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LESSON 14 / 16 · TOPIC 10.4

Bound charges reduce the field for fixed free charge

You will be able to: Explain dielectric polarization and distinguish free plate charge from bound surface charge.

Calculus-based electrostaticsFree study resourceReview editionTeacher review pending

What changes when an insulating material fills the gap?

Slide an insulating slab between charged plates after disconnecting the battery. Charges in its molecules shift slightly or orient; they do not travel freely through the material as conduction electrons do in metal.

A useful starting point: State what stays fixed before moving the plates →

Words and symbols before equations

Dielectric
An insulating material that polarizes in an applied electric field.
Dielectric constant κ
Dimensionless ratio ε/ε₀; this ideal linear model uses κ ≥ 1.
Free charge
Charge placed on the metal plates by an external source.
Bound charge
Polarization charge associated with the dielectric’s microscopic charge displacement.
Free and bound surface charges+ free− free− bound+ boundE = 1 N/CBound magnitude = 26.55 pC/m²; gap sketch not to scale
Read this model snapshot. E_free 4 + E_bound -3 = E_total 1 N/C at fixed free charge. κ = 4.
What this picture assumes

Fully filled narrow parallel-plate gap; homogeneous linear ideal dielectric. Free plate charge stays fixed. Positive field direction points from positive toward negative plate. Bound charges oppose the free-charge field; no leakage or breakdown is modeled.

Read the picture in three steps

  1. Locate the labeled sources, system boundary or graph axes. Read the units before comparing values.
  2. E_free 4 + E_bound -3 = E_total 1 N/C at fixed free charge. κ = 4.
  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

For a uniform linear dielectric completely filling a narrow plate gap at fixed free surface density σ_f, E = σ_f/(κε₀). Polarization contributes an opposing field, reducing the total from the vacuum value E₀ = σ_f/ε₀.

The polarization magnitude is P = ε₀(κ−1)E. Its surface bound-charge magnitude is σ_b = P = σ_f(1−1/κ). Bound charge next to the positive plate is negative; near the negative plate it is positive.

The resulting capacitance is C = κC₀ = κε₀A/d. The relation E = E₀/κ compares fixed free charge, not a fixed-voltage battery case. Real dielectrics can have leakage, nonlinear behavior and breakdown outside this ideal model.

A worked example, step by step

An isolated capacitor has σ_f = 35.4 pC/m². A κ = 4 dielectric fills the gap. Find E and the bound surface-charge magnitude.

  1. Vacuum field E₀ = σ_f/ε₀ = 4 N/C.
  2. Total field becomes E = E₀/κ = 1 N/C.
  3. σ_b = σ_f(1−1/κ) = 26.55 pC/m².
  4. The polarization field is −3 N/C relative to the positive E₀ direction, leaving a net +1 N/C.
Common mix-up

Bound charge does not mean electrons freely crossed the dielectric. Field reduction by κ assumes fixed free charge.

CHECK THE IDEA

Does polarization completely cancel E for finite κ?

Compare with an explanation

No. It reduces the field; ideal equilibrium metal is a different limiting model.

Now investigate one change Explore →

Predict. Change one thing. Explain.

Increase κ while holding free surface density fixed. Compare the free-charge field, opposing polarization field and total. Check that κ = 1 recovers vacuum.

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

Free and bound surface charges+ free− free− bound+ boundE = 1 N/CBound magnitude = 26.55 pC/m²; gap sketch not to scale

E_free 4 + E_bound -3 = E_total 1 N/C at fixed free charge. κ = 4.

Field superposition along +xN/C · same scale for all bars0Free-charge field4Bound-charge field-3Total field1

Fully filled narrow parallel-plate gap; homogeneous linear ideal dielectric. Free plate charge stays fixed. Positive field direction points from positive toward negative plate. Bound charges oppose the free-charge field; no leakage or breakdown is modeled.

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. A dielectric primarily responds through…

Show answer and reasoning

polarization. Microscopic charges shift or orient while remaining bound.

2. At fixed free charge, κ = 3 changes E to…

Show answer and reasoning

E₀/3. The total permittivity is three times ε₀.

Original written challenge

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

For fixed σ_f = 53.1 pC/m², insert a fully filling κ = 2 dielectric. Find vacuum field, total field, bound density and its orientation.

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

Compare with the answer and four-point rubric
  1. 1 point: E₀ = σ_f/ε₀ = 6 N/C.
  2. 1 point: E = E₀/2 = 3 N/C.
  3. 1 point: σ_b = σ_f/2 = 26.55 pC/m².
  4. 1 point: Negative bound charge faces the positive plate; its field opposes the original field.

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 does κ measure?

Permittivity relative to vacuum.

RECALL 2Which charge stays fixed in this investigation?

Free charge on the isolated plates.

RECALL 3Where are bound surface charges?

At dielectric boundaries, with signs that oppose the applied field.

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

Bound charges reduce the field for fixed free charge

  • ε = κε₀.
  • Fixed σ_f: E = σ_f/(κε₀).
  • σ_b = σ_f(1−1/κ) for a fully filled ideal plate gap.

Remember: Bound charge does not mean electrons freely crossed the dielectric. Field reduction by κ assumes fixed free charge.

Conditions: Fully filled narrow parallel-plate gap; homogeneous linear ideal dielectric. Free plate charge stays fixed. Positive field direction points from positive toward negative plate. Bound charges oppose the free-charge field; no leakage or breakdown is modeled.

Refresh Kid · AP Physics C: Electricity and Magnetism Unit 3 (official Unit 10) · Objectives 10.4.A · Review edition

Framework, scope and review status

Mapped to College Board CED, Topic 10.4, objectives 10.4.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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