Dielectrics

AP Physics C: Electricity and Magnetism· AP Physics C: E&M CED — Conductors, Capacitors, Dielectrics· 14 min read

1. What Are Dielectrics?★☆☆☆☆⏱ 2 min

Dielectrics are insulating (non-conducting) materials inserted between capacitor plates to modify electrical properties. This topic makes up 15-20% of AP Physics C: E&M Unit 2, or approximately 3-6% of the total exam score, and appears regularly on both multiple choice and free response sections.

📘 Definition

Dielectric Constant

$κ$ (Greek lowercase kappa)

Dimensionless quantity greater than or equal to 1 that describes how much a material reduces the net electric field inside it. $κ = 1$ for vacuum, and $κ > 1$ for all practical dielectric materials.

Example:

$κ \approx 1$ for air, $κ = 80$ for human body tissue.

Allow higher operating voltages before dielectric breakdown (arcing) between plates
Reduce the physical size of capacitors for a given target capacitance
Increase capacitance for a fixed plate charge or voltage

2. Dielectric Polarization and Capacitance Scaling★★★☆☆⏱ 4 min

When an external electric field is applied to a dielectric, bound charges within the material polarize: positive bound charges shift slightly toward the negative capacitor plate, and negative bound charges shift toward the positive plate. This creates a thin layer of induced surface charge that produces an induced electric field opposing the original field from free charge on the plates.

The net electric field inside the dielectric is reduced by a factor of $κ$ , so for a given amount of free charge $Q$ on the plates, the potential difference $V = E d$ between plates is also reduced by $κ$ . Since capacitance is defined as $C = Q / V$ , the new capacitance of a dielectric-filled capacitor becomes:

C = \kappa C_0

Where $C_{0}$ is the capacitance of the same geometry without the dielectric. For a fully filled parallel plate capacitor with area $A$ and separation $d$ , this simplifies to:

C = \frac{\kappa \epsilon_0 A}{d}

📐 Worked Example

A parallel plate air capacitor has an initial capacitance of $12 pF$ , with plate separation $2.0 mm$ . A dielectric slab with $κ = 3.5$ and thickness $1.5 mm$ is inserted between the plates, filling the full plate area. Calculate the new capacitance of the system.

1
This configuration is equivalent to two capacitors in series: one dielectric-filled of thickness $d_{1} = 1.5 mm$ , one air-filled of thickness $d_{2} = 0.5 mm$ . We use $C_{0} = ϵ_{0} A / d = 12 pF$ , so $ϵ_{0} A = C_{0} d$ to avoid recalculating constants.
2
$C_1 = \frac{\kappa \epsilon_0 A}{d_1} = \frac{\kappa C_0 d}{d_1} = \frac{3.5 (12\ \text{pF}) (2.0\ \text{mm})}{1.5\ \text{mm}} = 56\ \text{pF}, \quad C_2 = \frac{C_0 d}{d_2} = \frac{(12\ \text{pF}) (2.0\ \text{mm})}{0.5\ \text{mm}} = 48\ \text{pF}$
3
Apply the series capacitance combination rule:
4
$\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2}$
5
Substitute values and solve for $C_{eq}$ :
6
$\frac{1}{C_{\text{eq}}} = \frac{1}{56\ \text{pF}} + \frac{1}{48\ \text{pF}} = \frac{104}{2688}\ \text{pF}^{-1} \implies C_{\text{eq}} \approx 26\ \text{pF}$

Exam tip:

If the dielectric does not fill the entire volume of the capacitor, always decompose the system into combinations of fully filled or vacuum-filled capacitors in series or parallel before calculating equivalent capacitance, instead of relying on a memorized off-format formula.

3. Gauss's Law in Dielectrics★★★★☆⏱ 4 min

When working with dielectrics, Gauss's law can be rewritten to automatically account for bound induced charge, eliminating the need to calculate bound charge explicitly. Permittivity of a dielectric is defined as $ϵ = κ ϵ_{0}$ , and the electric displacement vector is $D = ϵ E = κ ϵ_{0} E$ .

Gauss's law in dielectrics takes the form:

\oint \vec{D} \cdot d\vec{A} = Q_{\text{free, enclosed}}

Only free charge (charge placed on conducting plates, not induced bound charge from polarization) is included on the right-hand side. For symmetric geometries, solve for $D$ first, then find $E = D / (κ ϵ_{0})$ , then calculate potential difference and capacitance.

📐 Worked Example

A spherical capacitor has an inner conducting shell of radius $a$ with free charge $+ Q$ , and an outer conducting shell of radius $b$ with free charge $- Q$ . The space between the shells is filled with a non-uniform dielectric with dielectric constant $κ (r) = κ_{0} \frac{a}{r}$ for $a < r < b$ . Derive an expression for the capacitance of this device.

