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Polarization and bound charge, the displacement $\vec{D} = \varepsilon_0\vec{E} + \vec{P}$ with $\nabla \cdot \vec{D} = \rho_f$, linear dielectrics and capacitors filled with them.
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By the end of this lesson you will be able to find fields, polarization and bound charge in linear dielectrics, and compute the capacitance of capacitors filled or partly filled with dielectric.
You know that molecules have permanent or induced dipole moments, that dipoles align in a field, and you have Gauss's law and the energy of capacitors from earlier lessons. From Physics C you know that a dielectric raises a capacitor's capacitance by $\kappa$. This lesson explains why, by adding up the fields of countless tiny dipoles inside matter.
| Term | What it means |
|---|---|
| Polarization | $\vec{P}$, the dipole moment per unit volume of a material. |
| Bound charge | Charge from uncancelled ends of molecular dipoles: $\sigma_b = \vec{P} \cdot \hat{n}$, $\rho_b = -\nabla \cdot \vec{P}$. |
| Free charge | Charge that can be placed or moved at will, such as on capacitor plates. |
| Electric displacement | $\vec{D} = \varepsilon_0\vec{E} + \vec{P}$, with $\nabla \cdot \vec{D} = \rho_f$. |
| Linear dielectric | A material with $\vec{P} = \varepsilon_0\chi_e\vec{E}$, so $\vec{D} = \varepsilon\vec{E}$. |
| Dielectric constant | $\kappa = \varepsilon/\varepsilon_0 = 1 + \chi_e$. |
| Dielectric strength | The largest field a material withstands before breaking down. |
In a field, the molecules of an insulator become dipoles, induced or aligned, and the material acquires a polarization $\vec{P}$, the dipole moment per unit volume. Adding up the potentials of all these dipoles gives the same result as two kinds of bound charge:
$$\sigma_b = \vec{P} \cdot \hat{n}, \qquad \rho_b = -\nabla \cdot \vec{P}.$$
In a uniformly polarized slab, the dipoles cancel inside, leaving a layer of positive charge on one face and negative on the other. Their field opposes the applied one, which is why fields inside dielectrics are weaker.
Gauss's law counts all charge, $\varepsilon_0\nabla \cdot \vec{E} = \rho_f + \rho_b$. Moving the bound part to the left side defines the electric displacement:
$$\vec{D} = \varepsilon_0\vec{E} + \vec{P}, \qquad \nabla \cdot \vec{D} = \rho_f.$$
For most materials in ordinary fields, $\vec{P}$ is proportional to $\vec{E}$, $\vec{P} = \varepsilon_0\chi_e\vec{E}$, so $\vec{D} = \varepsilon\vec{E}$ with $\varepsilon = \kappa\varepsilon_0$ and $\kappa = 1 + \chi_e$. In a capacitor filled with such a material, $D = \sigma_f$, the field is $\sigma_f/\kappa\varepsilon_0$, $\kappa$ times weaker than without it, and the capacitance is $\kappa$ times larger.
Another way: picture
Picture a crowd of tiny arrows inside the insulator, each a molecular dipole, all turned partly along the field. Inside, every arrow's head sits next to the next arrow's tail, and their charges cancel. At the surfaces, arrowheads poke out on one side and tails on the other: thin sheets of bound charge. Those sheets face the capacitor plates with opposite charge, cancelling part of the plates' field.
Another way: steps
Checks. The bound charge on a face must be opposite in sign to the free charge on the neighboring plate and smaller in magnitude, $\sigma_f(1 - 1/\kappa)$. Total bound charge on an isolated piece of dielectric must be zero. With $\kappa = 1$, every result must reduce to vacuum. And capacitance must increase with any added dielectric.
The potential of a single dipole is $k\,\vec{p} \cdot \hat{\mathscr{r}}/\mathscr{r}^2$. For a polarized material, add up $k\,\vec{P} \cdot \hat{\mathscr{r}}/\mathscr{r}^2\,d\tau$ over the volume. An integration by parts rewrites the integral as $k\oint\sigma_b\,da/\mathscr{r} + k\int\rho_b\,d\tau/\mathscr{r}$ with $\sigma_b = \vec{P} \cdot \hat{n}$ and $\rho_b = -\nabla \cdot \vec{P}$: exactly the potential of surface and volume charge.
Physically, where polarization is uniform, the head of each dipole cancels the tail of the next; only at the surfaces or where $\vec{P}$ varies is there net charge. The bound charge is as real as any other in its effects on the field — it is simply tied to the molecules and cannot be removed or conducted away.
