Energy stored in separated charge
You will be able to: Calculate capacitor energy and explain the factor one half.
Why does charging require increasing work?
As you transfer charge from one capacitor plate to the other, the voltage difference grows. Later increments require more work than earlier ones. The stored energy is the accumulated work, not simply final charge times final voltage.
A useful starting point: Capacitance depends on the geometry →
Words and symbols before equations
- Stored electric energy U_C
- Energy associated with the capacitor’s electric field, in J.
- Final voltage V
- Magnitude of the final plate potential difference in these formulas.
- Charge-voltage graph
- For ideal fixed C, voltage rises linearly with transferred charge.
- Average charging voltage
- V_final/2 when an initially uncharged ideal fixed-C capacitor is charged quasistatically.
What this picture assumes
Fixed ideal C=2 μF, initially uncharged. The V-versus-transferred-charge triangle gives capacitor field energy only, not all energy supplied by a dissipative charging circuit. μC·V=μJ.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- C=2 μF; Q=20 μC; U_C=100 μJ. Triangle area=½QV. At fixed C, voltage doubling quadruples energy.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
For fixed C starting uncharged, V=q/C during charging. The area under the V-versus-q line is a triangle: U_C=(1/2)QV=(1/2)CV²=Q²/(2C). Each form represents the same stored energy.
Choose the form that matches what is fixed. At fixed voltage, increasing C increases stored energy. At fixed charge, increasing C decreases stored energy. The source or mechanical interaction must be included to balance changes.
In an ordinary battery-resistor charging process the battery can supply more energy than is finally stored; some goes to the surroundings. The formula gives capacitor field energy, not automatically all energy delivered by a source.
A worked example, step by step
A 2 μF capacitor is charged to 10 V. Find Q and stored energy.
- Q=CV=(2×10⁻⁶)(10)=20 μC.
- U_C=(1/2)CV²=0.5(2×10⁻⁶)(100)=1.0×10⁻⁴ J.
- Using (1/2)QV gives the same result.
- Doubling V at fixed C quadruples U_C, while doubling Q.
QV uses final voltage for every increment and overcounts stored energy by a factor two in this ideal charging model.
Why is stored energy not QV?
Compare with an explanation
Voltage grows during charging; the average voltage for this linear process is V/2.
Predict. Change one thing. Explain.
Change voltage at fixed capacitance. Read the triangle under V versus q. Compare the effects of doubling V on charge and stored energy.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
C=2 μF; Q=20 μC; U_C=100 μJ. Triangle area=½QV. At fixed C, voltage doubling quadruples energy.
Fixed ideal C=2 μF, initially uncharged. The V-versus-transferred-charge triangle gives capacitor field energy only, not all energy supplied by a dissipative charging circuit. μC·V=μJ.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant charge, field 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.
Original written challenge
4 points · self-check · not an official AP questionA 4 μF capacitor has Q=20 μC. (a) Find V. (b) Find U_C. (c) Find U_C if Q doubles at fixed C. (d) Explain why the stored-energy expression has a half.
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Compare with the answer and four-point rubric
- 1 point: V=Q/C=5 V.
- 1 point: U_C=0.5(20×10⁻⁶)(5)=50 μJ.
- 1 point: U_C=200 μJ.
- 1 point: The charging voltage rises linearly from zero to its final value, giving a triangular area.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1Three equivalent energy forms?
½CV², Q²/(2C), and ½QV.
RECALL 2Which form helps at fixed Q?
Q²/(2C).
RECALL 3Does stored energy count every charging loss?
No. Include the rest of the circuit in a complete energy account.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Energy stored in separated charge
- U_C=(1/2)CV²=Q²/(2C)=(1/2)QV.
- At fixed C, U_C is proportional to V².
- Stored energy need not equal total energy supplied by the charging circuit.
Remember: QV uses final voltage for every increment and overcounts stored energy by a factor two in this ideal charging model.
Conditions: Fixed ideal C=2 μF, initially uncharged. The V-versus-transferred-charge triangle gives capacitor field energy only, not all energy supplied by a dissipative charging circuit. μC·V=μJ.
Refresh Kid · AP Physics 2 Unit 2 (official Unit 10) · Objectives 10.6.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 10.6, objectives 10.6.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. Refresh Kid calls this the second AP Physics 2 unit; College Board numbers it Unit 10; the first unit in this course is official Unit 9. 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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