Electrostatics

Contents

The electrostatic field
Methods for obtaining the potential
Electrostatic potential energy
Charge distributions in external fields
Electrostatics in macroscopic media

Electrostatics is the theory of the fields produced by charges at rest. The chapter follows the order of the source: the Coulomb force is postulated, the electrostatic field and its potential are constructed by continuous superposition, the local field equations — Gauss's law and the vanishing of the curl — are derived, and the boundary-value methods for obtaining the potential are developed. The Coulomb constant is kept throughout as $k_E$, so that every formula remains valid in any system of units; the worked examples, ported from the companion problem collection, are written in SI units, where $k_E = 1/(4\pi\epsilon_0)$.

The electrostatic field

Fundamentals

Point charge

Definition 58.1 (Point charge).

A point charge is a point of space to which one assigns a constant real scalar quantity $q$ called its charge.

The Coulomb force

We state as a fundamental principle:

Postulate 58.2 (Coulomb's law).

The force that a charge $q'$ located at $\vect{x}'$ exerts on a charge $q$ located at $\vect{x}$ is proportional to the product of the charges and inversely proportional to the square of the distance separating them. That is,

\begin{equation}\tag{58.1} \vect{F} = k_E\,\frac{qq'}{\abs{\vect{x}-\vect{x}'}^{2}}\, \frac{\vect{x}-\vect{x}'}{\abs{\vect{x}-\vect{x}'}}\ec \end{equation}

where $k_E$ is a positive real constant.

Remark 58.3 (Unit systems).

The constant $k_E$ selects the system of units: in Gaussian units $k_E = 1$, while in SI units $k_E = 1/(4\pi\epsilon_0)$, with $\epsilon_0$ the vacuum permittivity [Jackson:1999] [BIPM:2019]. The worked examples of this chapter are written in SI units.

Continuous charge distributions

Definition 58.4 (Continuous charge distribution).

Consider a set of infinitesimal charges $\dd q'$ located at points $\vect{x}'$ (with respect to a fixed coordinate origin) which together form a continuum occupying a volume $V'$. We call this continuum a continuous charge distribution.

Consider further a charge $q$ at $\vect{x}$. The force felt by $q$ due to a continuous charge distribution is the continuous sum of the forces that the individual charges $\dd q'$ exert on it. By Coulomb's law,

\begin{equation}\tag{58.2} \vect{F} = k_E\,q\int_{V'}\dd q'\, \frac{\vect{x}-\vect{x}'}{\abs{\vect{x}-\vect{x}'}^{3}}\ep \end{equation}

Charge densities

Definition 58.5 (Charge densities).

The volume charge density of a distribution is

\begin{equation}\tag{58.3} \rho(\vect{x}') = \frac{\dd q'}{\dd V'}\ep \end{equation}

Equivalently one defines the surface charge density

\begin{equation}\tag{58.4} \sigma(\vect{x}') = \frac{\dd q'}{\dd s'} \end{equation}

on a surface $S'$, and the line charge density

\begin{equation}\tag{58.5} \lambda(\vect{x}') = \frac{\dd q'}{\dd l'} \end{equation}

on a curve $C'$.

Substituting Equation (58.3) into Equation (58.2), the force felt by a charge $q$ due to a continuous distribution of density $\rho$ is

\begin{equation}\tag{58.6} \vect{F} = k_E\,q\int_{V'}\rho(\vect{x}')\, \frac{\vect{x}-\vect{x}'}{\abs{\vect{x}-\vect{x}'}^{3}}\,\dd V'\ec \end{equation}

and analogously for the surface and line densities,

\begin{equation}\tag{58.7} \vect{F} = k_E\,q\int_{S'}\sigma(\vect{x}')\, \frac{\vect{x}-\vect{x}'}{\abs{\vect{x}-\vect{x}'}^{3}}\,\dd s'\ec \qquad \vect{F} = k_E\,q\int_{C'}\lambda(\vect{x}')\, \frac{\vect{x}-\vect{x}'}{\abs{\vect{x}-\vect{x}'}^{3}}\,\dd l'\ep \end{equation}

Comparing Equation (58.6) with Equation (58.1), we see that a point charge $q$ located at $\vect{x}'$ is the particular distribution whose density is

\begin{equation}\tag{58.8} \rho(\vect{x}) = q\,\delta(\vect{x}-\vect{x}')\ep \end{equation}

The electrostatic field

Definition 58.6 (Electrostatic field).

