Electrostatics: the Field, the Potential and the Multipole Expansion

Electrostatics is the theory of the electric field produced by charges at rest. Everything in it follows from one experimental law — Coulomb's inverse-square force between point charges — combined with the principle of superposition. This chapter rebuilds electrostatics as a field theory in the manner of Griffiths' Chapters 2 and 3: from Coulomb's law to Gauss's law in integral and differential form, to the scalar potential and the Poisson and Laplace equations, to the electrostatic energy, and finally to the multipole expansion, which tells you what any localised charge distribution looks like from far away.

The organising idea is that the electrostatic field carries two pieces of local information, and each is captured by a differential operator. The divergence of $\vec{E}$ measures the charge density: $\nabla\cdot\vec{E} = \rho/\varepsilon_0$ . The curl of $\vec{E}$ is zero: $\nabla\times\vec{E} = \vec{0}$ . The first is Gauss's law; the second is what makes the potential exist. Once you have $V$ with $\vec{E} = -\nabla V$ , the whole subject reduces to solving one second-order partial differential equation, $\nabla^2 V = -\rho/\varepsilon_0$ , subject to boundary conditions — which is the business of the next chapter.

How to read this chapter. Keep two results in front of you: the pair $\nabla\cdot\vec{E} = \rho/\varepsilon_0$ , $\nabla\times\vec{E} = \vec{0}$ (Sections 2–4), and the multipole potential of Section 8. Every later chapter quotes the first pair without comment. The multipole expansion is the single most reused calculation in the course — the dipole term reappears in dielectrics (Chapter 3), in radiation (Chapter 7 onwards) and in quantum mechanics. Derive it once, carefully, here.

1. Coulomb's Law and the Electric Field

The force on a point charge $q$ at $\vec{r}$ due to a point charge $q'$ at $\vec{r}\,'$ is

\vec{F} = \frac{1}{4\pi\varepsilon_0}\,\frac{q\,q'}{\eta^2}\,\hat{\eta}, \qquad \vec{\eta} \equiv \vec{r} - \vec{r}\,', \quad \eta = |\vec{\eta}|

with $\varepsilon_0 = 8.854\times10^{-12}\ \mathrm{C^2\,N^{-1}\,m^{-2}}$ the permittivity of free space. The force is along the line joining the charges, repulsive for like signs, and falls off as the inverse square of the separation. The electric field is the force per unit test charge,

\vec{E}(\vec{r}) = \frac{1}{4\pi\varepsilon_0}\sum_i \frac{q_i}{\eta_i^2}\,\hat{\eta}_i \;\xrightarrow{\text{continuum}}\; \frac{1}{4\pi\varepsilon_0}\int \frac{\rho(\vec{r}\,')}{\eta^2}\,\hat{\eta}\;d\tau'

The sum-to-integral step is the principle of superposition — fields add linearly — and it is an independent experimental fact, not a consequence of Coulomb's law alone. Superposition is what lets every subsequent problem be built out of point-charge solutions.

2. Gauss's Law in Integral Form

The flux of $\vec{E}$ through a closed surface $\mathcal{S}$ equals the enclosed charge over $\varepsilon_0$ :

\oiint_{\mathcal{S}} \vec{E}\cdot d\vec{a} = \frac{Q_{\text{enc}}}{\varepsilon_0}

The proof for a single point charge is geometric: the flux of $\hat{\eta}/\eta^2$ through any surface enclosing the charge is the solid angle it subtends, $4\pi$ , independent of the surface's shape or the charge's position inside it; a charge outside contributes zero because every field line that enters also leaves. Superposition then extends the result to any distribution. Gauss's law is exact and completely general, but it determines $\vec{E}$ only when symmetry (spherical, cylindrical, or planar) lets you pull $\vec{E}$ out of the integral.

Worked example — uniformly charged sphere. A ball of radius $R$ carries total charge $Q$ spread uniformly, $\rho = 3Q/4\pi R^3$ . A concentric Gaussian sphere of radius $r > R$ gives $E\,(4\pi r^2) = Q/\varepsilon_0$ , so $\vec{E} = \dfrac{Q}{4\pi\varepsilon_0 r^2}\,\hat{r}$ — the external field is that of a point charge at the centre. For $r < R$ the enclosed charge is $Q(r/R)^3$ , so $\vec{E} = \dfrac{Q\,r}{4\pi\varepsilon_0 R^3}\,\hat{r}$ : the interior field grows linearly from zero at the centre. The two expressions agree at $r = R$ , as they must because there is no surface charge.

