A parallel-plate capacitor has plates of area separated by

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 of area A=0.030 m2A = 0.030\,\mathrm{m}^2 separated by d=2.0 mmd = 2.0\,\mathrm{mm}. With the gap evacuated it is charged by a 400 V400\,\mathrm{V} supply and then disconnected, so its charge is fixed. Neon gas is then admitted, filling the gap at a number density n=2.5×1025 m−3n = 2.5\times10^{25}\,\mathrm{m^{-3}}. A neon atom has polarisability volume α/4πε0=0.40×10−30 m3\alpha/4\pi\varepsilon_0 = 0.40\times10^{-30}\,\mathrm{m}^3, with p=αE\mathbf{p} = \alpha\mathbf{E}, and every atom is polarised alike. Edge effects are ignored, the gas is a linear dielectric with no free charge, and because the gas alters the field by only about one part in 10410^4 you may use the vacuum value of EE when computing the atomic moments. ((a)) Find the field in the gap and the induced dipole moment of one neon atom. ((b)) Find the polarisation PP of the gas, its susceptibility χe\chi_e and its relative permittivity εr\varepsilon_r. ((c)) Find the change in the capacitor's stored energy when the gas is admitted, stating whether it rises or falls.

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