The list below gives

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

The Boundary-Value Problem, Ch Molecular polarisability, permanent dipoles and the Clausius-Mossotti relation

Problem

The list below gives, for four liquids at T=293 KT=293\,\mathrm{K} (carbon tetrachloride is abbreviated CCl4_4), the number density NvN_v, the electronic polarisability volume α/4πε0\alpha/4\pi\varepsilon_0, the permanent dipole moment pp (zero for a non-polar molecule) and the measured relative permittivity εr\varepsilon_r. CCl4_4: Nv=6.24×1027 m−3N_v=6.24\times10^{27}\,\mathrm{m^{-3}}, α/4πε0=11.2×10−30 m3\alpha/4\pi\varepsilon_0=11.2\times10^{-30}\,\mathrm{m^3}, p=0p=0, measured εr=2.24\varepsilon_r=2.24 cyclohexane: Nv=5.58×1027 m−3N_v=5.58\times10^{27}\,\mathrm{m^{-3}}, α/4πε0=10.9×10−30 m3\alpha/4\pi\varepsilon_0=10.9\times10^{-30}\,\mathrm{m^3}, p=0p=0, measured εr=2.02\varepsilon_r=2.02 acetone: Nv=8.20×1027 m−3N_v=8.20\times10^{27}\,\mathrm{m^{-3}}, α/4πε0=6.4×10−30 m3\alpha/4\pi\varepsilon_0=6.4\times10^{-30}\,\mathrm{m^3}, p=9.61×10−30 C mp=9.61\times10^{-30}\,\mathrm{C\,m}, measured εr=20.7\varepsilon_r=20.7 acetonitrile: Nv=11.5×1027 m−3N_v=11.5\times10^{27}\,\mathrm{m^{-3}}, α/4πε0=4.4×10−30 m3\alpha/4\pi\varepsilon_0=4.4\times10^{-30}\,\mathrm{m^3}, p=13.08×10−30 C mp=13.08\times10^{-30}\,\mathrm{C\,m}, measured εr=37.5\varepsilon_r=37.5 ((a)) For each liquid, compute y=Nv3ε0(α+p23kBT).y = \frac{N_v}{3\varepsilon_0}\left(\alpha + \frac{p^2}{3k_BT}\right). ((b)) Where y<1y<1, find the predicted εr\varepsilon_r from εr=(1+2y)/(1−y)\varepsilon_r = (1+2y)/(1-y) and compare with the measured value. Where y≥1y\ge1, say what this means. ((c)) Sort the four liquids into those for which the Clausius-Mossotti relation is a good working approximation and those for which it fails, and explain, using the assumption behind the Lorentz local field, why acetone fails despite forming no hydrogen bonds.

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