$\vec{D}$ is useful because its sources are the free charges, which we control and know, while bound charges depend on the field we are trying to find. In symmetric problems, Gauss's law for $\vec{D}$ gives $\vec{D}$ directly from the free charge, and dividing by $\varepsilon$ gives $\vec{E}$ without ever computing the bound charge.
But $\vec{D}$ is not simply a field with free charge as its source: its curl need not vanish where dielectrics meet at an angle, so $\vec{D}$ is not determined by free charge alone without symmetry. Treat it as a tool, not as the true field. The force on a charge is always $q\vec{E}$, and $\vec{E}$ remains the fundamental quantity.
At an interface between two dielectrics with no free charge on it, a pillbox gives $D_{1,\perp} = D_{2,\perp}$, and a small loop gives $E_{1,\parallel} = E_{2,\parallel}$. So the normal component of $\vec{E}$ jumps by the ratio of the dielectric constants, $\kappa_1E_{1,\perp} = \kappa_2E_{2,\perp}$, and field lines bend at the interface.
This bending is the electrostatic version of refraction. It concentrates fields in low-$\kappa$ regions: an air bubble inside a high-voltage insulator carries a field several times higher than the surrounding material. That is why bubbles and voids cause partial discharges and eventual breakdown in cables and transformers, and why manufacturers vacuum-impregnate high-voltage insulation.
A dielectric raises capacitance by weakening the field for a given charge. It also raises the voltage a capacitor can hold, because most solid dielectrics have breakdown fields far above air's $3$ MV/m: polypropylene film withstands about $600$ MV/m. Both effects multiply the energy a capacitor can store.
Ceramic capacitors use barium titanate, whose dielectric constant can exceed $1000$; that is why a tiny surface-mount capacitor on a circuit board can hold microfarads. Electrolytic capacitors use an oxide layer only nanometers thick. The billions of capacitors made each year all rely on the same physics: the bound charge of aligned dipoles cancelling most of the field from the plates.
In a linear dielectric the energy density is $\tfrac{1}{2}\vec{D} \cdot \vec{E}$, which for a filled capacitor gives $\tfrac{1}{2}CV^2$ with the enhanced $C$. At fixed charge, inserting a dielectric lowers the energy by $\kappa$, so the field pulls the dielectric in: a slab held at the edge of a charged capacitor is drawn into the gap by the fringing field.
At fixed voltage, with a battery attached, the energy stored rises by $\kappa$, yet the slab is still pulled in; the battery supplies twice the energy increase, and half goes into mechanical work. This is the principle of capacitive level sensors in fuel tanks and of some micro-actuators, where the pull on a dielectric moves a mirror or a switch.
The linear law $\vec{P} = \varepsilon_0\chi_e\vec{E}$ fails in strong fields and in special materials. Ferroelectrics such as barium titanate have spontaneous polarization that can be flipped by a field, showing hysteresis like a ferromagnet; they are used in nonvolatile memory. Piezoelectrics such as quartz polarize when squeezed and deform when polarized, which is how quartz watches keep time and ultrasound probes send and receive sound.
Near breakdown, any material's response becomes nonlinear as electrons are torn from atoms. In optics, the nonlinear part of the polarization at the intense fields of lasers produces new colors of light — green laser pointers double the frequency of infrared light in a nonlinear crystal. Linear dielectric theory is the starting point for all of these.
A material's dielectric constant reflects which of its charges can respond to a field and how quickly. In nonpolar materials such as polyethylene, only electron clouds shift, giving values near two. In polar liquids such as water, whole molecules rotate, giving much larger values, about eighty for water at room temperature. In ionic crystals, positive and negative ions shift relative to one another, adding a further contribution.
Each mechanism has a speed limit. Molecular rotation in water cannot follow fields oscillating much faster than about ten gigahertz, and ions cannot follow optical frequencies, so at the frequency of visible light only the electrons respond. That is why water's dielectric constant is eighty for static fields but its refractive index is only $1.33$, whose square, $1.77$, is the dielectric constant at optical frequencies. Measuring how the dielectric constant changes with frequency, dielectric spectroscopy, reveals which charges move in a material and how freely.
A transistor's gate controls current through the channel below it by capacitive coupling across a thin insulator. For decades the insulator was silicon dioxide, $\kappa = 3.9$, thinned with each generation to keep the capacitance up as transistors shrank. By the early 2000s it was about $1.2$ nm thick, only five atomic layers, and electrons tunneled straight through it, wasting power.
