU=50\times10^{-6}\,\mathrm F\cdot12\,\mathrm V=600\,\mu\mathrm C$ , matching the given charge. Halving voltage halves Q in the same linear state, rather than C.

What changes for a capacitor disconnected from its battery?

In the following example the disconnected electrodes remain isolated without charge leakage while the dielectric changes capacitance. Energy stored in the capacitor must therefore be compared at fixed charge magnitude; fixed voltage would be a different boundary condition.[1][3]

Fill the gap of a disconnected capacitor

Given: An ideal parallel-plate capacitor with $C_0=120\,\mathrm{pF}$ is charged to $U_0=60\,\mathrm V$ and disconnected. A linear dielectric with $\varepsilon_r=3$ completely fills the gap; leakage is neglected.
Find: Find the new capacitance and voltage and the change in stored energy.

Keep free plate charge fixed and change the material capacitance.
$\begin{aligned}Q&=C_0U_0\\C_1&=\varepsilon_rC_0\end{aligned}$
Calculate charge and new capacitance.
$\begin{aligned}Q&=120\times10^{-12}\,\mathrm F\cdot60\,\mathrm V=7{,}2\,\mathrm{nC}\\C_1&=3\cdot120\,\mathrm{pF}=360\,\mathrm{pF}\end{aligned}$
Use the same charge to find the new voltage.
$U_1=\frac{Q}{C_1}=\frac{7{,}2\,\mathrm{nC}}{360\,\mathrm{pF}}=20\,\mathrm V$
Compare linear field energies at fixed Q.
$\begin{aligned}W_0&=\frac{Q^2}{2C_0}=216\,\mathrm{nJ}\\W_1&=\frac{Q^2}{2C_1}=72\,\mathrm{nJ}\\\Delta W&=-144\,\mathrm{nJ}\end{aligned}$

Result: $C_1=360\,\mathrm{pF}$ , $U_1=20\,\mathrm V$ . Field energy decreases by $144\,\mathrm{nJ}$ .

Check: At unchanged charge, tripling capacitance reduces voltage and field energy to one third. The energy difference must appear in a full balance involving mechanical work and possible losses; it does not vanish. With the battery connected, voltage would instead be fixed.

What changes when capacitors are connected

\begin{aligned}C_{\parallel}&=C_1+C_2\\C_{\mathrm{Reihe}}&=\frac{C_1C_2}{C_1+C_2}\end{aligned}

Parallel components share voltage; in a simple initially uncharged series combination they have equal charge magnitude.[2]

Rated capacitance alone does not describe a switching event. First identify connected nodes, whether a voltage source remains attached and the charge already present on isolated nodes. The voltage on an individual component may then change while its linear capacitance remains fixed. Terminal orientations and initial charges matter particularly for precharged series combinations.[2][4]

Explore further: nonlinear characteristic

For a reversible quasistatic characteristic, charging work is the integral of voltage over transported charge. The following numerical model is an original illustrative input, not a measured component.[5][3]

A nonlinear model characteristic

Given: A quasistatic lossless model has $Q(U)=aU+bU^3$ , $a=2.00\,\mathrm{\mu F}$ , $b=0.100\,\mathrm{\mu C/V^3}$ , and $Q(0)=0$ . Consider $U=2.00\,\mathrm V$ , with no hysteresis.
Find: Compare ratio, differential capacitance and charging work.

Approach: Total charge gives the ratio; a small voltage change follows the derivative. Charging work integrates voltage over transported charge.
$\begin{aligned}C_{\mathrm{sec}}&=\frac{Q(U)}U=a+bU^2\\C_{\mathrm{diff}}&=\frac{\mathrm dQ}{\mathrm dU}=a+3bU^2\\W&=\int_0^U u\frac{\mathrm dQ}{\mathrm du}\,\mathrm du=\frac{aU^2}2+\frac{3bU^4}4\end{aligned}$
Substitution: The two capacitance values answer different questions.
$\begin{aligned}Q(2\,\mathrm V)&=4.80\,\mathrm{\mu C}\\C_{\mathrm{sec}}&=2.40\,\mathrm{\mu F}\\C_{\mathrm{diff}}&=3.20\,\mathrm{\mu F}\\W&=5.20\,\mathrm{\mu J}\end{aligned}$

Result: Local slope exceeds $\frac{Q}{U}$ . Actual charging work is $5.20\,\mathrm{\mu J}$ , rather than $\frac12QU=4.80\,\mathrm{\mu J}$ .