The electrostatic field due to a continuous charge distribution is the vector field obtained from the force exerted on a test charge $q$ at $\vect{x}$ in the limit in which $q$ tends to zero — so that the field generated by the test charge itself does not interfere with the field generated by the distribution:

\begin{equation}\tag{58.9} \vect{E}(\vect{x}) = \lim_{q\to 0}\frac{\vect{F}}{q}\ep \end{equation}

From this definition and Equation (58.6), the electrostatic field due to a continuous charge distribution is

\begin{equation}\tag{58.10} \vect{E}(\vect{x}) = k_E\int_{V'}\rho(\vect{x}')\, \frac{\vect{x}-\vect{x}'}{\abs{\vect{x}-\vect{x}'}^{3}}\,\dd V'\ep \end{equation}

Conversely, consider a charge $q$ at $\vect{x}$ subject to an external electrostatic field $\vect{E}$ produced by some distribution of density $\rho$: the field is given by Equation (58.10) and the force it exerts on $q$ by Equation (58.6). Comparing the two expressions, the force felt by a charge in an external electrostatic field is

\begin{equation}\tag{58.11} \vect{F} = q\,\vect{E}\ep \end{equation}

The electrostatic potential

Lemma 58.7.

For $\vect{x}\neq\vect{x}'$,

\begin{equation}\tag{58.12} \pp_i\left(\frac{1}{\abs{\vect{x}-\vect{x}'}}\right) = -\frac{x_i-x'_i}{\abs{\vect{x}-\vect{x}'}^{3}}\ep \end{equation}

Derivation pending.

Gradient identity Equation (58.12): direct differentiation of $\abs{\vect{x}-\vect{x}'}^{-1}$ (asserted without proof in the source).

From Equation (58.10) we have, in components,

\begin{align} E_i(\vect{x}) &= k_E\int_{V'}\rho(\vect{x}')\, \frac{x_i-x'_i}{\abs{\vect{x}-\vect{x}'}^{3}}\,\dd V' \nonumber\\ &= -k_E\int_{V'}\rho(\vect{x}')\, \pp_i\left(\frac{1}{\abs{\vect{x}-\vect{x}'}}\right)\dd V' \tag{58.13}\\ &= -\pp_i\left(k_E\int_{V'} \frac{\rho(\vect{x}')}{\abs{\vect{x}-\vect{x}'}}\,\dd V'\right)\ec \tag{58.14} \end{align}

where Equation (58.12) was used in the second step and the derivative, which acts on $\vect{x}$ only, was taken outside the integral in the third.

Definition 58.8 (Electrostatic potential).

The electrostatic potential of a continuous charge distribution is

\begin{equation}\tag{58.15} \phi(\vect{x}) = k_E\int_{V'}\frac{\rho(\vect{x}')}{\abs{\vect{x}-\vect{x}'}}\, \dd V'\ep \end{equation}

With this definition, Equation (58.14) reads

\begin{equation}\tag{58.16} \vect{E}(\vect{x}) = -\vect{\nabla}\phi(\vect{x})\ep \end{equation}

Gauss's law

Lemma 58.9.

In the sense of distributions,

\begin{equation}\tag{58.17} \pp_i\pp_i\left(\frac{1}{\abs{\vect{x}-\vect{x}'}}\right) = -4\pi\,\delta(\vect{x}-\vect{x}')\ep \end{equation}

Derivation pending.

Distributional Laplacian of $\abs{\vect{x}-\vect{x}'}^{-1}$, Equation (58.17) (asserted without proof in the source).

Theorem 58.10 (Gauss's law).

