Insulators

Created by Samiksha Sriram Fall 2026

The Main Idea

An insulator is a material that slows down or stops the flow of electricity, heat, or sound. They are typically characterized by their poor conductivity of electromagnetic energy, which sets them apart from both conductors and semiconductors. Insulators contain atoms whose electrons are tightly bound, unlike conductors, which have free electrons. Because the atoms are bound to each other within the mass and cannot be reconfigured through electrochemical manipulation, insulators are characterized as having high resistivity.

Insulating materials, also known as dielectric materials, can exist in all states of matter. Although a perfect ideal insulator would conduct no electricity, all materials conduct electricity to some degree. These insulators can vary in physical characteristics since some materials may be hard/rigid while others can be soft/flexible.

Inductors can also be inside an electric field. Insulators lack free electrons; microscopic dipoles are created via polarization when an external electric field is applied. This internal polarization creates a countering field that only reduces rather than cancels it, and the net electric field inside the inductor is weaker than the external field. Because of this, electric fields exist inside the insulator, whereas inside a conductor, the internal electric field is always zero when at electrostatic equilibrium. Diving deeper into polarization, when it occurs, it causes positive and negative charges to shift or align in opposite directions. This separation creates tiny electric dipoles or aligns molecules that already have permanent dipoles. Overall, the material stays neutral; however, there are distinct positive and negative regions inside the material.

(Model by Samiksha Sriram)

Polarization of an insulated material occurs when molecules with pre-existing dipole moments rotate, or the dipole moments in the individual molecules are induced. This can be seen through the equation [math]\displaystyle{ \mathbf{D} = \varepsilon_0 \mathbf{E} + \mathbf{P} }[/math] where D is the electric displacement field, or electric flux density. This value is in the units of [math]\displaystyle{ \text{C/m}^2 }[/math] and accounts for both free charges and the effect of polarization inside the material. [math]\displaystyle{ \varepsilon_0 }[/math]​ is the permittivity of free space and is a physical constant that represents the value [math]\displaystyle{ 8.85 \times 10^{-12} \text{ C}^2 / (\text{N} \cdot \text{m}^2) }[/math]. This value describes how electric fields behave in a vacuum. E is the electric field and is in the units N/C. Finally, P is the polarization of the medium and represents the value of the dipole moment per unit volume and is in the units [math]\displaystyle{ \text{C/m}^2 }[/math]. Because of this equation, we know that [math]\displaystyle{ \mathbf{D} \gt \varepsilon_0 \mathbf{E} }[/math] inside the insulating material.

Not all materials polarize equally. Some materials are far more susceptible to polarization than others, and the ratio of the polarization of the medium to [math]\displaystyle{ \varepsilon_0 \mathbf{E} }[/math] is called the electric susceptibility, [math]\displaystyle{ \chi_e }[/math]​, of the medium: [math]\displaystyle{ \mathbf{P} = \chi_e \varepsilon_0 \mathbf{E} }[/math]. Materials that have a larger [math]\displaystyle{ \chi_e }[/math] value are more polarized because their charges are more easily displaced by an external electric field compared to a material with a smaller [math]\displaystyle{ \chi_e }[/math].

Factors Affecting Insulating Behavior

Material Composition is one example of what can affect an insulator. One commonly used dielectric material used in electronics and power systems is ceramics.

• One example of a ceramic used in electronic circuits is barium titanate, which is a ferroelectric perovskite that exhibits a high dielectric constant and underpins multilayer ceramic capacitors (MLCCs). Other examples are aluminum nitride and silicon carbide, which have high dielectric strength and good thermal conductivity, meaning that they are suitable for power device substrates and high-voltage insulators.
• Glass dielectric materials, such as borosilicate and fused silica, allow for very little dielectric loss and high dimensional stability. They also act as the dielectric layer in glass-based capacitors and as substrates in microwave packaging.
• The largest volume class of dielectric materials by weight is represented by organic polymers. Polyethylene and cross-linked polyethylene insulate power cables at high voltages and are important because of their low loss tangent and good dielectric strength. Polypropylene, polystyrene, and polytetrafluoroethylene are used in capacitors, coaxial cables, and microwave substrates. Epoxy resins can insulate circuit boards, transformer windings, and switchgear. However, polymers are still limited because of their susceptibility to partial discharge, thermal aging, and moisture absorption.

Materials that are classed as insulators are liable to decrease their resistance when the temperature of the material increases. Thus, a visible drop in resistance can only be seen when these insulators are at incredibly high temperatures. Since there are so few free electrons in the material, almost all of the electrons in an insulator are tightly bound within the atom. However, when heat is brought into the picture, the material vibrates the atoms, and if it is heated enough, the atoms violently vibrate to the point where captive electrons shake free, allowing these electrons to carry current. Thus, when the temperature is high, the insulator's resistance can fall. This means that most insulating materials chosen for real-world purposes will have a very low, negative temperature coefficient over their working range of temperature.