3. The Divergence of E: Gauss's Law in Differential Form

Applying the divergence theorem to Gauss's law, $\displaystyle\int_{\mathcal{V}}(\nabla\cdot\vec{E})\,d\tau = \frac{1}{\varepsilon_0}\int_{\mathcal{V}}\rho\,d\tau$ for every volume, so the integrands are equal:

\boxed{\;\nabla\cdot\vec{E} = \frac{\rho}{\varepsilon_0}\;}

This is the first of Maxwell's equations. It can also be got directly from the Coulomb integral using $\nabla\cdot\!\left(\dfrac{\hat{\eta}}{\eta^2}\right) = 4\pi\,\delta^3(\vec{\eta})$ : the divergence of the field integral picks out $\rho(\vec{r})$ at the field point. The delta-function identity is the rigorous statement that the inverse-square field is divergence-free everywhere except at its source, where all of its flux originates.

4. The Curl of E and the Scalar Potential

For a single point charge at the origin, $\vec{E} = \dfrac{q}{4\pi\varepsilon_0}\dfrac{\hat{r}}{r^2} = -\nabla\!\left(\dfrac{q}{4\pi\varepsilon_0 r}\right)$ , and the curl of a gradient is zero. Superposition gives, for any static distribution,

\nabla\times\vec{E} = \vec{0}, \qquad \oint \vec{E}\cdot d\vec{\ell} = 0

A curl-free field on a simply connected region is the gradient of a scalar. Define the electric potential

V(\vec{r}) = -\int_{\mathcal{O}}^{\vec{r}} \vec{E}\cdot d\vec{\ell}, \qquad \vec{E} = -\nabla V

with $\mathcal{O}$ a chosen reference point (infinity, for a localised distribution). The line integral is path-independent precisely because $\nabla\times\vec{E} = \vec{0}$ . For a charge distribution,

V(\vec{r}) = \frac{1}{4\pi\varepsilon_0}\int \frac{\rho(\vec{r}\,')}{\eta}\,d\tau'

a single scalar integral with no direction to track — which is why one always computes $V$ first and differentiates afterward.

Worked example — potential of a line charge. An infinite line with uniform density $\lambda$ has $\vec{E} = \dfrac{\lambda}{2\pi\varepsilon_0 s}\,\hat{s}$ by Gauss's law (cylindrical Gaussian surface). Integrating, $V(s) = -\dfrac{\lambda}{2\pi\varepsilon_0}\ln(s/s_0)$ , with reference distance $s_0$ : infinity cannot be the reference here because the charge distribution is not localised. This is a first warning that the choice of reference point is part of the definition of $V$ .

5. Poisson's and Laplace's Equations

Combining $\vec{E} = -\nabla V$ with $\nabla\cdot\vec{E} = \rho/\varepsilon_0$ :

\boxed{\;\nabla^2 V = -\frac{\rho}{\varepsilon_0}\;}\qquad\text{(Poisson)}, \qquad \nabla^2 V = 0 \quad\text{(Laplace, where }\rho = 0\text{)}

All of electrostatics is the boundary value problem for these equations. The formal solution $V = \frac{1}{4\pi\varepsilon_0}\int \rho/\eta\,d\tau'$ solves Poisson's equation because $\nabla^2(1/\eta) = -4\pi\,\delta^3(\vec{\eta})$ — the Green's function of the Laplacian is $-1/4\pi\eta$ . In practice the integral is unusable whenever conductors or dielectrics impose conditions on surfaces rather than prescribing $\rho$ everywhere, and one solves the PDE directly. Chapter 2 is devoted to the techniques.

