Past exam of the physics B course of the University of Cambridge / 2013 / Paper 2 / Section B / B6

In vacuum, displacement current contributes the additional term ${ϵ}_0\partial \mathbf{E}/\partial t$:Taking the divergence and using Gauss law gives $\nabla \cdot \mathbf{J}+\partial \rho /\partial t=0$, the local conservation of charge. Without displacement current the curl equation would incorrectly require a divergence-free current even during charge accumulation.

For a charging capacitor, a surface cutting the wire carries conduction current while a surface passing through the gap carries changing electric flux. The displacement current term makes the circulation independent of the spanning surface. It is not transport of free charge across the vacuum gap. In matter the macroscopic equation is $\nabla \times \mathbf{H}={\mathbf{J}}_f+\partial \mathbf{D}/\partial t$, with electric polarization included in $\mathbf{D}$. Together with Faraday's law of induction, the vacuum equations give electromagnetic waves of speed $1/\sqrt{{\mu }_0{ϵ}_0}$.

An applied electric field displaces bound charges or reorients permanent dipoles, producing electric polarization $\mathbf{P}$, the dipole moment per unit volume. It gives bound charges ${\rho }_b=-\nabla \cdot \mathbf{P}$ and ${\sigma }_b=\mathbf{P}\cdot \mathbf{n}$. Definefor a linear isotropic dielectric. Then $\nabla \cdot \mathbf{D}={\rho }_f$ separates free charge from the bound-charge response. Electronic, ionic and orientational electric polarization respond on different timescales; real permittivity is frequency dependent, and its dissipative part describes dielectric loss.

A uniform dielectric filling a parallel-plate capacitor raises its capacitance to $ϵA/d$. At fixed charge this reduces field energy $Q^2/(2C)$; at fixed voltage the stored energy $C{V}^2/2$ increases because the voltage source supplies charge and energy. Insulation and energy storage are useful applications, limited by loss and dielectric breakdown. Static interface conditions are continuous tangential $\mathbf{E}$ and a jump in normal $\mathbf{D}$ equal to free surface charge.

A current in one circuit links magnetic flux through another. It produces magnetic flux linkage through another. With fixed geometry and a linear reciprocal medium, choose consistent winding orientations and writeThe shared coefficient is mutual inductance; reversal of one winding reverses its sign. Faraday's law of induction gives the mutual induced emf ${\mathcal{E}}_2=-M{\dot{I}}_1$ for fixed $M$. The minus sign expresses Lenz's law.

The magnetic energy isNonnegative energy for every current pair requires $|M|\le \sqrt{L_1{L}_2}$, motivating the coupling coefficient $M/\sqrt{L_1{L}_2}$. Transformers use strong common flux to exchange energy between circuits; leakage flux reduces coupling. A shared core increases linkage, but saturation, hysteresis and eddy-current losses limit the linear description. A moving geometry also contributes $-I_1\dot{M}$ to the mutual emf.

Find the potential from $\mathbf{E}=-\nabla \varphi$ and $\nabla \cdot (ϵ\nabla \varphi )=-{\rho }_f$. One must specify boundary data: a potential on each Dirichlet boundary, a normal displacement on each Neumann boundary, or an appropriate mixture. Conductors are equipotentials; at dielectric interfaces $\varphi$ is continuous and the normal displacement jumps by the specified free surface charge. With no singular dipole sheet, tangential electric field is continuous.

To establish uniqueness of Poisson equation, subtract two solutions with the same charge and boundary data, writing their difference as $u$. Integration by parts givessince $\nabla \cdot (ϵ\nabla u)=0$. The right side vanishes for identical Dirichlet data, identical Neumann data, or their usual mixed combination. Interface terms cancel under the homogeneous difference conditions. With positive permittivity, $\nabla u=0$: the field is unique, and the potential is unique if any suitable reference potential is fixed. For a pure Neumann problem an additive constant remains, and the prescribed outward displacement flux must equal the enclosed free charge. This theorem justifies image-charge solutions once their boundary data have been checked.