Moisture can also pose serious risks in insulation. Heat is conducted 20 times better by water than by air. Increased thermal conductivity can occur when moisture seeps into insulation materials, which means that the insulating material has a reduced ability to maintain desired temperatures. Additionally, when the humidity of the air is high, the material absorbs more water, increasing the conductance and reducing the insulation resistance value.

If an electric field exceeds a material's dielectric breakdown strength ([math]\displaystyle{ \mathbf{E} \gt \mathbf{E}_{\text{breakdown}} }[/math]), the insulator loses its resistance and turns into a temporary or permanent conductor. Every insulating material has its own defined maximum electric field it can withstand before failure. The dielectric strength depends on material composition, degree of polarization under an electric field, presence of impurities or defects, and environmental/thermal conditions. By understanding dielectric breakdown analysis, catastrophic insulation failures can be prevented, maintenance of the material can be prioritized, life-extension strategies can be supported, and maintenance that is condition-based rather than time-based maintenance can be enabled. Because breakdowns mean that the physical/chemical structure of the insulation is altered, the breakdown can be permanent.

A Microscopic Model

A microscopic model of an electrical insulator shows that the electrons are tightly bound to the atoms and cannot move freely through the material: (Model by Samiksha Sriram)

In a typical insulator, the valence bands are filled with electrons. A wide space called a band gap sits right above these filled levels. Normal voltage cannot supply the large amount of extra energy needed to jump across this gap into the conduction band; thus, electrical current cannot flow. In the band insulator model, the electrons are treated as independent particles moving in a periodic crystal lattice, whereas in the Mott-Hubbard model, it explains that materials should conduct electricity based on their band structure, and insulators do not because of their strong electrical repulsion between electrons on the same atom.

A Mathematical Model

Polarization density in an insulator is the dipole moment per unit volume created when an external electric field distorts atoms or aligns polar molecules.

The electric dipole moment consists of a positive (+q) charge and a negative (-q) charge, which are separated by a displacement vector. The equation that represents this concept is [math]\displaystyle{ \mathbf{p} = q \mathbf{d} }[/math] and it uses the units Coulomb-meters (C*m). In this equation, the vector distance is directed from the negative to the positive charge. Furthermore, because the vector distance is generally smaller than an atom's size, it is small compared to any macroscopic dimension of interest.

Polarization inside a material creates microscopic charge separations, macroscopically described as bound charges: the bound volume charge density is equivalent to -change times P. The bound surface charge density is equivalent to P times the unit vector of n.

In linear, isotropic, and homogeneous dielectric media, polarization is proportional to the electric field: [math]\displaystyle{ \mathbf{P} = \varepsilon_0 \chi_e \mathbf{E} }[/math]', where [math]\displaystyle{ \chi_e }[/math] is the electric susceptibility of the material, and [math]\displaystyle{ \varepsilon_0 }[/math] is the vacuum permittivity. It links the electric field E and the electric displacement field D via: [math]\displaystyle{ \mathbf{D} = \varepsilon_0 \mathbf{E} + \mathbf{P} = \varepsilon_0 (1 + \chi_e) \mathbf{E} = \varepsilon \mathbf{E} }[/math].

Examples

Simple

A solid insulating sphere of radius [math]\displaystyle{ R = 0.05 \text{ m} }[/math] has a total positive charge [math]\displaystyle{ Q = 3.0 \times 10^{-6} \text{ C} }[/math] distributed uniformly throughout its volume. Find the magnitude of the electric field at a distance [math]\displaystyle{ r = 0.02 \text{ m} }[/math] from the center of the sphere.

Solution

The magnitude of the electric field inside the insulating sphere at [math]\displaystyle{ r = 0.02 \text{ m} }[/math] is [math]\displaystyle{ 4.32 \times 10^6 \text{ N/C} }[/math].

Intermediate

A solid conducting sphere of radius [math]\displaystyle{ a }[/math] carries a net positive charge [math]\displaystyle{ 2Q }[/math]. It is surrounded by a concentric, thick insulating spherical shell of inner radius [math]\displaystyle{ b }[/math] and outer radius [math]\displaystyle{ c }[/math] (where [math]\displaystyle{ a \lt b \lt c }[/math]). The insulating shell has a uniform volume charge density [math]\displaystyle{ \rho = - \rho_0 }[/math] (a negative constant). Find the magnitude of the electric field [math]\displaystyle{ \mathbf{E} }[/math] in the region inside the insulator ([math]\displaystyle{ b \lt r \lt c }[/math]).

