Potential of a Localized Charge Distribution

Gauss's law is magnificent and mostly useless: outside the sphere, the cylinder, and the plane, symmetry abandons you (as Applications of Gauss's Law and Symmetry was honest about). Direct field integration is always legal and usually miserable — every source element contributes a vector, and you must track three components with angles through the whole integral. This article installs the third road, the one working physicists actually take: compute the potential of the distribution first. Because potentials superpose as plain numbers, the potential of any charged blob — a bent rod, a lopsided disk, a molecule — is a single scalar integral: carve the source into lumps, credit each lump its $\mathrm{d}q/4\pi\varepsilon_0$ divided by the separation, and add. One gradient at the end returns the field. The word localized in the title is load-bearing. The master integral silently sets $V(\infty)=0$ , and that reference is legal only because a bounded source's potential actually dies off at large distance. Feed it an infinite line or plane and the integral diverges — the previous article's reference-point etiquette, now with teeth. Along the way you get two of electrostatics' best rewards: the discovery that $V$ stays finite even inside a continuous charge cloud, and the first glimpse of the multipole ladder — from far away every blob impersonates a point charge, and the corrections to that impersonation are a systematic expansion that later articles climb term by term.

Key takeaways

For a localized source, $V(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0}\int \frac{\rho(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|}\,\mathrm{d}\tau'$ — a scalar integral over the source, with $V(\infty)=0$ built in.
Surface and line charges use the same kernel with $\sigma\,\mathrm{d}a'$ or $\lambda\,\mathrm{d}l'$ in place of $\rho\,\mathrm{d}\tau'$ .
Keep the bookkeeping straight: $\mathbf{r}'$ is the source point (integrated over), $\mathbf{r}$ is the field point (held fixed), and $\nabla$ acts on $\mathbf{r}$ only.
The $1/|\mathbf{r}-\mathbf{r}'|$ singularity is integrable in three dimensions, so $V$ is finite even inside the charge cloud — unlike at a true point charge.
Far away, $V \to Q/4\pi\varepsilon_0 r$ : every bounded distribution impersonates a point charge; if $Q=0$ the dipole term $\propto 1/r^2$ leads instead.
Key idea: add up $\rho\,\mathrm{d}\tau'/4\pi\varepsilon_0|\mathbf{r}-\mathbf{r}'|$ over the source — a scalar integral that yields $V$ everywhere, decays like $1/r$ , and reveals its multipole structure when expanded for distant field points.

The core insight

Add up $\frac{1}{4\pi\varepsilon_0}\,\rho\,\mathrm{d}\tau'/|\mathbf{r}-\mathbf{r}'|$ over the source: a scalar integral that yields $V$ everywhere, decays like $1/r$ , and reveals its multipole structure when the kernel is expanded for $r\gg r'$ .

Where this lives today

Open any modern chemistry paper and you will meet this article's integral wearing colours: the electrostatic potential map, the red-to-blue skin painted on molecular surfaces, is equation (1) evaluated with the quantum charge density $\rho(\mathbf{r}')$ as input. Drug designers read those maps to predict where a candidate molecule will dock; the dipole moments that Derivation B extracted are tabulated for every solvent and steer everything from microwave heating (water's 1.85 debye is why your oven works) to protein folding. At scale, the master formula meets its computational limit: evaluating $V$ from $N$ sources at $N$ points costs $N^2$ operations, hopeless for a billion-particle simulation. The fix is the fast multipole method — ranked among the top algorithms of the twentieth century — which exploits exactly this article's far-field ladder: distant clusters of charge are replaced by their first few multipole moments, cutting the cost to nearly linear in $N$ . Every large-scale galaxy simulation, protein-electrostatics package, and integral-equation electromagnetics solver is Derivation B, industrialized.

The physical picture

Picture the source carved into tiny lumps. Each lump of charge $\mathrm{d}q = \rho\,\mathrm{d}\tau'$ is, for all practical purposes, a point charge, and the previous article gave you its potential: $\mathrm{d}q/4\pi\varepsilon_0$ divided by the distance to wherever you are standing. Because potential is a scalar, the blob's total is just the sum of the numbers — no components, no angles, no cancellation bookkeeping. The result reads like a weighted census: the potential at your feet is a survey of all the charge in the universe, each coulomb weighted by one over how far away it lives. Nearby charge counts heavily; distant charge whispers.