The divergence of the electrostatic field due to a distribution of charge density $\rho$ is

\begin{equation}\tag{58.18} \vect{\nabla}\cdot\vect{E}(\vect{x}) = 4\pi k_E\,\rho(\vect{x})\ep \end{equation}

Derivation. Applying $\pp_i$ to Equation (58.13),

\begin{align*} \pp_iE_i(\vect{x}) &= -k_E\int_{V'}\rho(\vect{x}')\, \pp_i\pp_i\left(\frac{1}{\abs{\vect{x}-\vect{x}'}}\right)\dd V'\\ &= 4\pi k_E\int_{V'}\rho(\vect{x}')\, \delta(\vect{x}-\vect{x}')\,\dd V'\\ &= 4\pi k_E\,\rho(\vect{x})\ec \end{align*}

where Equation (58.17) was used in the second step.

∎

Integrating Equation (58.18) over an arbitrary volume $V$ enclosing a total charge $q_V$, and using that by Equation (58.3) the volume integral of the density is the enclosed charge,

\begin{align*} \int_{V}\pp_iE_i\,\dd V &= 4\pi k_E\int_{V}\rho\,\dd V = 4\pi k_E\,q_V\ec \end{align*}

and converting the left-hand side into a surface integral by the divergence theorem, we obtain the flux of the electrostatic field through a closed surface:

\begin{equation}\tag{58.19} \oint_{\pp V}\vect{E}\cdot\dd\vect{s} = 4\pi k_E\,q_V\ec \end{equation}

a result known as Gauss's law in integral form.

Curl of the electrostatic field

Corollary 58.11.

The electrostatic field satisfies

\begin{equation}\tag{58.20} \vect{\nabla}\times\vect{E}(\vect{x}) = \vect{0}\ep \end{equation}

Proof.

By Equation (58.14) there exists a scalar function such that the electrostatic field can be written as the gradient Equation (58.16); that is, the electrostatic field is a conservative vector field. Since the curl of a gradient vanishes identically, Equation (58.20) follows.

∎

Geometry of the field

Electrostatic field lines

Definition 58.12 (Field lines).

The electrostatic field lines are the curves $\vect{x}=\vect{x}(\lambda)$ to which the electrostatic field is tangent at every point:

\begin{equation*} \dv{\vect{x}}{\lambda}(\lambda) = \alpha\,\vect{E}\bigl(\vect{x}(\lambda)\bigr)\ep \end{equation*}

The constant $\alpha$ can be removed by the reparametrization $\lambda'=\alpha\lambda$, since

\begin{align*} \dv{\vect{x}}{\lambda}(\lambda) &= \alpha\,\vect{E}\bigl(\vect{x}(\lambda)\bigr)\ec\\ \dv{\vect{x}}{\lambda'}\,\dv{\lambda'}{\lambda} &= \alpha\,\vect{E}\bigl(\vect{x}(\lambda')\bigr)\ec\\ \dv{\vect{x}}{\lambda'}\,\alpha &= \alpha\,\vect{E}\bigl(\vect{x}(\lambda')\bigr)\ec\\ \dv{\vect{x}}{\lambda'} &= \vect{E}\bigl(\vect{x}(\lambda')\bigr)\ec \end{align*}

or, renaming the parameter,

\begin{equation}\tag{58.21} \dv{\vect{x}}{\lambda}(\lambda) = \vect{E}\bigl(\vect{x}(\lambda)\bigr)\ec \end{equation}

which is called the field-line equation.

Closed line integrals of the electrostatic field

Proposition 58.13.

The line integral of the electrostatic field around any closed curve vanishes:

\begin{equation}\tag{58.22} \oint_{\pp S}\vect{E}\cdot\dd\vect{l} = 0\ep \end{equation}

Derivation. By Stokes's theorem, for an arbitrary compact surface $S$ with closed boundary $\pp S$,

\begin{equation*} \oint_{\pp S}\vect{E}\cdot\dd\vect{l} = \int_{S}\bigl(\vect{\nabla}\times\vect{E}\bigr)\cdot\dd\vect{s}\ec \end{equation*}

which vanishes by Equation (58.20).

∎

Proposition 58.14.

There are no closed electrostatic field lines.