The solution was a material with a higher dielectric constant. In 2007 Intel introduced hafnium-based gate dielectrics, $\kappa \approx 25$, in its $45$ nm processors. A hafnium oxide layer $3$ nm thick has the same capacitance as silicon dioxide only $3 \times 3.9/25 = 0.47$ nm thick — its equivalent oxide thickness — while being thick enough to cut tunneling leakage by orders of magnitude. Every modern processor, including those fabricated at new plants in Arizona and Ohio, uses such high-κ dielectrics.
Film capacitors made of polypropylene store energy in inverters for electric vehicles, solar farms and wind turbines, smoothing the voltage of the DC link between power stages. Polypropylene's dielectric constant is only $2.2$, but it withstands fields of several hundred megavolts per meter and heals itself after small breakdowns, so capacitors can run at high field and store energy densities of a few joules per cubic centimeter.
Researchers at national laboratories and universities are developing polymer nanocomposites and ceramic films that combine higher $\kappa$ with high breakdown strength, since stored energy density scales as $\tfrac{1}{2}\kappa\varepsilon_0E_{\max}^2$. Doubling $\kappa$ doubles the energy, but doubling the breakdown field quadruples it, which is why dielectric strength is the most prized property in these materials.
Inserting a dielectric raises capacitance, which can sound as if the dielectric supplies charge. It does not: its bound charges come in equal and opposite layers, and the material stays neutral. What changes is the field. The bound layers partly cancel the plates' field, so for the same free charge the voltage is lower, which is exactly what a larger capacitance means.
A second error is to treat $\vec{D}$ as a field determined by free charge alone, as $\vec{E}$ is by all charge. $\nabla \cdot \vec{D} = \rho_f$, but the curl of $\vec{D}$ need not vanish, so without symmetry $\vec{D}$ cannot be found from free charge by Coulomb's law. Use it where symmetry lets Gauss's law do the work.
Plates carry $\pm\sigma_f$ with a dielectric of constant $\kappa$ filling the gap. Find $D$.
$D = \sigma_f$
Gauss's law for $\vec{D}$ with a pillbox through one plate.
Find the field.
$E = \dfrac{D}{\kappa\varepsilon_0} = \dfrac{\sigma_f}{\kappa\varepsilon_0}$
Weaker by $\kappa$.
Find the voltage.
$V = Ed = \dfrac{\sigma_fd}{\kappa\varepsilon_0}$
Uniform field across the gap.
Find the capacitance.
$C = \dfrac{\sigma_fA}{V} = \dfrac{\kappa\varepsilon_0A}{d}$
Larger by $\kappa$.
Find the bound surface charge.
$\sigma_b = P = D - \varepsilon_0E = \sigma_f\left(1 - \dfrac{1}{\kappa}\right)$
Opposite to the adjacent plate.
A sphere of radius $R$ has uniform polarization $\vec{P} = P\hat{z}$. Find the bound volume charge.
$\rho_b = -\nabla \cdot \vec{P} = 0$
Uniform polarization has no divergence.
Find the bound surface charge.
$\sigma_b = \vec{P} \cdot \hat{r} = P\cos\theta$
Positive on top, negative on the bottom.
Recognize the field inside.
$\vec{E}_{\text{in}} = -\dfrac{\vec{P}}{3\varepsilon_0}$
A $\cos\theta$ surface charge produces a uniform field inside.
Recognize the field outside.
$\text{a pure dipole with } \vec{p} = \tfrac{4}{3}\pi R^3\vec{P}$
The total dipole moment of the sphere.
Check the surface boundary condition.
$E_{\text{out},r} - E_{\text{in},r} = \dfrac{\sigma_b}{\varepsilon_0}$
At $\theta = 0$: $\tfrac{2P}{3\varepsilon_0} + \tfrac{P}{3\varepsilon_0} = \tfrac{P}{\varepsilon_0}$.
Interpret the inside field.
$\vec{E}_{\text{in}} \text{ opposes } \vec{P}$
The depolarizing field, which limits how strongly a sphere polarizes.
Plates of area $100$ cm² are $2.0$ mm apart; a slab with $\kappa = 4$ and thickness $1.0$ mm lies against one plate. Find $D$.
$D = \sigma_f$
The same in air and slab.
Find the field in the air.
$E_{\text{air}} = \dfrac{\sigma_f}{\varepsilon_0}$
No polarization in air, nearly.