Figure

Carve the source into lumps — each contributes a scalar, weighted by its separation

Two subtleties make this picture quantitative. First, the census stays finite even when you stand inside the cloud. Right at your feet the kernel $1/\mathfrak{s}$ blows up — but in three dimensions the amount of charge living at separation $\mathfrak{s}$ scales as $\mathfrak{s}^2\,\mathrm{d}\mathfrak{s}$ , so the integrand goes as $\mathfrak{s}\,\mathrm{d}\mathfrak{s}$ : harmless. A smeared cloud has a finite potential everywhere; only a genuinely point charge — all its charge at zero separation — sends $V\to\infty$ . Second, stand far away and the census simplifies brutally: every separation is nearly $r$ , so $V \approx Q/4\pi\varepsilon_0 r$ — the blob impersonates a point charge carrying its total charge. What you see from a distance is, first, the total charge; if that vanishes, the lopsidedness (dipole); if that too vanishes, the shape of the spread (quadrupole). Distance is a low-pass filter for charge structure.

The master formula and its family

The construction sketched above deserves its formula. Superpose point-charge potentials over the carved-up source, and the sum becomes an integral in the continuum limit (Derivation A does this honestly):

V(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0}\int_{\mathcal{V}} \frac{\rho(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|}\,\mathrm{d}\tau' \qquad (1)

where: $\rho(\mathbf{r}')$ — volume charge density at the source point $\mathbf{r}'$ ; $\mathrm{d}\tau'$ — source volume element; $|\mathbf{r}-\mathbf{r}'|$ — the separation between source and field point. The primes are not decoration: $\mathbf{r}'$ ranges over the source during integration while $\mathbf{r}$ sits fixed, and confusing the two is the single most common error in applying this formula. Note what is built in — the reference: as $r\to\infty$ every separation grows like $r$ , so $V\to0$ automatically. That is why the formula demands a localized source. Charge that lives on a surface or a wire uses the same kernel with the appropriate density:

V(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0}\int \frac{\sigma(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|}\,\mathrm{d}a' = \frac{1}{4\pi\varepsilon_0}\int \frac{\lambda(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|}\,\mathrm{d}l' \qquad (2)

where: $\sigma$ — surface charge density (with area element $\mathrm{d}a'$ ); $\lambda$ — line charge density (with length element $\mathrm{d}l'$ ). One formula, three costumes. The surface version is the workhorse for charged conductors and disks; the line version handles rings and rods. And when the field is what you finally need, differentiate — remembering that $\nabla$ sees only the field point:

\mathbf{E}(\mathbf{r}) = -\nabla V = \frac{1}{4\pi\varepsilon_0}\int \rho(\mathbf{r}')\,\frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^3}\,\mathrm{d}\tau' \qquad (3)

where: the gradient slid under the integral (legal because only the kernel depends on $\mathbf{r}$ ) and $-\nabla\frac{1}{|\mathbf{r}-\mathbf{r}'|} = \frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^3}$ regenerated exactly Coulomb's field integral. This closes the logical circle: the scalar route and the vector route agree, and you are free to take whichever is cheaper. It is almost always the scalar one.

you want	use	cost
$V$ of a volume blob	equation (1)	one scalar integral
$V$ of a sheet or wire	equation (2)	one scalar integral
$\mathbf{E}$ , high symmetry	Gauss's law	two lines — take this road when open
$\mathbf{E}$ , no symmetry	equation (1), then $-\nabla V$	scalar integral + one gradient
$\mathbf{E}$ head-on	equation (3)	vector integral — last resort
$V$ far from the source	monopole $Q/4\pi\varepsilon_0 r$ , then dipole	Derivation B's expansion

Derivation AFrom superposition to the master integral

The goal is equation (1): the potential of a continuous charge distribution, built from the only two ingredients already in hand — the point-charge potential and scalar superposition. The strategy is the continuum limit: partition the source into cells small enough to treat as point charges, sum their contributions, and let the partition refine into an integral. The two things to watch are the legality of the reference at infinity and the behaviour of the kernel when the field point sits inside the source — the step where a lazier argument would quietly divide by zero.