Check: Independent check using $W=\int_0^Q U(q)\,\mathrm dq$ : integration by parts gives $UQ-\int_0^U Q(u)\,\mathrm du$ , yielding $9.60-4.40=5.20\,\mathrm{\mu J}$ . Setting $b=0$ recovers the linear energy expression.

Five exercises with distinct reasoning

Calculate · Foundation

A linear capacitor has $C=220\,\mathrm{nF}$ and $U=9.0\,\mathrm V$ . What charge magnitude does one plate carry?

Show hint

Solve for Q and convert nanofarads to farads.

Learning goal: Calculate charge from capacitance and voltage using compatible prefixes.

Show solution and calculation

Approach: $Q=CU$ .
The charge magnitude is $1.98\,\mu\mathrm C$ .
The other plate carries the opposite charge; the ideal component has zero net charge.

Q=CU=220\times10^{-9}\,\mathrm F\cdot9{,}0\,\mathrm V=1{,}98\times10^{-6}\,\mathrm C

[1]

Rearrange · Foundation

For $C=15\,\mu\mathrm F$ , one plate carries $Q=45\,\mu\mathrm C$ . Find the voltage.

Show hint

Solve $Q=CU$ for U; identical prefixes cancel.

Learning goal: Determine voltage from charge and verify it by substitution.

Show solution and calculation

Approach: $U=\frac QC$ .
The voltage is $3.0\,\mathrm V$ .
Verification: $15\,\mu\mathrm F\cdot3\,\mathrm V=45\,\mu\mathrm C$ .

U=\frac{Q}{C}=\frac{45\,\mu\mathrm C}{15\,\mu\mathrm F}=3{,}0\,\mathrm V

[1]

Interpret · Application

A charge curve is a straight line through the origin between 0 and 4 V. It contains $(2\,\mathrm V,6\,\mu\mathrm C)$ and $(4\,\mathrm V,12\,\mu\mathrm C)$ . What capacitance does it show? Does it also establish safe operation at 40 V?

Show hint

Read the slope; distinguish behaviour in the measured range from a component rating.

Learning goal: Interpret capacitance as a slope and respect the measured range.

Show solution and calculation

Approach: $C=\frac{\Delta Q}{\Delta U}$ .
The slope is $3\,\mu\mathrm F$ .
Both point ratios agree.
These data do not establish a 40 V rating; linearity beyond the interval and the breakdown limit are unknown.

C=\frac{\Delta Q}{\Delta U}=\frac{12-6}{4-2}\,\mu\mathrm F=3\,\mu\mathrm F

[1][4]

Error analysis · Application

Two plates carry $+4\,\mu\mathrm C$ and $-4\,\mu\mathrm C$ , with a voltage of 8 V. A solution inserts zero net charge into $C=\frac QU$ and claims $C=0$ . Locate and correct the error.

Show hint

Which charge does Q mean in the definition?

Learning goal: Distinguish plate charge from the capacitor’s total net charge.

Show solution and calculation

Approach: $C=\frac{|Q_{\mathrm{plate}}|}{U}$ .
Q is the magnitude on one plate rather than the sum.
The correct capacitance is $0.50\,\mu\mathrm F$ ; zero net charge is consistent with this.

C=\frac{4\,\mu\mathrm C}{8\,\mathrm V}=0{,}50\,\mu\mathrm F

[1]

Transfer · Challenge

Capacitor $C_1=2\,\mu\mathrm F$ is charged to 12 V, while $C_2=4\,\mu\mathrm F$ is uncharged. Matching terminal sides are connected in parallel without a battery. Neglect leakage after equilibration. Find final voltage and the field-energy change.

Show hint

Charge on the connected nodes is conserved; capacitances add.

Learning goal: Distinguish conserved charge in an isolated parallel connection from its energy balance.

