Fermi energy of electron gas with electrostatic interaction I have been given the following exam question and am unsure how I would go about solving it:

Consider the case of a one-dimensional metal, consisting of a chain of $N$
positive charges $+q$ separated by a distance $2R$ and immersed in a neutralizing background of electrons with density per unit length $n_e$. The electrostatic energy due to the interaction of the electrons with the ion cores and among themselves is:
$$\mathcal{E}_{el}=-\frac{\alpha q^{2}}{4\pi \epsilon_{0}}\frac{N}{R}$$
Assuming that the electrons form a non-interacting Fermi gas, calculate the Fermi energy. Write down an expression for the total kinetic energy of the electrons.

Now, ordinarily I would compute the Fermi energy as follows: First write down the density of states for a gas of Fermions:
$$g(\epsilon)=\frac{V m^{3/2}}{\sqrt{2}\pi^{2}\hbar^3}\epsilon^{1/2}$$
Then we have:
$$N = \int_{0}^{\infty}g(\epsilon)n_{F}(\epsilon)\:\mathrm{d}\epsilon = \int_{0}^{\infty}g(\epsilon)\Theta(E_{F}-\epsilon)\:\mathrm{d}\epsilon$$
Where $\Theta(\epsilon)$ is the Heaviside step function. We can thus calculate:
$$N=\int_{0}^{E_{F}}g(\epsilon)\:\mathrm{d}\epsilon = \frac{Vm^{3/2}}{\sqrt{2}\pi^{2}\hbar^{3}}\frac{2}{3}\epsilon_{F}^{3/2}$$
So we have the Fermi energy:
$$\epsilon_{F} = \left(\frac{N}{V}\frac{2\sqrt{2}\pi^{2}\hbar^{3}}{3m^{3/2}}\right)^{2/3}=\frac{2\hbar^{2}}{m}\left(\frac{\pi^{2}}{3}\right)^{2/3}n_{e}^{2/3}$$
To calculate the total energy, we have:
$$E = \int_{0}^{E_{F}}\epsilon g(\epsilon)\:\mathrm{d}\epsilon$$
But none of this takes into account the electrostatic energy, so I fear that I have terribly misunderstood something.
Thanks!
 A: Kinetic energy in 1D, method 1. Free electrons. Assume no potential energy at the moment. Zero temperature.
\begin{equation}
n_e=\int_0^{E_F}g(\epsilon)d\epsilon=\int_0^{E_F}\frac{1}{\pi\hbar}\sqrt{\frac{m}{2\epsilon}}d\epsilon=\frac{\sqrt{2mE_F}}{\pi\hbar}
\end{equation}
\begin{equation}
E_F=\frac{\pi^2\hbar^2n_e^2}{2m}
\end{equation}
Kinetic energy, method 2. Number of filled states
\begin{equation}
N=\frac{kL}{\pi}
\end{equation}
where $L=\frac{N}{n_e}$ is the chain length. This gives the wave vector
\begin{equation}
k=\frac{\pi N}{L}=\pi n_e
\end{equation}
and thus the kinetic energy
\begin{equation}
E_{kin}=\frac{\hbar^2}{2m}k^2=\frac{\hbar^2\pi^2n_e^2}{2m}
\end{equation}
similarly to method 1. The total (Fermi) energy is the sum of kinetic and potential (electrostatic) energy:
\begin{equation}
E_F=E_{kin}+E_{el}=\frac{\hbar^2\pi^2n_e^2}{2m}-\frac{\alpha q^2}{4\pi\epsilon_0}\frac{N}{R}
\end{equation}
Hope this make sense. In your solution, DOS is taken 3D, but the problem says the system is 1D metal. This, at least, should be corrected.  
