Physics · Electricity

Stored energy in a capacitor

The energy stored in the electric field of a capacitor grows quadratically with the voltage.

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Formula

E = \frac{1}{2} C U^{2}

E in joules [J] · C in farads [F] · U in volts [V]

Variables & units

Variables & units
Symbol	Meaning	Unit
$E$	Stored electric energy	J
$C$	Capacitance	F
$U$	Charging voltage	V

LaTeX

E = \frac{1}{2} C U^2

QUANTA STUDYquanta-study.de

Linear voltage-charge curve. Only the area between initial charge 28.2 mC and final charge 84.6 mC is shaded as additional energy; it is 6.768 J. — An already charged 470 µF capacitor is charged from 60 V to 180 V. Additional stored energy is the trapezoidal area between 28.2 and 84.6 mC: 6.768 J.

Capacitor energy calculator

Find the stored energy, capacitance or voltage of a capacitor, or how the stored energy changes when the voltage rises or falls.

Calculation model

Find

Capacitance C

Capacitance of the capacitor, usually printed on it, e.g. 200 µF.

Voltage magnitude |U|

Voltage across the terminals as a magnitude, e.g. 300 V in a camera flash.

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

Working · rearrange for E, substitute, evaluate

1. Rearrange for E
$E=\frac12 CU^2$
2. Substitute values (in SI units)
- C = 200 µF = 0.0002 F
$E=\frac12\cdot 0.0002\cdot 300^2=9$
3. Result
E = 9 J

Result: Stored energy E

E = 9 J

With C = 200 µF and |U| = 300 V, E = 9 J.

Quick tips

Double the voltage, four times the energy: E grows with U², but only linearly with the capacitance C.
Unit check: 1 F · 1 V² = 1 C/V · V² = 1 C·V = 1 J. Convert µF to F first, so multiply by 10⁻⁶.

Assumptions and limits

Linear capacitor with constant C > 0: E = ½CU². Voltage is a magnitude. The result is stored field energy, not automatically the work supplied by a charging source.

Derivation & background

During reversible charging, voltage rises with charge. Charging work is the area under voltage as a function of charge. For constant C, $E=\frac12 CU^2=\frac12 QU=\frac{Q^2}{2C}$ . Here E denotes energy, not field strength. With initial charge, energy increase is the difference between storage states. Source work and resistor heat must be accounted for separately.

Exam blueprint

Validity range

Linear constant C and a defined charge state. Energy increase is final energy minus initial energy. The formula gives stored energy, not automatically source work or useful output. Nonlinear reversible characteristics must be integrated.

Derivation steps

Voltage grows during charging; integrating it over charge explains half the final product.

1Consider a linear capacitor with constant C, initially uncharged.
2At instantaneous charge q, its voltage is $U(q)=\frac qC$ .
3Reversibly adding a small charge requires instantaneous voltage multiplied by that charge increment.
4Integrate this work from the uncharged starting state to final charge Q.
5The straight line gives $W_C=\frac{Q^2}{2C}$ , geometrically a triangular area on the charge–voltage graph.
6Using $Q=CU$ gives $W_C=\frac12 CU^2=\frac12 QU$ .
7With initial charge, subtract initial energy; source work and losses require extra accounting terms.

Rearrangements

Voltage from the energy

U = \sqrt{\frac{2E}{C}}

For double the energy you only need $\sqrt{2}$ times the voltage.

Form with charge

E = \frac{Q^2}{2C}

Practical when the charge is given instead of the voltage.

Capacitance from energy and voltage

C = \frac{2E}{U^2}

This is how buffer capacitors are sized for an energy demand.

Task variant

The charging voltage of a capacitor is doubled. What happens to the energy?

$E \propto{} U^{2}$ : the stored energy quadruples.

A defibrillator capacitor $(C = 150 \mu{}F)$ is charged to 2,000 V. Find E.

$E = \frac{1}{2} \times{} 1.5\times{}10^{-4} \times{} (2\,000)^{2} = \frac{1}{2} \times{} 1.5\times{}10^{-4} \times{} 4\times{}10^{6} = 300 J$ .

Common mistakes

Dropping the factor ½ and computing $E = CU^{2}$ .

For an initially uncharged linear capacitor, voltage averaged over charge is half the final voltage. This simple averaging rule does not apply to precharge or a nonlinear characteristic.

Confusing $E = \frac{1}{2}CU^{2}$ with $E = QU$ .

$E = QU$ holds for charge through a fixed voltage; on a capacitor U rises along.

Not squaring U.

The energy grows quadratically with voltage, which is why flash capacitors are charged to high voltages.

Exam context

Exam classics: flash unit and defibrillator, energy comparison before/after inserting a dielectric, redistributing charge between two capacitors.

Formula cluster

Field energy

The energy resides in the electric field between the plates.

C = Q/U

Capacitance of a Capacitor

P = U·I

Electric Power

Epot = mgh

Potential Energy (Gravitational)

Worked example

A flash model with $C=1000\,\mathrm{\mu F}$ and $U=300\,\mathrm V$ stores $E=\frac12(10^{-3}\,\mathrm F)(300\,\mathrm V)^2=45\,\mathrm J$ . Energy actually delivered to a load also depends on the discharge process.

Applications

Camera flash, defibrillator, buffer storage in electronics, supercapacitors (regenerative braking)

How do you calculate with Energy Stored in a Capacitor?

