Parallel capacitors add charge at a common voltage
You will be able to: Use shared nodes to calculate individual and total stored charge.
Why do parallel capacitances add?
Two capacitors connected across the same battery have the same voltage. Each stores its own plate-charge magnitude, so the battery moves the sum onto the positive-terminal plates.
A useful starting point: Series capacitors share charge under a neutral-junction condition →
Words and symbols before equations
- Common voltage
- The potential difference between the two shared nodes.
- Total supplied charge
- Sum of charges transferred onto the positive-terminal plates.
- Parallel equivalent
- A capacitor storing that same total charge at the shared voltage.
What this picture assumes
Capacitors across a common ideal source in the final settled state. Q_total is charge supplied to one terminal group, not net charge of both plate groups.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- C_eq = 6 μF; supplied charge = 72 μC. Q₁ = 24, Q₂ = 48 μC; V₁ = 12, V₂ = 12 V. Total stored U = 432 μJ. Settled state: no steady capacitor-branch current.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the physics
Q₁ = C₁V and Q₂ = C₂V because both capacitors span the same two nodes.
The supplied charge is Q_total = (C₁+C₂)V, so C_eq = C₁+C₂. The larger capacitor stores more charge at the common voltage.
Stored energies also add: U_total = ½(C₁+C₂)V². These are settled endpoint states with a fixed source; connecting precharged capacitors without a battery is a different charge-sharing problem.
A worked example, step by step
A 2 μF and a 4 μF capacitor are in parallel at 12 V. Find charges and total stored energy.
- C_eq = 6 μF.
- Q₁ = 24 μC and Q₂ = 48 μC, totaling 72 μC.
- U_total = ½(6 μF)(12 V)² = 432 μJ.
- Each capacitor has 12 V across it; the charges need not match.
Do not use the resistor parallel formula for capacitors.
Does total charge here mean net charge of both plates together?
Compare with an explanation
No. It is the amount transferred to one terminal’s plate group; the opposite group has the negative amount.
Predict. Change one thing. Explain.
Change one capacitance at fixed battery voltage. Identify which individual charge remains unchanged and how total charge grows.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
C_eq = 6 μF; supplied charge = 72 μC. Q₁ = 24, Q₂ = 48 μC; V₁ = 12, V₂ = 12 V. Total stored U = 432 μJ. Settled state: no steady capacitor-branch current.
Capacitors across a common ideal source in the final settled state. Q_total is charge supplied to one terminal group, not net charge of both plate groups.
Explain what you noticed: Which quantity changed? Which stayed fixed? Use the relevant current, circuit topology, charge or energy conservation, or RC 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 questionFind C_eq, each plate-charge magnitude and total energy for 3 μF and 5 μF in parallel at 4 V.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: C_eq = 8 μF.
- 1 point: Charges are 12 μC and 20 μC.
- 1 point: Total supplied charge is 32 μC.
- 1 point: Energy is ½(8)(16) = 64 μJ.
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 1What defines parallel capacitors?
The same two terminal nodes.
RECALL 2How does Q divide?
In proportion to C at common voltage.
RECALL 3What is the whole pair’s net charge?
Zero if the plate groups are equal and opposite.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Parallel capacitors add charge at a common voltage
- C_eq = C₁+C₂.
- Q_i = C_iV.
- U_total = ½C_eqV².
Remember: Do not use the resistor parallel formula for capacitors.
Conditions: Capacitors across a common ideal source in the final settled state. Q_total is charge supplied to one terminal group, not net charge of both plate groups.
Refresh Kid · AP Physics C: Electricity and Magnetism Unit 4 (official Unit 11) · Objectives 11.8.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 11.8, objectives 11.8.A. CED effective Fall 2024, current PDF ©2026; checked September 16, 2026. This is E&M Unit 4: Electric Circuits, numbered Unit 11 in the official combined Physics C sequence. Topics 11.1–11.8 retain their official identifiers. Models include signed charge flow, prescribed current-density and resistivity integrals, DC resistor networks, nonideal batteries and meters, capacitor combinations and finite-resistance RC transients. Circuit schematics use conventional-current references and explicit node connectivity. Initial capacitor voltage and positive time constants are stated. Unequal ideal sources are never directly wired in parallel. The optional spatial wire view supplements a complete 2D current-density explanation. 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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