Electric vectors E, P and D

The three vectors

• Electric field intensity E: force per unit positive test charge; it is the total field in the dielectric, produced by all (free and bound) charges. Unit: V/m.
• Polarization P: electric dipole moment per unit volume of the dielectric. Unit: C/m². For a linear isotropic dielectric P = ε₀ χₑ E, where χₑ is the electric susceptibility.
• Electric displacement D: the vector whose flux depends only on free charge; defined as D = ε₀E + P. Unit: C/m².

Relation D = ε₀E + P using a parallel-plate capacitor

   + + + + + + + + +   plate, free charge +q (σf = q/A)
   - - - - - - - - -   bound charge -σp (dielectric surface)
   [   dielectric   ]
   + + + + + + + + +   bound charge +σp
   - - - - - - - - -   plate, free charge -q

1. Plates carry free surface charge density ±σf. A dielectric slab fills the gap. It becomes polarized, and bound charges of surface density −σp (near the positive plate) and +σp (near the negative plate) appear on its faces.
2. The bound-charge density equals the polarization: σp = P.
3. Apply Gauss' law with a pillbox enclosing the surface of the positive plate and part of the dielectric. The net charge per unit area enclosed is σf − σp: ε₀ E = σf − σp = σf − P
4. Rearrange: σf = ε₀E + P
5. Define D by the Gauss' law for free charge, ∮ D·dA = q_free, so D = σf. Therefore

D = ε₀E + P

Dielectric behaviour in the capacitor

• With P = ε₀χₑE: D = ε₀(1 + χₑ)E = ε₀ε_r E = εE, where ε_r = 1 + χₑ is the dielectric constant.
• The field falls to E = E₀/ε_r (E₀ is the field without the dielectric) because the bound charges oppose the free charges.
• At the same charge, the potential difference falls by ε_r, so the capacitance increases to C = ε_r C₀.