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LESSON 12 / 16 · TOPIC 10.3

Charging work is the area under the voltage curve

You will be able to: Integrate charging work and connect capacitor energy with field energy density.

Calculus-based electrostaticsFree study resourceReview editionTeacher review pending

Why is capacitor energy one half of QΔV?

The first bit of charge is easy to move onto an uncharged capacitor. Later charge must be moved against a larger voltage. Charging work adds all those changing costs, rather than using the final voltage for every bit.

A useful starting point: A cylindrical gap produces a logarithmic voltage →

Words and symbols before equations

Increment dq
An infinitesimally small transferred charge.
Charging work dW
Incremental external work V(q)dq in reversible charging.
Stored energy U
Energy of the electric field/configuration, in joules.
Energy density u
Energy per volume, in J/m³.
Charging work: area under V(q)Instantaneous V (V)Transferred q (pC)005002.7510005.515008.25200011
Read this model snapshot. Final Q = 2000 pC; stored U = 10 nJ. Area in pC·V is pJ, so divide by 1000 for nJ.
What this picture assumes

Reversible charging of an ideal capacitor with fixed C from q = 0. The triangular area under V(q) gives stored energy, not necessarily total energy from a real battery and charging circuit.

Read the picture in three steps

  1. Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
  2. Final Q = 2000 pC; stored U = 10 nJ. Area in pC·V is pJ, so divide by 1000 for nJ.
  3. Check what the picture assumes below. Use the Explore task to predict one change before moving a control.

Connect the picture to the physics

At intermediate charge q, voltage is V(q) = q/C for fixed capacitance. Integrate dW = V(q)dq from 0 to Q: U = ∫₀^Q(q/C)dq = Q²/(2C). The triangular area under the V-versus-q graph explains the factor one half.

Using Q = CΔV gives equivalent forms U = ½QΔV = ½C(ΔV)². Choose a form that matches what is held fixed. These formulas describe energy stored, not automatically all energy supplied by a battery during a real charging process.

For vacuum plates, substitute C = ε₀A/d and ΔV = Ed: U = ½ε₀E²Ad. Dividing by field volume Ad gives u = ½ε₀E². The volume model neglects fringe fields; energy density is nonnegative.

A worked example, step by step

A 200 pF capacitor is charged reversibly from zero to 10 V. Find Q and stored energy.

  1. Q = CΔV = 2000 pC = 2 nC.
  2. U = ½C(ΔV)² = ½(200×10⁻¹²)(100).
  3. U = 10 nJ.
  4. The final QΔV is 20 nJ; averaging the linearly increasing voltage supplies the factor one half.
Common mix-up

QΔV is twice the stored energy for charging a fixed linear capacitor from zero. The battery and external apparatus require their own energy accounting.

CHECK THE IDEA

If voltage reverses, does stored energy become negative?

Compare with an explanation

No. Energy depends on voltage squared for an ideal capacitor.

Now investigate one change Explore →

Predict. Change one thing. Explain.

Change the final voltage at fixed C and predict the graph area. Double voltage and compare stored energy; then vary C at fixed voltage.

On narrow screens, swipe or scroll diagrams sideways to read all labels.

Charging work: area under V(q)Instantaneous V (V)Transferred q (pC)005002.7510005.515008.25200011

Final Q = 2000 pC; stored U = 10 nJ. Area in pC·V is pJ, so divide by 1000 for nJ.

Reversible charging of an ideal capacitor with fixed C from q = 0. The triangular area under V(q) gives stored energy, not necessarily total energy from a real battery and charging circuit.

Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant conductor equilibrium, charge conservation, capacitance or energy relationship to justify your prediction.

Apply the idea to a fresh problem Practice →

Show what you understand.

Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.

1. Doubling voltage at fixed C changes U by…

Show answer and reasoning

4. U is proportional to voltage squared.

2. C = 100 pF and ΔV = 20 V gives U…

Show answer and reasoning

20 nJ. ½(100×10⁻¹²)(20²) = 20×10⁻⁹ J.

Original written challenge

4 points · self-check · not an official AP question

Derive U = Q²/(2C) from incremental work. For vacuum plates, use it to derive the electric-field energy density.

This response is not submitted or saved. Copy it before leaving.

Compare with the answer and four-point rubric
  1. 1 point: At intermediate q, dW = (q/C)dq.
  2. 1 point: Integrating gives U = Q²/(2C).
  3. 1 point: Use Q = CΔV, C = ε₀A/d and ΔV = Ed to obtain U = ½ε₀E²Ad.
  4. 1 point: Divide by Ad to obtain u = ½ε₀E² with units J/m³.

Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.

Recall the ideas without notes Review →

Retrieve it before you reveal it.

RECALL 1Why the factor one half?

Voltage grows linearly from zero during charging at fixed C.

RECALL 2Which graph area gives stored energy?

Voltage versus transferred charge.

RECALL 3Is stored energy necessarily all battery work?

No. Other transfers or dissipation may occur in the charging apparatus.

Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.

Charging work is the area under the voltage curve

  • U = ∫₀^Q(q/C)dq = Q²/(2C).
  • U = ½QΔV = ½C(ΔV)².
  • Vacuum field energy density: u = ½ε₀E².

Remember: QΔV is twice the stored energy for charging a fixed linear capacitor from zero. The battery and external apparatus require their own energy accounting.

Conditions: Reversible charging of an ideal capacitor with fixed C from q = 0. The triangular area under V(q) gives stored energy, not necessarily total energy from a real battery and charging circuit.

Refresh Kid · AP Physics C: Electricity and Magnetism Unit 3 (official Unit 10) · Objectives 10.3.A · Review edition

Framework, scope and review status

Mapped to College Board CED, Topic 10.3, objectives 10.3.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. This is E&M Unit 3: Conductors and Capacitors, numbered Unit 10 in the official combined Physics C sequence. Topics 10.1–10.4 retain their official identifiers. Models state the electrostatic conditions, geometry approximations and whether charge or voltage stays fixed. Capacitor geometries include parallel plates, concentric spheres and long coaxial cylinders. Dielectric comparisons assume a fully filling ideal linear material. The optional 3D plate view uses explicitly different gap and lateral scales to show the small separation. Checked with the Fall 2026 clarifications. The lesson breakdown and questions are original Refresh Kid work, not official topic subdivisions.

Implementation and automated checks are separate from independent teacher review and observation of students. Both human review stages remain pending. This is a review edition, not a certified or validated assessment.

Optional further resource: College Board’s released questions and scoring guides. Papers can combine units; this link is an archive, not an assignment of every question to this lesson.

Our learn, explore, practice and recall sequence is informed by the IES learning guide. The exact Refresh Kid implementation has not been evaluated for learning effectiveness.

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