Show solution and calculation

Approach: $U_f=\frac{C_1U_1+C_2U_2}{C_1+C_2}$ .
Final voltage is $4\,\mathrm V$ .
Field energy decreases from $144\,\mu\mathrm J$ to $48\,\mu\mathrm J$ , a decrease of $96\,\mu\mathrm J$ .
During real equilibration this becomes heat and possibly electromagnetic radiation.
A perfectly lossless oscillatory model would not spontaneously reach the assumed stationary final state.

\begin{aligned}U_f&=\frac{C_1U_1}{C_1+C_2}=\frac{2\cdot12}{2+4}\,\mathrm V=4\,\mathrm V\\W_i&=\frac12 C_1U_1^2=144\,\mu\mathrm J\\W_f&=\frac12(C_1+C_2)U_f^2=48\,\mu\mathrm J\end{aligned}

[1][2][3][4]

When a single capacitance value is insufficient

C_{\mathrm{diff}}(U)=\frac{dQ}{dU}

For a nonlinear charge curve the local slope matters for small voltage changes; it need not equal $\frac{Q}{U}$ .[5]

Constant capacitance assumes the selected linear material and operating range. With a voltage-dependent characteristic, hysteresis or changed geometry, the ratio can vary. Temperature, measurement frequency and circuit surroundings may also matter for a real component. Then the characteristic and conditions belong to the specification. Energy follows from charging work in general; $W=\frac12CU^2$ cannot be extended to any nonlinear characteristic using an arbitrary single ratio.[5][3][4]

A design also needs separate consideration of voltage rating, leakage, losses and terminal inductance. The pure capacitance model describes charge separation and quasistatic field storage. It describes neither current limitation during sudden connection nor the safe design of a high-voltage store. The RC time constant adds a time-dependent model for an explicitly resistive charging path; capacitance alone does not specify the charging curve.[4][6]

In brief

First identify one plate’s charge and the terminal voltage. In a fixed linear model, $C$ is constant; a source or isolation determines whether $U$ or $Q$ stays fixed. Capacitance alone determines neither stored energy nor permitted operating voltage.[1][3]

Sources for this explanation

Quanta · Updated October 1, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

Editorial responsibility, sources and methods

Change history

October 1, 2026Made plate charge, rearrangements, farads/coulombs, isolation and differential capacitance discoverable through learning questions; retained nonlinear and dielectric calculations and added contextual geometry, energy and RC-time-model links.
October 1, 2026Added an original DE/EN learning article with model limits, two worked examples and five distinct solved exercises.
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 Capacitor. Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/kondensator-kapazitaet

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Frequently asked questions about Capacitance of a Capacitor

How do you calculate with the capacitor formula

C =\frac{ Q}{U}

Use the charge magnitude on one plate and the voltage between terminals, rather than the net charge of the whole component. In the linear model, $C=\frac QU$ . For $Q=600\,\mathrm{\mu C}$ and $U=12\,\mathrm V$ , this gives $C=50\,\mathrm{\mu F}$ . Back-substitution $CU=600\,\mathrm{\mu C}$ confirms the result. When capacitance is known, use $Q=CU$ or $U=\frac QC$ . A $100\,\mathrm{\mu F}$ capacitor at $9\,\mathrm V$ , for example, carries $900\,\mathrm{\mu C}$ on one plate. Convert differing prefixes explicitly and check coulombs per volt equals farads. The ratio cannot be evaluated at zero voltage, but the linear proportionality factor or slope remains meaningful. This calculation does not determine the maximum permitted operating voltage.

What does the capacitance of a capacitor mean intuitively?+

Capacitance describes charge separation per volt. A linear component with greater capacitance carries more charge on each plate at the same voltage. It is the slope of the charge-voltage line: for $C=3\,\mathrm{\mu F}$ , one additional volt means an additional $3\,\mathrm{\mu C}$ on the selected plate. A tank analogy can illustrate proportionality, but does not replace the electric field or opposite electrode charges. The component can remain neutral overall. Capacitance is also not its present charge level: charged and uncharged capacitors can have the same value. For a nonlinear characteristic, ratio $\frac QU$ and local slope $\frac{\mathrm dQ}{\mathrm dU}$ differ, so the operating range must be specified.

What does the capacitance of a parallel-plate capacitor depend on?+

For large parallel plates with a uniform linear isotropic material filling the whole gap, approximately $C=\varepsilon_0\varepsilon_r\frac Ad$ . Use the overlap of one plate, rather than the sum of both areas; measure separation normally to the plates. Doubling area doubles capacitance, while doubling separation halves it, with unchanged material and still negligible fringe fields. For $A=0.010\,\mathrm{m^2}$ , $d=1.0\,\mathrm{mm}$ and approximate air, the result is about $88.5\,\mathrm{pF}$ . A dielectric raises capacitance within the applicable model, but its value depends on material and operating range. Layers along the field add voltage drops, so a simple arithmetic material average is insufficient.

