PHYSICS 2 • ELECTRIC POTENTIAL

Energy Stored in Capacitors — Compute energy stored in a capacitor

Understand how capacitors store electrostatic energy and master the equations that quantify it.

Historical Context & Motivation

The story of energy storage in capacitors begins with one of the earliest electrical devices ever constructed. In the mid-eighteenth century, experimenters were captivated by the ability of certain glass-and-metal arrangements to accumulate static electricity and release it in a dramatic spark. These devices, known as Leyden jars, were the first practical capacitors, and their ability to deliver a sudden jolt of energy raised a fundamental question: exactly how much energy does a charged conductor store, and what governs that quantity? Answering this question required centuries of theoretical development, linking the concepts of charge, voltage, and the electric field into a coherent mathematical framework.

1745
Invention of the Leyden Jar
Pieter van Musschenbroek and Ewald Georg von Kleist independently develop the Leyden jar, a glass jar coated with metal foil inside and out. It demonstrated that electrical charge could be accumulated and stored for later discharge, sparking widespread interest in electrical phenomena across Europe.
1800
Volta's Pile & Quantitative Electrostatics
Alessandro Volta's invention of the voltaic pile provided the first source of continuous current and established the concept of electromotive force (voltage). This enabled systematic studies of how charge, voltage, and stored energy relate in capacitive systems.
1837
Faraday's Dielectric Studies
Michael Faraday introduced the concept of the dielectric constant, showing that inserting an insulating material between capacitor plates increases the capacitance. His work revealed that energy is stored in the electric field permeating the dielectric, not merely on the plates themselves.
1861
Maxwell's Field Energy Density
James Clerk Maxwell formalized the idea that the electric field itself carries energy with a volumetric energy density of ½ε₀E². This unified the energy stored in a capacitor with the broader electromagnetic theory and provided the theoretical underpinning for all modern energy-storage calculations.
1950s–Present
Modern Supercapacitors
The development of supercapacitors (also called ultracapacitors) using porous carbon electrodes and advanced dielectrics has pushed capacitive energy storage into practical engineering applications, from regenerative braking systems to power grid stabilization.

The central question that this lesson addresses is deceptively simple: if a capacitor of capacitance C has been charged to a potential difference V, how much electrostatic potential energy is stored in the device? We will see that the answer involves an integral because the voltage across the capacitor changes as charge accumulates, and this nuance distinguishes capacitor energy from the simpler relationship U = qV for a point charge moved through a constant potential difference.

Core Principles & Definitions

Before diving into the energy formula, it is essential to establish the foundational concepts that underpin capacitor energy storage. A capacitor is any arrangement of two conductors separated by an insulator (or vacuum) that can store charge and the associated electric field energy. The most common idealized geometry is the parallel-plate capacitor, but the energy relationships we derive are entirely general and apply to any capacitor regardless of geometry. The key to understanding why the stored energy is ½CV² rather than simply CV² lies in recognizing that charging a capacitor is an incremental process: each additional bit of charge must be moved against a progressively increasing voltage.

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Capacitance (C)

Capacitance is defined as the ratio C = Q/V, where Q is the magnitude of charge on either plate and V is the potential difference across the plates. Measured in farads (F), capacitance is a geometric property that depends on plate area, separation, and the dielectric material between the plates.
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Charge–Voltage Linearity

For a linear capacitor (the standard case), Q and V are directly proportional: Q = CV. This linear relationship is what makes the energy integral straightforward. As charge is added incrementally, the voltage rises linearly from 0 to V, and the energy integral becomes a simple polynomial.
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Work Done by the Battery

When a battery charges a capacitor, it does total work W = QV on the charges. However, only half of this work ends up as stored energy in the capacitor; the other half is dissipated as heat in the circuit resistance. This factor of ½ arises naturally from the integration.
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Energy Resides in the Electric Field

Physically, the energy is not stored on the metal plates but rather in the electric field that fills the space between them. The energy density is u = ½ε₀E², and integrating this over the volume between the plates recovers U = ½CV². This field-based perspective generalizes to all electromagnetic energy storage.
KEY TAKEAWAY
Think of charging a capacitor like filling a spring-loaded reservoir. At first the spring offers little resistance and it is easy to push water in, but as the reservoir fills, the spring pushes back harder, requiring more work per unit volume. The total energy stored is the integral of this increasing resistance — analogous to the area under a linear force–displacement curve, which gives ½kx². For a capacitor the analogous result is U = ½CV², where the factor of ½ reflects the fact that voltage grows linearly with charge during the charging process.