Derivation. Suppose there were a closed field line $C$ with parameter $\lambda$. Then, using the field-line equation Equation (58.21),

\begin{align*} \oint_{C}\vect{E}\cdot\dd\vect{x} &= \oint_{C}\vect{E}\bigl(\vect{x}(\lambda)\bigr)\cdot \dv{\vect{x}}{\lambda}\,\dd\lambda\\ &= \oint_{C}\dv{\vect{x}}{\lambda}\cdot \dv{\vect{x}}{\lambda}\,\dd\lambda\\ &= \oint_{C}\abs{\dv{\vect{x}}{\lambda}}^{2}\dd\lambda \neq 0\ec \end{align*}

since along an actual curve

\begin{equation*} \abs{\dv{\vect{x}}{\lambda}}^{2} > 0\ep \end{equation*}

This contradicts Proposition 58.13, by which every closed line integral of the electrostatic field vanishes. Hence no closed field line exists.

∎

Equipotential surfaces

Definition 58.15 (Equipotential surface).

Consider a charge distribution generating an electrostatic potential $\phi$. An equipotential surface is a surface given by the equation

\begin{equation}\tag{58.23} \phi(\vect{x}) = C \end{equation}

with $C$ constant.

Methods for obtaining the potential

Boundary conditions

Line integral of the electrostatic field

Consider two points $P_i$ and $P_f$ joined by a curve $C$ described by $\vect{x}$. From Equation (58.16), integrating along $C$,

\begin{align*} \int_{C}\vect{E}\cdot\dd\vect{x} &= -\int_{C}\vect{\nabla}\phi\cdot\dd\vect{x} = -\int_{C}\dd\phi\ec \end{align*}

so that

\begin{equation*} \phi(P_f)-\phi(P_i) = -\int_{C}\vect{E}\cdot\dd\vect{x}\ep \end{equation*}

Taking $P_i$ to be the point of coordinates $\vect{x}$ and $P_f$ a point located at $\vect{x}_0$, we obtain the following expression for the potential:

\begin{equation}\tag{58.24} \phi(\vect{x}) = \phi(\vect{x}_0) + \int_{C}\vect{E}\cdot\dd\vect{l}\ep \end{equation}

Continuity condition; interface conditions; ideal conductors

[Headings reserved in the source outline.] The worked examples below make use of the standard boundary conditions of electrostatics [Jackson:1999]: the potential is continuous across any interface; the tangential component of $\vect{E}$ is continuous across an interface; the jump of the normal component of the field across a charged surface is fixed by the surface density, so that on the surface of an ideal conductor (in SI units)

\begin{equation}\tag{58.25} \sigma = \epsilon_0\,\vect{E}\cdot\hat{\vect{n}}\big|_{\pp V}\ec \end{equation}

with $\hat{\vect{n}}$ the outward normal of the conductor; the potential is constant throughout an ideal conductor, and vanishes on a grounded one.

Derivation pending.

Continuity and interface conditions of the electrostatic field and the ideal-conductor boundary conditions, including Equation (58.25): the source reserves these headings without content, while its worked problems use the results.

The method of images

[Heading reserved in the source outline.] The method replaces a boundary-value problem in the presence of conductors by an equivalent charge configuration — the original sources plus fictitious image charges placed outside the physical region — chosen so that the boundary conditions of Section 58.2.1 are satisfied. The source develops the method only through the following worked problems, both of which rely on the ideal-dipole potential recorded in Section 58.4.4.

Example 58.16 (Ideal dipole inclined at $$45^\circ$$ before a grounded conducting plane).

An ideal dipole of dipole moment $\vect{p}$ is located at a distance $d$ from an infinite conducting plane connected to ground, and forms an angle of $45^\circ$ with the normal to the plane. Compute the electrostatic potential at every point and the surface charge density induced on the plane.

