Past exam of the physics 2 paper of the University of Oxford / 2015 resit / 10 / c / 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,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 isexactly the negative-sheet boundary condition. The ideal sheet itself has distinct one-sided fields rather than one common field value.

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.