Here is how to work through a typical Energy Stored in a Capacitor (E = ½CU²) task step by step:

1
Task
The charging voltage of a capacitor is doubled. What happens to the energy?
Solution path
$E \propto{} U^{2}$ : the stored energy quadruples.
2
Task
A defibrillator capacitor $(C = 150 \mu{}F)$ is charged to 2,000 V. Find E.
Solution path
$E = \frac{1}{2} \times{} 1.5\times{}10^{-4} \times{} (2\,000)^{2} = \frac{1}{2} \times{} 1.5\times{}10^{-4} \times{} 4\times{}10^{6} = 300 J$ .

Understand and apply the equation

Charging a capacitor builds up its voltage. Each additional increment of charge therefore encounters a different voltage from the previous one. This response determines stored field energy. Read it as an area, compare precharged and final states, and distinguish storage from source work. Connecting two stores shows why charge conservation does not guarantee that their combined capacitor energy stays unchanged.

How do you calculate stored capacitor energy?

Capacitance of a capacitor relates plate charge to voltage; energy measures work contained in the stored state. An ideal linear capacitor connects them through a straight charge–voltage relation. Decide whether you need energy of a state, its change or total work supplied by a charger. Energy already present must not be counted again as new input.[1][2]

W_C=\frac12 CU^2=\frac{Q^2}{2C}=\frac12 QU

$W_C$ denotes energy, U voltage magnitude and Q charge magnitude on one plate; $C>0$ is constant.[1]

The introductory formula calls energy E, while the field article uses E for electric field strength. $W_C$ helps distinguish them here. Energy is nonnegative; reversing plate polarity leaves stored energy unchanged at equal voltage magnitude. The three forms are equivalent only for linear $Q=CU$ with unchanged C. A variable or hysteretic charge-voltage curve needs its actual charging work.[1]

Why does the energy formula contain one half?

The horizontal axis counts added charge; the vertical axis shows instantaneous capacitor voltage. A narrow area strip equals voltage times added charge. Starting uncharged gives a triangle; precharging makes the newly added area a trapezoid. A constant source voltage traces a different curve and therefore has a different work area.[1]

\begin{aligned}dW&=U(q)\,dq\\U(q)&=\frac qC\\W_C&=\int_0^Q\frac qC\,dq=\frac{Q^2}{2C}\end{aligned}

Instantaneous capacitor voltage rather than always its final value determines the next charge increment.[1]

For an initially uncharged linear capacitor, the U-q line starts at the origin. Its triangular area is $\frac12QU$ . Transporting a test charge across an already fixed voltage instead has work $qU$ and does not build that voltage first. With initial voltage $U_0$ the change is $\Delta W_C=\frac C2(U_f^2-U_0^2)$ . Initial stored energy is already present and must not be counted again as newly supplied energy.[1]

Charge–voltage area is energy

QUANTA STUDYquanta-study.de

Endpoints come from the retained interpretation exercise. A rectangle at final voltage over the entire charge would be 32 μJ and double stored energy.[1][2]

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Farads times volts squared gives joules

\begin{aligned}[CU^2]&=\frac{\mathrm C}{\mathrm V}\mathrm{V^2}=\mathrm J\\U&=\sqrt{\frac{2W_C}{C}}\end{aligned}

The square root yields voltage magnitude, not polarity.

$1\,\mathrm{\mu F}=10^{-6}\,\mathrm F$ , whereas $1\,\mathrm{mF}=10^{-3}\,\mathrm F$ ; confusing them changes calculated energy by 1000. At fixed C, doubling voltage quadruples stored energy. At fixed Q, larger C instead reduces $\frac{Q^2}{2C}$ . Saying greater capacitance means greater energy therefore requires the boundary condition, fixed voltage or fixed charge. A flash model with 1000 µF at 300 V stores 45 J.[1]

Calculating precharge, source work and charge redistribution

Energy increase of an already charged store

Given: An ideal linear capacitor $C=470\,\mathrm{\mu F}$ is charged from 60.0 V to 180 V. Capacitance unchanged; calculate the difference in stored energy.
Find: Find initial, final and added stored energy.

Approach: The same constant capacitance describes both states. Only final energy minus energy already stored initially is additional storage.
$\begin{aligned}W_0&=\frac12CU_0^2\\W_f&=\frac12CU_f^2\\\Delta W_C&=W_f-W_0\end{aligned}$
Convert capacitance and evaluate both states separately.
$\begin{aligned}W_0&=\frac12(470\times10^{-6}\,\mathrm F)(60\,\mathrm V)^2\\&=0.846\,\mathrm J\\W_f&=\frac12(470\times10^{-6}\,\mathrm F)(180\,\mathrm V)^2\\&=7.614\,\mathrm J\end{aligned}$
Only the difference is newly stored energy.
$\Delta W_C=7.614-0.846=6.768\,\mathrm J$
Charge endpoints give the same area calculation.
$\begin{aligned}Q_0&=28.2\,\mathrm{mC}\\Q_f&=84.6\,\mathrm{mC}\\\Delta W_C&=\frac12(U_0+U_f)(Q_f-Q_0)\end{aligned}$

Result: Initial 0.846 J, final 7.614 J and added storage 6.768 J. These values do not yet specify losses in a particular charger.

