Electric Fields in Material Space
Until now the fields lived in empty space. Real engineering happens inside matter — copper that conducts, glass that insulates, ceramics that store charge. This chapter asks what a material does to a field, and what a field does to the material.
- How materials are classed by conductivity \(\sigma\), and the two kinds of current density.
- Ohm's law in point form \(\vec{J}=\sigma\vec{E}\), and conductors in electrostatic equilibrium.
- The continuity equation and the relaxation time of a conductor.
- Polarization \(\vec{P}\), bound charge, and how a dielectric weakens an internal field.
- Permittivity, susceptibility and the dielectric constant \(\varepsilon_r\).
- The electrostatic boundary conditions on \(\vec{E}\) and \(\vec{D}\) across an interface.
Materials & Current Density
Materials are graded by how freely charge moves through them — their conductivity \(\sigma\) (S/m). Conductors (copper, \(\sigma\approx 5.8\times10^7\)) have vast numbers of free electrons; dielectrics or insulators (glass, mica) have almost none; semiconductors sit between. Moving charge constitutes a current density \(\vec{J}\) (A/m²), the current per unit cross-sectional area.
Two mechanisms produce \(\vec{J}\). Convection current is charge \(\rho_v\) physically carried along at velocity \(\vec{u}\) — like a charged beam in vacuum; it needs no conductor and does not obey Ohm's law. Conduction current is the drift of free charges inside a conductor under a field, and it does.
Conductors & Ohm's Law
Inside a conductor an applied field drives free electrons until they reach a steady drift, giving a conduction current proportional to the field. This is Ohm's law in point form — the local statement behind the familiar \(V=IR\):
In electrostatic equilibrium — no steady current — a conductor must have \(\vec{E}=0\) inside. If any field remained, it would keep pushing charges; they stop only when the field vanishes. The consequences are sharp and worth memorising:
Excess charge flees to the outer surface, the interior is field-free (the basis of electrostatic shielding), and because no work moves a charge along a field-free conductor, the whole body sits at one potential. The external field meets the surface at right angles.
The Continuity Equation
Charge is conserved: it can move, but never appear or vanish. The current flowing out of a closed surface must equal the rate at which the enclosed charge drops. In point form, via the divergence theorem, this is the continuity equation:
Combining continuity with Ohm's law and Gauss's law shows that any charge placed inside a conductor decays exponentially to the surface with a characteristic relaxation time \(T_r = \varepsilon/\sigma\). For copper this is about \(10^{-19}\) seconds — effectively instantaneous, which is why we treat conductors as reaching equilibrium at once.
Polarization in Dielectrics
A dielectric has no free charge, but its bound charges shift. An applied field nudges each molecule's positive and negative centres apart, turning every molecule into a tiny dipole. Summed per unit volume, these dipoles give the polarization \(\vec{P}\) (C/m²). The aligned dipoles create their own field opposing the applied one, so the net field inside a dielectric is weaker.
Polarization adds to \(\varepsilon_0\vec{E}\) so that the flux density picks up the material's response. This generalises \(\vec{D}=\varepsilon_0\vec{E}\) to matter:
Permittivity & Susceptibility
For most materials the polarization is proportional to the field: \(\vec{P}=\varepsilon_0\chi_e\vec{E}\), where \(\chi_e\) is the electric susceptibility. Substituting collapses everything back into the simple form \(\vec{D}=\varepsilon\vec{E}\), now with the material's permittivity:
The dimensionless dielectric constant \(\varepsilon_r\) (relative permittivity) tells you how many times stronger the material's response is than vacuum. It is always greater than one — vacuum is \(\varepsilon_r=1\):
| Material | \(\varepsilon_r\) (approx.) | Dielectric strength (MV/m) |
|---|---|---|
| Vacuum / air | 1.0 | 3 (air) |
| Paper | 2–4 | 15 |
| Mica | 6 | 200 |
| Glass | 4–10 | 30 |
| Distilled water | 81 | — |
Dielectric Strength
A dielectric insulates only up to a point. Raise the field high enough and electrons are torn free, the material suddenly conducts, and it is often destroyed — this is dielectric breakdown. The maximum field a material withstands is its dielectric strength. This single number sets the voltage rating of every capacitor and the spacing of every high-voltage insulator, linking directly to the breakdown work studied across high-voltage engineering.
Boundary Conditions
When a field crosses from one material into another, it bends. Two rules — derived by applying \(\oint\vec{E}\cdot d\vec{l}=0\) to a thin loop and \(\oint\vec{D}\cdot d\vec{S}=Q\) to a thin pillbox straddling the interface — govern the change. The tangential \(\vec{E}\) is continuous; the normal \(\vec{D}\) jumps only by any free surface charge present:
With no free charge on the interface (\(\rho_S=0\)), the normal component of \(\vec{D}\) is continuous too, and the field lines refract according to the ratio of permittivities — a "law of refraction" for field lines:
Since \(\vec{E}=0\) inside a conductor, the field just outside is purely normal to the surface and equals the surface charge density divided by \(\varepsilon\). No field line ever runs along a conductor's surface — it always leaves perpendicular.
