momentum $k_{z}$ of an electron in the electrosphere of the quark stars is given by $$k_{z}=\sqrt{m_{e}^{2}+k_{\perp }^{2}-\left[ \omega -V\left(
z\right) \right] ^{2}},$$ where $k_{\perp }^{2}=k_{x}^{2}+k_{y}^{2}$ [@So82; @Kim02]. The mean particle production number $\left\langle N\right\rangle$ is given by summing over all modes [@Gr85], $$\left\langle N\right\rangle =\int e^{-2\sigma }\frac{dtd\omega }{2\pi }\frac{dxdk_{x}}{2\pi }\frac{dydk_{y}}{2\pi }.$$

In order to estimate the electron-positron pair production rate in the electrosphere, we have to find first the thickness of the classically forbidden zone, or, equivalently, the limits of integration $z_{\pm }$ in the transmission probability. They can be obtained as solutions of the equation $k_{z}\left( z\right) =0$ and are given by $$z_{\pm }=\frac{1}{2}\sqrt{\frac{3}{\alpha \pi }}\frac{1}{T}{\rm arcsinh}\left(\frac{\sqrt{2}\pi T}{\omega \mp \sqrt{m_{e}^{2}+k_{\perp
}^{2}}}\right)-z_{0}.$$

Therefore, by taking into account the form of the electrostatic potential in the electrosphere we obtain for $\sigma $ the expression $$\sigma =\int_{z_{-}}^{z_{+}}\sqrt{m_{e}^{2}+k_{\perp }^{2}-\left\{ \omega -\frac{\sqrt{2}\pi T}{\sinh \left[ 2\sqrt{\frac{\alpha \pi
}{3}}T\left( z+z_{0}\right) \right] }\right\} ^{2}}dz.$$

With the help of the transformation $\eta =\left(
1/\sqrt{m_{e}^{2}+k_{\perp
}^{2}}\right) \left\{ \omega -\sqrt{2}\pi T/\sinh \left[ 2\sqrt{\alpha \pi /3}T\left( z+z_{0}\right) \right] \right\} $ we obtain the following integral representation for $\sigma $: $$\label{sigma}
\sigma =\sqrt{\frac{3\pi }{2\alpha }}\frac{m_{e}^{2}+k_{\perp
}^{2}}{\omega
^{2}}\int_{-1}^{1}\frac{\sqrt{1-\eta ^{2}}d\eta }{\left( \frac{\sqrt{m_{e}^{2}+k_{\perp }^{2}}}{\omega }\eta -1\right) \sqrt{\frac{2\pi ^{2}T^{2}}{\omega ^{2}}+\left( \frac{\sqrt{m_{e}^{2}+k_{\perp
}^{2}}}{\omega }\eta -1\right) ^{2}}}.$$

The 