Dielectrics: An Atomic View — Question 2

Question 2

Number density connects molecular polarizability to susceptibility

Use $\epsilon_0=8.854\times10^{-12}$ F/m. A dilute nonpolar material has number density $n=2.00\times10^{28}$ m$^{-3}$ and relative permittivity 1.50. Neglect local-field corrections and use $p=\alpha E$, $P=np=\epsilon_0\chi E$.

1. Infer the molecular polarizability.

2. Find P and the induced moment per molecule at E=$2.00\times10^5$ V/m.

Question 2 - Solution

Number density connects molecular polarizability to susceptibility

(a) Method and calculation

$$\chi=\kappa-1=.500,\quad n\alpha=\epsilon_0\chi.$$$$\alpha=\frac{8.854\times10^{-12}(.500)}{2.00\times10^{28}}=2.21350\times10^{-40}\,\mathrm{C\,m^2/V}.$$

(b) Interpretation and further calculation

$$P=\epsilon_0\chi E=8.85400\times10^{-7}\,\mathrm{C/m^2}.$$$$p=\alpha E=P/n=4.42700\times10^{-35}\,\mathrm{C\,m}.$$

Physical check

The number density converts a moment per molecule into dipole moment per volume; the resulting unit is C per square meter.