We have two large plates, each of area A, separated by a distance d. The charge on the plates is ±Q, corresponding to the charge density ±σ (with σ = Q/A). When there is vacuum between the plates,
Consider next a dielectric inserted between the plates fully occupying the intervening region. The dielectric is polarised by the field, the effect is equivalent to two charged sheets (at the surfaces of the dielectric normal to the field) with surface charge densities σ p and –σ p.
The electric field in the dielectric then corresponds to the case when the net surface charge density on the plates is ±(σ– σ p ). That is,
So that the potential difference across the plates is
For linear dielectrics, we expect σ p to be proportional to ε0, i.e., to σ. Thus, (σ– σ p ) is proportional to σ and we can write