Estimating the (average local field)/(Applied field) in water

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Homework Statement

At room temperature the relative permittivity (ε) of water is 80. The dipole moment of a water molecule is 6.2×10^-30 coulomb meters. What is the average value of E_local/E for a water molecule? (In working out this problem, neglect the contribution to the relative permittivity from induced dipole moments.)

Homework Equations

$$\textbf{P}=\varepsilon_0 \chi_E \textbf{E}$$
$$\textbf{p}=(\chi_E/N)_{gas} \varepsilon_0 \textbf{E}_{local}$$
$$E_{local}/E=(N / \chi_E)_{gas} (\chi_E/N)_{liquid}$$
where N is the molecular density. Problem with the latter formula is that I don't know $\varepsilon_{gas}$. I guess there is more reasoning needed than equations.

The Attempt at a Solution

I don't know why you shouldn't be able to use the Clausius-Mossotti formula for this. It gives a very incorrect result, but anyway, here is the reasoning i thought was correct:
Imagine a spherical cavity in the dielectric. Here lies the molecule that is under the influence of the local field. Now there will be negative charges in the direction of the field and positive in the other because of the polarization of the dielectric. The field from these boundary charges creates a field in the same direction as the applied field and the problem is to estimate it.

If anyone wants me to show how this is done I can of course spend an hour in deriving it. But this will not do much good since it yields the wrong answer. The end result (the derivation is correct) is:
$$E_{local}/E \simeq 1+ \chi_E/3=27.3$$
But the correct result is 6,6. Completely wrong. What is wrong in my assumptions here? How can I reason about the dipole moment?

The book talks about it: "Even in a liquid, dielectric behaviour is complicated when polar molecules are present, since very large fields are generated by the permanent dipole moment." So why leave it to the student to solve such a difficult problem? Is there something i have missed?

(I Heard, by the way, that this is a special problem. I solved all the other problems in the chapter, but here I don't even know where to start. I have reread the whole chapter and parts of it resulting in better understanding, but still...)

 

[TSny]

Have you covered the Langevin (Debye) formula which relates the polarization $P$, the local field $E_{local}$, the permanent dipole moment of a molecule, the number of molecules per unit volume, and the temperature? See for example equation 1.9.11 here.

If you combine the Langevin formula with the formula $P = \epsilon_o \chi_E E$ I think you can get the result of 6.6 for the ratio of the local and applied fields.

 

[Order]

Have you covered the Langevin (Debye) formula which relates the polarization $P$, the local field $E_{local}$, the permanent dipole moment of a molecule, the number of molecules per unit volume, and the temperature? See for example equation 1.9.11 here.

If you combine the Langevin formula with the formula $P = \epsilon_o \chi_E E$ I think you can get the result of 6.6 for the ratio of the local and applied fields.

Yes, I have covered the orientational polarizability induced by dipoles aligning with the Electric field, but with a simpler derivation than done in your linked presentation. As they hint I let the distortional polarizability be zero and then I get the quotient 6,6.