Check: Trapezoidal area $120\,\mathrm V\cdot56.4\,\mathrm{mC}=6.768\,\mathrm J$ independently matches the energy difference. Reporting final 7.614 J as additional charging work would include energy already present.

Empty capacitor on a fixed ideal source through a resistor

Given: $C=220\,\mathrm{\mu F}$ is initially uncharged and fully charged through $R=330\,\Omega$ by an ideal constant source $U_s=90.0\,\mathrm V$ . No other stores or losses; ideal RC circuit.
Find: Find final storage, source work and resistor heat.

Approach: The constant ideal source supplies its voltage for every charge increment while the initially empty capacitor builds its voltage. Remaining source work becomes resistor heat.
$\begin{aligned}W_s&=CU_s^2\\W_C&=\frac12CU_s^2\\W_R&=W_s-W_C\end{aligned}$
The source supplies its fixed voltage for each charge increment.
$\begin{aligned}Q_f&=CU_s=19.8\,\mathrm{mC}\\W_s&=U_sQ_f=CU_s^2=1.782\,\mathrm J\end{aligned}$
The capacitor builds its voltage gradually.
$\begin{aligned}W_C&=\frac12 CU_s^2=0.891\,\mathrm J\\W_R&=W_s-W_C=0.891\,\mathrm J\end{aligned}$
Resistance changes the time constant, not the fully charged final balance.
$\begin{aligned}\tau&=RC=0.0726\,\mathrm s\\i(t)&=\frac{U_s}{R}e^{-\frac{t}{RC}}\end{aligned}$

Result: Half of 1.782 J source work remains stored and 0.891 J becomes resistor heat. This half fraction belongs to the stated initially empty RC case, not every charging process.

Check: Independently, $W_R=\int_0^\infty R i(t)^2\,\mathrm dt$ . Substituting the exponential current and integrating gives $W_R=R\left(\frac{U_s}{R}\right)^2\frac{RC}{2}=\frac{CU_s^2}{2}=0.891\,\mathrm J$ . Different precharge or a controlled variable source changes the loss fraction. Simply setting $R=0$ makes the RC transient model singular.

For connected capacitors, common final voltage and conserved total charge must be used together. The following original example derives its balance from these relations and the energy expression. Positive connecting resistance is an explicitly identified heat path, not disappearing energy. The RC time constant additionally describes how quickly redistribution proceeds along a resistive path; it does not replace charge and final-energy accounting.[2][1][3][5]

Charge redistribution between two isolated stores

Given: An ideal 100 μF capacitor is charged to 12 V; a second 200 μF capacitor is uncharged. They are joined with matching polarities through a positive resistor, without an external source. Capacitances stay constant; only the resistor dissipates. Consider the settled final state.
Find: Find common final voltage, final stored energy and resistor heat.

Approach: charge on the connected positive plates is conserved. Both capacitors have the same final voltage; account for energy separately.
$\begin{aligned}U_f&=\frac{C_1U_1+C_2U_2}{C_1+C_2}\\W_i&=\frac12C_1U_1^2+\frac12C_2U_2^2\\W_f&=\frac12(C_1+C_2)U_f^2\end{aligned}$
Redistribute charge and calculate energies.
$\begin{aligned}Q_{\mathrm{tot}}&=1.20\,\mathrm{mC}\\U_f&=\frac{1.20\,\mathrm{mC}}{300\,\mathrm{\mu F}}=4.0\,\mathrm V\\W_i&=7.20\,\mathrm{mJ}\\W_f&=2.40\,\mathrm{mJ}\end{aligned}$
The reduction in field energy is heat in the stated resistor model.
$\begin{aligned}W_R&=W_i-W_f=4.80\,\mathrm{mJ}\\W_R&=\frac12\frac{C_1C_2}{C_1+C_2}(U_1-U_2)^2\end{aligned}$

Result: Both stores settle at 4 V. Field energy is 2.40 mJ and 4.80 mJ became heat. Charge conservation does not require unchanged field energy.

Check: Final charges 0.40 mC and 0.80 mC add to 1.20 mC. The independent loss expression also gives 4.80 mJ. Resistance affects settling time; an ideal zero-resistance connection would be a singular transition model and is not assumed.

Explore further: Field energy and mechanical boundary conditions

\begin{aligned}W_C&=\int u_E\,d^3r\\u_E&=\frac12\varepsilon_0E^2\end{aligned}

In an ideal vacuum plate gap, energy is distributed spatially in the field.[1]

For constant field throughout volume $Ad$ , integration gives $\frac12\varepsilon_0E^2Ad$ . Using $E=\frac{U}{d}$ and $C=\varepsilon_0\frac Ad$ yields the same component energy $\frac12CU^2$ . The ideal model counts the gap; finite plates also have fringe fields. In a linear lossless dielectric the corresponding energy density depends on the constitutive relation and includes polarization within the model. Electric-field energy density in vacuum describes the local energy contribution; total stored energy follows only from the appropriate volume integral.[1][5]

Increasing plate separation in an isolated capacitor keeps Q fixed and lowers C; energy rises when external mechanical work separates attracting plates. At fixed connected voltage the source participates in accounting and field energy falls as d grows. The energy expression alone determines neither source work nor mechanical work without this boundary condition.[5][4]

Distinguish energy, power and device purpose

A capacitor can release stored energy rapidly, but its energy formula determines neither pulse duration nor peak output power. Those need discharge path, resistance, inductance and component tolerances. Flash units or medical pulse systems illustrate storage purpose, not a prescription for suitable operating voltage. Equal $W_C$ can arise from different C and U and does not imply identical discharge behaviour.[1][3]

Without further input, an ideal discharge can supply at most initial stored energy. With remaining voltage $U_f$ , released energy is $\frac C2(U_0^2-U_f^2)$ . Energy divided by duration estimates mean power rather than the time-pulse shape.[1]

Five storage calculations

Calculate · Foundation

$C=100\,\mathrm{\mu F}$ , $U=40.0\,\mathrm V$ . Find $W_C$ .

