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LESSON 15 / 18 · TOPIC 10.6

Dielectrics: what stays fixed matters

You will be able to: Compare dielectric insertion into isolated and fixed-voltage capacitors.

Official College Board Unit 10Free study resourceReview editionTeacher review pending

Does inserting a dielectric always lower voltage?

Two identical charged capacitors start at the same voltage. Disconnect one from its battery and leave the other connected. Inserting the same dielectric increases both capacitances, but only the disconnected capacitor must keep its free plate charge fixed.

A useful starting point: Energy stored in separated charge →

Words and symbols before equations

Dielectric
An insulating material that polarizes in the field.
Fixed Q
An isolated capacitor cannot exchange free charge with a battery.
Fixed V
An ideal connected battery maintains its terminal potential difference.
Fully filled model
Uniform linear dielectric occupies the whole gap; geometry stays fixed.
Capacitor energy before and after insertionμJ · same scale for all bars0Initially0.5After insertion0.125
Read this model snapshot. Isolated: fixed Q. κ=4; C=400 pF, Q=10 nC, V=25 V, U=0.125 μJ. Include mechanics in a complete energy account.
What this picture assumes

Initial vacuum C₀=100 pF and V₀=100 V; geometry unchanged and linear dielectric completely fills the gap. Isolated means fixed free Q; connected means fixed V. Mechanical work and battery exchanges are not part of the displayed capacitor-only energy.

Read the picture in three steps

  1. Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
  2. Isolated: fixed Q. κ=4; C=400 pF, Q=10 nC, V=25 V, U=0.125 μJ. Include mechanics in a complete energy account.
  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

Capacitance becomes κC₀ in both cases. For isolated plates, Q stays fixed, V=Q/C falls by 1/κ, and E=V/d falls by 1/κ. Stored energy Q²/(2C) also falls by 1/κ.

With a connected ideal battery, V stays fixed and Q=CV rises by κ. For unchanged d the total gap field V/d stays fixed, even though the dielectric creates an opposing contribution: the battery supplies additional free charge. Stored energy ½CV² rises by κ.

Energy is not lost from the universe in the isolated case. Mechanical work, motion and other transfers during insertion complete the account. In the connected case the battery exchanges energy and charge. Do not compare only the capacitor and assume it is isolated.

Fully inserting dielectric at unchanged geometry
QuantityIsolated: fixed QBattery: fixed V
CapacitanceMultiplies by κMultiplies by κ
Free plate chargeUnchangedMultiplies by κ
Voltage and gap fieldDivide by κUnchanged
Capacitor energyDivides by κMultiplies by κ

A worked example, step by step

C₀=100 pF and V₀=100 V. Insert κ=4. Compare isolated and connected results.

  1. Initially Q₀=10 nC and U₀=0.5 μJ. Both final capacitances are 400 pF.
  2. Isolated: Q=10 nC, V=25 V, U=0.125 μJ.
  3. Connected: V=100 V, Q=40 nC, U=2.0 μJ.
  4. The different constraints explain the opposite energy trends.
Common mix-up

Fixed charge and fixed voltage are different experiments. State the battery connection before calculating.

CHECK THE IDEA

Why can the fixed-voltage gap field stay the same despite polarization?

Compare with an explanation

The battery supplies additional free charge, balancing the dielectric’s opposing contribution while V/d stays fixed.

Now investigate one change Explore →

Predict. Change one thing. Explain.

Switch between isolated plates and a connected battery, then change κ. Predict Q, V and U before comparing the readouts.

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

Capacitor energy before and after insertionμJ · same scale for all bars0Initially0.5After insertion0.125

Isolated: fixed Q. κ=4; C=400 pF, Q=10 nC, V=25 V, U=0.125 μJ. Include mechanics in a complete energy account.

Initial vacuum C₀=100 pF and V₀=100 V; geometry unchanged and linear dielectric completely fills the gap. Isolated means fixed free Q; connected means fixed V. Mechanical work and battery exchanges are not part of the displayed capacitor-only energy.

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.

1. In an isolated capacitor, dielectric insertion leaves fixed…

Show answer and reasoning

Free plate Q. No conducting path permits charge exchange.

2. With an ideal battery connected and d fixed, the total gap E…

Show answer and reasoning

Stays fixed. The battery maintains V, so E=V/d is unchanged.

Original written challenge

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

A capacitor is charged, disconnected, and filled with κ=3 dielectric. (a) Give C’s factor. (b) Give Q’s factor. (c) Give V’s factor. (d) Give stored energy’s factor and identify a missing part of a complete energy account.

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

Compare with the answer and four-point rubric
  1. 1 point: C multiplies by 3.
  2. 1 point: Q stays fixed.
  3. 1 point: V becomes one third.
  4. 1 point: U becomes one third; include mechanical energy/work during insertion.

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 1What stays fixed when isolated?

Free charge on the plates.

RECALL 2What stays fixed with ideal battery attached?

Voltage.

RECALL 3Can you predict energy without a constraint?

No. Fixed-Q and fixed-V cases give different trends.

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

Dielectrics: what stays fixed matters

  • C_new=κC₀.
  • Fixed Q: V,E,U scale as 1/κ.
  • Fixed V: Q,U scale as κ; E unchanged for fixed d.

Remember: Fixed charge and fixed voltage are different experiments. State the battery connection before calculating.

Conditions: Initial vacuum C₀=100 pF and V₀=100 V; geometry unchanged and linear dielectric completely fills the gap. Isolated means fixed free Q; connected means fixed V. Mechanical work and battery exchanges are not part of the displayed capacitor-only energy.

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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