What happens to charge and voltage when you disconnect the capacitor from the source?+

After disconnection, free plate charge remains approximately fixed provided neither leakage nor a conducting connection permits charge exchange. If capacitance is unchanged, voltage therefore initially stays unchanged too. If capacitance subsequently changes, voltage follows $U=\frac QC$ . Fully inserting a linear dielectric with $\varepsilon_r=4$ quadruples capacitance and reduces voltage to one quarter. With an ideal source still connected, voltage instead stays fixed and the source supplies extra charge according to $Q=CU$ . These boundary conditions also determine energy accounting. Plate movement or material insertion can involve mechanical work and losses. Stored field energy alone therefore does not always describe the complete energy exchange.

How do capacitances add in series and parallel circuits?+

Parallel capacitors share a voltage and their charge contributions add, giving $C_{\parallel}=C_1+C_2$ . Two components of $100\,\mathrm{\mu F}$ each give $200\,\mathrm{\mu F}$ . In a simple initially uncharged series combination with a neutral isolated intermediate node, plate-charge magnitudes are equal while voltages add. Thus $\frac1{C_{\mathrm{series}}}=\frac1{C_1}+\frac1{C_2}$ ; the same components give $50\,\mathrm{\mu F}$ . For precharged components, initial charges and terminal orientations need additional attention. Series connection does not automatically provide a reliably usable higher voltage rating: real leakage, tolerances and transient voltage division can require suitable balancing circuitry. Equivalent capacitance initially describes the idealised circuit relation rather than complete component behaviour or operating limits.

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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 capacitor

Capacitance relates the charge magnitude on one capacitor plate to the voltage between plates. For linear material and unchanged geometry, C is constant.

AdvancedExam-relevantUpper secondaryUniversity/collegeBachelor foundations

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

Symbol

Meaning

Unit

C

Capacitance in the chosen linear model

Q

Charge magnitude on one plate, not total net charge of both plates

U

Voltage magnitude between plates

Derivation & background

Exam blueprint

Validity range

Derivation steps

In the linear electrostatic model, plate charge, field and voltage scale together; their ratio defines capacitance.

1Consider two conductors with equal opposite charge and fixed geometry. Q is the magnitude on one plate; the net charge of both plates can be zero.
2For linear material, doubling charge doubles the electric field, provided geometry and boundary conditions remain unchanged.
3Voltage is the field integral between conductors and then scales with Q as well.
4 $C=\frac{Q}{U}$ defines the constant proportionality factor for positive U; $Q=CU$ also describes the uncharged state.
5 $1\,\mathrm F=\frac{1\,\mathrm C}{1\,\mathrm V}$ . Material and geometry determine C in this model; a nonlinear characteristic needs separate specification.

How do you calculate with Capacitance of a Capacitor?

Here is how to work through a typical Capacitance of a Capacitor (C = Q/U) task step by step:

Task

A 100 µF capacitor is at $U=9\,\mathrm V$ . How much charge does it store?

Solution path

$Q=CU=(100\times10^{-6}\,\mathrm F)(9\,\mathrm V)=0.9\,\mathrm{mC}$

Task

$C=150\,\mathrm{\mu F}$ carries $Q=3\times10^{-3}\,\mathrm C$ . Find the voltage.

Solution path

$U=\frac{Q}{C}=\frac{3\times10^{-3}\,\mathrm C}{150\times10^{-6}\,\mathrm F}=20\,\mathrm V$

Understand and apply the equation

Capacitance is not a charge gauge

Which plate charge and voltage belong in C = Q/U?

\begin{aligned}C&=\frac QU\\Q&=CU\\U&=\frac QC\end{aligned}

Q is the positive charge magnitude on one plate; U is the magnitude of the potential difference between the plates.[1]

Same voltage, different plate charge

QUANTA STUDYquanta-study.de

U=4\,\mathrm V

, the selected plates carry

8\,\mathrm{\mu C}

and

20\,\mathrm{\mu C}

. Both components can remain neutral overall. Slope shows capacitance, not net charge.[1]

Free with a Quanta account. Your download stays ready after registration.

