= Solution
The signed normal-field profile is constant and negative below $z=-d$, rises linearly through the slab, jumps down at the negative sheet, then remains constant and positive above. More explicitly,
$$
E_z(-d^-)=E_z(-d^+)=\frac{\sigma-\rho d}{2\epsilon_0},\quad E_z(0^-)=\frac{\rho d+\sigma}{2\epsilon_0},\quad E_z(0^+)=\frac{\rho d-\sigma}{2\epsilon_0}.
$$
There is no surface charge at the slab's lower face, so the field is continuous there; only its slope changes. At the ideal sheet, the field jump is
$$
\boxed{E_z(0^+)-E_z(0^-)=-\frac\sigma{\epsilon_0}},
$$
exactly the negative-sheet boundary condition. The ideal sheet itself has distinct one-sided fields rather than one common field value.
\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2015-resit-charged-sheet-slab-field.png]
{title=Signed normal electric field for a negative sheet above a positively charged slab}
The graph uses the illustrative ratio $\sigma/(\rho d)=1/2$, clearly separated from the symbolic general result. The zero occurs inside the lower half of the slab, and the two outside fields have equal magnitude and opposite sign.
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