If one presumes that there exists a true charge density ρ_{true},
a polarization or bound charge density ρ_{bound},
and a total charge density ρ_{total}, such
that ρ_{true }+ ρ_{bound
}= ρ_{total}, 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**= ρ_{total}/ε_{0},**∇**×**E**= 0, (SI units).

**∇∙E**= ρ_{total}/ε_{0}is the source equation for**E**.

Let ρ_{true }+ ρ_{bound }= ρ_{total}, then**∇∙E**= (ρ_{true }+ ρ_{bound})/ε_{0}= ρ_{true}/ε_{0}-**∇∙P**/ε_{0}.

**∇∙P**= -ρ_{bound }is the source equation for**P**.

**∇∙**(ε_{0}**E**+**P**) = ρ_{true}.**D**= ε_{0}**E**+**P**,**∇∙D**= ρ_{true}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 }ρ_{true}dV = Q_{free inside}.

For lih materials:**P**= ε_{0}χ_{e}**E**,**D**= ε_{0}(1 + χ_{e})**E**= ε_{0}κ_{e}**E**= ε**E**.

Equations of macroscopic electrostatics:**∇∙D**= ρ_{true, }**∇**×**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**= ε**E**. Unfavorable external conditions, however, can destroy those special properties.

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:

ρ_{true }+ ρ_{bound }= ρ_{total},**∇∙P**= -ρ_{bound},**P**·**n**= σ_{bound},**D**= ε_{0}**E**+**P**This is not a linear, isotropic material.

- Details of the calculation:

The bound charge density on upper face is σ_{bound}=**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/ε_{0})**k**= -(P/ε_{0})cosθ**k.**The electric displacement inside the material is

**D**= ε_{0}**E**+**P**= −P cosθ**k**+ P(sinθ**i**+ cosθ**k**) = P sinθ**i**Outside,

**E**=**P**= 0 so**D**= 0.

Note**D**_{inside}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.

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**= ε_{0}**E**+**P**, ρ_{true }+ ρ_{bound }= ρ_{total},**∇∙P**= -ρ_{bound},**P**·**n**= σ_{bound} - 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**D**:**D**= σ**i**inside both slabs.(b)

**E**_{1}=**D**/ε_{1}= σ/(2ε_{0})**i**,**E**_{2}=**D**/ε_{2}= 2σ/(3ε_{0})**i**.

(c)**P**=**D**- ε_{0}**E**.**P**_{1}= σ/2ε**i**,**P**_{2}= σ/3**i**,

(d) ΔV = V(0) - V(2s) = E_{1}s + E_{2}s = 7σs/(6ε_{0}).

(e) ρ_{bound}= 0 in each slab, the volume charge density is zero in each slab.

Surface charge σ_{bound}on surface of dielectric 1 at x = 0:

E_{1}= σ_{total}/ε_{0}= (σ + σ_{bound})/ε_{0}= σ/(2ε_{0}). σ_{bound}= -σ/2.

Surface charge σ_{bound}on surface of dielectric 1 at x = s: σ_{bound}= σ/2.

Surface charge σ_{bound}on surface of dielectric 2 at x = 2s:

E_{2}= -σ_{total}/ε_{0}= (-σ + σ_{bound})/ε_{0}= -2σ/(3ε_{0}). σ_{bound}= σ/3.

Surface charge σ_{bound}on surface of dielectric 1 at x = s: σ_{bound}= -σ/3.

Total surface charge at the boundary at x = s: σ_{bound}= σ/6.

(f) Gauss' law for**E**:

E_{1}= σ_{total}/ε_{0}**i**= (σ - σ/2)/ε_{0}**i**= σ/(2ε_{0})**i**.**E**_{2}= (-σ + σ/3)/ε_{0}(-**i**) = 2σ/(3ε_{0})**i**.

The results agree with what we found in (b).