Why is the ratio constant in the linear model?

\begin{aligned}Q&\longmapsto kQ\\U&\longmapsto kU\\C&=\frac{kQ}{kU}=\frac{Q}{U}\end{aligned}

A common scaling factor cancels in the ratio; at zero voltage the line slope is used.[1]

\begin{aligned}E&=\frac{Q}{\varepsilon A}\\U&=Ed\\C&=\frac{Q}{U}=\varepsilon\frac{A}{d}\end{aligned}

For large parallel plates the uniform field explains the geometry relation; material and boundary conditions remain part of the model.[1][4]

Calculating capacitance, charge and voltage: farads and coulombs

\begin{aligned}1\,\mathrm F&=\frac{1\,\mathrm C}{1\,\mathrm V}\\1\,\mu\mathrm F&=10^{-6}\,\mathrm F\\1\,\mathrm{nF}&=10^{-9}\,\mathrm F\end{aligned}

The variable C denotes capacitance; the upright unit symbol C instead denotes coulombs.[1]

How do you calculate capacitance from charge and voltage?

Find capacitance from plate charge and voltage

Given: A linear capacitor carries $Q=6.0\times10^{-4}\,\mathrm C$ on one plate; the voltage between the plates is $U=12\,\mathrm V$ .
Find: Find capacitance in farads and microfarads.

Use the ratio of one plate’s charge to terminal voltage.
$C=\frac{Q}{U}$
Substitute SI values and check the unit.
$C=\frac{6{,}0\times10^{-4}\,\mathrm C}{12\,\mathrm V}=5{,}0\times10^{-5}\,\mathrm F$
Convert farads to microfarads.
$C=50\,\mu\mathrm F$

Result: The capacitance is $50\,\mu\mathrm F$ .

Check: Back-substitution: $CU=50\times10^{-6}\,\mathrm F\cdot12\,\mathrm V=600\,\mu\mathrm C$ , matching the given charge. Halving voltage halves Q in the same linear state, rather than C.

What changes for a capacitor disconnected from its battery?

Fill the gap of a disconnected capacitor

Given: An ideal parallel-plate capacitor with $C_0=120\,\mathrm{pF}$ is charged to $U_0=60\,\mathrm V$ and disconnected. A linear dielectric with $\varepsilon_r=3$ completely fills the gap; leakage is neglected.
Find: Find the new capacitance and voltage and the change in stored energy.

Keep free plate charge fixed and change the material capacitance.
$\begin{aligned}Q&=C_0U_0\\C_1&=\varepsilon_rC_0\end{aligned}$
Calculate charge and new capacitance.
$\begin{aligned}Q&=120\times10^{-12}\,\mathrm F\cdot60\,\mathrm V=7{,}2\,\mathrm{nC}\\C_1&=3\cdot120\,\mathrm{pF}=360\,\mathrm{pF}\end{aligned}$
Use the same charge to find the new voltage.
$U_1=\frac{Q}{C_1}=\frac{7{,}2\,\mathrm{nC}}{360\,\mathrm{pF}}=20\,\mathrm V$
Compare linear field energies at fixed Q.
$\begin{aligned}W_0&=\frac{Q^2}{2C_0}=216\,\mathrm{nJ}\\W_1&=\frac{Q^2}{2C_1}=72\,\mathrm{nJ}\\\Delta W&=-144\,\mathrm{nJ}\end{aligned}$

Result: $C_1=360\,\mathrm{pF}$ , $U_1=20\,\mathrm V$ . Field energy decreases by $144\,\mathrm{nJ}$ .

What changes when capacitors are connected

\begin{aligned}C_{\parallel}&=C_1+C_2\\C_{\mathrm{Reihe}}&=\frac{C_1C_2}{C_1+C_2}\end{aligned}

Parallel components share voltage; in a simple initially uncharged series combination they have equal charge magnitude.[2]

Explore further: nonlinear characteristic

A nonlinear model characteristic

Given: A quasistatic lossless model has $Q(U)=aU+bU^3$ , $a=2.00\,\mathrm{\mu F}$ , $b=0.100\,\mathrm{\mu C/V^3}$ , and $Q(0)=0$ . Consider $U=2.00\,\mathrm V$ , with no hysteresis.
Find: Compare ratio, differential capacitance and charging work.

