Dielectrics — AP Physics C: E&M Study Guide

For: AP Physics C: E&M candidates sitting AP Physics C: E&M.

Covers: Dielectric polarization, capacitance modification by dielectrics, Gauss’s law in dielectrics, energy storage in dielectric-filled capacitors, and calculation of equivalent capacitance for mixed dielectric configurations, aligned to AP CED requirements.

You should already know: Capacitance of parallel plate capacitors in vacuum. Gauss's law for electric fields in free space. Energy stored in vacuum-filled capacitors.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics C: E&M style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Dielectrics?

Dielectrics are insulating (non-conducting) materials inserted between the plates of a capacitor to modify its electrical properties. This topic makes up approximately 15-20% of Unit 2 (Conductors, Capacitors, Dielectrics) for AP Physics C: E&M, translating to roughly 3-6% of the total exam score. Dielectrics appear regularly on both multiple choice (MCQ) and free response (FRQ) sections, often combined with other topics like energy storage or RC circuit behavior. Synonyms sometimes used include "insulating materials for capacitors", though dielectrics are specifically selected for their polarization properties rather than just lack of conductivity.

Notation convention: the dielectric constant is always written as $\kappa$ (Greek lowercase kappa), a dimensionless quantity greater than or equal to 1 ($\kappa =1$ for vacuum, $\kappa >1$ for all practical dielectric materials). Dielectrics serve three key practical purposes: they allow higher operating voltages before dielectric breakdown (arcing) between plates, they reduce the physical size of capacitors for a given capacitance, and they increase capacitance for a fixed plate charge or voltage.

2. Dielectric Polarization and Capacitance Scaling

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 on opposite faces of the dielectric, which produces an induced electric field that opposes the original field from the free charge on the capacitor plates.

The net electric field inside the dielectric is reduced by a factor of $\kappa$, so for a given amount of free charge $Q$ on the plates, the potential difference $V=Ed$ between plates is also reduced by $\kappa$. 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 air, ${\kappa }_{\text{air}}\approx 1$, so $C_0$ is approximately equal to the capacitance in air). For the common case of a fully filled parallel plate capacitor with area $A$ and separation $d$, this simplifies to: $$C=\frac{\kappa {ϵ}_{0}A}{d}$$

Worked Example

A parallel plate air capacitor has an initial capacitance of $12\text{pF}$, with plate area $1.36\times 10^{-2}{\text{m}}^2$ and plate separation $2.0\text{mm}$. A dielectric slab with $\kappa =3.5$ and thickness $1.5\text{mm}$ 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 with thickness $d_1=1.5\text{mm}$ filled with dielectric, and one with thickness $d_2=0.5\text{mm}$ filled with air ($\kappa \approx 1$). We use the original capacitance relation $C_0={ϵ}_{0}A/d=12\text{pF}$, so ${ϵ}_{0}A=C_{0}d$ to avoid recalculating constants.
2. Calculate individual capacitances: $C_1=\frac{\kappa {ϵ}_{0}A}{d_1}=\frac{\kappa C_{0}d}{d_1}=\frac{3.5(12\text{pF})(2.0\text{mm})}{1.5\text{mm}}=56\text{pF}$, and $C_2=\frac{C_{0}d}{d_2}=\frac{(12\text{pF})(2.0\text{mm})}{0.5\text{mm}}=48\text{pF}$.
3. Apply series capacitance combination: $\frac{1}{C_{\text{eq}}}=\frac{1}{C_1}+\frac{1}{C_2}$.
4. Substitute values: $\frac{1}{C_{\text{eq}}}=\frac{1}{56\text{pF}}+\frac{1}{48\text{pF}}=\frac{104}{2688}{\text{pF}}^{-1}$, so $C_{\text{eq}}\approx 26\text{pF}$.

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

When working with dielectrics, Gauss's law can be rewritten to simplify calculations by automatically accounting for bound induced charge, eliminating the need to calculate bound charge explicitly. For any dielectric, the permittivity is defined as $ϵ=\kappa {ϵ}_0$, and the electric displacement vector $\vec{D}=ϵ\vec{E}=\kappa {ϵ}_0\vec{E}$.

