A parallel-plate capacitor has plates that are very large compared with the gap

PHYSICS · Multi-part · Fermi University Physics · Electricity, Magnetism & Electromagnetism · Electric fields in matter · Electric fields in matter

Electricity, Magnetism and Electromagnetism, Ch Dielectrics: polarisation, bound charge and the displacement field

Problem

A parallel-plate capacitor has plates that are very large compared with the gap, so the fields between them are uniform and edge effects are ignored. The free surface charge density on the positive plate is

\sigma_f = 2.0\times10^{-6}\,\mathrm{C\,m^{-2}}

and the equal and opposite free charge sits on the negative plate. The gap is completely filled by a homogeneous, isotropic, linear dielectric of relative permittivity

\varepsilon_r = 4.0

. ((a)) Use Gauss's law for the displacement field with a pillbox straddling the positive plate to find the magnitude of

\mathbf{D}

in the gap. ((b)) Find the magnitude of

\mathbf{E}

inside the dielectric. ((c)) Find the magnitude of the polarisation

\mathbf{P}

, and the bound surface charge density

\sigma_b

on the face of the dielectric that touches the positive plate (give its sign). ((d)) State the electric susceptibility

\chi_e

of the dielectric.

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