Series capacitors share charge under a neutral-junction condition
You will be able to: Derive equivalent capacitance and voltage division for an initially neutral series pair.
Why is series capacitance smaller than either capacitor?
Connect two initially uncharged capacitors in series to a battery. The isolated conductor between them cannot acquire net charge, so its two facing plates gain equal opposite amounts.
A useful starting point: Use one node voltage to connect several loops →
Words and symbols before equations
- Floating junction
- An internal conductor with no external conducting path.
- Plate charge Q
- Magnitude on a plate, not the net zero charge of the whole capacitor.
- Series equivalent C_eq
- Single capacitor with the same outer-terminal Q-versus-V relation.
What this picture assumes
Initially uncharged series capacitors, neutral isolated internal conductor, final settled state. Equal plate-charge magnitudes depend on that initial condition; no steady conduction through the dielectric.
Read the picture in three steps
- Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
- C_eq = 1.333 μF; supplied charge = 16 μC. Q₁ = 16, Q₂ = 16 μC; V₁ = 8, V₂ = 4 V. Total stored U = 96 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
For initially uncharged capacitors with a neutral floating junction, both have the same charge magnitude Q. Their voltages are Q/C₁ and Q/C₂.
The source voltage is their sum, giving 1/C_eq = 1/C₁+1/C₂. The smaller capacitor receives the larger voltage at common Q.
Equal series charge relies on the stated initial floating-node charge. A precharged internal node can change the individual charge relation. This investigation uses initially uncharged capacitors and the final settled state.
| Property | Series, neutral inner node | Parallel |
|---|---|---|
| Shared quantity | Charge magnitude | Voltage |
| Equivalent C | Reciprocal sum | Sum of capacitances |
| Division | Voltage inversely with C | Charge proportional to C |
A worked example, step by step
Capacitors 2 μF and 4 μF are in series across 12 V. Find equivalent capacitance, charge and voltages.
- C_eq = 2(4)/(2+4) = 4/3 μF.
- Q = C_eqV = 16 μC.
- V₁ = 16/2 = 8 V and V₂ = 16/4 = 4 V.
- The voltages add to 12 V, and C_eq is below 2 μF.
Series capacitor voltages are generally unequal; the smaller C receives the larger drop under the neutral-junction condition.
Does a series capacitor have a steady conduction current through its dielectric?
Compare with an explanation
No. In this ideal settled state the branch current is zero.
Predict. Change one thing. Explain.
Increase C₂ while holding C₁ and the battery fixed. Track equivalent capacitance, common charge and the two voltages.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
C_eq = 1.333 μF; supplied charge = 16 μC. Q₁ = 16, Q₂ = 16 μC; V₁ = 8, V₂ = 4 V. Total stored U = 96 μJ. Settled state: no steady capacitor-branch current.
Initially uncharged series capacitors, neutral isolated internal conductor, final settled state. Equal plate-charge magnitudes depend on that initial condition; no steady conduction through the dielectric.
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 questionInitially uncharged 3 μF and 6 μF capacitors are placed in series across 9 V. Find C_eq, charge and voltage on each.
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Compare with the answer and four-point rubric
- 1 point: C_eq = 3(6)/(3+6) = 2 μF.
- 1 point: Q = 2(9) = 18 μC.
- 1 point: Voltages are 6 V and 3 V.
- 1 point: The neutral internal conductor supports equal opposite inner charges and the drops sum to 9 V.
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 1Which quantity matches in this series pair?
Plate-charge magnitude, under the stated initial condition.
RECALL 2Why is C_eq small?
The required total voltage is the sum of individual Q/C drops.
RECALL 3What assumption is often left unstated?
The floating internal node starts with zero net charge.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Series capacitors share charge under a neutral-junction condition
- 1/C_eq = 1/C₁+1/C₂.
- Neutral internal junction: |Q₁| = |Q₂|.
- V_i = Q/C_i.
Remember: Series capacitor voltages are generally unequal; the smaller C receives the larger drop under the neutral-junction condition.
Conditions: Initially uncharged series capacitors, neutral isolated internal conductor, final settled state. Equal plate-charge magnitudes depend on that initial condition; no steady conduction through the dielectric.
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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