density at the surface and hence it can also significantly reduce the intensity of the electric field and the electromagnetic emissivity of the electrons in the electrosphere. Therefore, in order to describe the radiation properties of the electrosphere we assume that $V_q\neq 0$.

The Poisson equation for the electrostatic potential $V\left(
z,T\right) $ generated by the finite temperature electron distribution reads [@Al86; @Ke95] $$\frac{d^{2}V}{dz^{2}}=\frac{4\alpha }{3\pi }\left[ \left(
V^{3}-V_{q}^{3}\right) +\pi ^{2}\left( V-V_{q}\right) T^{2}\right]
,z\leq 0, \label{p2}$$ $$\frac{d^{2}V}{dz^{2}}=\frac{4\alpha }{3\pi }\left( V^{3}+\pi
^{2}T^{2}V\right) ,z\geq 0,  \label{3}$$ where $T$ is the temperature of the electron layer, which can be taken as a constant, since we assume the electrons are in thermodynamic equilibrium with the constant temperature quark matter. In Eqs. (\[p2\])-(\[3\]), $z$ is the space coordinate measuring height above the quark surface, $\alpha $ is the fine structure constant and $V_{q}/3\pi ^{2}$ is the quark charge density inside the quark matter. The boundary conditions for Eqs. (\[p2\])-(\[3\]) are $V\rightarrow V_{q}$ as $z\rightarrow -\infty $ and $V\rightarrow
0$ for $z\rightarrow \infty $. In the case of the zero temperature electron distribution at the boundary $z=0$ we have the condition $V(0)=(3/4)V_{q}$ [@Al86].

The general solution of Eq. (\[3\]) is given by [@ChHa03] $$V\left( z,T\right) =\frac{\sqrt{2}\pi T}{\sinh 