6.6.2. The dielectric constant of helium at temperature $0^\circ C$ and pressure 1 atm is $1.000074$. Find the dipole moment of a helium atom in a uniform electric field of strength 300 V/cm.
Solution
1st Solution Option.
0. Write down what we have.
$$\varepsilon = 1.000074$$
$$t = 0^\circ C \Leftrightarrow T = 273\ \text{K}$$
$$p = \alpha E_{\text{loc}}$$ where $\alpha$ is the polarizability of the atom, which can be found using the Clausius-Mossotti formula, and $$E_{\text{loc}} = E + E_{\text{l}} = \frac{\varepsilon + 2}{3}E$$ Since the fraction $\approx 1$, we have $E_{\text{loc}} \approx E$.
$E_{\text{l}}$ is the electric field of the surroundings, created by polarization outside the Lorentz sphere. (An explanation of this formula will be given before the final answer, in case you are encountering it for the first time.)
2. Finding the polarizability of the atom.
Write down the Clausius-Mossotti formula, then express $\alpha$:
$$\alpha = \frac{3\varepsilon_{0}}{n}\left(\frac{\varepsilon - 1}{\varepsilon + 2}\right)$$ where $n$ is the concentration of helium atoms, which we can find from the ideal gas equation of state.
$$P = nKT \Rightarrow n = \frac{P}{KT}$$ where $K$ is the Boltzmann constant.
The Clausius--Mossotti formula describes the relationship between the static dielectric constant of a dielectric and the polarizability of its constituent particles. It was derived independently by Ottaviano F. Mossotti in 1850 and by Rudolf J. E. Clausius in 1879. In cases where the substance consists of particles of one kind, in the Gaussian system of units the formula is: