= Solution
Choose $z=0$ at the sheet, positive $z$ upwards, so the uniformly charged slab occupies $-d<z<0$. Use the usual isolated-plane superposition convention, with no imposed uniform background field. The slab alone produces $+\rho d/(2\epsilon_0)\hat{\mathbf z}$ above it, while the negatively charged sheet produces $-\sigma/(2\epsilon_0)\hat{\mathbf z}$ there. Consequently
$$
\boxed{\mathbf E(z>0)=\frac{\rho d-\sigma}{2\epsilon_0}\hat{\mathbf z}}.
$$
Since $\rho d>\sigma$, the field is upward, away from the net positive charge distribution, with magnitude $(\rho d-\sigma)/(2\epsilon_0)$. The equal halves in the separate infinite-sheet fields follow from their reflection symmetry before superposition.
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