3.4.5 The fundamental equations and the potential in material media

Let us collect the two fundamental equations of electrostatics, now valid in any medium, in both their differential and their integral forms. The integral forms follow from the differential ones by Stokes's theorem and by the divergence theorem (Section 2.4): integrate $\nabla \times E = 0$ over any open surface bounded by a contour $C$ , and $\nabla \cdot D = ρ_{v}$ over any volume $v$ bounded by a closed surface $S$ .

The two fundamental equations

	differential form	integral form
First fundamental equation	$\nabla \times E = 0$	$\oint_{C} E \cdot d l = 0$
Second fundamental equation	$\nabla \cdot D = ρ_{v}$	$\oint_{S} D \cdot d S = \int_{v} ρ_{v} d v$

The integral form of the second equation is Gauss's law in a medium: the outward flux of $D$ through any closed surface equals the free charge enclosed. In problems with symmetry it gives $D$ directly, without our having to know anything about the bound charges, and $E = D / ε$ follows afterwards. Exercises 3.4.2 and 3.4.3 use exactly this route.

The first equation tells us that the electrostatic field in a medium is still conservative. Everything we learnt about the potential in Section 3.3 therefore carries over unchanged:

E = - \nabla V, V_{B} - V_{A} = - \int_{A}^{B} E \cdot d l, (3.4.12)

the integral being independent of the path. Substituting $E = - \nabla V$ into the second fundamental equation with $D = ε E = - ε \nabla V$ gives Poisson's equation in a medium,

\nabla \cdot (ε \nabla V) = - ρ_{v} . (3.4.13)

In a simple medium $ε$ is a constant and can be taken out of the divergence:

Poisson's and Laplace's equations in a simple medium

\nabla^{2} V = - \frac{ρ _{v}}{ε} and, where ρ_{v} = 0, \nabla^{2} V = 0. (3.4.14)

Laplace's equation is the same as in vacuum. Hence all the solutions of Laplace's equation found in Section 3.3 (Problems 3.3.1 to 3.3.3) remain solutions in a homogeneous dielectric. What changes is only the relation between the potential and the free charge on the conductors, in which $ε_{0}$ is replaced by $ε$ . Problem 3.4.1 is the first example. In a nonhomogeneous medium, however, Eq. (3.4.13) must be used, and where two different media meet the solution must be patched together with the boundary conditions that we derive next.

Common mistake ·

ε

outside the derivative or the integral in an inhomogeneous medium

If $ε$ varies in space, $\nabla \cdot (ε \nabla V)$ is not $ε \nabla^{2} V$ : the product rule gives $\nabla \cdot (ε \nabla V) = ε \nabla^{2} V + \nabla ε \cdot \nabla V$ . In the same way $\int E \cdot d l$ with $E = D / ε (x)$ , or the energy $\frac{1}{2} \int ε E^{2} d v$ , must keep $ε$ inside the integral (Exercise 3.4.4). Only in a simple medium may $ε$ be pulled out.

Quick check

A parallel-plate capacitor connected to a fixed voltage $V$ is filled with oil ( $ε_{r} = 2.2$ ). Does the potential $V (x)$ between the plates change? Does the charge on the plates change?

$V (x)$ does not change: it satisfies Laplace's equation, which does not contain $ε$ , with the same plate potentials. The charge does change: $ρ_{S} = D_{n} = ε E_{n}$ grows by the factor $2.2$ (Problem 3.4.1).