1
By symmetry, $D$ points radially outward and has constant magnitude on any spherical Gaussian surface of radius $r$ between $a$ and $b$ . Apply Gauss's law for dielectrics:
2
$\oint \vec{D} \cdot d\vec{A} = D (4\pi r^2) = Q_{\text{free, enclosed}} = Q$
3
Solve for $D$ , then find $E (r)$ using $E = D / (κ (r) ϵ_{0})$ :
4
$D = \frac{Q}{4\pi r^2} \implies E(r) = \frac{Q}{4\pi r^2 \epsilon_0 \left(\kappa_0 \frac{a}{r}\right)} = \frac{Q}{4\pi \epsilon_0 \kappa_0 a r}$
5
Calculate the potential difference between the shells by integrating $E (r)$ :
6
$V = \int_a^b E(r) dr = \frac{Q}{4\pi \epsilon_0 \kappa_0 a} \int_a^b \frac{1}{r} dr = \frac{Q}{4\pi \epsilon_0 \kappa_0 a} \ln\left(\frac{b}{a}\right)$
7
Use the definition of capacitance $C = Q / V$ to get the final result:
8
$C = \frac{4\pi \epsilon_0 \kappa_0 a}{\ln\left(\frac{b}{a}\right)}$

Exam tip:

Always remember that Gauss's law in dielectrics only counts free charge in the enclosed term; bound induced charge is already accounted for by the $κ$ in the relation $D = κ ϵ_{0} E$ .

4. Energy Storage in Dielectric-Filled Capacitors★★★☆☆⏱ 3 min

The general formula for stored energy in any capacitor remains $U = \frac{1}{2} C V^{2} = \frac{Q ^{2}}{2 C} = \frac{1}{2} Q V$ regardless of whether a dielectric is present. The change in stored energy after inserting a dielectric depends entirely on whether the capacitor is connected to a battery (fixed potential difference) or isolated (fixed free charge):

Battery connected (fixed $V$ ): $C$ increases by $κ$ , so $U = κ U_{0}$ . Energy increases, because the battery does work to add extra charge to maintain constant voltage.
Isolated (fixed $Q$ ): $C$ increases by $κ$ , so $U = U_{0} / κ$ . Energy decreases, because the attractive force between induced surface charges pulls the dielectric inward, reducing total stored energy.

The energy density (energy per unit volume) in a dielectric is:

u = \frac{1}{2} \kappa \epsilon_0 E^2 = \frac{1}{2}DE

📐 Worked Example

An isolated parallel plate capacitor has a vacuum capacitance of $5.0 μ F$ , charged to a potential difference of $100 V$ by a battery, then disconnected from the battery. A dielectric with $κ = 2.5$ is fully inserted between the plates. Calculate the final stored energy in the capacitor after insertion.

1
The capacitor is isolated, so charge $Q$ is constant. Calculate initial charge:
2
$Q = C_0 V_0 = (5.0 \times 10^{-6}\ \text{F})(100\ \text{V}) = 5.0 \times 10^{-4}\ \text{C}$
3
Find new capacitance after insertion:
4
$C = \kappa C_0 = 2.5 (5.0 \times 10^{-6}\ \text{F}) = 12.5 \times 10^{-6}\ \text{F}$
5
Use the constant-charge energy formula to solve for $U$ :
6
$U = \frac{Q^2}{2C} = \frac{(5.0 \times 10^{-4}\ \text{C})^2}{2 (12.5 \times 10^{-6}\ \text{F})} = 0.010\ \text{J}$
7
Check with the scaling rule for fixed $Q$ : initial energy $U_{0} = 0.025 J$ , so $U = U_{0} / κ = 0.025/2.5 = 0.010 J$ , which matches.

Exam tip:

Always check if the capacitor is connected to a battery (fixed $V$ ) or isolated (fixed $Q$ ) before calculating energy change after inserting a dielectric — the change in energy has opposite signs for the two cases.

5. AP-Style Practice Problems★★★☆☆⏱ 5 min

✓ Quick check

Test your understanding with this multiple choice question:

An isolated parallel plate capacitor with vacuum between the plates has stored energy $U_{0}$ , charge $Q_{0}$ , capacitance $C_{0}$ . A dielectric with $κ = 2$ is fully inserted between the plates, with the capacitor remaining isolated. What is the new stored energy?
- $U_{0} /4$
- $U_{0} /2$
- $U_{0}$
- $2 U_{0}$
Reveal answer
1 —
An isolated capacitor has constant free charge, so $C = 2 C_{0}$ , and $U = Q^{2} / (2 C) = U_{0} /2$ . If the capacitor were connected to a battery, the answer would be $2 U_{0}$ , but that is not the case here.

