# Electrostatics of Microscopic Media

## Suceptibility

$$\vec{p}_{mol} = \varepsilon_0\gamma_{mol}\vec{E}$$. C.f. $$\vec{P} = \varepsilon_0\chi_e\vec{E}$$ for macroscopic media.

Consider a polar molecule in an electric field. The polar molecule tries to align along the field and experiences a torque. $$\langle \vec{p}\rangle$$ becomes more along the $z$-axis, assuming $$\vec{E}$$ is along the $z$-axis, over time due to the torque felt by the polar molecules.

$$p_i\propto \exp(-E_i/kT)$$. Thus, lower energies are most likely. The leading constant is, $$1=\sum p_i = A\sum_i \exp(-E_i/kT) = AZ$$. Thus, $$p_i = \frac{\exp(-E_i/kT)}{Z}$$.

$$H = H_0 - \vec{p}_0\cdot\vec{E}$$. $$\vec{p} = q\vec{d},\vec{\tau} = \vec{r}\times\vec{F} = q\frac{\vec{d}}{2}\times\vec{E} = \vec{p}_0\times\vec{E} = -\hat{y}P_0E\sin\theta = -\frac{\partial U}{\partial\theta}$$. So, $$U = -p_0E\cos\theta$$.

In the classical case, we have a continuous energy spectra, so $$Z = \iiint\iiint \exp(-E(p,q)/kT)d^3pd^3q$$.

Using $$p_z = p_0\cos\theta$$, $$\langle p_{mol}\rangle = \frac{\int p_0\cos\theta \exp\left(\frac{-H}{kT}\right)d\Omega}{Z} = p_0(\coth\alpha - 1/\alpha),\alpha=p_0E/kT=p_0E\beta$$. Note that $$\alpha$$ is the ratio between the thermal and interaction energies.

For $$T\to 0$$, $$\langle p_{mol}\rangle \approx p_0$$. Thus, the particles are very well aligned.

For $$kT\gg p_0 E$$, $$\alpha\ll 1$$. So, $$\coth\alpha\approx \frac{1}{\alpha}\frac{(1+\frac{1}{2}\alpha^2)}{(1+\frac{1}{6}\alpha^2)}\approx \frac{1}{\alpha}(1+\frac{1}{2}\alpha^2)(1-\frac{1}{6}\alpha^2) \approx \frac{1}{\alpha}\left(1+\frac{1}{3}\alpha^2\right)$$. Then, $$\langle p_{mol}\approx p_0\frac{1}{3}\alpha = \frac{1}{3}\frac{p_0^2E}{kT} = \varepsilon_0\gamma_{mol}E$$. So, $$\gamma_{mol} \approx \frac{p_0^2E}{3\varepsilon_0 kT}$$.

Created: 2023-06-25 Sun 02:33

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