The potential of an ideal dipole located at $\vect{x}'$ is given by

\begin{equation*} \phi(\vect{x}) = \frac{1}{4\pi\epsilon_0}\, \frac{\vect{x}-\vect{x}'}{\abs{\vect{x}-\vect{x}'}^{3}} \cdot\vect{p}\ep \end{equation*}

Using the method of images, we place an image dipole $\vect{p}_I$ on the other side of the plane, at a distance $d_I$; by the symmetry of the problem we may take $\abs{\vect{p}_I}=\abs{\vect{p}}$. The potential at every point is then

\begin{equation*} \phi(\vect{x}) = \frac{1}{4\pi\epsilon_0}\left( \frac{\vect{x}-\vect{x}'}{\abs{\vect{x}-\vect{x}'}^{3}} \cdot\vect{p} + \frac{\vect{x}-\vect{x}_I}{\abs{\vect{x}-\vect{x}_I}^{3}} \cdot\vect{p}_I\right)\ep \end{equation*}

Placing the origin at the point of the plane closest to the dipole, with $\hat{\vect{x}}$ along the normal,

\begin{align*} \vect{x} &= x\,\hat{\vect{x}}+y\,\hat{\vect{y}}+z\,\hat{\vect{z}}\ec\\ \vect{x}' &= d\,\hat{\vect{x}}\ec\qquad \vect{x}_I = d_I\,\hat{\vect{x}}\ec\\ \vect{p} &= p\left(\cos45^\circ\,\hat{\vect{x}} +\sin45^\circ\,\hat{\vect{y}}\right) = \frac{p}{\sqrt{2}}\left(\hat{\vect{x}}+\hat{\vect{y}}\right)\ec\\ \vect{p}_I &= p\left(\cos45^\circ\,\hat{\vect{x}} -\sin45^\circ\,\hat{\vect{y}}\right) = \frac{p}{\sqrt{2}}\left(\hat{\vect{x}}-\hat{\vect{y}}\right)\ep \end{align*}

We must now impose the boundary conditions of the problem to determine the distance $d_I$. Since the conducting plane is grounded, the potential must vanish on it; evaluating at the origin,

\begin{align*} \left.\frac{1}{4\pi\epsilon_0}\frac{p}{\sqrt{2}}\left( \frac{x-d+y}{\left((x-d)^2+y^2+z^2\right)^{3/2}} +\frac{x-d_I-y}{\left((x-d_I)^2+y^2+z^2\right)^{3/2}} \right)\right|_{\vect{x}=\vect{0}} &= 0\ec\\ \frac{-d}{(d^2)^{3/2}}+\frac{-d_I}{(d_I^2)^{3/2}} &= 0\ec \end{align*}

so that

\begin{equation*} d_I = -d\ec\qquad \vect{x}_I = -d\,\hat{\vect{x}}\ep \end{equation*}

Therefore

\begin{equation}\tag{58.26} \phi(\vect{x}) = \frac{p}{4\sqrt{2}\,\pi\epsilon_0}\left( \frac{x-d+y}{\left((x-d)^2+y^2+z^2\right)^{3/2}} +\frac{x+d-y}{\left((x+d)^2+y^2+z^2\right)^{3/2}}\right)\ec \end{equation}

and with these values the two numerators cancel on the whole plane $x=0$, so the boundary condition is satisfied everywhere on the conductor.

With the potential in hand we can compute the electric field. In particular,

\begin{align*} E_x &= -\pdv{\phi}{x}\\ &= -\frac{p}{4\sqrt{2}\,\pi\epsilon_0}\Bigl( \left((x-d)^2+y^2+z^2\right)^{-3/2} -3(x-d)(x-d+y)\left((x-d)^2+y^2+z^2\right)^{-5/2}\Bigr.\\ &\qquad\Bigl. +\left((x+d)^2+y^2+z^2\right)^{-3/2} -3(x+d)(x+d-y)\left((x+d)^2+y^2+z^2\right)^{-5/2}\Bigr)\ep \end{align*}

