\left[
2\sqrt{\frac{\alpha \pi }{3}}T\left( z+z_{0}\right) \right] },$$ where $z_{0}$ is a constant of integration. Its value can be obtained from the condition of the continuity of the potential across the star’s surface, requiring $V_{q}(0,T)=V\left(
0,T\right) $, where $V_{q}\left( z,T\right) $ is the value of the electrostatic potential in the region $z\leq 0$, described by Eq. (\[p2\]). Therefore $$z_{0}=\frac{1}{2}\sqrt{\frac{3}{\alpha \pi }}\frac{1}{T} {\rm
arcsinh}\left[ \frac{\sqrt{2}\pi T}{V_{q}\left( 0,T\right)
}\right] ,$$

The number density distribution $n_{e}$ of the electrons at the quark star surface can be obtained from $n_{e}\left( z,T\right)
=V^{3}/3\pi ^{2}+VT^{2}/3$ [@Ke95; @ChHa03] and is given by $$n_{e}\left( z,T\right) =\frac{\sqrt{2}\pi }{3}\frac{1+\cosh ^{2}\left[ 2\sqrt{\frac{\alpha \pi }{3}}T\left( z+z_{0}\right) \right] }{\sinh
^{3}\left[ 2\sqrt{\frac{\alpha \pi }{3}}T\left( z+z_{0}\right)
\right] }T^{3}.$$

In the limit of zero temperature, $T\rightarrow 0$ we obtain $V(z)=a_0/(z+b)$, where $a_0=\sqrt{3\pi /2\alpha }$ and $b$ is an integration constant. $b$ can be determined from the boundary condition $V(0)=(3/4)V_{q}$, which gives $b=\left(
4a_0/3V_{q}\right) $. Therefore, in this case we find for the electron particle number distribution the expression: $n_{e}(z)=(1/3\pi ^{2})a^{3}_{0}/ \left(z+b\right) ^{3}$.

In the absence of a crust of the quark star, the electron layer can extend to several thousands fermis outside the star’s surface.

The strength of the electric field $E$ outside the quark star surface 