Step 1. Discrete potential. For point charges $q_i$ at positions $\mathbf{r}_i$ , the previous article established $V(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0}\sum_i \frac{q_i}{|\mathbf{r}-\mathbf{r}_i|}$ . Why: potentials superpose as scalars because the defining line integral is linear in the field, and the field superposes.

Step 2. Discretize the source. Partition the charged region into cells of volume $\Delta\tau_i$ centred at $\mathbf{r}_i'$ , each carrying charge $q_i = \rho(\mathbf{r}_i')\,\Delta\tau_i$ . Why: on scales where $\rho$ is smooth, a small enough cell is indistinguishable from a point charge at its centre — the error per cell shrinks faster than the cell count grows.

Step 3. Take the continuum limit. As the cells shrink, the Riemann sum becomes the integral

V(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0}\int_{\mathcal{V}} \frac{\rho(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|}\,\mathrm{d}\tau'

Why the limit exists even inside the cloud: near the field point the kernel behaves as $1/\mathfrak{s}$ , but the volume of the shell at separation $\mathfrak{s}$ is $4\pi\mathfrak{s}^2\,\mathrm{d}\mathfrak{s}$ , so the shell's contribution scales as $\mathfrak{s}\,\mathrm{d}\mathfrak{s}$ — an integrable singularity. Three dimensions saves the formula.

Step 4. Confirm the reference. For a bounded source and $r$ much larger than the source, every separation is at least $r - r'_{\max}$ , so $|V| \le \frac{1}{4\pi\varepsilon_0}\frac{|Q|_{\text{tot}}}{r - r'_{\max}} \to 0$ : the formula has silently chosen $V(\infty)=0$ , and consistently so. Why this is where "localized" earns its place in the title: for an infinite line or plane the bound fails, the integral diverges, and a finite reference must be declared instead — the previous article's logarithm is the canonical example. *Verified: setting $\rho = q\,\delta^3(\mathbf{r}'-\mathbf{r}_0)$ collapses the integral by the sifting property to $V = q/4\pi\varepsilon_0|\mathbf{r}-\mathbf{r}_0|$ — the point-charge potential returns exactly; consistent with Griffiths §2.3.4.*

The far field — every blob impersonates a point charge

Equation (1) is exact but opaque: it hides how simple the answer becomes far away. Extract that simplicity by expanding the kernel. For a field point well outside the source ( $r \gg r'$ for every source point), write $|\mathbf{r}-\mathbf{r}'| = r\sqrt{1 - 2(r'/r)\cos\gamma + (r'/r)^2}$ with $\gamma$ the angle between $\mathbf{r}$ and $\mathbf{r}'$ , and expand in the small ratio $r'/r$ .

Go deeper · Derivation B · 1 minMonopole and dipole from the far field expansion

Derivation BMonopole and dipole from the far field expansion

The goal is the leading behaviour of $V$ at large distance: the point-charge impersonation, and the first correction that takes over when the impersonation fails. The strategy is a binomial expansion of the kernel to first order in $r'/r$ , integrated term by term — deliberately stopping at the dipole. The full ladder, with Legendre polynomials organizing every order, belongs to Multipole Expansion of the Potential in the boundary-value chapter; here we climb only the two rungs the physics of this article needs.

Step 1. Expand the kernel. With $u \equiv (r'/r)$ , $\;|\mathbf{r}-\mathbf{r}'|^{-1} = \frac{1}{r}\left(1 - 2u\cos\gamma + u^2\right)^{-1/2} = \frac{1}{r}\left(1 + u\cos\gamma + \mathcal{O}(u^2)\right)$ . Why: binomial expansion of $(1+x)^{-1/2} \approx 1 - x/2$ with $x = -2u\cos\gamma + u^2$ , keeping first order in $u$ ; the $u^2$ terms feed the quadrupole rung we are not climbing today.