6. Conductors and Boundary Conditions

In electrostatic equilibrium a conductor satisfies: $\vec{E} = \vec{0}$ inside; $\rho = 0$ inside, so all net charge resides on the surface; the surface is an equipotential; and just outside, $\vec{E}$ is perpendicular to the surface with magnitude $\sigma/\varepsilon_0$ . The last is a special case of the general matching conditions at any surface carrying surface charge $\sigma$ :

E_\perp^{\text{above}} - E_\perp^{\text{below}} = \frac{\sigma}{\varepsilon_0}, \qquad \vec{E}_\parallel^{\text{above}} = \vec{E}_\parallel^{\text{below}}, \qquad V^{\text{above}} = V^{\text{below}}

The normal component jumps by $\sigma/\varepsilon_0$ (from Gauss on a pillbox); the tangential component is continuous (from $\oint\vec{E}\cdot d\vec{\ell} = 0$ on a thin loop); $V$ itself is continuous because $\vec{E}$ is finite.

7. Electrostatic Energy

The work to assemble a charge distribution, brought in from infinity against the field of what is already there, is

W = \frac{1}{2}\int \rho\,V\,d\tau = \frac{\varepsilon_0}{2}\int_{\text{all space}} E^2\,d\tau

The second form follows by writing $\rho = \varepsilon_0\nabla\cdot\vec{E}$ , integrating by parts, and letting the surface term vanish at infinity. It says the energy is stored in the field, with density $u = \tfrac12\varepsilon_0 E^2$ , everywhere the field is non-zero. The factor $\tfrac12$ avoids double-counting each pair of charge elements. Note the two forms disagree about where the energy is (the field form is positive everywhere; the $\rho V$ form has contributions only where there is charge) but agree on the total — and the field form is the one that generalises to radiation.

Worked example — energy of a charged sphere. For the uniformly charged ball of Section 2, integrate $u = \tfrac12\varepsilon_0 E^2$ over all space: the exterior contributes $\dfrac{Q^2}{8\pi\varepsilon_0}\displaystyle\int_R^\infty \dfrac{dr}{r^2} = \dfrac{Q^2}{8\pi\varepsilon_0 R}$ , the interior (with $E\propto r$ ) contributes $\dfrac{Q^2}{40\pi\varepsilon_0 R}$ , for a total $W = \dfrac{3}{5}\dfrac{Q^2}{4\pi\varepsilon_0 R}$ . The self-energy diverges as $R\to0$ : a true point charge has infinite electrostatic energy, a problem classical electrodynamics never fully resolves.

8. The Multipole Expansion

Far from a localised charge distribution confined near the origin, expand $1/\eta$ in powers of $r'/r$ using the generating function of the Legendre polynomials:

\frac{1}{\eta} = \frac{1}{\sqrt{r^2 + r'^2 - 2rr'\cos\alpha}} = \frac{1}{r}\sum_{n=0}^{\infty}\left(\frac{r'}{r}\right)^{n} P_n(\cos\alpha)

where $\alpha$ is the angle between $\vec{r}$ and $\vec{r}\,'$ . Substituting into $V$ ,

V(\vec{r}) = \frac{1}{4\pi\varepsilon_0}\sum_{n=0}^{\infty}\frac{1}{r^{\,n+1}}\int (r')^{n} P_n(\cos\alpha)\,\rho(\vec{r}\,')\,d\tau'

The first three terms are the physically important ones:

Monopole ( $n=0$ ): $V_{\text{mono}} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r}$ , with $Q = \int\rho\,d\tau'$ the total charge. Dominates at large $r$ unless $Q = 0$ .
Dipole ( $n=1$ ): $V_{\text{dip}} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\vec{p}\cdot\hat{r}}{r^2}$ , with dipole moment $\vec{p} = \int \vec{r}\,'\rho(\vec{r}\,')\,d\tau'$ . Dominates when $Q = 0$ .
Quadrupole ( $n=2$ ): falls as $1/r^3$ , described by the traceless quadrupole tensor $Q_{ij} = \int(3x'_i x'_j - r'^2\delta_{ij})\rho\,d\tau'$ .

The monopole moment is origin-independent; the dipole moment is origin-independent only if $Q = 0$ ; each higher moment is well-defined only once all lower ones vanish.

