Question

For dense insulating materials, the dielec- tric response function is expressed in terms of the microscopic polarizability, a, as no ēr = 1 + že=1+- 1- nea/3 (a) Solve this for a in terms of ēr to obtain the Clausius-Mossotti equation ēr – 1 Ēr - 2 - nea 3 (b) Under what condition does this Clausius-Mossotti relationship reduce to the relationship between thc dielectric constant and the electric susceptibility given for a gas, ēr=1+nea?

Please show all work

Answer 1

Given

nea Er =1+1 n,a/3

1-nea/3+nea 1- nea/3

Er(1 – nea/3) = 1 - " + na

vºuz + 1 = du*3-4

- - ]= 2' קע"ט

Er – 1 = (Ēr + 2)",0

$\boxed {\frac{\bar\varepsilon_r- 1}{\bar\varepsilon_r +2}=\frac {n_e\alpha}3}$

Note: This is the correct Clausius-Mossitti equation, there is a positive sign in the denominator of LHS.

(b)

For a gas, number density is low and the condition is we have

nea << 1

1-nea/ 31

This reduces equation (1) to

E1+nea