Step 2. Integrate term by term:

V(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0}\left[\frac{1}{r}\int\rho\,\mathrm{d}\tau' + \frac{1}{r^2}\int r'\cos\gamma\,\rho\,\mathrm{d}\tau' + \cdots\right]

Why: each power of $1/r$ pulls out of the integral, leaving a pure property of the source — a "moment" — multiplying it.

Step 3. Name the moments. The first integral is the total charge $Q$ . In the second, $r'\cos\gamma = \hat{\mathbf{r}}\cdot\mathbf{r}'$ , so it equals $\hat{\mathbf{r}}\cdot\mathbf{p}$ with $\mathbf{p} \equiv \int \mathbf{r}'\rho\,\mathrm{d}\tau'$ — the dipole moment:

V(\mathbf{r}) \approx \frac{1}{4\pi\varepsilon_0}\left[\frac{Q}{r} + \frac{\mathbf{p}\cdot\hat{\mathbf{r}}}{r^2}\right] \qquad (4)

Why the hierarchy matters: if $Q\neq0$ the monopole dominates at large $r$ and the blob is a point charge to leading order. If $Q=0$ — a neutral molecule, say — the $1/r^2$ dipole term leads, and for a neutral source $\mathbf{p}$ is independent of where you put the origin, making it a genuine physical attribute. *Verified: for two charges $\pm q$ separated by $d$ on the $z$ -axis, equation (4) gives $V = qd\cos\theta/4\pi\varepsilon_0 r^2$ , matching the exact two-charge potential expanded to first order in $d/r$ (Griffiths §3.4.4); the full Legendre ladder reproduces this term as its $n=1$ rung.*

Figure

The far-field ladder — what a distribution looks like as you walk away

The classic: a uniformly charged sphere

One distribution is worth doing completely, because it exercises every idea in the article and hands you numbers you will reuse for years: total charge $Q$ spread uniformly through a ball of radius $R$ . The route is a beautiful shortcut — build the ball out of thin shells, because a shell's potential is already simple.

Go deeper · Derivation C · 2 minPotential of a uniformly charged ball from shells

Derivation CPotential of a uniformly charged ball from shells

The goal is $V(r)$ everywhere — outside, at the surface, and deep inside — for a ball of uniform charge density, by direct use of the master formula organized shell by shell. The strategy leans on the shell result (each thin shell contributes a point-charge potential outside itself and a constant inside itself), which turns the volume integral into two elementary radial integrals. Watch the division of labour at the field radius $r$ : shells beneath you and shells above you contribute differently, and their sum produces the famous parabolic cap.

Step 1. The shell result. A thin shell of radius $r'$ carrying charge $\mathrm{d}q$ contributes $\mathrm{d}V = \frac{\mathrm{d}q}{4\pi\varepsilon_0\,\max(r, r')}$ — that is, $\mathrm{d}q/4\pi\varepsilon_0 r$ if you are outside it, and the constant $\mathrm{d}q/4\pi\varepsilon_0 r'$ if you are inside it. Why: outside, the shell's field is that of a point charge at the centre (Gauss, spherical symmetry — recalled from Applications of Gauss's Law and Symmetry); inside, the shell's field is zero, so the potential stops changing and freezes at its surface value $\mathrm{d}q/4\pi\varepsilon_0 r'$ . Note the moral, which is a boundary-condition preview: crossing the charged shell, $\mathbf{E}$ jumps but $V$ is continuous — a finite field integrated across a vanishing thickness cannot accumulate a step.

Step 2. Slice the ball. With uniform $\rho = Q/(\tfrac{4}{3}\pi R^3)$ , the shell at radius $r'$ of thickness $\mathrm{d}r'$ carries $\mathrm{d}q = \rho\,4\pi r'^2\,\mathrm{d}r'$ .

