Answer:
a) The induced surface charge densities on the inner and outer surfaces of each plate are zero.
b) The new free charges on the surfaces of the plates remain the same as before the introduction of the dielectric. Only the induced charges change due to the presence of the dielectric.
Step-by-step explanation:
(a) To determine the induced surface charge densities on the inner and outer surfaces of each plate, we can consider the electric field between the plates.
Since the plates are equipotentials, the electric field between them must be zero. However, in the presence of the sheet of charge with a uniform surface charge density σ at z = b, there will be an electric field generated.
Let's consider the electric field at a point between the plates, close to the inner surface of the lower plate. At this point, the electric field is due to the electric field from the induced charges on both plates and the electric field from the sheet of charge.
The electric field due to the sheet of charge is given by E_sheet = σ/(2ε₀), where ε₀ is the vacuum permittivity.
The electric field due to the induced charges on the inner surface of the lower plate (which is at z = 0) will be directed upwards, and we'll call it E_inner. The electric field due to the induced charges on the outer surface of the upper plate (which is at z = s) will be directed downwards, and we'll call it E_outer.
Since the electric field between the plates is zero, we have:
E_sheet + E_inner - E_outer = 0.
However, since the distance s between the plates is much smaller than their lateral dimensions, we can neglect the effect of the electric field from the induced charges and consider that the electric field due to the sheet of charge is the only significant contribution.
Therefore, we have:
E_sheet = σ/(2ε₀) = 0.
This equation implies that the induced surface charge densities on the inner and outer surfaces of each plate are zero.
(b) When a dielectric slab of thickness t < b and relative permittivity εr is introduced and placed on top of the lower plate, the presence of the dielectric will cause polarization and induce bound charges on the surfaces of the dielectric and the plates.
Let's consider the three plates: top, middle (dielectric slab), and bottom.
For the top plate (which coincides with the plane z = s), the induced surface charge density on the upper surface will be zero because it's connected to the outer surface of the upper plate (which is an equipotential) and there is no electric field between the plates. The induced surface charge density on the lower surface will also be zero because the dielectric is on top of the lower plate.
For the middle plate (dielectric slab), the induced surface charge density on the upper surface will be zero because it's connected to the lower surface of the top plate, which has a zero induced charge density. The induced surface charge density on the lower surface of the middle plate will be given by σ_induced = -σ * (εr - 1), where σ is the surface charge density of the lower plate.
For the bottom plate (which coincides with the xˆ − yˆ plane), the induced surface charge density on the upper surface will be given by σ_induced = -σ * (εr - 1), where σ is the surface charge density of the lower plate. The induced surface charge density on the lower surface will be zero because it's connected to the dielectric slab.
The new free charges on the surfaces of the plates remain the same as before the introduction of the dielectric. Only the induced charges change due to the presence of the dielectric.