Approach: Total charge gives the ratio; a small voltage change follows the derivative. Charging work integrates voltage over transported charge.
$\begin{aligned}C_{\mathrm{sec}}&=\frac{Q(U)}U=a+bU^2\\C_{\mathrm{diff}}&=\frac{\mathrm dQ}{\mathrm dU}=a+3bU^2\\W&=\int_0^U u\frac{\mathrm dQ}{\mathrm du}\,\mathrm du=\frac{aU^2}2+\frac{3bU^4}4\end{aligned}$
Substitution: The two capacitance values answer different questions.
$\begin{aligned}Q(2\,\mathrm V)&=4.80\,\mathrm{\mu C}\\C_{\mathrm{sec}}&=2.40\,\mathrm{\mu F}\\C_{\mathrm{diff}}&=3.20\,\mathrm{\mu F}\\W&=5.20\,\mathrm{\mu J}\end{aligned}$

Result: Local slope exceeds $\frac{Q}{U}$ . Actual charging work is $5.20\,\mathrm{\mu J}$ , rather than $\frac12QU=4.80\,\mathrm{\mu J}$ .

Five exercises with distinct reasoning

Calculate · Foundation

A linear capacitor has $C=220\,\mathrm{nF}$ and $U=9.0\,\mathrm V$ . What charge magnitude does one plate carry?

Show hint

Solve for Q and convert nanofarads to farads.

Learning goal: Calculate charge from capacitance and voltage using compatible prefixes.

Show solution and calculation

Approach: $Q=CU$ .
The charge magnitude is $1.98\,\mu\mathrm C$ .
The other plate carries the opposite charge; the ideal component has zero net charge.

Q=CU=220\times10^{-9}\,\mathrm F\cdot9{,}0\,\mathrm V=1{,}98\times10^{-6}\,\mathrm C

[1]

Rearrange · Foundation

For $C=15\,\mu\mathrm F$ , one plate carries $Q=45\,\mu\mathrm C$ . Find the voltage.

Show hint

Solve $Q=CU$ for U; identical prefixes cancel.

Learning goal: Determine voltage from charge and verify it by substitution.

Show solution and calculation

Approach: $U=\frac QC$ .
The voltage is $3.0\,\mathrm V$ .
Verification: $15\,\mu\mathrm F\cdot3\,\mathrm V=45\,\mu\mathrm C$ .

U=\frac{Q}{C}=\frac{45\,\mu\mathrm C}{15\,\mu\mathrm F}=3{,}0\,\mathrm V

[1]

Interpret · Application

A charge curve is a straight line through the origin between 0 and 4 V. It contains $(2\,\mathrm V,6\,\mu\mathrm C)$ and $(4\,\mathrm V,12\,\mu\mathrm C)$ . What capacitance does it show? Does it also establish safe operation at 40 V?

Show hint

Read the slope; distinguish behaviour in the measured range from a component rating.

Learning goal: Interpret capacitance as a slope and respect the measured range.

Show solution and calculation

Approach: $C=\frac{\Delta Q}{\Delta U}$ .
The slope is $3\,\mu\mathrm F$ .
Both point ratios agree.
These data do not establish a 40 V rating; linearity beyond the interval and the breakdown limit are unknown.

C=\frac{\Delta Q}{\Delta U}=\frac{12-6}{4-2}\,\mu\mathrm F=3\,\mu\mathrm F

[1][4]

Error analysis · Application

Two plates carry $+4\,\mu\mathrm C$ and $-4\,\mu\mathrm C$ , with a voltage of 8 V. A solution inserts zero net charge into $C=\frac QU$ and claims $C=0$ . Locate and correct the error.

Show hint

Which charge does Q mean in the definition?

Learning goal: Distinguish plate charge from the capacitor’s total net charge.

Show solution and calculation

Approach: $C=\frac{|Q_{\mathrm{plate}}|}{U}$ .
Q is the magnitude on one plate rather than the sum.
The correct capacitance is $0.50\,\mu\mathrm F$ ; zero net charge is consistent with this.

C=\frac{4\,\mu\mathrm C}{8\,\mathrm V}=0{,}50\,\mu\mathrm F

[1]

Transfer · Challenge

Capacitor $C_1=2\,\mu\mathrm F$ is charged to 12 V, while $C_2=4\,\mu\mathrm F$ is uncharged. Matching terminal sides are connected in parallel without a battery. Neglect leakage after equilibration. Find final voltage and the field-energy change.