Gauss's law in dielectrics takes the form: $$\oint \vec{D}\cdot d\vec{A}=Q_{\text{free, enclosed}}$$ Only free charge (the charge placed on the conducting plates, not induced bound charge from polarization) is included on the right-hand side. For symmetric geometries (parallel plates, spherical, cylindrical capacitors), we first solve for $D$ using Gauss's law, then find the net electric field as $E=D/(\kappa {ϵ}_0)$, then calculate potential difference and capacitance from $E$.

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 $\kappa (r)={\kappa }_0\frac{a}{r}$ for $a<r<b$. Derive an expression for the capacitance of this device.

1. By symmetry, $\vec{D}$ 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: $\oint \vec{D}\cdot d\vec{A}=D(4\pi r^2)=Q_{\text{free, enclosed}}=Q$.
2. Solve for $D=\frac{Q}{4\pi r^2}$, then substitute into $E=D/(\kappa (r){ϵ}_0)$: $E(r)=\frac{Q}{4\pi r^2{ϵ}_0\left({\kappa }_0\frac{a}{r}\right)}=\frac{Q}{4\pi {ϵ}_0{\kappa }_{0}ar}$.
3. Calculate the potential difference between the shells: $V={\int }_a^{b}E(r)dr=\frac{Q}{4\pi {ϵ}_0{\kappa }_{0}a}{\int }_a^b\frac{1}{r}dr=\frac{Q}{4\pi {ϵ}_0{\kappa }_{0}a}\ln \left(\frac{b}{a}\right)$.
4. Capacitance is $C=Q/V$, so $C=\frac{4\pi {ϵ}_0{\kappa }_{0}a}{\ln \left(\frac{b}{a}\right)}$.

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 $\kappa$ in the relation $D=\kappa {ϵ}_{0}E$.

4. Energy Stored in Dielectric-Filled Capacitors

The general formula for stored energy in any capacitor remains $U=\frac{1}{2}C{V}^2=\frac{Q^2}{2C}=\frac{1}{2}QV$ 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 $V$) or isolated (fixed free charge $Q$):

• Battery connected (fixed $V$): $C$ increases by $\kappa$, so $U=\frac{1}{2}(\kappa C_0)V^2=\kappa U_0$. Energy increases by a factor of $\kappa$, because the battery does work to add extra charge to the plates to maintain constant voltage.
• Isolated (fixed $Q$): $C$ increases by $\kappa$, so $U=\frac{Q^2}{2\kappa C_0}=U_0/\kappa$. Energy decreases by a factor of $\kappa$, because no extra charge can be added, and 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 $u=\frac{1}{2}\kappa {ϵ}_0{E}^2=\frac{1}{2}DE$, which generalizes the vacuum energy density formula.

Worked Example

An isolated parallel plate capacitor has a vacuum capacitance of $5.0\mu \text{F}$, charged to a potential difference of $100\text{V}$ by a battery, then disconnected from the battery. A dielectric with $\kappa =2.5$ 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: $Q=C_0{V}_0=(5.0\times 10^{-6}\text{F})(100\text{V})=5.0\times 10^{-4}\text{C}$.
2. New capacitance after insertion: $C=\kappa C_0=2.5(5.0\times 10^{-6}\text{F})=12.5\times 10^{-6}\text{F}$.
3. Use the constant-charge energy formula: $U=\frac{Q^2}{2C}=\frac{(5.0\times 10^{-4}\text{C}{)}^2}{2(12.5\times 10^{-6}\text{F})}=0.010\text{J}$.
4. Check: Initial energy was $U_0=\frac{1}{2}{C}_0{V}_0^2=0.025\text{J}$, so $U=U_0/\kappa =0.025/2.5=0.010\text{J}$, 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. Common Pitfalls (and how to avoid them)

• Wrong move: Assuming inserting a dielectric always increases stored energy, regardless of battery connection. Why: Students memorize that $C$ always increases, so they assume $U=1/2C{V}^2$ 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 $Q_{\text{enclosed}}$ 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 the dielectric and vacuum regions share the same potential difference, so they are in parallel.
• Wrong move: Using $C=\kappa {ϵ}_{0}A/d$ 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 $\kappa$ as having units of farads per meter. Why: Students mix up $\kappa$ with permittivity $ϵ=\kappa {ϵ}_0$, leading to unit errors in final capacitance. Correct move: Confirm $\kappa$ is dimensionless, so $C$ will always have units of farads when using the formula correctly.

