← Part I: Electrostatics
Chapter 4

Conductors & Dielectrics

Electrostatic induction, capacitance, polarization, and the displacement field.

4.1 Conductors in Electrostatic Equilibrium

In a perfect conductor, charges redistribute until the electric field is zero everywhere inside. The key electrostatic properties of a conductor in equilibrium are:

$\mathbf{E} = 0$
inside the conductor
$\rho = 0$
no free charge inside; charge resides on surfaces
$V = \text{const}$
the conductor is an equipotential
$\mathbf{E} \perp \text{surface}$
field is normal to the surface just outside
$E_{\rm outside} = \sigma/\epsilon_0$
surface field proportional to surface charge density

4.1.1 Boundary Conditions

At the interface between two media, the boundary conditions follow from integrating Maxwell's equations across the interface. With surface charge $\sigma$ and surface current $\mathbf{K}$:

Normal component (from Gauss)

$$\epsilon_2 E_{2}^\perp - \epsilon_1 E_1^\perp = \sigma_f$$

Tangential component (from Faraday)

$$E_1^\parallel = E_2^\parallel$$

4.2 Capacitance

The capacitance $C$ of a conductor (or pair of conductors) is defined as the ratio of stored charge to potential:

$$C = \frac{Q}{V} \qquad \text{[Farads = C/V]}$$

Parallel plates

$$C = \frac{\epsilon_0 A}{d}$$

Spherical capacitor

$$C = 4\pi\epsilon_0\frac{ab}{b-a}$$

Cylindrical capacitor

$$C = \frac{2\pi\epsilon_0 L}{\ln(b/a)}$$

The energy stored in a capacitor is:

$$W = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{1}{2}QV$$

4.3 Dielectric Materials

In a dielectric, an applied electric field induces a polarization $\mathbf{P}$— a dipole moment per unit volume. For a linear, isotropic dielectric:

$$\mathbf{P} = \epsilon_0 \chi_e \mathbf{E}$$

where $\chi_e$ is the electric susceptibility. The displacement field $\mathbf{D}$ is:

$$\mathbf{D} = \epsilon_0 \mathbf{E} + \mathbf{P} = \epsilon_0(1 + \chi_e)\mathbf{E} = \epsilon_0 \epsilon_r \mathbf{E} = \epsilon \mathbf{E}$$

where $\epsilon_r = 1 + \chi_e$ is the relative permittivity (dielectric constant). Gauss's law in terms of $\mathbf{D}$:

$$\nabla \cdot \mathbf{D} = \rho_f, \qquad \oint_S \mathbf{D} \cdot d\mathbf{a} = Q_{f,\rm enc}$$

where $\rho_f$ is the free charge density (excluding bound charges in the dielectric). A parallel-plate capacitor filled with dielectric has $C = \epsilon_r \epsilon_0 A/d$.

Simulation: Capacitor with Dielectric Slab

Solves Laplace's equation by finite differences for a parallel-plate capacitor partially filled with a dielectric ($\epsilon_r = 4$), showing how the potential bends at the interface.

Parallel-Plate Capacitor with Dielectric

Finite-difference solution of Laplace's equation in a capacitor with a dielectric slab (ε_r = 4).

Click Run to execute the Python code

First run will download Python environment (~15MB)