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.