Finally, by the interface condition Equation (58.25) with $\hat{\vect{n}}=\hat{\vect{x}}$, the surface charge density induced on the plane is

\begin{align*} \sigma(y,z) &= \left.\epsilon_0\,\vect{E}\cdot\hat{\vect{n}}\right|_{\pp V} = \epsilon_0\,E_x(0,y,z)\\ &= -\frac{p}{4\sqrt{2}\,\pi}\Bigl( \left(d^2+y^2+z^2\right)^{-3/2} +3d\,(y-d)\left(d^2+y^2+z^2\right)^{-5/2}\Bigr.\\ &\qquad\Bigl. +\left(d^2+y^2+z^2\right)^{-3/2} -3d\,(d-y)\left(d^2+y^2+z^2\right)^{-5/2}\Bigr)\\ &= \frac{p}{2\sqrt{2}\,\pi}\left( 3d\,(d-y)\left(d^2+y^2+z^2\right)^{-5/2} -\left(d^2+y^2+z^2\right)^{-3/2}\right)\ep \end{align*}

Example 58.17 (Ideal dipole normal to a grounded conducting plane).

An ideal dipole of dipole moment $\vect{p}$ is oriented along the normal at a distance $d$ from an infinite conducting plane connected to ground. Compute the potential at every point and the force exerted on the dipole.

The image construction is the same as in Example 58.16: an image dipole $\vect{p}_I$ with $\abs{\vect{p}_I}=\abs{\vect{p}}$ is placed at $\vect{x}_I=d_I\,\hat{\vect{x}}$, and now

\begin{equation*} \vect{x}' = d\,\hat{\vect{x}}\ec\qquad \vect{p} = p\,\hat{\vect{x}}\ec\qquad \vect{p}_I = p\,\hat{\vect{x}}\ep \end{equation*}

Imposing that the potential vanish on the grounded plane — evaluating at the origin as before —

\begin{equation*} \frac{-d}{(d^2)^{3/2}}+\frac{-d_I}{(d_I^2)^{3/2}} = 0 \quad\implies\quad d_I = -d\ec\qquad \vect{x}_I = -d\,\hat{\vect{x}}\ep \end{equation*}

Therefore

\begin{equation}\tag{58.27} \phi(\vect{x}) = \frac{p}{4\pi\epsilon_0}\left( \frac{x-d}{\left((x-d)^2+y^2+z^2\right)^{3/2}} +\frac{x+d}{\left((x+d)^2+y^2+z^2\right)^{3/2}}\right)\ec \end{equation}

which vanishes on the plane $x=0$, as required.

The force felt by an ideal dipole $\vect{p}$ in an external field $\vect{E}$ is given by

\begin{equation*} \vect{F} = \vect{\nabla}\left(\vect{p}\cdot\vect{E}\right) \end{equation*}

(cf. Section 58.4.4). In this case the external field is the one generated by the plane — equivalently, by the image dipole,

\begin{equation*} \vect{E}(\vect{x}) = \frac{1}{4\pi\epsilon_0}\, \frac{3(\vect{p}_I\cdot\hat{\vect{r}})\hat{\vect{r}}-\vect{p}_I} {r^{3}}\ec \end{equation*}

with $r$ measured from the image dipole. The dipole sits at distance $r=2d$ from its image, on the axis $\theta=0$ (spherical coordinates centred on the image, polar axis $\hat{\vect{x}}$). Thus, with $\vect{p}_I=\vect{p}$,

\begin{align*} \vect{F} &= \left.\vect{\nabla}\left(\vect{p}\cdot \frac{1}{4\pi\epsilon_0}\, \frac{3(\vect{p}\cdot\hat{\vect{r}})\hat{\vect{r}}-\vect{p}} {r^{3}}\right)\right|_{r=2d,\;\theta=0}\\ &= \left.\frac{1}{4\pi\epsilon_0}\,\vect{\nabla}\left( \frac{3(\vect{p}\cdot\hat{\vect{r}})^{2}-p^{2}}{r^{3}} \right)\right|_{r=2d,\;\theta=0}\\ &= \left.\frac{p^{2}}{4\pi\epsilon_0}\,\vect{\nabla}\left( 3\,\frac{\cos^{2}\theta}{r^{3}}-\frac{1}{r^{3}} \right)\right|_{r=2d,\;\theta=0}\\ &= \frac{p^{2}}{4\pi\epsilon_0}\left( 3\left(-3\,\frac{\cos^{2}\theta}{r^{4}}\right) +\frac{3}{r^{4}}\right)\bigg|_{r=2d,\;\theta=0}\, \hat{\vect{r}}\\ &= -\frac{1}{4\pi\epsilon_0}\,\frac{6p^{2}}{(2d)^{4}}\, \hat{\vect{x}} = -\frac{3p^{2}}{32\pi\epsilon_0\,d^{4}}\,\hat{\vect{x}}\ec \end{align*}

