Consider a capacitor half filled with a dielectric as shown in the figure.
I understand that due to the polarization of the dielectric, there will be some surface charge $\sigma_B$ on the surface of the dielectric on the left side. Meanwhile, the plates are conductors so the charge density on the plate, $\sigma_F$ must be uniform.
My question is why doesn't the bound charge $\sigma_B$ induce a charge $-\sigma_B$ on the left side of the parallel plate? I would have thought one would see $\sigma_F - \sigma_B$ on the plate over the dielectric and just $\sigma_F$ in the area withou the dielectric.