Show hint

Convert μF to farads first.

Learning goal: Find energy of a linear capacitor from capacitance and voltage.

Show solution and calculation

Approach: capacitance is constant; use the energy expression involving the given voltage.
$W_C=0.5\cdot10^{-4}\cdot1600\,\mathrm J=0.0800\,\mathrm J$ .
Charge is $Q=4.00\,\mathrm{mC}$ ; $\frac12QU$ gives the same value.

\begin{aligned}W_C&=\frac12CU^2\\&=\frac12(100\times10^{-6}\,\mathrm F)(40\,\mathrm V)^2\\&=0.0800\,\mathrm J\end{aligned}

Rearrange · Application

A linear capacitor $C=2.00\,\mathrm{mF}$ must store 0.250 J. Find voltage magnitude.

Show hint

Solve the quadratic relation for U.

Learning goal: Reconstruct voltage magnitude from an energy target.

Show solution and calculation

Approach: rearrange the quadratic energy relation for voltage magnitude.
$\begin{aligned}U&=\sqrt{\frac{2\cdot0.250\,\mathrm J}{0.00200\,\mathrm F}}\\{}&=\sqrt{250}\,\mathrm V\approx15.8114\,\mathrm V\text{.}\end{aligned}$ Both polarities store the same energy.

U\approx15.8114\,\mathrm V

Transfer · Application

An isolated linear capacitor retains $Q=6.00\,\mathrm{\mu C}$ . Its capacitance rises from 2.00 to 6.00 μF. Find initial and final energy.

Show hint

Q is fixed here, not U.

Learning goal: Find energy change from changed capacitance at isolated fixed charge.

Show solution and calculation

Approach: isolated charge remains fixed, so use the energy expression involving charge and capacitance.
$W_0=\frac{Q^2}{2C_0}=9.00\,\mathrm{\mu J}$ and $W_f=3.00\,\mathrm{\mu J}$ .
In an actual capacitance change, the 6.00 µJ energy difference needs an accounted mechanical or other energy path.

\begin{aligned}W_0&=9.00\,\mathrm{\mu J}\\W_f&=3.00\,\mathrm{\mu J}\end{aligned}

Interpret · Application

For an ideal linear capacitor, voltage U plotted against charge q passes through the origin and (q=8 µC,U=4 V). What stored energy corresponds to the area from q=0 to 8 µC? Would charging to −8 µC instead of +8 µC at the same capacitance give negative energy?

Show hint

In this graph the energy integral is a triangle’s area. For negative charge, integration direction and voltage are both negative.

Learning goal: Interpret charge–voltage area as energy and account for voltage polarity.

Show solution and calculation

Approach: find the area under the linear voltage curve.
It is 16 µJ and capacitance is 2 µF.
Reversing polarity also makes voltage negative, but stored energy stays positive and unchanged.

Check: the squared-charge expression gives 16 µJ for either sign; the rectangle of final voltage times final charge would be twice as large.

\begin{aligned}C&=\frac QU=\frac{8\,\mu\mathrm C}{4\,\mathrm V}=2\,\mu\mathrm F\\W&=\int_0^Q U(q)\,\mathrm dq=\frac12QU=16\,\mu\mathrm J\\W(-Q)&=\frac{(-Q)^2}{2C}=W(Q)\end{aligned}

[1][2]

Error analysis · Application

An initially uncharged ideal 5 µF capacitor is connected through a resistor to an ideal constant 10 V source and fully charged. A report calls the entire source work of 500 µJ the stored capacitor energy. Correct this and find heat in the resistor; other losses are excluded.

Show hint

The source always supplies 10 V per unit charge. Capacitor voltage instead rises from zero to 10 V.

Learning goal: Separate battery work, stored energy and resistor heating during RC charging.

Show solution and calculation

Approach: find final charge first, then calculate source work and stored energy separately.
The source indeed supplies 500 µJ, but only 250 µJ is stored.
The remaining 250 µJ becomes heat in the resistor.

Check: source work equals stored-energy increase plus heat. The factor of two follows from constant source voltage versus increasing capacitor voltage.

\begin{aligned}Q_f&=CU_0=50\,\mu\mathrm C\\W_{\mathrm{source}}&=U_0Q_f=500\,\mu\mathrm J\\W_C&=\frac12CU_0^2=250\,\mu\mathrm J\\Q_R&=W_{\mathrm{source}}-W_C=250\,\mu\mathrm J\end{aligned}

[1][3]

When does the energy formula fail for nonlinear capacitance?