6. Practice Questions (AP Physics C: E&M Style)

Question 1 (Multiple Choice)

An isolated parallel plate capacitor with vacuum between the plates has stored energy $U_0$, charge $Q_0$, capacitance $C_0$. A dielectric with $\kappa =2$ is fully inserted between the plates, with the capacitor remaining isolated. What is the new stored energy? (A) $U_0/4$ (B) $U_0/2$ (C) $U_0$ (D) $2U_0$

Worked Solution: An isolated capacitor has constant free charge, so $Q=Q_0$ after inserting the dielectric. Capacitance increases to $C=\kappa C_0=2C_0$. Stored energy for constant charge is $U=Q^2/(2C)=Q_0^2/(2(2C_0))=U_0/2$. If the capacitor were connected to a battery, energy would double, but that is not the case here. The correct answer is (B).

Question 2 (Free Response)

A parallel plate capacitor has square plates of side length $L$, plate separation $d$. Half the volume is filled with two dielectrics: dielectric 1 (${\kappa }_1$) fills the left half of the plate area, full separation $d$, and dielectric 2 (${\kappa }_2$) fills the right half of the plate area, full separation $d$. (a) Derive an expression for the equivalent capacitance of this system in terms of given quantities and ${ϵ}_0$. (b) The capacitor is connected to a battery of voltage $V$. What is the total free charge on each plate, in terms of given quantities and ${ϵ}_0$? (c) The battery is disconnected, then the two dielectrics are removed. What is the new potential difference between the plates, in terms of the original voltage $V$?

Worked Solution: (a) Each dielectric fills half the plate area, so $A_1=A_2=L^2/2$, and both span the full separation $d$. They share the same potential difference, so they are in parallel. Individual capacitances: $C_1=\frac{{\kappa }_1{ϵ}_0(L^2/2)}{d}$, $C_2=\frac{{\kappa }_2{ϵ}_0(L^2/2)}{d}$. Adding parallel capacitances gives $C_{\text{eq}}=\frac{{ϵ}_0{L}^2({\kappa }_1+{\kappa }_2)}{2d}$. (b) Total charge $Q=C_{\text{eq}}V=\frac{{ϵ}_0{L}^2({\kappa }_1+{\kappa }_2)V}{2d}$. (c) Charge is constant after disconnecting the battery. New capacitance with vacuum is $C_0=\frac{{ϵ}_0{L}^2}{d}$. New voltage $V^{\prime }=Q/C_0=\frac{({\kappa }_1+{\kappa }_2)}{2}V$.

Question 3 (Application / Real-World Style)

A touch sensor in a smartphone uses a parallel plate capacitor where your finger (which acts as a dielectric with $\kappa =80$, matching the dielectric constant of body tissue) replaces the air gap between the plates when you touch the screen. The sensor capacitor has area $2.0{\text{mm}}^2$, plate separation $0.50\text{mm}$, and is normally filled with air (${\kappa }_{\text{air}}\approx 1$). When you touch the screen, your finger fully replaces the air between the plates. The capacitor is connected to a $3.3\text{V}$ logic circuit. What is the change in charge on the capacitor when you touch the sensor?

Worked Solution: Original capacitance with air: $C_0=\frac{{\kappa }_{\text{air}}{ϵ}_{0}A}{d}=\frac{(1)(8.85\times 10^{-12}\text{F/m})(2.0\times 10^{-6}{\text{m}}^2)}{0.50\times 10^{-3}\text{m}}\approx 3.5\times 10^{-14}\text{F}=35\text{fF}$. New capacitance with finger: $C=\kappa C_0=80(35\text{fF})=2800\text{fF}=2.8\times 10^{-12}\text{F}$. Change in charge (constant $V=3.3\text{V}$): $\Delta Q=V(C-C_0)\approx 9.1\times 10^{-12}\text{C}=9.1\text{pC}$. This large, easily measurable change in charge is detected by the phone's control circuit to register that a touch has occurred at that sensor location.

7. Quick Reference Cheatsheet

8. What's Next

This chapter on dielectrics 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 with mixed dielectrics, 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.

• RC Circuits
• Electric Fields in Materials
• Maxwell's Equations