The plane is the flat face of a large block of linear dielectric of relative permittivity

PHYSICS · Multi-part · Fermi University Physics · Electricity, Magnetism & Electromagnetism · Electric fields in matter · Maxwell's equations and conservation laws

Electricity, Magnetism and Electromagnetism, Ch Displacement current, Maxwell's equations and boundary conditions

Problem

The plane z=0z = 0 is the flat face of a large block of linear dielectric of relative permittivity εr=4.0\varepsilon_r = 4.0, which fills the region z<0z < 0; the region z>0z > 0 is vacuum. The fields are static. A measurement gives, just above and just below the face,Evac=(300, 0, 200) V m−1,Ediel=(300, 0, 50) V m−1.\mathbf{E}_{\text{vac}} = (300,\,0,\,200)\ \mathrm{V\,m^{-1}},\qquad \mathbf{E}_{\text{diel}} = (300,\,0,\,50)\ \mathrm{V\,m^{-1}}. Take the unit normal n^=+z^\hat{\mathbf{n}} = +\hat{\mathbf{z}}, pointing out of the dielectric into the vacuum, and use the pillbox convention of this chapter, in which the jump of D⋅n^\mathbf{D}\cdot\hat{\mathbf{n}} on going along n^\hat{\mathbf{n}} across the face equals the free surface charge density σf\sigma_f. ((a)) Check that the tangential components are consistent with Faraday's law applied to a thin loop straddling the face, and find the free surface charge density σf\sigma_f on the face. ((b)) Find the bound surface charge density σb\sigma_b on the face of the dielectric. ((c)) Verify your value of σb\sigma_b against the measured jump in the normal component of E\mathbf{E}, by stating that jump in V m−1\mathrm{V\,m^{-1}}.

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