W_C=\int_0^Q U(q^{\prime})\,dq^{\prime}

For reversible construction the charging curve matters; constant-C factor ½ follows only from a straight line.[5][6]

Time-, voltage- or temperature-dependent capacitance may make a single rated capacitance insufficient for accurate energy. The actual charging curve and initial conditions must be known. Hysteresis, leakage and chemical processes in real stores require additional models; supplied work may differ from reversibly stored field energy. Discharge power also depends on resistance, wiring and the load.[1][6][3]

In brief

Stored energy follows from the charge–voltage area. At constant C it is quadratic and independent of polarity. Changes require an initial state; source work, resistor heat and mechanical work must be accounted for separately within one consistent system boundary.

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 internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

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 stored energy, the factor one half, precharge, source work and charge redistribution accessible through existing direct answers; retained all energy and heat balances and linked capacitance, RC-time model and local field-energy density in context.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Added an accessible learning sequence, an additional model example, a quantitative diagram and a closing summary; retained existing examples, exercises and sources.

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Quanta. Energy Stored in a Capacitor. Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/kondensator-energie

When citing a specific scientific statement, also cite the specialist source linked to that statement.

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Frequently asked questions about Energy Stored in a Capacitor

How do you calculate the energy stored in a capacitor?+

For a linear capacitor with constant capacitance, stored energy is half the capacitance multiplied by voltage squared. Equivalent forms use charge and capacitance or the product of charge and voltage. They refer to the same final state. At 1000 µF and 300 V, the result is 45 J. If the capacitor was already charged initially, its energy increase is final energy minus initial energy. Total work supplied by a charging source can be larger because resistors or other losses absorb energy. Nonlinear capacitance requires the actual reversible charging curve rather than an assumed constant rated value.

Where does the factor ½ in

E = \frac{1}{2}CU^{2}

come from?+

As charge builds up, voltage on a linear capacitor rises from zero to its final value. Adding a small amount of charge therefore does not always require the same work: voltage is small early and larger later. Integrating voltage over added charge gives the triangular area under a straight line, producing the factor one half. Final voltage multiplied by all charge instead gives twice the stored energy. With an already charged initial state, additional energy corresponds to a trapezoidal area. For a nonlinear charging curve, area remains decisive, but the simple half factor is no longer automatic.

Why are flash units and defibrillators charged to high voltages?+

At fixed capacitance, stored energy grows with voltage squared. Tripling voltage therefore provides nine times the energy. A compact capacitor can consequently supply a brief energy pulse for a flash unit or a specifically designed medical pulse generator. The formula initially describes only the electrical storage state. Energy actually delivered to a load, pulse duration and current also depend on discharge circuitry, wiring resistance and the load itself. Increasing voltage does not replace component limits or engineering design. Capacitance and voltage alone therefore establish neither an appropriate medical use nor the effectiveness of any particular pulse.

Where is the energy in a capacitor actually located?+

In the ideal vacuum model, electric energy resides in the field. Between large parallel plates, field magnitude is approximately uniform; integrate local electric energy density over the gap. This matches the linear capacitor-energy expression. Fringe fields also contribute for real finite plates. A dielectric additionally changes polarization and energy accounting, so unchanged vacuum density is not the complete material-energy description there. Charge and voltage are practical quantities for calculating that same idealised storage state. Electrical power delivered during discharge is a separate question requiring a time-dependent discharge process and a model of its load and circuit.

Why is half the energy lost when charging through a resistor?+

The half result applies to an initially uncharged linear capacitor charged fully through a positive resistor from an ideal constant-voltage source. The source supplies its voltage multiplied by final charge; only half remains stored in the capacitor. The other portion is dissipated in the resistor. For 220 µF and 90 V, the source supplies 1.782 J; storage and heat are each 0.891 J. Resistance changes charging time, not this complete final accounting. Initial charge, variable source voltage, additional energy stores or other charging methods alter the division. Half loss is therefore not a universal property of every transfer of energy into a capacitor.

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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, §8.3: Energy Stored in a Capacitor

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\mathcal E=-N\frac{\mathrm d\Phi_B}{\mathrm dt}

Faraday’s law of induction

\oint_{\partial V} \mathbf{E}\cdot d\mathbf{A} = \frac{Q_{\mathrm{innen}}}{\varepsilon_0}

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Symbol

Meaning

Unit

E

Stored electric energy

C

Capacitance

U

Charging voltage

Derivation & background

Exam blueprint

Validity range

Derivation steps

Voltage grows during charging; integrating it over charge explains half the final product.

1Consider a linear capacitor with constant C, initially uncharged.
2At instantaneous charge q, its voltage is $U(q)=\frac qC$ .
3Reversibly adding a small charge requires instantaneous voltage multiplied by that charge increment.
4Integrate this work from the uncharged starting state to final charge Q.
5The straight line gives $W_C=\frac{Q^2}{2C}$ , geometrically a triangular area on the charge–voltage graph.
6Using $Q=CU$ gives $W_C=\frac12 CU^2=\frac12 QU$ .
7With initial charge, subtract initial energy; source work and losses require extra accounting terms.

How do you calculate with Energy Stored in a Capacitor?

Here is how to work through a typical Energy Stored in a Capacitor (E = ½CU²) task step by step:

Task

The charging voltage of a capacitor is doubled. What happens to the energy?

Solution path

$E \propto{} U^{2}$ : the stored energy quadruples.

Task

A defibrillator capacitor $(C = 150 \mu{}F)$ is charged to 2,000 V. Find E.

