Part 2 · Chapter 7

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.

Electromagnetic Field Theory Prof. Mithun Mondal Reading time ≈ 55 min
i What you'll learn
  • 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.
Section 7-1

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.

Current from current density
\[ I = \int_S \vec{J}\cdot d\vec{S}, \qquad \vec{J} = \rho_v \vec{u} \]

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.

Section 7-2

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 \(V=IR\):

Point form of Ohm's law
\[ \vec{J} = \sigma\vec{E} \]

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:

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A conductor in equilibrium
\(\vec{E}=0\) inside; all charge on the surface; the surface is an equipotential

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.

E = 0 inside ++++ +++++
Excess charge resides on the surface; the interior field is zero
Section 7-3

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:

Continuity equation
\[ \nabla\cdot\vec{J} = -\frac{\partial \rho_v}{\partial t} \]

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.

Section 7-4

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.

no field field applied → E applied
Random molecular dipoles (left) align with an applied field (right), producing P

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:

Flux density in a dielectric
\[ \vec{D} = \varepsilon_0\vec{E} + \vec{P} \]
Section 7-5

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:

Permittivity, susceptibility, dielectric constant
\[ \vec{D} = \varepsilon\vec{E}, \qquad \varepsilon = \varepsilon_0\varepsilon_r, \qquad \varepsilon_r = 1 + \chi_e \]

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 / air1.03 (air)
Paper2–415
Mica6200
Glass4–1030
Distilled water81
Section 7-6

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.

Why a dielectric helps twice. Slipping a dielectric between capacitor plates does two good things at once: its \(\varepsilon_r\) raises the stored charge for a given voltage, and its dielectric strength lets the capacitor survive a higher voltage than air would. Both effects reappear quantitatively in Chapter 8.
Section 7-7

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:

Dielectric–dielectric boundary conditions
\[ E_{1t} = E_{2t}, \qquad D_{1n} - D_{2n} = \rho_S \]

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:

Refraction of field lines
\[ \frac{\tan\theta_1}{\tan\theta_2} = \frac{\varepsilon_1}{\varepsilon_2} \]
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At a conductor–dielectric boundary
\[ E_t = 0, \qquad D_n = \rho_S \]

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.

Section 7-8

Worked Examples

1 Current density to current

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, \(I = JA\):

Working
\[ I = (5\times10^{6})(2\times10^{-6}) = 10\ \text{A} \]
2 Point form of Ohm's law

Problem. Copper (\(\sigma=5.8\times10^{7}\)) carries \(\vec{E}=1\ \text{mV/m}\). Find \(\vec{J}\).

Solution. Apply \(\vec{J}=\sigma\vec{E}\):

Working
\[ J = (5.8\times10^{7})(10^{-3}) = 5.8\times10^{4}\ \text{A/m}^2 \]
3 Relaxation time

Problem. Find the relaxation time of mica (\(\sigma=10^{-15}\ \text{S/m}\), \(\varepsilon_r=6\)).

Solution. Use \(T_r=\varepsilon/\sigma\):

Working
\[ T_r = \frac{6(8.854\times10^{-12})}{10^{-15}} = 5.3\times10^{4}\ \text{s} \approx 15\ \text{hours} \]

A good insulator holds its charge for hours — the opposite extreme from copper.

4 D in a dielectric

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 \(\vec{P}=\varepsilon_0(\varepsilon_r-1)\vec{E}\):

Working
\[ D = 4\varepsilon_0(2\times10^{5}) = 7.08\ \mu\text{C/m}^2, \qquad P = 3\varepsilon_0(2\times10^{5}) = 5.31\ \mu\text{C/m}^2 \]
5 Crossing a boundary

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 \(\tan\theta_1/\tan\theta_2 = \varepsilon_1/\varepsilon_2\):

Working
\[ \tan\theta_2 = \frac{5}{2}\tan 30^\circ = 1.443 \;\Rightarrow\; \theta_2 = 55.3^\circ \]
6 Field at a conductor surface

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:

Working
\[ E = \frac{20\times10^{-9}}{8.854\times10^{-12}} = 2.26\ \text{kV/m (normal to surface)} \]
Review

Chapter Summary

Current density

\(I=\int_S\vec{J}\cdot d\vec{S}\); convection \(\vec{J}=\rho_v\vec{u}\); conduction \(\vec{J}=\sigma\vec{E}\).

Conductors

In equilibrium: \(\vec{E}=0\) inside, charge on surface, body is one equipotential.

Continuity

\(\nabla\cdot\vec{J}=-\partial\rho_v/\partial t\); relaxation time \(T_r=\varepsilon/\sigma\).

Polarization

\(\vec{D}=\varepsilon_0\vec{E}+\vec{P}\); aligned dipoles weaken the internal field.

Permittivity

\(\vec{D}=\varepsilon\vec{E}\), \(\varepsilon=\varepsilon_0\varepsilon_r\), \(\varepsilon_r=1+\chi_e\); breakdown limits the field.

Boundaries

\(E_{1t}=E_{2t}\), \(D_{1n}-D_{2n}=\rho_S\); at a conductor \(E_t=0\), \(D_n=\rho_S\).

Practice

Problems

For each item, identify the material involved and whether the question is about conduction, polarization, or a boundary. Difficulty rises down the list.

  1. A current of \(15\ \text{A}\) flows uniformly through a wire of radius \(1\ \text{mm}\). Find \(\vec{J}\).
  2. Aluminium has \(\sigma=3.5\times10^{7}\ \text{S/m}\). Find \(\vec{E}\) needed to drive \(\vec{J}=10^{6}\ \text{A/m}^2\).
  3. Compute the relaxation time of distilled water (\(\sigma=10^{-4}\), \(\varepsilon_r=81\)).
  4. A dielectric (\(\varepsilon_r=3\)) has \(\vec{D}=6\ \mu\text{C/m}^2\). Find \(\vec{E}\) and \(\vec{P}\).
  5. Find the susceptibility \(\chi_e\) of a material with \(\varepsilon_r=5.5\).
  6. A field in glass (\(\varepsilon_r=6\)) meets air at \(40^\circ\) to the normal. Find the angle in air.
  7. A conductor surface in a dielectric (\(\varepsilon_r=2.5\)) carries \(\rho_S=30\ \text{nC/m}^2\). Find \(\vec{E}\) just outside.
  8. 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.
  9. Show that the relaxation time explains why a charged conductor neutralises almost instantly while an insulator does not.
  10. A parallel-plate gap is half air, half mica. Argue which layer is more likely to break down first under rising voltage.
  11. 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.
  12. Explain physically why the tangential field is continuous but the normal flux density can jump.
Tip: at any interface, sort the field into tangential and normal parts before doing anything else. Tangential \(\vec{E}\) always passes through unchanged; normal \(\vec{D}\) only jumps if free surface charge sits on the boundary. Almost every "field crossing a boundary" problem is just these two rules applied to the two components separately — exactly the decomposition that powers the capacitor and Laplace problems ahead.