Visual Explanation — The Charging Process

The plot above shows voltage v(q) = q/C as a linear function of charge q. The shaded triangular area beneath the line equals the total energy stored: U = ½QV. The incremental work dU = v · dq (shown in amber) represents the energy added when a tiny charge dq is moved across the instantaneous voltage v.

The diagram makes the origin of the factor of ½ visually transparent. If the voltage were constant at V throughout the charging process (as it would be if you could somehow maintain V while adding charge), the energy would be QV — the area of the full rectangle. However, because the voltage starts at zero and rises linearly to V, the actual energy is the area of the triangle, which is exactly half the rectangle's area. This geometric argument is equivalent to performing the integral U = ∫₀Q (q/C) dq = Q²/(2C) = ½CV². The incremental energy element dU = v · dq is the thin horizontal strip highlighted in amber; summing all such strips from q = 0 to q = Q recovers the triangle.

Mathematical Framework — Deriving the Energy Formula

We now derive the energy stored in a capacitor rigorously from first principles. Consider a capacitor initially uncharged. At some intermediate stage of charging, let q denote the charge already deposited on the positive plate. The instantaneous potential difference across the capacitor is v(q) = q/C. To move an additional infinitesimal charge dq from the negative plate to the positive plate requires work dW = v · dq = (q/C) dq. The total work done in charging the capacitor from 0 to final charge Q is the integral of these incremental contributions.

DERIVATION INTEGRAL
U = ∫₀Q (q / C) dq = (1/C) · [q² / 2]₀Q = Q² / (2C)
U = energy stored (joules), Q = total charge on either plate (coulombs), C = capacitance (farads). The integral evaluates as a simple power rule since C is constant during charging.

Using the fundamental relation Q = CV, we can re-express this result in three equivalent and equally important forms. Each form is useful depending on which quantities are known or held constant in a particular problem.

THREE EQUIVALENT FORMS
U = ½CV² = ½QV = Q² / (2C)
All three expressions give the same energy. Use U = ½CV² when C and V are known; use U = ½QV when both Q and V are given; use U = Q²/(2C) when Q and C are specified but V is not directly available.

Connection to Field Energy Density

For a parallel-plate capacitor with plate area A and separation d, the uniform electric field between the plates is E = V/d, and the capacitance is C = ε₀A/d (in vacuum). Substituting into U = ½CV² yields U = ½(ε₀A/d)(Ed)² = ½ε₀E²(Ad). Since Ad is the volume of the region between the plates, we identify the energy density (energy per unit volume) as u = ½ε₀E². This result, though derived for a parallel-plate geometry, holds for any electrostatic field configuration and is one of the most important results in electromagnetism.

ENERGY DENSITY
u = ½ε₀E²
u = energy per unit volume (J/m³), ε₀ = permittivity of free space (8.854 × 10⁻¹² F/m), E = electric field magnitude (V/m). In a dielectric medium, replace ε₀ with ε = κε₀ where κ is the dielectric constant.
Where Does the Other Half Go?
When a battery of EMF V charges a capacitor to voltage V, the battery delivers total energy W = QV. But the capacitor stores only U = ½QV. The remaining ½QV is dissipated as heat in the resistance of the circuit — no matter how small that resistance is. This is a fundamental and somewhat surprising result: exactly half the battery's energy is always lost during RC charging of a capacitor from zero to full voltage.

Energy at Constant Charge vs. Constant Voltage

One of the most insightful aspects of the three equivalent energy formulas becomes apparent when we ask what happens to the stored energy if we change the capacitance — for instance by pulling the plates apart or inserting a dielectric. The answer depends critically on whether the capacitor is connected to a battery (constant voltage) or isolated (constant charge). These two scenarios give opposite results, and mastering this distinction is essential for exam success and physical intuition.