Solution path

$E = \frac{1}{2} \times{} 1.5\times{}10^{-4} \times{} (2\,000)^{2} = \frac{1}{2} \times{} 1.5\times{}10^{-4} \times{} 4\times{}10^{6} = 300 J$ .

Understand and apply the equation

How do you calculate stored capacitor energy?

W_C=\frac12 CU^2=\frac{Q^2}{2C}=\frac12 QU

$W_C$ denotes energy, U voltage magnitude and Q charge magnitude on one plate; $C>0$ is constant.[1]

Why does the energy formula contain one half?

\begin{aligned}dW&=U(q)\,dq\\U(q)&=\frac qC\\W_C&=\int_0^Q\frac qC\,dq=\frac{Q^2}{2C}\end{aligned}

Instantaneous capacitor voltage rather than always its final value determines the next charge increment.[1]

Charge–voltage area is energy

QUANTA STUDYquanta-study.de

Endpoints come from the retained interpretation exercise. A rectangle at final voltage over the entire charge would be 32 μJ and double stored energy.[1][2]

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

Farads times volts squared gives joules

\begin{aligned}[CU^2]&=\frac{\mathrm C}{\mathrm V}\mathrm{V^2}=\mathrm J\\U&=\sqrt{\frac{2W_C}{C}}\end{aligned}

The square root yields voltage magnitude, not polarity.

Calculating precharge, source work and charge redistribution

Energy increase of an already charged store

Given: An ideal linear capacitor $C=470\,\mathrm{\mu F}$ is charged from 60.0 V to 180 V. Capacitance unchanged; calculate the difference in stored energy.
Find: Find initial, final and added stored energy.

Approach: The same constant capacitance describes both states. Only final energy minus energy already stored initially is additional storage.
$\begin{aligned}W_0&=\frac12CU_0^2\\W_f&=\frac12CU_f^2\\\Delta W_C&=W_f-W_0\end{aligned}$
Convert capacitance and evaluate both states separately.
$\begin{aligned}W_0&=\frac12(470\times10^{-6}\,\mathrm F)(60\,\mathrm V)^2\\&=0.846\,\mathrm J\\W_f&=\frac12(470\times10^{-6}\,\mathrm F)(180\,\mathrm V)^2\\&=7.614\,\mathrm J\end{aligned}$
Only the difference is newly stored energy.
$\Delta W_C=7.614-0.846=6.768\,\mathrm J$
Charge endpoints give the same area calculation.
$\begin{aligned}Q_0&=28.2\,\mathrm{mC}\\Q_f&=84.6\,\mathrm{mC}\\\Delta W_C&=\frac12(U_0+U_f)(Q_f-Q_0)\end{aligned}$

Result: Initial 0.846 J, final 7.614 J and added storage 6.768 J. These values do not yet specify losses in a particular charger.

Empty capacitor on a fixed ideal source through a resistor

Given: $C=220\,\mathrm{\mu F}$ is initially uncharged and fully charged through $R=330\,\Omega$ by an ideal constant source $U_s=90.0\,\mathrm V$ . No other stores or losses; ideal RC circuit.
Find: Find final storage, source work and resistor heat.

Approach: The constant ideal source supplies its voltage for every charge increment while the initially empty capacitor builds its voltage. Remaining source work becomes resistor heat.
$\begin{aligned}W_s&=CU_s^2\\W_C&=\frac12CU_s^2\\W_R&=W_s-W_C\end{aligned}$
The source supplies its fixed voltage for each charge increment.
$\begin{aligned}Q_f&=CU_s=19.8\,\mathrm{mC}\\W_s&=U_sQ_f=CU_s^2=1.782\,\mathrm J\end{aligned}$
The capacitor builds its voltage gradually.
$\begin{aligned}W_C&=\frac12 CU_s^2=0.891\,\mathrm J\\W_R&=W_s-W_C=0.891\,\mathrm J\end{aligned}$
Resistance changes the time constant, not the fully charged final balance.
$\begin{aligned}\tau&=RC=0.0726\,\mathrm s\\i(t)&=\frac{U_s}{R}e^{-\frac{t}{RC}}\end{aligned}$

Result: Half of 1.782 J source work remains stored and 0.891 J becomes resistor heat. This half fraction belongs to the stated initially empty RC case, not every charging process.

Charge redistribution between two isolated stores

Given: An ideal 100 μF capacitor is charged to 12 V; a second 200 μF capacitor is uncharged. They are joined with matching polarities through a positive resistor, without an external source. Capacitances stay constant; only the resistor dissipates. Consider the settled final state.
Find: Find common final voltage, final stored energy and resistor heat.

Approach: charge on the connected positive plates is conserved. Both capacitors have the same final voltage; account for energy separately.
$\begin{aligned}U_f&=\frac{C_1U_1+C_2U_2}{C_1+C_2}\\W_i&=\frac12C_1U_1^2+\frac12C_2U_2^2\\W_f&=\frac12(C_1+C_2)U_f^2\end{aligned}$
Redistribute charge and calculate energies.
$\begin{aligned}Q_{\mathrm{tot}}&=1.20\,\mathrm{mC}\\U_f&=\frac{1.20\,\mathrm{mC}}{300\,\mathrm{\mu F}}=4.0\,\mathrm V\\W_i&=7.20\,\mathrm{mJ}\\W_f&=2.40\,\mathrm{mJ}\end{aligned}$
The reduction in field energy is heat in the stated resistor model.
$\begin{aligned}W_R&=W_i-W_f=4.80\,\mathrm{mJ}\\W_R&=\frac12\frac{C_1C_2}{C_1+C_2}(U_1-U_2)^2\end{aligned}$

Result: Both stores settle at 4 V. Field energy is 2.40 mJ and 4.80 mJ became heat. Charge conservation does not require unchanged field energy.

