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PU-102 / L14 · lecture

Dielectrics, Polarization and the D Field

~8 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)

BEFORE YOU START

You should be comfortable with Gauss's law for the electric field \(\vec{E}\) and the concept of electric flux. Familiarity with vector calculus operators, specifically the divergence \(\nabla \cdot\), is required to follow the charge density relations.

Why this matters

The \(\vec{D}\) field simplifies electrostatics by isolating the source charges you control from the complex response of the material.

In the previous lecture, we treated conductors as regions where charges move freely to cancel internal fields. Now we turn to dielectrics, insulating materials where charges are bound to atoms but can shift slightly under an external field. This shift creates microscopic dipoles, generating a macroscopic polarization \(\vec{P}\). Calculating the total electric field \(\vec{E}\) in the presence of such materials is difficult because the bound charges depend on the field itself. By introducing the auxiliary field \(\vec{D}\), we can apply a form of Gauss's law that depends only on the free charges we place on our capacitor plates, bypassing the need to calculate the intricate distribution of bound charges within the material.

The Physical Picture: Polarization

Imagine a dielectric material as a collection of neutral atoms. In the absence of an external electric field, the positive nucleus and the electron cloud are centered on each other. When an external field is applied, the positive charges are pulled slightly in the direction of the field, while the negative charges are pulled slightly against it. This separation of charge centers creates an electric dipole moment for each atom.

When we look at the material on a macroscopic scale, we define the polarization vector \(\vec{P}\) as the net electric dipole moment per unit volume. If the material is uniform and the field is constant, \(\vec{P}\) is constant throughout the bulk. However, at the surface of the dielectric, the dipoles are not perfectly cancelled by their neighbors. This results in a net accumulation of charge on the surface, known as bound surface charge. Inside the material, if the polarization is uniform, the bound charges cancel out perfectly. But if the polarization varies with position, there is a net accumulation of bound charge within the volume as well. This is the physical origin of the "bound charge" that complicates the \(\vec{E}\) field calculation.

Bound Charge Densities

To quantify the effect of polarization, we derive the charge densities associated with the dipoles. Consider a small volume element within the dielectric. The net charge inside this volume due to polarization is determined by how much dipole moment flows out of the volume. Mathematically, this is expressed as the negative divergence of the polarization field.

\[ \rho_b = -\nabla \cdot \vec{P} \]

This equation tells us that if the polarization is uniform (\(\nabla \cdot \vec{P} = 0\)), there is no bound volume charge. However, at the boundary of the dielectric, the polarization drops to zero outside the material. This discontinuity creates a surface charge density. The bound surface charge density is given by the component of the polarization vector normal to the surface.

\[ \sigma_b = \vec{P} \cdot \hat{n} \]

Here, \(\hat{n}\) is the unit normal vector pointing outward from the dielectric surface. These two equations, \(\rho_b\) and \(\sigma_b\), allow us to calculate the total charge distribution if we know the polarization. However, since \(\vec{P}\) itself depends on \(\vec{E}\), solving for \(\vec{E}\) directly is often a circular and difficult problem.

Defining the D Field

To resolve this difficulty, we define a new field, the electric displacement field \(\vec{D}\). The definition is constructed specifically to absorb the polarization term. We define \(\vec{D}\) as the sum of the vacuum permittivity times the electric field plus the polarization.

\[ \vec{D} = \varepsilon_0 \vec{E} + \vec{P} \]

This definition is not arbitrary; it is chosen so that the divergence of \(\vec{D}\) yields only the free charge density. Let us verify this by taking the divergence of both sides. Using the linearity of the divergence operator, we get \(\nabla \cdot \vec{D} = \varepsilon_0 (\nabla \cdot \vec{E}) + \nabla \cdot \vec{P}\). From Gauss's law for the total electric field, we know that \(\varepsilon_0 (\nabla \cdot \vec{E}) = \rho_{\text{total}} = \rho_f + \rho_b\), where \(\rho_f\) is the free charge density and \(\rho_b\) is the bound charge density. Substituting the expression for bound volume charge \(\rho_b = -\nabla \cdot \vec{P}\), we find that the \(\nabla \cdot \vec{P}\) terms cancel out exactly.