Show hint

Charge on the connected nodes is conserved; capacitances add.

Learning goal: Distinguish conserved charge in an isolated parallel connection from its energy balance.

Show solution and calculation

Approach: $U_f=\frac{C_1U_1+C_2U_2}{C_1+C_2}$ .
Final voltage is $4\,\mathrm V$ .
Field energy decreases from $144\,\mu\mathrm J$ to $48\,\mu\mathrm J$ , a decrease of $96\,\mu\mathrm J$ .
During real equilibration this becomes heat and possibly electromagnetic radiation.
A perfectly lossless oscillatory model would not spontaneously reach the assumed stationary final state.

\begin{aligned}U_f&=\frac{C_1U_1}{C_1+C_2}=\frac{2\cdot12}{2+4}\,\mathrm V=4\,\mathrm V\\W_i&=\frac12 C_1U_1^2=144\,\mu\mathrm J\\W_f&=\frac12(C_1+C_2)U_f^2=48\,\mu\mathrm J\end{aligned}

[1][2][3][4]

When a single capacitance value is insufficient

C_{\mathrm{diff}}(U)=\frac{dQ}{dU}

For a nonlinear charge curve the local slope matters for small voltage changes; it need not equal $\frac{Q}{U}$ .[5]

In brief

Sources for this explanation

Quanta · Updated October 1, 2026

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An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 1, 2026Made plate charge, rearrangements, farads/coulombs, isolation and differential capacitance discoverable through learning questions; retained nonlinear and dielectric calculations and added contextual geometry, energy and RC-time-model links.
October 1, 2026Added an original DE/EN learning article with model limits, two worked examples and five distinct solved exercises.
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 Capacitor. Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/kondensator-kapazitaet

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Frequently asked questions about Capacitance of a Capacitor

How do you calculate with the capacitor formula

C =\frac{ Q}{U}

What does the capacitance of a capacitor mean intuitively?+

What does the capacitance of a parallel-plate capacitor depend on?+

What happens to charge and voltage when you disconnect the capacitor from the source?+

How do capacitances add in series and parallel circuits?+

Capacitance of a capacitor

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How do you calculate with Capacitance of a Capacitor?

Capacitance is not a charge gauge

Which plate charge and voltage belong in C = Q/U?

Same voltage, different plate charge

Why is the ratio constant in the linear model?

Calculating capacitance, charge and voltage: farads and coulombs

How do you calculate capacitance from charge and voltage?

Find capacitance from plate charge and voltage

What changes for a capacitor disconnected from its battery?

Fill the gap of a disconnected capacitor

What changes when capacitors are connected

Explore further: nonlinear characteristic

A nonlinear model characteristic

Five exercises with distinct reasoning

A linear capacitor has C=220 nFC=220\,\mathrm{nF}C=220nF and U=9.0 VU=9.0\,\mathrm VU=9.0V. What charge magnitude does one plate carry?

For C=15 μFC=15\,\mu\mathrm FC=15μF, one plate carries Q=45 μCQ=45\,\mu\mathrm CQ=45μC. Find the voltage.

A charge curve is a straight line through the origin between 0 and 4 V. It contains (2 V,6 μC)(2\,\mathrm V,6\,\mu\mathrm C)(2V,6μC) and (4 V,12 μC)(4\,\mathrm V,12\,\mu\mathrm C)(4V,12μC). What capacitance does it show? Does it also establish safe operation at 40 V?

Two plates carry +4 μC+4\,\mu\mathrm C+4μC and −4 μC-4\,\mu\mathrm C−4μC, with a voltage of 8 V. A solution inserts zero net charge into C=QUC=\frac QUC=UQ​ and claims C=0C=0C=0. Locate and correct the error.

Capacitor C1=2 μFC_1=2\,\mu\mathrm FC1​=2μF is charged to 12 V, while C2=4 μFC_2=4\,\mu\mathrm FC2​=4μF is uncharged. Matching terminal sides are connected in parallel without a battery. Neglect leakage after equilibration. Find final voltage and the field-energy change.