Explore further: Field energy and mechanical boundary conditions

\begin{aligned}W_C&=\int u_E\,d^3r\\u_E&=\frac12\varepsilon_0E^2\end{aligned}

In an ideal vacuum plate gap, energy is distributed spatially in the field.[1]

Distinguish energy, power and device purpose

Five storage calculations

Calculate · Foundation

$C=100\,\mathrm{\mu F}$ , $U=40.0\,\mathrm V$ . Find $W_C$ .

Show hint

Convert μF to farads first.

Learning goal: Find energy of a linear capacitor from capacitance and voltage.

Show solution and calculation

Approach: capacitance is constant; use the energy expression involving the given voltage.
$W_C=0.5\cdot10^{-4}\cdot1600\,\mathrm J=0.0800\,\mathrm J$ .
Charge is $Q=4.00\,\mathrm{mC}$ ; $\frac12QU$ gives the same value.

\begin{aligned}W_C&=\frac12CU^2\\&=\frac12(100\times10^{-6}\,\mathrm F)(40\,\mathrm V)^2\\&=0.0800\,\mathrm J\end{aligned}

Rearrange · Application

A linear capacitor $C=2.00\,\mathrm{mF}$ must store 0.250 J. Find voltage magnitude.

Show hint

Solve the quadratic relation for U.

Learning goal: Reconstruct voltage magnitude from an energy target.

Show solution and calculation

Approach: rearrange the quadratic energy relation for voltage magnitude.
$\begin{aligned}U&=\sqrt{\frac{2\cdot0.250\,\mathrm J}{0.00200\,\mathrm F}}\\{}&=\sqrt{250}\,\mathrm V\approx15.8114\,\mathrm V\text{.}\end{aligned}$ Both polarities store the same energy.

U\approx15.8114\,\mathrm V

Transfer · Application

An isolated linear capacitor retains $Q=6.00\,\mathrm{\mu C}$ . Its capacitance rises from 2.00 to 6.00 μF. Find initial and final energy.

Show hint

Q is fixed here, not U.

Learning goal: Find energy change from changed capacitance at isolated fixed charge.

Show solution and calculation

Approach: isolated charge remains fixed, so use the energy expression involving charge and capacitance.
$W_0=\frac{Q^2}{2C_0}=9.00\,\mathrm{\mu J}$ and $W_f=3.00\,\mathrm{\mu J}$ .
In an actual capacitance change, the 6.00 µJ energy difference needs an accounted mechanical or other energy path.

\begin{aligned}W_0&=9.00\,\mathrm{\mu J}\\W_f&=3.00\,\mathrm{\mu J}\end{aligned}

Interpret · Application

For an ideal linear capacitor, voltage U plotted against charge q passes through the origin and (q=8 µC,U=4 V). What stored energy corresponds to the area from q=0 to 8 µC? Would charging to −8 µC instead of +8 µC at the same capacitance give negative energy?

Show hint

In this graph the energy integral is a triangle’s area. For negative charge, integration direction and voltage are both negative.

Learning goal: Interpret charge–voltage area as energy and account for voltage polarity.

Show solution and calculation

Approach: find the area under the linear voltage curve.
It is 16 µJ and capacitance is 2 µF.
Reversing polarity also makes voltage negative, but stored energy stays positive and unchanged.

Check: the squared-charge expression gives 16 µJ for either sign; the rectangle of final voltage times final charge would be twice as large.

\begin{aligned}C&=\frac QU=\frac{8\,\mu\mathrm C}{4\,\mathrm V}=2\,\mu\mathrm F\\W&=\int_0^Q U(q)\,\mathrm dq=\frac12QU=16\,\mu\mathrm J\\W(-Q)&=\frac{(-Q)^2}{2C}=W(Q)\end{aligned}

[1][2]

Error analysis · Application

An initially uncharged ideal 5 µF capacitor is connected through a resistor to an ideal constant 10 V source and fully charged. A report calls the entire source work of 500 µJ the stored capacitor energy. Correct this and find heat in the resistor; other losses are excluded.

Show hint

The source always supplies 10 V per unit charge. Capacitor voltage instead rises from zero to 10 V.

Learning goal: Separate battery work, stored energy and resistor heating during RC charging.

Show solution and calculation

Approach: find final charge first, then calculate source work and stored energy separately.
The source indeed supplies 500 µJ, but only 250 µJ is stored.
The remaining 250 µJ becomes heat in the resistor.

Check: source work equals stored-energy increase plus heat. The factor of two follows from constant source voltage versus increasing capacitor voltage.

\begin{aligned}Q_f&=CU_0=50\,\mu\mathrm C\\W_{\mathrm{source}}&=U_0Q_f=500\,\mu\mathrm J\\W_C&=\frac12CU_0^2=250\,\mu\mathrm J\\Q_R&=W_{\mathrm{source}}-W_C=250\,\mu\mathrm J\end{aligned}

[1][3]

When does the energy formula fail for nonlinear capacitance?