📐 Worked Example

A parallel plate capacitor has square plates of side length $L$ , plate separation $d$ . Half the volume is filled with two dielectrics: dielectric 1 ( $κ_{1}$ ) fills the left half of the plate area (full separation $d$ ), and dielectric 2 ( $κ_{2}$ ) fills the right half of the plate area (full separation $d$ ). (a) Derive an expression for equivalent capacitance. (b) The capacitor is connected to a battery of voltage $V$ . What is the total free charge on each plate? (c) The battery is disconnected, then the dielectrics are removed. What is the new potential difference?

1
Part (a): Each dielectric fills half the plate area, and both span the full separation. They share the same potential difference, so they are in parallel:
2
$C_1 = \frac{\kappa_1 \epsilon_0 (L^2/2)}{d}, \quad C_2 = \frac{\kappa_2 \epsilon_0 (L^2/2)}{d} \implies C_{\text{eq}} = C_1 + C_2 = \frac{\epsilon_0 L^2 (\kappa_1 + \kappa_2)}{2d}$
3
Part (b): Total charge is $Q = C_{eq} V$ :
4
$Q = \frac{\epsilon_0 L^2 (\kappa_1 + \kappa_2) V}{2d}$
5
Part (c): Charge is constant after disconnection. New vacuum capacitance is $C_{0} = ϵ_{0} L^{2} / d$ , so:
6
$V' = \frac{Q}{C_0} = \frac{(\kappa_1 + \kappa_2)}{2} V$

6. Common Pitfalls

Wrong move:

Assuming inserting a dielectric always increases stored energy, regardless of battery connection

Why:

Students memorize that $C$ always increases, so they assume $U = \frac{1}{2} C V^{2}$ always means $U$ increases, forgetting $V$ is not constant for disconnected capacitors

Correct move:

Explicitly identify if $V$ is fixed (battery connected) or $Q$ is fixed (battery disconnected) before writing the energy expression

Wrong move:

Including bound induced charge in the $Q_{enclosed}$ term when using Gauss's law for dielectrics

Why:

Students confuse the original all-charge Gauss's law with the dielectric form, leading to incorrect $E$ values

Correct move:

For $ \oint \vec{D} \cdot d\vec{A} = Q_{\text{free, enclosed}}$, only include charge placed on the conductors, not induced polarization charge

Wrong move:

Treating a dielectric that fills half the area (full plate separation) as two capacitors in series

Why:

Students mix up series vs parallel for partial volume fills

Correct move:

If the dielectric extends across the entire plate separation, both regions share the same potential difference, so they are in parallel

Wrong move:

Using $C = κ ϵ_{0} A / d$ for a capacitor that is only partially filled with dielectric

Why:

Students memorize the fully filled formula and apply it to all geometries

Correct move:

Decompose any partially filled capacitor into fully filled or vacuum sub-capacitors, then combine with series/parallel rules

Wrong move:

Treating dielectric constant $κ$ as having units of farads per meter

Why:

Students mix up $κ$ with permittivity $ϵ = κ ϵ_{0}$ , leading to unit errors in final capacitance

Correct move:

Confirm $κ$ is dimensionless, so $C$ will always have correct units of farads when using the formula properly

7. Quick Reference Cheatsheet

Category	Formula	Notes
Dielectric capacitance scaling	$C = κ C_{0}$	$κ \geq 1$ , $κ = 1$ for vacuum, $C_{0}$ = vacuum capacitance
Fully filled parallel plate	$C = \frac{κ ϵ _{0} A}{d}$	Only applies when dielectric fills entire volume between plates
Gauss's law in dielectrics	$\oint D \cdot d A = Q_{free, enclosed}$	Only free charge counted; bound charge is automatically accounted for
Displacement-Electric relation	$D = κ ϵ_{0} E$	Relates $D$ from Gauss's law to net electric field $E$
General stored energy	$U = \frac{1}{2} C V^{2} = \frac{Q ^{2}}{2 C}$	Same form for any capacitor, with or without dielectric
Energy change (fixed $V$ , battery connected)	$U = κ U_{0}$	Energy increases; battery adds charge to maintain voltage
Energy change (fixed $Q$ , isolated)	$U = U_{0} / κ$	Energy decreases; attractive force pulls dielectric inward
Energy density in dielectric	$u = \frac{1}{2} κ ϵ_{0} E^{2} = \frac{1}{2} D E$	Energy per unit volume in any dielectric medium

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Energy change after dielectric insertion
2022 · FRQ
Capacitance of non-uniform dielectric

Going deeper

unit overviewAP Physics C: E&M Unit 2 Overview

What's Next

This subtopic completes the fundamentals of capacitors needed for the rest of the AP Physics C: E&M course. Next you will apply your understanding of capacitance (including dielectrics) to analyze RC circuits, where capacitors charge and discharge through resistors. Without mastering how to calculate equivalent capacitance for systems with dielectrics, you will not be able to correctly find the time constant or solve for voltage/current in RC circuits, which are common on AP FRQs. Dielectrics also lay the groundwork for understanding electric fields in materials, a core concept for analyzing Maxwell's equations later in the course.