When a single capacitance value is insufficient

In brief

Sources for this explanation

Frequently asked questions about Capacitance of a Capacitor

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More formulas: Electricity and magnetism

Retain Capacitance of a Capacitor for exams

Capacitance of a capacitor

Variables & units

Capacitance calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

The capacitor

Worked example

Applications

How do you calculate with Capacitance of a Capacitor?

Capacitance is not a charge gauge

Which plate charge and voltage belong in C = Q/U?

Same voltage, different plate charge

Why is the ratio constant in the linear model?

Calculating capacitance, charge and voltage: farads and coulombs

How do you calculate capacitance from charge and voltage?

Find capacitance from plate charge and voltage

What changes for a capacitor disconnected from its battery?

Fill the gap of a disconnected capacitor

What changes when capacitors are connected

Explore further: nonlinear characteristic

A nonlinear model characteristic

Five exercises with distinct reasoning

A linear capacitor has C=220 nFC=220\,\mathrm{nF}C=220nF and U=9.0 VU=9.0\,\mathrm VU=9.0V. What charge magnitude does one plate carry?

For C=15 μFC=15\,\mu\mathrm FC=15μF, one plate carries Q=45 μCQ=45\,\mu\mathrm CQ=45μC. Find the voltage.

A charge curve is a straight line through the origin between 0 and 4 V. It contains (2 V,6 μC)(2\,\mathrm V,6\,\mu\mathrm C)(2V,6μC) and (4 V,12 μC)(4\,\mathrm V,12\,\mu\mathrm C)(4V,12μC). What capacitance does it show? Does it also establish safe operation at 40 V?

Two plates carry +4 μC+4\,\mu\mathrm C+4μC and −4 μC-4\,\mu\mathrm C−4μC, with a voltage of 8 V. A solution inserts zero net charge into C=QUC=\frac QUC=UQ​ and claims C=0C=0C=0. Locate and correct the error.

Capacitor C1=2 μFC_1=2\,\mu\mathrm FC1​=2μF is charged to 12 V, while C2=4 μFC_2=4\,\mu\mathrm FC2​=4μF is uncharged. Matching terminal sides are connected in parallel without a battery. Neglect leakage after equilibration. Find final voltage and the field-energy change.

When a single capacitance value is insufficient

In brief

Sources for this explanation

Frequently asked questions about Capacitance of a Capacitor

A linear capacitor has $C=220\,\mathrm{nF}$ and $U=9.0\,\mathrm V$ . What charge magnitude does one plate carry?

For $C=15\,\mu\mathrm F$ , one plate carries $Q=45\,\mu\mathrm C$ . Find the voltage.

A charge curve is a straight line through the origin between 0 and 4 V. It contains $(2\,\mathrm V,6\,\mu\mathrm C)$ and $(4\,\mathrm V,12\,\mu\mathrm C)$ . What capacitance does it show? Does it also establish safe operation at 40 V?

Two plates carry $+4\,\mu\mathrm C$ and $-4\,\mu\mathrm C$ , with a voltage of 8 V. A solution inserts zero net charge into $C=\frac QU$ and claims $C=0$ . Locate and correct the error.

Capacitor $C_1=2\,\mu\mathrm F$ is charged to 12 V, while $C_2=4\,\mu\mathrm F$ is uncharged. Matching terminal sides are connected in parallel without a battery. Neglect leakage after equilibration. Find final voltage and the field-energy change.

A linear capacitor has $C=220\,\mathrm{nF}$ and $U=9.0\,\mathrm V$ . What charge magnitude does one plate carry?

For $C=15\,\mu\mathrm F$ , one plate carries $Q=45\,\mu\mathrm C$ . Find the voltage.

A charge curve is a straight line through the origin between 0 and 4 V. It contains $(2\,\mathrm V,6\,\mu\mathrm C)$ and $(4\,\mathrm V,12\,\mu\mathrm C)$ . What capacitance does it show? Does it also establish safe operation at 40 V?

Two plates carry $+4\,\mu\mathrm C$ and $-4\,\mu\mathrm C$ , with a voltage of 8 V. A solution inserts zero net charge into $C=\frac QU$ and claims $C=0$ . Locate and correct the error.

Capacitor $C_1=2\,\mu\mathrm F$ is charged to 12 V, while $C_2=4\,\mu\mathrm F$ is uncharged. Matching terminal sides are connected in parallel without a battery. Neglect leakage after equilibration. Find final voltage and the field-energy change.