Worked Examples
Problem. A wire of cross-section \(2\ \text{mm}^2\) carries uniform \(\vec{J}=5\times10^{6}\,\hat{a}_z\ \text{A/m}^2\). Find \(I\).
Solution. For uniform \(\vec{J}\) normal to the area, \(I = JA\):
Problem. Copper (\(\sigma=5.8\times10^{7}\)) carries \(\vec{E}=1\ \text{mV/m}\). Find \(\vec{J}\).
Solution. Apply \(\vec{J}=\sigma\vec{E}\):
Problem. Find the relaxation time of mica (\(\sigma=10^{-15}\ \text{S/m}\), \(\varepsilon_r=6\)).
Solution. Use \(T_r=\varepsilon/\sigma\):
A good insulator holds its charge for hours — the opposite extreme from copper.
Problem. A dielectric with \(\varepsilon_r=4\) has \(\vec{E}=2\times10^{5}\,\hat{a}_x\ \text{V/m}\). Find \(\vec{D}\) and \(\vec{P}\).
Solution. Use \(\vec{D}=\varepsilon_0\varepsilon_r\vec{E}\) and \(\vec{P}=\varepsilon_0(\varepsilon_r-1)\vec{E}\):
Problem. Field in region 1 (\(\varepsilon_{r1}=2\)) makes \(30^\circ\) with the normal. Find the angle in region 2 (\(\varepsilon_{r2}=5\)), no free charge.
Solution. Apply the refraction law \(\tan\theta_1/\tan\theta_2 = \varepsilon_1/\varepsilon_2\):
Problem. A conductor in air carries surface charge \(\rho_S = 20\ \text{nC/m}^2\). Find \(\vec{E}\) just outside.
Solution. At a conductor surface \(E_n = \rho_S/\varepsilon_0\), purely normal:
Chapter Summary
\(I=\int_S\vec{J}\cdot d\vec{S}\); convection \(\vec{J}=\rho_v\vec{u}\); conduction \(\vec{J}=\sigma\vec{E}\).
In equilibrium: \(\vec{E}=0\) inside, charge on surface, body is one equipotential.
\(\nabla\cdot\vec{J}=-\partial\rho_v/\partial t\); relaxation time \(T_r=\varepsilon/\sigma\).
\(\vec{D}=\varepsilon_0\vec{E}+\vec{P}\); aligned dipoles weaken the internal field.
\(\vec{D}=\varepsilon\vec{E}\), \(\varepsilon=\varepsilon_0\varepsilon_r\), \(\varepsilon_r=1+\chi_e\); breakdown limits the field.
\(E_{1t}=E_{2t}\), \(D_{1n}-D_{2n}=\rho_S\); at a conductor \(E_t=0\), \(D_n=\rho_S\).
Problems
For each item, identify the material involved and whether the question is about conduction, polarization, or a boundary. Difficulty rises down the list.
- A current of \(15\ \text{A}\) flows uniformly through a wire of radius \(1\ \text{mm}\). Find \(\vec{J}\).
- Aluminium has \(\sigma=3.5\times10^{7}\ \text{S/m}\). Find \(\vec{E}\) needed to drive \(\vec{J}=10^{6}\ \text{A/m}^2\).
- Compute the relaxation time of distilled water (\(\sigma=10^{-4}\), \(\varepsilon_r=81\)).
- A dielectric (\(\varepsilon_r=3\)) has \(\vec{D}=6\ \mu\text{C/m}^2\). Find \(\vec{E}\) and \(\vec{P}\).
- Find the susceptibility \(\chi_e\) of a material with \(\varepsilon_r=5.5\).
- A field in glass (\(\varepsilon_r=6\)) meets air at \(40^\circ\) to the normal. Find the angle in air.
- A conductor surface in a dielectric (\(\varepsilon_r=2.5\)) carries \(\rho_S=30\ \text{nC/m}^2\). Find \(\vec{E}\) just outside.
- Region 1 (\(z>0\), \(\varepsilon_{r1}=4\)) has \(\vec{E}_1=3\hat{a}_x+5\hat{a}_z\). Find \(\vec{E}_2\) in region 2 (\(z<0\), \(\varepsilon_{r2}=2\)), no free charge.
- Show that the relaxation time explains why a charged conductor neutralises almost instantly while an insulator does not.
- A parallel-plate gap is half air, half mica. Argue which layer is more likely to break down first under rising voltage.
- Derive the boundary condition \(E_{1t}=E_{2t}\) by applying \(\oint\vec{E}\cdot d\vec{l}=0\) to a vanishingly thin loop across the interface.
- Explain physically why the tangential field is continuous but the normal flux density can jump.