Past exam of the physics B course of the University of Cambridge / 2011 / Paper 2 / Section B / B6

Apply Faraday's law to a shrinking rectangular loop crossing the interface. For bounded fields the magnetic-flux term vanishes with its area, giving ${\mathbf{E}}_{t,+}={\mathbf{E}}_{t,-}$. A shrinking Gaussian pillbox similarly gives $D_{n,+}-D_{n,-}={\sigma }_{\mathrm{free}}$. Thus normal $D$ is continuous when no free surface charge is present, even though a bound polarization charge may remain. These are the assumptions needed for the stated dielectric-interface boundary conditions.

Write $ϵ={ϵ}_r{ϵ}_0$, $r_{+}=\sqrt{(x-a{)}^2+y^2+z^2}$ and $r_{-}=\sqrt{(x+a{)}^2+y^2+z^2}$. Use each region's own permittivity in its Coulomb potential:At $x=0$, continuity of potential and normal displacement givesSolving,This permittivity convention for transmitted image charges matters: if the vacuum Coulomb prefactor is used even for ${\varphi }_{-}$, its combined effective source is $2Q/(1+{ϵ}_r)$ instead. The physical potential is the same. The two numbered leaves identify how these sources satisfy the equations in their respective regions.

The real charge feels only the image contribution after removing its own singular self-field. The image at $(-a,0,0)$ is separated from it by $2a$, soFor the bound charge, the outward normal of the dielectric is $+\hat{x}$ and $\mathbf{P}={ϵ}_0({ϵ}_r-1){\mathbf{E}}_{-}$. Differentiating ${\varphi }_{-}$ at $x=0$ gives $E_{-,x}=-a(Q+Q^{\prime })/[4\pi ϵ(a^2+y^2+z^2{)}^{3/2}]$. ThereforeIts integral over the plane is $Q^{\prime \prime }$, consistent with the external image representation. For ${ϵ}_r=1$ both the image force and induced surface charge vanish.

For the dielectric half-space the two formal sources coincide at $(a,0,0)$, outside the region being represented. Their sum is $Q+Q^{\prime }=2{ϵ}_{r}Q/(1+{ϵ}_r)$ in the dielectric Coulomb kernel, henceThere is no charge singularity inside this half-space. The expression satisfies Laplace's equation there, decays at infinity, and matches the vacuum solution's tangential electric field and normal electric displacement. The coincident image is a mathematical representation, not an extra physical charge in the vacuum region.

For the vacuum half-space keep the actual charge $Q$ at $(a,0,0)$ and put its image at the reflected point $(-a,0,0)$. ThenThe reflected singularity lies outside this region. The real singularity has the correct charge strength, and the interface coefficients obey the two matching equations in the parent Solution at every $y,z$. With decay at infinity, uniqueness of an electrostatic dielectric boundary-value problem establishes the construction throughout both half-spaces.