\[ \nabla \cdot \vec{D} = \rho_f \]

This is the differential form of Gauss's law for the \(\vec{D}\) field. It states that the source of the \(\vec{D}\) field is only the free charge, which is the charge we explicitly place on conductors or electrodes. This is a powerful simplification because \(\rho_f\) is usually known or easily calculated, whereas \(\rho_b\) is a complex function of the field.

Integral Form and Boundary Conditions

By applying the divergence theorem to the differential equation, we obtain the integral form of Gauss's law for \(\vec{D}\). This form is particularly useful for calculating fields in symmetric geometries, such as parallel plate capacitors or spherical dielectrics.

\[ \oint \vec{D} \cdot d\vec{a} = Q_{f,\text{enc}} \]

Here, \(Q_{f,\text{enc}}\) is the total free charge enclosed by the Gaussian surface. Notice that bound charges do not appear on the right-hand side. This means that when using a Gaussian surface that cuts through a dielectric, we only need to account for the free charges on the plates, ignoring the induced bound charges on the dielectric surfaces.

At the interface between two different dielectric materials, or between a dielectric and a vacuum, the \(\vec{D}\) field obeys specific boundary conditions. If there is no free surface charge at the interface, the normal component of \(\vec{D}\) is continuous. That is, \(D_{1n} = D_{2n}\). If there is a free surface charge density \(\sigma_f\) at the interface, the discontinuity in the normal component is equal to that free charge density: \(D_{2n} - D_{1n} = \sigma_f\). These conditions are essential for solving problems involving multiple dielectric layers.

The derivation

◆ DERIVATION EMBED · D-048 · not re-derived here

Dielectrics, Bound Charge and the D Field

A dielectric with polarization field \(\vec{P}(\vec{r})\) (electric dipole moment per unit volume) is electrostatically equivalent to a bound volume charge density \(\rho_b=-\nabla\cdot\vec{P}\) together with a bound surface charge density \(\sigma_b=\vec{P}\cdot\hat{n}\). Consequently the auxiliary field \(\vec{D}\equiv\varepsilon_0\vec{E}+\vec{P}\) obeys Gauss's law sourced by the free charge alone, \(\nabla\cdot\vec{D}=\rho_f\), with integral form \(\oint\vec{D}\cdot d\vec{a}=Q_{f,\text{enc}}\).

\[ \rho_b=-\nabla\cdot\vec{P},\quad \sigma_b=\vec{P}\cdot\hat{n},\quad \vec{D}=\varepsilon_0\vec{E}+\vec{P},\quad \nabla\cdot\vec{D}=\rho_f,\quad \oint\vec{D}\cdot d\vec{a}=Q_{f,\text{enc}} \]
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Tier 3 ยท watch out

A common misconception is that \(\vec{D}\) is a more "fundamental" field than \(\vec{E}\). It is \(\vec{E}\) that exerts forces: the force on a charge \(q\) is \(q\vec{E}\), not \(q\vec{D}\). The definition \(\vec{D} = \varepsilon_0\vec{E} + \vec{P}\) and \(\nabla\cdot\vec{D} = \rho_f\) hold in any material, but the simple relation \(\vec{D} = \varepsilon\vec{E}\) holds only for linear isotropic dielectrics; in anisotropic crystals \(\varepsilon\) becomes a tensor and in non-linear materials \(\vec{P}\) is not proportional to \(\vec{E}\). Note too that \(\nabla\times\vec{D}\) need not be zero, so \(\vec{D}\) is not determined by the free charge alone. C

Worked example

Problem. A parallel plate capacitor has plates of area \(A = 0.02\ \mathrm{m^2}\) separated by a distance \(d = 5.0\ \mathrm{mm}\). The space between the plates is filled with a linear dielectric material with relative permittivity \(\varepsilon_r = 4.0\). A free charge \(Q_f = 1.0\ \mu\mathrm{C}\) is placed on the positive plate. Calculate the magnitude of the electric field \(\vec{E}\) inside the dielectric and the bound surface charge density \(\sigma_b\) on the dielectric surface adjacent to the positive plate.