W_C=\int_0^Q U(q^{\prime})\,dq^{\prime}

For reversible construction the charging curve matters; constant-C factor ½ follows only from a straight line.[5][6]

In brief

Sources for this explanation

Quanta · Updated October 1, 2026

Editorial responsibility: Quanta Study editorial team

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An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

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 stored energy, the factor one half, precharge, source work and charge redistribution accessible through existing direct answers; retained all energy and heat balances and linked capacitance, RC-time model and local field-energy density in context.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Added an accessible learning sequence, an additional model example, a quantitative diagram and a closing summary; retained existing examples, exercises and sources.

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Quanta. Energy Stored in a Capacitor. Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/kondensator-energie

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Frequently asked questions about Energy Stored in a Capacitor

How do you calculate the energy stored in a capacitor?+

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E = \frac{1}{2}CU^{2}

come from?+

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Stored energy in a capacitor

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How do you calculate with Energy Stored in a Capacitor?

How do you calculate stored capacitor energy?

Why does the energy formula contain one half?

Charge–voltage area is energy

Farads times volts squared gives joules

Calculating precharge, source work and charge redistribution

Energy increase of an already charged store

Empty capacitor on a fixed ideal source through a resistor

Charge redistribution between two isolated stores

Explore further: Field energy and mechanical boundary conditions

Distinguish energy, power and device purpose

Five storage calculations

C=100 μFC=100\,\mathrm{\mu F}C=100μF, U=40.0 VU=40.0\,\mathrm VU=40.0V. Find WCW_CWC​.

A linear capacitor C=2.00 mFC=2.00\,\mathrm{mF}C=2.00mF must store 0.250 J. Find voltage magnitude.

An isolated linear capacitor retains Q=6.00 μCQ=6.00\,\mathrm{\mu C}Q=6.00μC. Its capacitance rises from 2.00 to 6.00 μF. Find initial and final energy.

For an ideal linear capacitor, voltage U plotted against charge q passes through the origin and (q=8 µC,U=4 V). What stored energy corresponds to the area from q=0 to 8 µC? Would charging to −8 µC instead of +8 µC at the same capacitance give negative energy?

An initially uncharged ideal 5 µF capacitor is connected through a resistor to an ideal constant 10 V source and fully charged. A report calls the entire source work of 500 µJ the stored capacitor energy. Correct this and find heat in the resistor; other losses are excluded.

When does the energy formula fail for nonlinear capacitance?

In brief

Sources for this explanation

Frequently asked questions about Energy Stored in a Capacitor

Subject references and further reading

More formulas: Electricity and magnetism

Retain Energy Stored in a Capacitor for exams

Stored energy in a capacitor

Variables & units

Capacitor energy calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Field energy

Worked example

Applications

How do you calculate with Energy Stored in a Capacitor?

How do you calculate stored capacitor energy?

Why does the energy formula contain one half?

Charge–voltage area is energy

Farads times volts squared gives joules

Calculating precharge, source work and charge redistribution

Energy increase of an already charged store

Empty capacitor on a fixed ideal source through a resistor

Charge redistribution between two isolated stores

Explore further: Field energy and mechanical boundary conditions

Distinguish energy, power and device purpose

Five storage calculations

C=100 μFC=100\,\mathrm{\mu F}C=100μF, U=40.0 VU=40.0\,\mathrm VU=40.0V. Find WCW_CWC​.

A linear capacitor C=2.00 mFC=2.00\,\mathrm{mF}C=2.00mF must store 0.250 J. Find voltage magnitude.

An isolated linear capacitor retains Q=6.00 μCQ=6.00\,\mathrm{\mu C}Q=6.00μC. Its capacitance rises from 2.00 to 6.00 μF. Find initial and final energy.

For an ideal linear capacitor, voltage U plotted against charge q passes through the origin and (q=8 µC,U=4 V). What stored energy corresponds to the area from q=0 to 8 µC? Would charging to −8 µC instead of +8 µC at the same capacitance give negative energy?

An initially uncharged ideal 5 µF capacitor is connected through a resistor to an ideal constant 10 V source and fully charged. A report calls the entire source work of 500 µJ the stored capacitor energy. Correct this and find heat in the resistor; other losses are excluded.

When does the energy formula fail for nonlinear capacitance?

In brief

Sources for this explanation

Frequently asked questions about Energy Stored in a Capacitor

Subject references and further reading

More formulas: Electricity and magnetism

Retain Energy Stored in a Capacitor for exams

$C=100\,\mathrm{\mu F}$ , $U=40.0\,\mathrm V$ . Find $W_C$ .

A linear capacitor $C=2.00\,\mathrm{mF}$ must store 0.250 J. Find voltage magnitude.

An isolated linear capacitor retains $Q=6.00\,\mathrm{\mu C}$ . Its capacitance rises from 2.00 to 6.00 μF. Find initial and final energy.

$C=100\,\mathrm{\mu F}$ , $U=40.0\,\mathrm V$ . Find $W_C$ .

A linear capacitor $C=2.00\,\mathrm{mF}$ must store 0.250 J. Find voltage magnitude.

An isolated linear capacitor retains $Q=6.00\,\mathrm{\mu C}$ . Its capacitance rises from 2.00 to 6.00 μF. Find initial and final energy.