In symmetric situations, electric fields in linear dielectrics can simply be treated as reducing the electric field to one of a medium with permittivity $\epsilon$.
Charges within the atoms of the dielectric are not free to move, but align to reduce electric field
$\mathbf P$ is the polarization density, which is the dipole moment per unit volume (can model it as a cube of dipole moment $\mu$ and side length $d$, so the polarisation density is $P=\mu/d^3$. But any reasonable model would work, like a cuboid)
An equivalent model with bound charges on the surface and volume can produce an equivalent electric field. Note the subscript $b$ which represents bound charges in the model.
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$$ \textrm{Surface: }\sigma_b= \mathbf P \cdot \mathbf{\hat n} \qquad \textrm{Volume: }\rho_b=-\nabla \cdot \mathbf P $$
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Dipole moment density is proportional to net electric field. Result of id-id interactions
$$ \mathbf P = \epsilon_0\chi_e\mathbf E $$
$\chi_e$ is the electric susceptibility and $\kappa=1+\chi_e$ is the dielectric constant
If we take the divergence of that,
$$ -\rho_b=\chi_e \rho \implies \rho=\rho_f/\kappa $$
This is nice as we can easily get the bound charge density, but the problem arises when there are changes in $\kappa$, as the divergence argument has to include the divergence in the susceptibility. Solving this is equivalent to solving the Laplacian, where the boundary conditions in this case are that just beneath the surface of a discontinuity, the electric field normal to the surface creates polarisation that leads to the right surface bound charge.
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$$ \textrm{Electrical energy density: } \frac{1}{2}\epsilon E^2 \\ \textrm{Magnetic energy density: } \frac{B^2}{2\mu} $$
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More that the energy density is a factor of 2 off from the energy density of an equivalent permanently polarised material
Similar to the electric case, in very symmetric situations, we will see that we can compute the $\mathbf H$ field using only the free currents, the B field changes to be equivalent to if the material has a permeability of $\mu$.