### Polarization and the electric displacement

#### Problem:

If one presumes that there exists a true charge density ρtrue, a polarization or bound charge density ρbound, and a total charge density ρtotal, such that ρtrue + ρbound = ρtotal, write the source equations for D, E, and P.  Explain the meaning of these equations.  Briefly address the question:  Which of the fields D or E might be considered the more fundamental field?  Why?  Write the equation(s) describing the relationships between the three field quantities.

Solution:

• Concepts:
Maxwell's equations for electrostatics
• Reasoning:
Maxwell's equations connect the source equations for E and P and yield the source equations for D.
• Details of the calculation:
Maxwell's equations for electrostatics are
∇∙E = ρtotal0×E = 0, (SI units).
∇∙E = ρtotal0 is the source equation for E.
Let ρtrue + ρbound = ρtotal, then ∇∙E = (ρtrue + ρbound)/ε0 = ρtrue0 - ∇∙P0.
∇∙P = -ρbound is the source equation for P.
∇∙0E + P) = ρtrueD = ε0E + P, ∇∙D = ρtrue is the source equation for D.
Space derivatives of E, P, and D must exist, otherwise use the integral form.
For example:   ∫V ∇·D dV = ∮A D·n dA = ∫V ρtrue dV = Qfree inside.

For lih materials:
P = ε0χeE, D = ε0(1 + χe)E =  ε0κeE = εE.
Equations of macroscopic electrostatics:  ∇∙D = ρtrue,  ×E = 0.

D should not be regarded as a fundamental field of the same status as E.  It is rather a mathematical construct related to the way in which we seek a macroscopic solution for E from the basic equations.  D is a useful construct if the material has special properties, for example if it is a lih dielectric.  Then D = εE.  Unfavorable external conditions, however, can destroy those special properties.

#### Problem:

A thin electrically insulating sheet of material has thickness d and lateral extent L × L, where d << L.  It is in a vacuum and isolated from external electric fields.  It has frozen-in polarization per unit volume P oriented in the (x,z) plane at an angle θ to the normal to the surfaces, which is in the z-direction.
Find the magnitudes and directions of the electric displacement D and the electric field E both inside and outside the material, stating clearly your reasoning.

Solution:

• Concepts:
Polarization and bound charge densities
• Reasoning:
ρtrue + ρbound = ρtotal,  ∇∙P = -ρboundP·n = σbound,   D = ε0E + P
This is not a linear, isotropic material.
• Details of the calculation:
The bound charge density on upper face is σbound = P·n = P cosθ
The bound charge density on lower face has the opposite sign.
From Gauss's law we find the electric field inside the material.
E = −(σ/ε0)k = -(P/ε0)cosθ k.
The electric displacement inside the material is
D = ε0 E + P = −P cosθ k +  P(sinθ i + cosθ k) = P sinθ i
Outside, E = P = 0 so D = 0.
Note Dinside is parallel to the boundary, ∇∙D = 0, Gauss' law for D holds, but the problem lacks the symmetry to determine D from Gauss' law alone.

#### Problem:

The space between the plates of a parallel-plate capacitor is filled with two slabs of linear dielectric material.  Each slab has thickness s, so the total distance between the plates is 2s.  Slab 1 has a dielectric constant of 2, and slab 2 has a dielectric constant of 1.5.  The free charge density on the top plate is σ and on the bottom plate −σ.  (The top plate is touching slab 1.)
(a)  Find the electric displacement D in each slab.
(b)  Find the electric field E in each slab.
(c)  Find the polarization P in each slab.
(d)  Find the potential difference between the plates.
(e)  Find the location and amount of all the bound charge.
(f)  Now that you know all the charge (free and bound), recalculate the field in each slab, and compare with your answer to (b).

Solution:

• Concepts:
Gauss' law for D,  E = D/ε,  D = ε0E + P,  ρtrue + ρbound = ρtotal,  ∇∙P = -ρboundP·n = σbound
• Reasoning:
Gauss' law can be used to find D.  Given D, we find E, P, V, and the charge densities.
• Details of the calculation:
(a)  Let x-axis point downward and let the upper plate be at x = 0.
Gauss' law for DD = σ i inside both slabs.
(b)  E1 = D1 = σ/(2ε0) iE2 = D2 = 2σ/(3ε0) i.
(c)  P = D - ε0EP1 = σ/2ε i,  P2 = σ/3 i,
(d)  ΔV = V(0) - V(2s) = E1s + E2s = 7σs/(6ε0).
(e)  ρbound = 0 in each slab, the volume charge density is zero in each slab.
Surface charge σbound on surface of dielectric 1 at x = 0:
E1 = σtotal0 = (σ + σbound)/ε0 = σ/(2ε0).  σbound =  -σ/2.
Surface charge σbound on surface of dielectric 1 at x = s:  σbound = σ/2.
Surface charge σbound on surface of dielectric 2 at x = 2s:
E2 = -σtotal0 = (-σ + σbound)/ε0 = -2σ/(3ε0).  σbound =  σ/3.
Surface charge σbound on surface of dielectric 1 at x = s:  σbound = -σ/3.
Total surface charge at the boundary at x = s:  σbound = σ/6.
(f)  Gauss' law for E:
E
1 = σtotal0 i = (σ - σ/2)/ε0 i = σ/(2ε0) iE2 = (-σ + σ/3)/ε0 (-i) = 2σ/(3ε0) i.
The results agree with what we found in (b).