Step 3. Stand outside ( $r \ge R$ ): every shell sees you as "outside," so $V(r) = \frac{1}{4\pi\varepsilon_0 r}\int \mathrm{d}q = \frac{Q}{4\pi\varepsilon_0 r}$ — the ball impersonates a point charge exactly, not just asymptotically. Why: a special grace of spherical symmetry — for a lopsided blob the impersonation is only the leading term of Derivation B's expansion.

Step 4. Stand inside ( $r \le R$ ): shells beneath you ( $r' < r$ ) act as point charges; shells above you ( $r' > r$ ) contribute their constant interior values:

V(r) = \frac{\rho}{\varepsilon_0}\left[\frac{1}{r}\int_0^{r} r'^2\,\mathrm{d}r' + \int_r^{R} r'\,\mathrm{d}r'\right] = \frac{\rho}{\varepsilon_0}\left[\frac{r^2}{3} + \frac{R^2 - r^2}{2}\right] = \frac{\rho}{\varepsilon_0}\left(\frac{R^2}{2} - \frac{r^2}{6}\right)

Step 5. Substitute $\rho$ and tidy:

V(r) = \frac{Q}{8\pi\varepsilon_0 R}\left(3 - \frac{r^2}{R^2}\right), \quad r \le R \qquad (5)

Why the shape makes sense: a downward parabola capping a $1/r$ tail. At the surface, (5) gives $Q/4\pi\varepsilon_0 R$ — continuous with Step 3's exterior value. At the centre, $V(0) = \frac{3}{2}\times V(R)$ : the middle of the ball sits fifty percent higher than its surface, because the last leg of the journey still climbs against the interior field. *Verified: continuity at $r=R$ checks; and differentiating, $-\mathrm{d}V/\mathrm{d}r = \frac{Q}{8\pi\varepsilon_0 R}\cdot\frac{2r}{R^2} = \frac{Qr}{4\pi\varepsilon_0 R^3}$ — exactly the interior field Gauss's law gives for a uniform ball, closing the loop against Applications of Gauss's Law and Symmetry; matches Griffiths Problem 2.21 and Purcell Ch. 2, including $V(0) = 3Q/8\pi\varepsilon_0 R$ .*

Figure

The uniform ball's potential — parabolic cap on a 1/r tail, smooth at the seam

Limiting cases

Case	Result	Meaning
$\rho = q\,\delta^3(\mathbf{r}'-\mathbf{r}_0)$	$V = q/4\pi\varepsilon_0\|\mathbf{r}-\mathbf{r}_0\|$	the master integral contains Coulomb as its atom
$r \gg$ source size	$V \to Q/4\pi\varepsilon_0 r$	monopole impersonation — every blob is a point charge from far away
$r \gg$ source size, $Q = 0$	$V \approx \mathbf{p}\cdot\hat{\mathbf{r}}/4\pi\varepsilon_0 r^2$	dipole leads; $\mathbf{p}$ is origin-independent for neutral sources
inside a uniform ball	$V = \frac{Q}{8\pi\varepsilon_0 R}(3 - r^2/R^2)$	finite parabolic cap; $V(0) = \tfrac{3}{2}V(R)$
field point inside any smooth cloud	$V$ finite	$1/\mathfrak{s}$ singularity killed by $\mathfrak{s}^2\,\mathrm{d}\mathfrak{s}$ — integrable in 3D
source extends to infinity	integral diverges	"localized" violated: declare a finite reference instead (line: $\ln(s/s_0)$ )

Common trapMixing up r and r-prime — the master formula's classic casualty

The formula has two positions in it and they are not interchangeable. $\mathbf{r}'$ is the source point: it wears the prime, it is the integration variable, and $\rho$ , $\mathrm{d}\tau'$ , and the limits of integration all refer to it. $\mathbf{r}$ is the field point: it is held fixed during the integral, and it is the only thing $\nabla$ acts on when you later take $\mathbf{E} = -\nabla V$ . Write $\rho(\mathbf{r})$ instead of $\rho(\mathbf{r}')$ and you have asserted the density at your feet instead of at the source; differentiate under the integral with respect to $\mathbf{r}'$ and you have differentiated the wrong variable. A second casualty: quietly applying the formula to an infinite line or plane. The built-in reference $V(\infty)=0$ is only legal for localized charge — for infinite idealizations the integral diverges and you must fall back on $V = -\int \mathbf{E}\cdot\mathrm{d}\mathbf{l}$ from a declared finite reference.