where the angular part of the gradient drops out because $\dd(\cos^{2}\theta)/\dd\theta=-2\cos\theta\sin\theta$ vanishes at $\theta=0$. The force is attractive: the dipole is pulled toward the plane.

The Poisson equation

Poisson equation

From Equation (58.16), taking the divergence of both sides and using Gauss's law Equation (58.18),

\begin{align*} \vect{E} &= -\vect{\nabla}\phi\ec &&\text{apply } \vect{\nabla}\cdot(\ )\ec\\ 4\pi k_E\,\rho &= -\nabla^{2}\phi\ec \end{align*}

so that

\begin{equation}\tag{58.28} \nabla^{2}\phi(\vect{x}) = -4\pi k_E\,\rho(\vect{x})\ec \end{equation}

a result known as the Poisson equation.

Laplace equation

In a region where the charge density vanishes, Equation (58.28) reduces to

\begin{equation}\tag{58.29} \nabla^{2}\phi(\vect{x}) = 0\ec \end{equation}

known as the Laplace equation.

Green functions

[Headings reserved in the source outline: the source function, the general solution of the Poisson equation, and the Dirichlet and Neumann boundary conditions are pending port from a completed source; see [Jackson:1999] for the standard treatment.]

Multipole expansion

[Headings reserved in the source outline: the expansion of the potential in inverse powers of the distance, the multipole moments of order $n$, and the ideal-dipole limit are pending port from a completed source. The ideal-dipole formulas used by the worked examples are recorded, as pending statements, in Section 58.4.4.]

Electrostatic potential energy

[Heading reserved in the source outline; no content to port. The energy functional of a linear dielectric medium, used by the worked examples of Electrodynamics in Matter, is recorded there.]

Charge distributions in external fields

Potential energy of a charge distribution in an external field

[Heading reserved in the source outline; no content to port.]

Force on a charge distribution in an external field

[Heading reserved in the source outline; no content to port.]

Torque on a charge distribution in an external field

[Heading reserved in the source outline; no content to port.]

The ideal dipole

The worked examples of Section 58.2.2 employ the following standard results for an ideal (point) dipole of moment $\vect{p}$ [Jackson:1999] [Griffiths:2018]. The potential of an ideal dipole located at $\vect{x}'$ is

\begin{equation}\tag{58.30} \phi(\vect{x}) = k_E\, \frac{\left(\vect{x}-\vect{x}'\right)\cdot\vect{p}} {\abs{\vect{x}-\vect{x}'}^{3}}\ec \end{equation}

and the corresponding field, at position $r\hat{\vect{r}}$ relative to the dipole, is

\begin{equation}\tag{58.31} \vect{E}(\vect{x}) = k_E\, \frac{3\left(\vect{p}\cdot\hat{\vect{r}}\right)\hat{\vect{r}} -\vect{p}}{r^{3}}\ep \end{equation}

In an external electrostatic field $\vect{E}$, the force felt by an ideal dipole is

\begin{equation}\tag{58.32} \vect{F} = \vect{\nabla}\left(\vect{p}\cdot\vect{E}\right)\ep \end{equation}

Derivation pending.

Ideal-dipole potential Equation (58.30), field Equation (58.31) and force Equation (58.32): to be derived from the multipole expansion of Section 58.2.5 and from the energy of a distribution in an external field, both of which the source reserves without content.

Electrostatics in macroscopic media

[Heading reserved in the source outline.] The macroscopic model of electrostatics in material media — polarization, the displacement field, and the linear dielectric constitutive relation — together with the two dielectric worked problems of the companion collection, is ported to Electrodynamics in Matter.