  1. Calculate the free charge density \(\sigma_f\) on the plate: \(\sigma_f = Q_f / A\).
  2. Use the integral form of Gauss's law for \(\vec{D}\) to find the magnitude of \(\vec{D}\). By symmetry, \(\vec{D}\) is uniform and perpendicular to the plates, so \(D A = Q_f\), giving \(D = \sigma_f\).
  3. Use the constitutive relation for a linear dielectric, \(\vec{D} = \varepsilon \vec{E} = \varepsilon_0 \varepsilon_r \vec{E}\), to find \(\vec{E}\).
  4. Calculate the polarization \(\vec{P}\) using \(\vec{P} = \vec{D} - \varepsilon_0 \vec{E}\).
  5. Calculate the bound surface charge density \(\sigma_b\) using \(\sigma_b = \vec{P} \cdot \hat{n}\). On the dielectric surface next to the positive plate, the outward normal of the dielectric points towards the plate, opposite to \(\vec{P}\), so \(\sigma_b\) is negative there.

Numbers: \(\sigma_f = 1.0\times10^{-6}/0.02 = 5.0\times10^{-5}\ \mathrm{C/m^2}\), so \(D = 5.0\times10^{-5}\ \mathrm{C/m^2}\); \(E = D/(\varepsilon_0\varepsilon_r) = 5.0\times10^{-5}/(4.0\times8.854\times10^{-12}) \approx 1.41\times10^6\ \mathrm{V/m}\); \(P = D - \varepsilon_0 E = \sigma_f(1-1/\varepsilon_r) = 3.75\times10^{-5}\ \mathrm{C/m^2}\).

\[ \boxed{E \approx 1.41 \times 10^6\ \mathrm{V/m}, \quad \sigma_b = -3.75 \times 10^{-5}\ \mathrm{C/m^2}} \]

Units check. \(D/\varepsilon\) has units \((\mathrm{C/m^2})/(\mathrm{F/m}) = \mathrm{C/(F\,m)} = \mathrm{V/m}\), and \(P\) has the units of \(D\), \(\mathrm{C/m^2}\), as a surface charge density should. The bound charge (\(-3.75\times10^{-5}\)) partly cancels the free charge (\(+5.0\times10^{-5}\)), leaving a net \(1.25\times10^{-5}\ \mathrm{C/m^2} = \varepsilon_0 E\).

CHECK YOURSELF

Three quick tests

  1. What is the physical meaning of the divergence of the polarization field \(\nabla \cdot \vec{P}\)?
    AnswerIt represents the negative of the bound volume charge density \(\rho_b\). A non-zero divergence indicates a net accumulation of bound charge within the volume.
  2. If a dielectric with \(\varepsilon_r = 2\) is placed between capacitor plates with free charge density \(\sigma_f\), what is the ratio of the electric field \(E\) to the displacement field \(D\)?
    AnswerThe ratio is \(E/D = 1/(\varepsilon_0 \varepsilon_r) = 1/(2\varepsilon_0)\). The field is half of what it would be in a vacuum for the same free charge.
  3. At a boundary between two dielectrics with no free surface charge, which component of \(\vec{D}\) is continuous?
    AnswerThe normal component of \(\vec{D}\) is continuous. The tangential component of \(\vec{E}\) is also continuous, but the tangential component of \(\vec{D}\) is not necessarily continuous if the permittivities differ.
PROBLEMS

Problem set for PU-102 →

Each derivation in this unit ends with five graded problems with full solutions.

NOTES

Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.

Derivations cited: D-048 · Threads: force, energy, fields