Left panel (constant Q): the capacitor is isolated. Inserting a dielectric increases C, so V = Q/C decreases, and U = Q²/(2C) decreases. Right panel (constant V): the battery maintains V. Inserting a dielectric increases C, more charge flows from the battery (Q = CV increases), and U = ½CV² increases.
Choosing the right energy formula based on constraints
ScenarioHeld ConstantBest FormulaEffect of Increasing C
Isolated capacitorQU = Q²/(2C)U decreases (energy leaves as work done by field pulling dielectric in)
Connected to batteryVU = ½CV²U increases (battery supplies additional charge and energy)

The lesson here is strategic: always use the form of the energy equation that features the quantity held constant. If the charge cannot change (isolated capacitor), use U = Q²/(2C); if the voltage is fixed by an external source, use U = ½CV². This choice ensures that the constant quantity stays in the numerator (or as a fixed factor), making it immediately clear how changes in C affect U.

Worked Example — Energy and Dielectric Insertion

A parallel-plate capacitor has capacitance C₀ = 5.0 μF and is charged to a voltage of V₀ = 12 V by a battery. The battery is then disconnected. A dielectric slab with dielectric constant κ = 3.0 is inserted between the plates, completely filling the gap. Find: (a) the initial energy stored, (b) the new capacitance, voltage, and charge after the dielectric is inserted, and (c) the final energy stored. Where did the lost energy go?

Dielectric Insertion at Constant Charge
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Step 1 — Identify Given Values and ConstraintsC₀ = 5.0 μF = 5.0 × 10⁻⁶ F, V₀ = 12 V, κ = 3.0. The battery is disconnected before the dielectric is inserted, so Q is constant (the charge has nowhere to go).
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Step 2 — Compute Initial Stored EnergyUsing U = ½CV²: U₀ = ½ × (5.0 × 10⁻⁶ F) × (12 V)² = ½ × (5.0 × 10⁻⁶) × 144 = 3.6 × 10⁻⁴ J.
U₀ = 3.6 × 10⁻⁴ J = 360 μJ
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Step 3 — Find New Capacitance After Dielectric InsertionInserting the dielectric multiplies the capacitance by κ: C = κC₀ = 3.0 × 5.0 μF = 15.0 μF.
C = 15.0 μF
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Step 4 — Find Charge and New VoltageThe charge is conserved: Q = C₀V₀ = (5.0 × 10⁻⁶)(12) = 6.0 × 10⁻⁵ C = 60 μC. The new voltage is V = Q/C = (6.0 × 10⁻⁵)/(15.0 × 10⁻⁶) = 4.0 V.
Q = 60 μC, V = 4.0 V
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Step 5 — Compute Final Energy and Account for Energy LossUsing U = Q²/(2C) or equivalently U = ½CV²: U = ½ × (15.0 × 10⁻⁶) × (4.0)² = ½ × 15.0 × 10⁻⁶ × 16 = 1.2 × 10⁻⁴ J = 120 μJ. The energy decreased from 360 μJ to 120 μJ, a loss of 240 μJ. This energy went into the mechanical work done by the electric field pulling the dielectric slab into the gap (the fringing fields at the edges of the plates exert a net inward force on the dielectric). In a real system, this kinetic energy would eventually be dissipated as heat through friction or oscillations.
U = 120 μJ; ΔU = −240 μJ (converted to mechanical work)

Capacitors vs. Batteries — Energy Storage Compared

Capacitors are not the only devices that store electrical energy — batteries and inductors also serve this purpose. Understanding the relative strengths and limitations of capacitive energy storage is crucial for selecting the right component in circuit design, power electronics, and energy systems. The table below highlights the key differences between capacitors and batteries, the two most common energy-storage elements in electrical systems.

Capacitors excel at rapid energy delivery; batteries excel at sustained energy supply
PropertyCapacitorBattery
Energy densityLow (0.01–0.05 Wh/kg for standard; up to ~10 Wh/kg for supercapacitors)High (100–265 Wh/kg for Li-ion)
Power densityVery high (10,000+ W/kg); can discharge in microsecondsModerate (250–1,000 W/kg); limited by reaction kinetics
Charge/discharge cyclesMillions (essentially unlimited)Hundreds to thousands before degradation
Energy storage mechanismElectrostatic field (no chemical changes)Electrochemical reactions
Voltage behavior during dischargeVoltage drops linearly with charge loss (V = Q/C)Voltage remains roughly constant until nearly depleted
Typical applicationsCamera flashes, defibrillators, power conditioning, regenerative brakingLaptops, electric vehicles, grid storage, portable electronics
KEY TAKEAWAY
Think of a capacitor as a sprinter and a battery as a marathon runner. A capacitor can release all of its stored energy in an extremely short burst — ideal for applications like camera flashes or defibrillator shocks — but it cannot carry much total energy. A battery stores far more energy but delivers it at a limited rate. In many modern systems, such as electric vehicles, capacitors and batteries work together: the battery provides sustained power for cruising, while supercapacitors handle transient power demands during acceleration and capture regenerative braking energy.

