Physic Labs

Electrodynamics

Electrostatic fields, conductors, and dielectrics

Static electric fields are sourced by charge; an equilibrium conductor has zero internal field, while a dielectric polarizes in an applied field.

Electrostatics is the time-independent limit of electrodynamics. Free charge sources electric displacement; bound charge is represented through material polarization.

∇⋅D=ρf,∇×E=0∇·D = ρ_f, ∇×E = 0

Definition: Quantities and model

In a linear isotropic medium, D = ε₀E + P = εE. In electrostatic equilibrium E = 0 inside a conductor; excess charge resides on its surface.

Adjust parameters and rotate the view to inspect field structure; this is illustrative, not a general Maxwell solver.

Interpretation and consequences

Boundary conditions: tangential E is continuous; the normal component of D jumps by the free surface-charge density σ_f.

Example: Quantitative example

A linear dielectric with ε_r = 4 is subject to E = 2.0×10⁵ V/m. Then D = ε₀ε_rE = 7.08×10⁻⁶ C/m².

Solution

Substitute into the stated relation, keep SI units consistent, and check the result dimensionally.

The conditions EtE_t continuous and D2n−D1n=σfD_{2n}-D_{1n}=\sigma_f determine fields at an interface. For a conductor, the interior field is zero, so just outside in vacuum En=σf/ε0E_n=\sigma_f/\varepsilon_0. This explains electrostatic shielding and why charge concentration at a sharp tip can produce a particularly intense local field.

In a dielectric, polarization PP produces bound volume charge ρb=−∇⋅P\rho_b=-\nabla\cdot P and surface charge σb=P⋅n^\sigma_b=P\cdot\hat n. Gauss's law for DD contains only free charge because the microscopic dipole response is already encoded in PP. This macroscopic bookkeeping differs from ∇⋅E=ρtotal/ε0\nabla\cdot E=\rho_{\rm total}/\varepsilon_0, which counts all charge.

A parallel-plate capacitor of area AA, gap dd, and uniform dielectric illustrates the distinction between EE and DD. Neglecting fringing, C=εA/dC=\varepsilon A/d, E=V/dE=V/d, and D=εE=Qf/AD=\varepsilon E=Q_f/A. Removing the dielectric from an isolated capacitor leaves QfQ_f fixed, so DD stays fixed while EE rises. With a voltage source attached, EE stays fixed and free charge changes instead. Comparisons must specify which electrical constraint is maintained.

Quick check

Which relation is correct in the idealized situation described?

What should be checked first when applying a field formula?

References

  1. David J. Griffiths (2017). Introduction to Electrodynamics