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Let $\xi$ denote geodesic distance from the charge; $R$ is the fixed curvature radius, not the radial variable. In round three-dimensional hyperbolic space, a sphere has areaUsing the convention that electric flux is $Q/{ϵ}_0$, Gauss's law gives the radial orthonormal electric fieldIts magnitude is $|Q|/A{ϵ}_0$, and a negative charge reverses the direction. In rationalized natural units set ${ϵ}_0=1$; Gaussian conventions instead use flux $4\pi Q$.

For $\xi \ll R$ this approaches $Q/(4\pi {ϵ}_0{\xi }^2)$. For $\xi \gg R$, $E\sim Q{e}^{-2\xi /R}/(\pi {ϵ}_0{R}^2)$. This hyperbolic-space Coulomb field falls rapidly because the sphere area grows exponentially; conserved electric flux is unchanged. A potential vanishing at infinity is $V=Q[\coth (\xi /R)-1]/(4\pi {ϵ}_{0}R)$.

The original PDF actually prints $\sin \theta$ rather than ${\sin }^2\theta$ in the angular metric tensor. The round angular factor used above is required by its description as hyperbolic space. If the printed angular metric tensor is taken literally, the sphere-area coefficient is insteadand a radial Gauss-law solution has $E=Q/[{ϵ}_0{\Omega }_{\ast }{R}^2{\sinh }^2(\xi /R)]$. That angular geometry is not a smooth round sphere, so the literal metric tensor does not describe the asserted homogeneous hyperbolic space. Both readings retain the same radial area-growth factor; the geometric repair is explicit.