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Consider a capacitor half filled with a dielectric as shown in the figure.enter image description here

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.

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The free charges on the metal plates induce (bound) charges on the dielectric.
The free charges on the metal plates cannot move onto the dielectric which is an insulator and the induced (bound) charges cannot move onto the metal plates. So the "effective" charge density in the region of the left-hand plates does become $\sigma_{\rm F}-\sigma_{\rm B} $ and so the electric field in that region is reduced by a factor equal to the relative permittivity of the dielectric.

If you replaced the dielectric with a conductor which just did not touch the metal plates then the induced charge density on the conductor would be equal to that on the metal plates and so the electric field inside the conductor would be zero ie the conductor is a material with an infinite permittivity.

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