If we assume a 1D world, then the divergence of the electric field would read
$\nabla\cdot\vec{E} = \partial_xE_x$,
since $\partial_yE_y = \partial_zE_z = 0$ in a 1D world. Therefore, Gauss Law reads
$\partial_xE_x(x) = \rho/\epsilon_0$, where $\rho$ is still defined in 3D. We can assume a box xyz in which our $\rho$ is and if we apply the Divergence theorem:
$\int \nabla\cdot\vec{E} dV = \int \vec{E}(x)\cdot d\vec{S} = E(x)\cdot y\cdot z = E(x)\cdot S = \frac{1}{\epsilon_0}\int\rho dV $, we might get finally
$E(x)\cdot S = \frac{1}{\epsilon_0}\int\rho\cdot S\cdot dx \to E(x) = \frac{1}{\epsilon_0}\int\rho dx = \frac{\sigma}{\epsilon_0}$.
This expression $E(x) = \sigma/\epsilon_0$ is exactly the same than the electric field created by a flat capacitor, which was answered here that this is in fact what charges look like in a 1D universe. However, if $E(x) = \sigma/\epsilon_0$ then $\partial_xE(x) = 0$ and does not this contradict Gauss Law from which we have started, i.e., $\partial_xE_x(x) = \rho/\epsilon_0$?
