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I believe that the steps in the posted question are (more or less) correct. The final connection between the first and last equations can be seen by integrating the first equation:

$$\int d\mathbf{x} \, n(\mathbf{x}) = \int d\mathbf{x} \, n_0 e^{-q\Phi(\mathbf{x}/kT}$$ $$N = n_0 \int d\mathbf{x} \, e^{-q\Phi(\mathbf{x})/kT}$$

Solving for $n_0$ gives

$$n_0 = \frac{N}{\int d\mathrm{x} \, \exp(-q\Phi(\mathbf{x})/kT)}$$

Comparing this to the last equation in the original question, we see that the last equation is in fact identical to the first equation.

As a side note, for $q\Phi(\mathbf{x}) \ll kT$ the density becomes $$n(\mathbf{x}) = n_0$$ If the source of $\Phi$ is at the origin, then this approximation is valid for $\mathbf{x} \to \infty$ For this reason, F.F. Chen (Introduction to Plasma Physics and Controlled Fusion) uses $n_\infty$ for $n_0$.

I would like to thank one of my colleagues for helping me with this. He could have posted the answer, but he doesn't have an account. Pity :)

I believe that the steps in the posted question are (more or less) correct. The final connection between the first and last equations can be seen by integrating the first equation:

$$\int d\mathbf{x} \, n(\mathbf{x}) = \int d\mathbf{x} \, n_0 e^{-q\Phi(\mathbf{x})/kT}$$ $$N = n_0 \int d\mathbf{x} \, e^{-q\Phi(\mathbf{x})/kT}$$

Solving for $n_0$ gives

$$n_0 = \frac{N}{\int d\mathrm{x} \, \exp(-q\Phi(\mathbf{x})/kT)}$$

Comparing this to the last equation in the original question, we see that the last equation is in fact identical to the first equation.

As a side note, for $q\Phi(\mathbf{x}) \ll kT$ the density becomes $$n(\mathbf{x}) = n_0$$ If the source of $\Phi$ is at the origin, then this approximation is valid for $\mathbf{x} \to \infty$ For this reason, F.F. Chen (Introduction to Plasma Physics and Controlled Fusion) uses $n_\infty$ for $n_0$.

Also, Chapter 4 of Statistical Mechanics by McQuarrie gives a very extensive discussion of how to transition from the probability that a system will be in a given state to the probability that a particle will be in a given state. This helps justify using a single particle in the original question. The assumptions are that the particles do not interact, temperature is high, and density is low. The first assumption of non-interacting particles is perhaps not so great for charged particles in a plasma. However, if the high temperature and low density conditions are met, then the assumption of non-interaction becomes better.

I would like to thank one of my colleagues for helping me with this. He could have posted the answer, but he doesn't have an account. Pity :)

I believe that the steps in the posted question are (more or less) correct. The final connection between the first and last equations can be seen by integrating the first equation:

$$\int d\mathbf{x} \, n(\mathbf{x}) = \int d\mathbf{x} \, n_0 e^{-q\Phi(\mathbf{x}/kT}$$ $$N = n_0 \int d\mathbf{x} \, e^{-q\Phi(\mathbf{x})/kT}$$

Solving for $n_0$ gives

$$n_0 = \frac{N}{\int d\mathrm{x} \, \exp(-q\Phi(\mathbf{x})/kT)}$$

Comparing this to the last equation in the original question, we see that the last equation is in fact identical to the first equation.

As a side note, for $q\Phi(\mathbf{x}) \ll kT$ the density becomes $$n(\mathbf{x}) = n_0 = \frac{N}{V}$$ If the source of $\Phi$ is at the origin, then this approximation is valid for $\mathbf{x} \to \infty$ For this reason, F.F. Chen (Introduction to Plasma Physics and Controlled Fusion) uses $n_\infty$ for $n_0$.

I would like to thank one of my colleagues for helping me with this. He could have posted the answer, but he doesn't have an account. Pity :)

I believe that the steps in the posted question are (more or less) correct. The final connection between the first and last equations can be seen by integrating the first equation:

$$\int d\mathbf{x} \, n(\mathbf{x}) = \int d\mathbf{x} \, n_0 e^{-q\Phi(\mathbf{x}/kT}$$ $$N = n_0 \int d\mathbf{x} \, e^{-q\Phi(\mathbf{x})/kT}$$

Solving for $n_0$ gives

$$n_0 = \frac{N}{\int d\mathrm{x} \, \exp(-q\Phi(\mathbf{x})/kT)}$$

Comparing this to the last equation in the original question, we see that the last equation is in fact identical to the first equation.

As a side note, for $q\Phi(\mathbf{x}) \ll kT$ the density becomes $$n(\mathbf{x}) = n_0$$ If the source of $\Phi$ is at the origin, then this approximation is valid for $\mathbf{x} \to \infty$ For this reason, F.F. Chen (Introduction to Plasma Physics and Controlled Fusion) uses $n_\infty$ for $n_0$.

I would like to thank one of my colleagues for helping me with this. He could have posted the answer, but he doesn't have an account. Pity :)

I believe that the steps in the posted question are (more or less) correct. The final connection between the first and last equations can be seen by integrating the first equation:

$$\int d\mathbf{x} \, n(\mathbf{x}) = \int d\mathbf{x} \, n_0 e^{-q\Phi(\mathbf{x}/kT}$$ $$N = n_0 \int d\mathbf{x} \, e^{-q\Phi(\mathbf{x})/kT}$$

Solving for $n_0$ gives

$$n_0 = \frac{N}{\int d\mathrm{x} \, \exp(-q\Phi(\mathbf{x})/kT)}$$

Comparing this to the last equation in the original question, we see that the last equation is in fact identical to the first equation.

As a side note, notice that for $q\Phi(\mathbf{x}) \ll kT$ the density becomes $$n(\mathbf{x}) = n_0 = \frac{N}{V}$$

I would like to thank one of my colleagues for helping me with this. He could have posted the answer, but he doesn't have an account. Pity :)

I believe that the steps in the posted question are (more or less) correct. The final connection between the first and last equations can be seen by integrating the first equation:

$$\int d\mathbf{x} \, n(\mathbf{x}) = \int d\mathbf{x} \, n_0 e^{-q\Phi(\mathbf{x}/kT}$$ $$N = n_0 \int d\mathbf{x} \, e^{-q\Phi(\mathbf{x})/kT}$$

Solving for $n_0$ gives

$$n_0 = \frac{N}{\int d\mathrm{x} \, \exp(-q\Phi(\mathbf{x})/kT)}$$

Comparing this to the last equation in the original question, we see that the last equation is in fact identical to the first equation.

As a side note, for $q\Phi(\mathbf{x}) \ll kT$ the density becomes $$n(\mathbf{x}) = n_0 = \frac{N}{V}$$ If the source of $\Phi$ is at the origin, then this approximation is valid for $\mathbf{x} \to \infty$ For this reason, F.F. Chen (Introduction to Plasma Physics and Controlled Fusion) uses $n_\infty$ for $n_0$.

I would like to thank one of my colleagues for helping me with this. He could have posted the answer, but he doesn't have an account. Pity :)