Worked example — the physical dipole. Two charges $\pm q$ separated by $\vec{d}$ (from $-q$ to $+q$ ) have $Q = 0$ and $\vec{p} = q\vec{d}$ . The exact potential is $V = \dfrac{q}{4\pi\varepsilon_0}\!\left(\dfrac{1}{\eta_+} - \dfrac{1}{\eta_-}\right)$ ; for $r \gg d$ this reduces to $\dfrac{1}{4\pi\varepsilon_0}\dfrac{p\cos\theta}{r^2}$ . Taking the gradient in spherical coordinates,

\vec{E}_{\text{dip}} = \frac{p}{4\pi\varepsilon_0 r^3}\left(2\cos\theta\,\hat{r} + \sin\theta\,\hat{\theta}\right)

the field of a pure (point) dipole, obtained in the limit $d\to0$ , $q\to\infty$ with $p = qd$ fixed. A pure dipole in a uniform external field $\vec{E}_0$ feels no net force but a torque $\vec{\tau} = \vec{p}\times\vec{E}_0$ and has energy $U = -\vec{p}\cdot\vec{E}_0$ .

9. Common Pitfalls

Using Gauss's law where there is no symmetry. Gauss's law is always true, but it gives $\vec{E}$ only when $\vec{E}$ can be pulled out of the flux integral — i.e. for spherical, infinite-cylindrical or infinite-planar symmetry. Otherwise use $V$ .
Forgetting that superposition is a separate postulate. Coulomb's law is about two charges; the field integral needs superposition as well.
Setting $V(\infty) = 0$ for a non-localised distribution. Infinite lines and planes require a finite reference distance; only a charge distribution of finite extent lets you use infinity.
Claiming $\vec{E}$ is discontinuous "because of the conductor". The normal component jumps by $\sigma/\varepsilon_0$ at any charged surface; the tangential component and $V$ are always continuous.
Equating the two energy expressions term by term. $\tfrac12\int\rho V\,d\tau$ and $\tfrac{\varepsilon_0}{2}\int E^2\,d\tau$ give the same total but localise the energy differently; the field form includes self-energy and is never negative.
Reading a dipole moment off a distribution with net charge. If $Q\ne0$ , $\vec{p}$ depends on the choice of origin and is not a meaningful intrinsic property.
Confusing a physical dipole with a pure dipole. The $1/r^3$ field is exact only for the point dipole; a physical dipole has higher-multipole corrections at finite $r$ .

Summary

Coulomb's law + superposition give $\vec{E}(\vec{r}) = \dfrac{1}{4\pi\varepsilon_0}\int \dfrac{\rho}{\eta^2}\hat{\eta}\,d\tau'$ and $V(\vec{r}) = \dfrac{1}{4\pi\varepsilon_0}\int \dfrac{\rho}{\eta}\,d\tau'$ .
The two field equations of electrostatics: $\nabla\cdot\vec{E} = \rho/\varepsilon_0$ (Gauss) and $\nabla\times\vec{E} = \vec{0}$ (which makes $V$ exist). Integral forms: $\oiint\vec{E}\cdot d\vec{a} = Q_{\text{enc}}/\varepsilon_0$ and $\oint\vec{E}\cdot d\vec{\ell} = 0$ .
Potential formulation: $\vec{E} = -\nabla V$ , $\nabla^2 V = -\rho/\varepsilon_0$ (Poisson), $\nabla^2 V = 0$ in charge-free regions (Laplace). Green's function: $\nabla^2(1/\eta) = -4\pi\delta^3(\vec{\eta})$ .
Boundary conditions: $E_\perp$ jumps by $\sigma/\varepsilon_0$ , $\vec{E}_\parallel$ and $V$ continuous. In a conductor $\vec{E} = 0$ , the surface is an equipotential, and $E_{\text{out}} = \sigma/\varepsilon_0$ .
Energy: $W = \tfrac12\int\rho V\,d\tau = \tfrac{\varepsilon_0}{2}\int E^2\,d\tau$ , field energy density $u = \tfrac12\varepsilon_0 E^2$ .
Multipole expansion: $V = \dfrac{1}{4\pi\varepsilon_0}\!\left[\dfrac{Q}{r} + \dfrac{\vec{p}\cdot\hat{r}}{r^2} + \cdots\right]$ with $\vec{p} = \int\vec{r}\,'\rho\,d\tau'$ . Pure dipole field $\vec{E}_{\text{dip}} = \dfrac{p}{4\pi\varepsilon_0 r^3}(2\cos\theta\,\hat{r} + \sin\theta\,\hat{\theta})$ ; in a uniform field, $\vec{\tau} = \vec{p}\times\vec{E}_0$ , $U = -\vec{p}\cdot\vec{E}_0$ .