Solution

Identify the Concept: Due to spherical symmetry, we apply Gauss's Law using a spherical Gaussian surface of radius [math]\displaystyle{ r }[/math] located within the insulating shell ([math]\displaystyle{ b \lt r \lt c }[/math]):

[math]\displaystyle{ \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_0} }[/math]

The left side evaluates to [math]\displaystyle{ E(4\pi r^2) }[/math] because the electric field is purely radial.

Determine the Enclosed Charge ([math]\displaystyle{ Q_{\text{enc}} }[/math]): The total charge enclosed by our Gaussian sphere consists of two parts: the entire charge on the inner conductor, and the portion of the negative insulating shell charge located between radius [math]\displaystyle{ b }[/math] and radius [math]\displaystyle{ r }[/math].

[math]\displaystyle{ Q_{\text{enc}} = Q_{\text{conductor}} + Q_{\text{insulator, enc}} }[/math]

[math]\displaystyle{ Q_{\text{enc}} = 2Q + \int_b^r \rho \, dV = 2Q + (-\rho_0) \cdot \left( \frac{4}{3}\pi r^3 - \frac{4}{3}\pi b^3 \right) }[/math]

[math]\displaystyle{ Q_{\text{enc}} = 2Q - \frac{4}{3}\pi \rho_0 (r^3 - b^3) }[/math]

Apply Gauss's Law to Solve for [math]\displaystyle{ E }[/math]: Substituting [math]\displaystyle{ Q_{\text{enc}} }[/math] into Gauss's Law:

[math]\displaystyle{ E(4\pi r^2) = \frac{1}{\varepsilon_0} \left[ 2Q - \frac{4}{3}\pi \rho_0 (r^3 - b^3) \right] }[/math]

Dividing both sides by [math]\displaystyle{ 4\pi r^2 }[/math] yields the electric field magnitude:

[math]\displaystyle{ E = \frac{2Q}{4\pi \varepsilon_0 r^2} - \frac{\rho_0}{3 \varepsilon_0 r^2} (r^3 - b^3) }[/math]

[math]\displaystyle{ E = \frac{1}{4\pi \varepsilon_0} \frac{2Q}{r^2} - \frac{\rho_0}{3 \varepsilon_0} \left( r - \frac{b^3}{r^2} \right) }[/math]

Difficult

Identify the Concept: Because the cylinder is infinitely long and the charge depends only on [math]\displaystyle{ r }[/math], the system exhibits cylindrical symmetry. We use Gauss's Law with a coaxial cylindrical Gaussian surface of radius [math]\displaystyle{ r }[/math] ([math]\displaystyle{ r \lt R }[/math]) and length [math]\displaystyle{ L }[/math]:

[math]\displaystyle{ \oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_0} }[/math]

The electric flux through the flat end caps is zero because the field points radially outward. The flux through the curved side wall evaluates to [math]\displaystyle{ E(2\pi r L) }[/math].

Calculate Enclosed Charge by Integration: Since the charge density [math]\displaystyle{ \rho(r) }[/math] is non-uniform, we must find [math]\displaystyle{ Q_{\text{enc}} }[/math] by integrating over a volume element [math]\displaystyle{ dV = 2\pi r' L \, dr' }[/math] from the axis [math]\displaystyle{ r'=0 }[/math] to the Gaussian radius [math]\displaystyle{ r'=r }[/math]:

[math]\displaystyle{ Q_{\text{enc}} = \int_0^r \rho(r') \, dV = \int_0^r \rho_0 \left(1 - \frac{r'}{R}\right) (2\pi r' L) \, dr' }[/math]

[math]\displaystyle{ Q_{\text{enc}} = 2\pi L \rho_0 \int_0^r \left(r' - \frac{(r')^2}{R}\right) dr' }[/math]

Evaluating the definite integral yields:

[math]\displaystyle{ Q_{\text{enc}} = 2\pi L \rho_0 \left[ \frac{(r')^2}{2} - \frac{(r')^3}{3R} \right]_0^r = 2\pi L \rho_0 \left( \frac{r^2}{2} - \frac{r^3}{3R} \right) }[/math]

Solve for the Electric Field [math]\displaystyle{ E }[/math]: Equating the electric flux to the enclosed charge expression divided by [math]\displaystyle{ \varepsilon_0 }[/math]:

[math]\displaystyle{ E(2\pi r L) = \frac{2\pi L \rho_0}{\varepsilon_0} \left( \frac{r^2}{2} - \frac{r^3}{3R} \right) }[/math]

Dividing both sides by [math]\displaystyle{ 2\pi r L }[/math] simplifies to:

[math]\displaystyle{ E = \frac{\rho_0}{\varepsilon_0} \left( \frac{r}{2} - \frac{r^2}{3R} \right) }[/math]