这是一份nottingham诺丁汉大学PHYS2004作业代写的成功案例

location is proportional to $\exp \left[-q \phi /\left(k_{B} T\right)\right]$. Then the charge density $\rho_{c}$ in the vicinity of an ion at $\boldsymbol{x}=0$ is
$$
\rho_{c}=Z e \delta(0)-n_{c} e \exp \left[\frac{e \phi}{k_{B} T}\right]+n_{i} e Z \exp \left[\frac{-e Z \phi}{k_{B} T}\right]
$$
If we assume that $|q \phi| \ll k_{B} T$ and that the plasma is quasi-neutral, then this becomes
$$
\rho_{c}=Z e \delta(0)-\frac{e^{2} \phi}{k_{B} T}\left(n_{e}+n_{i} Z^{2}\right)=Z e \delta(0)-\frac{\phi}{4 \pi \lambda_{D}^{2}}
$$
At this point we can write Poisson’s equation in spherical coordinates, assuming that the charges are distributed with spherical symmetry, as
$$
\frac{1}{r^{2}} \frac{\mathrm{d}}{\mathrm{d} r}\left(r^{2} \frac{\mathrm{d} \phi}{\mathrm{d} r}\right)=-4 \pi Z e \delta(0)+\frac{\phi}{\lambda_{D}^{2}}
$$
which (in cgs units) has the solution
$$
\phi=\frac{Z e}{r} \mathrm{e}^{-r / \lambda_{D}}
$$

PHYS2004 COURSE NOTES ：

This particular equation helps one see the physics of the wave we are finding. It is a purely longitudinal wave like an acoustic wave, in which the fluctuating electric field and compression by the electron pressure both cause the electron density to vary. The first term on the right-hand side can be evaluated from Poisson’s , which gives in this case
$$
\nabla \cdot \boldsymbol{E}{1}=4 \pi\left(Z e n{i o}-e n_{e o}-e n_{e 1}\right)
$$
in which the first two terms in parentheses cancel because the plasma is quasineutral. Then assuming the electrons behave as a polytropic gas with index $\gamma_{e}$, we obtain a wave equation
$$
\left(\frac{\partial^{2}}{\partial t^{2}}+\omega_{p e}^{2}-\frac{\gamma_{e} p_{e o}}{n_{e o} m_{e}} \nabla^{2}\right) n_{e 1}=0,
$$
in which we have introduced the electron plasma frequency,
$$
\omega_{p e}=\sqrt{4 \pi e^{2} n_{e o} / m_{e}}=5.64 \times 10^{4} \sqrt{n_{e o}} \mathrm{rad} / \mathrm{s},
$$