Connection to Advanced Theory — Field Energy & Electromagnetic Waves

The energy density u = ½ε₀E² that we derived for the electric field inside a capacitor generalizes far beyond static configurations. In the full theory of electromagnetism, the electromagnetic field carries energy with a density that includes contributions from both the electric and magnetic fields. The total electromagnetic energy density is u = ½ε₀E² + B²/(2μ₀), where B is the magnetic field and μ₀ is the permeability of free space. For a propagating electromagnetic wave, the time-averaged electric and magnetic energy densities are equal, each contributing half of the total wave energy. This connection — from a humble capacitor to the energy carried by light — is one of the great unifying themes in physics.

From capacitor energy to full electromagnetic energy theory
ConceptThis Lesson (Electrostatics)Advanced Extension (Electrodynamics)
Energy densityu = ½ε₀E² (electric field only)u = ½ε₀E² + B²/(2μ₀) (both fields)
Energy flowStatic: energy is localized in the gapPoynting vector S = (1/μ₀)(E × B) describes energy flux
Energy storage deviceCapacitor: stores electric field energyLC circuit: energy oscillates between capacitor (E-field) and inductor (B-field)
Governing equationU = ½CV² (lumped element)U = ∫(½ε₀E² + B²/(2μ₀)) dV (field integral over all space)

In courses on electromagnetic theory and optics, you will encounter situations where energy is not confined to a region between plates but propagates through space. The Poynting vector formalism extends the ideas developed here — quantifying energy stored per unit volume — into a description of energy flow per unit area per unit time. The seeds of that framework are planted right here in the capacitor energy formula.

Practice Problems

PROBLEM 1CONCEPTUAL
A capacitor is charged and then disconnected from the battery. The plates are then pulled farther apart. Does the stored energy increase, decrease, or remain the same? Explain your reasoning by identifying which quantity is held constant and selecting the appropriate energy formula.
PROBLEM 2BASIC CALCULATION
A 10 μF capacitor is charged to 24 V. Calculate the energy stored in the capacitor using U = ½CV².
PROBLEM 3INTERMEDIATE
Two capacitors, C₁ = 4.0 μF and C₂ = 6.0 μF, are connected in series across a 100 V battery. Find the total energy stored in the combination.
PROBLEM 4APPLIED
A cardiac defibrillator delivers 300 J of energy through a patient's chest. If its internal capacitor is rated at 32 μF, to what voltage must the capacitor be charged? Also estimate the charge stored on the capacitor at this voltage.
PROBLEM 5CRITICAL THINKING
A parallel-plate capacitor with plate area A = 0.02 m², separation d = 1.0 mm, and vacuum between the plates is charged to V = 500 V and then isolated. (a) Derive the stored energy using the field energy density u = ½ε₀E². (b) A dielectric slab (κ = 5.0) that fills half the gap (thickness d/2) is slid between the plates from one edge. Model this as two capacitors in series (one with dielectric, one with vacuum, each with gap d/2). Find the new equivalent capacitance and the new stored energy. (c) Was the dielectric pulled in or pushed out by the fringing fields? Justify your answer energetically.

Summary — Energy Stored in Capacitors

The energy stored in a capacitor arises from the work done to separate charge against an increasing voltage. Because the voltage builds linearly from zero to V as charge accumulates, the stored energy is U = ½CV² = ½QV = Q²/(2C) — exactly half what it would be if the full voltage were present throughout the charging process. The factor of ½ is geometric in origin, corresponding to the area of a triangle on a V-versus-Q plot. Physically, this energy resides in the electric field between the plates with an energy density u = ½ε₀E².

When solving problems, choose the formula that features the quantity held constant: use U = ½CV² for constant voltage (battery connected) and U = Q²/(2C) for constant charge (isolated capacitor). Inserting a dielectric increases C by a factor of κ; the effect on stored energy depends on the constraint. The capacitor energy formula is the electrostatic precursor to the general electromagnetic energy density u = ½ε₀E² + B²/(2μ₀), connecting this topic to the broader framework of Maxwell's equations and electromagnetic wave energy.

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