Find the field in the slab.
$E_{\text{slab}} = \dfrac{\sigma_f}{4\varepsilon_0}$
A quarter as strong.
Add the voltage drops.
$V = \dfrac{\sigma_f}{\varepsilon_0}\left(1.0 + \dfrac{1.0}{4}\right)\ \text{mm} = \dfrac{\sigma_f}{\varepsilon_0} \times 1.25\ \text{mm}$
An effective gap of $1.25$ mm.
Find the capacitance.
$C = \dfrac{8.85 \times 10^{-12} \times 0.010}{1.25 \times 10^{-3}} = 70.8\ \text{pF}$
Up from $44.3$ pF empty.
Compare with a full slab.
$C_{\text{full}} = 4 \times 44.3 = 177\ \text{pF}$
The remaining air gap limits the gain.
Find the bound charge on the slab.
$\sigma_b = \sigma_f\left(1 - \dfrac{1}{4}\right) = 0.75\sigma_f$
On both faces, with opposite signs.
Find where breakdown would start.
$E_{\text{air}} = 4E_{\text{slab}}$
The air gap carries the strongest field and breaks down first.
Write the bound charge formula.
$\sigma_b = \sigma_f\left(1 - \dfrac{1}{\kappa}\right)$
From $P = D - \varepsilon_0E$.
Substitute the values.
$\sigma_b = 20 \times \left(1 - \dfrac{1}{5}\right)$
In μC/m².
Evaluate the bound charge.
An isolated capacitor charged to $45$ V is disconnected, and a dielectric with $\kappa = 4$ is slid in to fill the gap. What is the new voltage?
Complete the worked solution: a dielectric with $\kappa = 5$ fills a capacitor whose plates carry $41$ μC/m². Find $\varepsilon_0E$ and $P$ in μC/m², and the susceptibility $\chi$.
Divide the displacement by $\kappa$.
$\varepsilon_0E = \dfrac{41}{5} =$ e
$D = \kappa\varepsilon_0E$.
Subtract to find the polarization.
$P = D - \varepsilon_0E =$ p
The rest of $D$ comes from aligned dipoles.
Find the susceptibility.
$\chi = \kappa - 1 =$ x
Check: $P = \chi\varepsilon_0E$.
Match each quantity to its expression.
| $\vec{P} \cdot \hat{n}$ | $-\nabla \cdot \vec{P}$ | $\varepsilon_0\vec{E} + \vec{P}$ | $\nabla \cdot \vec{D} = \rho_f$ | |
|---|---|---|---|---|
| bound surface charge | ||||
| bound volume charge | ||||
| the displacement | ||||
| Gauss's law in matter |
An isolated $66$ pF capacitor holds charge at $59$ V. A dielectric with $\kappa = 8$ fills the gap. Fill in the new capacitance, voltage and energy.
| with dielectric | |
|---|---|
| capacitance (pF) | |
| voltage (V) | |
| energy (pJ) |
The plates of a capacitor carry free charge $\pm32$ μC/m², and a dielectric with $\kappa = 5$ fills the gap. What is the magnitude of the bound charge on the dielectric's surface, in μC/m²?
Answer: μC/m²
A parallel-plate capacitor has plates of area $100$ cm² separated by $2$ mm. A slab of dielectric constant $\kappa = 4$ and thickness $1$ mm is placed against one plate. What is the capacitance, in pF? Use $\varepsilon_0 = 8.85 \times 10^{-12}$ F/m.
Answer: pF
Modern transistors replace the silicon dioxide gate insulator ($\kappa = 3.9$) with hafnium oxide ($\kappa = 25$). A hafnium oxide layer $4.4$ nm thick gives the same capacitance per area as what thickness of silicon dioxide, in nm?
Answer: nm
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
An isolated $41$ pF capacitor holds charge at $77$ V. A dielectric with $\kappa = 5$ fills the gap. Fill in the new capacitance, voltage and energy.
| with dielectric | |
|---|---|
| capacitance (pF) | |
| voltage (V) | |
| energy (pJ) |
You can analyze dielectrics in fields. Explain to someone how a dielectric raises a capacitor's capacitance without adding any charge.
19. Your turn: a capacitor with $\sigma_f = 20$ μC/m² is filled with a dielectric of $\kappa = 5$. What is the bound surface charge?, step 3
$\sigma_b = 16\ \mu\text{C/m}^2$
Cancelling eighty percent of the plate's field.