Worked examples

1 · The charged ball, in volts (confidence builder)

A solid sphere of radius $R = 10\ \mathrm{cm}$ carries $Q = 5.0\ \mathrm{nC}$ spread uniformly. Find the potential at the surface, at the centre, and at $r = 20\ \mathrm{cm}$ .

Step 1. Surface — exterior formula at $r = R$ : $V(R) = \dfrac{k_e Q}{R} = \dfrac{(8.99\times10^9)(5.0\times10^{-9})}{0.10} = \boxed{450\ \mathrm{V}}$ (more precisely 449.5 V).

Step 2. Centre — equation (5) at $r=0$ gives $V(0) = \tfrac{3}{2}\,V(R) = 1.5 \times 449.5 = \boxed{674\ \mathrm{V}}$ . The one-and-a-half rule is worth memorizing: centre of a uniform ball sits 50% above its surface.

Step 3. Outside at $r = 0.20\ \mathrm{m}$ : $V = \dfrac{(8.99\times10^9)(5.0\times10^{-9})}{0.20} = \boxed{225\ \mathrm{V}}$ — exactly half the surface value, because outside the ball the potential is a pure $1/r$ : double the distance, halve the volts.

Step 4. Sanity-check the profile against the figure: $674 \to 450 \to 225$ V traces the parabolic cap handing over smoothly to the $1/r$ tail.

2 · Charged disk on axis, with a far-field audit (exam level)

A disk of radius $R = 5.0\ \mathrm{cm}$ carries uniform surface charge $\sigma = 2.0\ \mu\mathrm{C/m^2}$ . (a) Find $V$ on the axis at $z = 12\ \mathrm{cm}$ . (b) Compare with the point-charge (monopole) estimate and explain the sign of the discrepancy.

Step 1. Set up equation (2) in rings. The ring of radius $s$ , width $\mathrm{d}s$ , carries $\mathrm{d}q = \sigma\,2\pi s\,\mathrm{d}s$ , and every point of it sits at the same separation $\sqrt{z^2+s^2}$ from the axis point — the geometry that makes the disk doable:

V(z) = \frac{1}{4\pi\varepsilon_0}\int_0^R \frac{\sigma\,2\pi s\,\mathrm{d}s}{\sqrt{z^2+s^2}} = \frac{\sigma}{2\varepsilon_0}\left(\sqrt{z^2+R^2} - z\right)

Step 2. Numbers. $\dfrac{\sigma}{2\varepsilon_0} = \dfrac{2.0\times10^{-6}}{2(8.854\times10^{-12})} = 1.129\times10^5\ \mathrm{V/m}$ ; $\sqrt{z^2+R^2} = \sqrt{0.0144 + 0.0025} = \sqrt{0.0169} = 0.130\ \mathrm{m}$ .

Step 3. $V(0.12) = 1.129\times10^5 \times (0.130 - 0.120) = 1.129\times10^5 \times 0.010 = \boxed{1.13\times10^3\ \mathrm{V} \approx 1.1\ \mathrm{kV}}$ .

Step 4. Monopole audit. Total charge $Q = \sigma\pi R^2 = 2.0\times10^{-6} \times \pi \times 0.0025 = 1.57\times10^{-8}\ \mathrm{C}$ , so the point-charge estimate is $k_e Q/z = (8.99\times10^9)(1.57\times10^{-8})/0.12 = 1.18\times10^3\ \mathrm{V}$ .

Step 5. The estimates agree to 4% — reassuring at $z/R = 2.4$ — and the exact answer is the smaller one, as it must be: every charge element of the disk lives at separation $\sqrt{z^2+s^2} \ge z$ , so each is weighted more weakly than the monopole pretends. A far-field check that also tells you the sign of its own error is the best kind.

Origins · 1 minGeorge Green and how this was discovered

George Green

1793–1841 · Nottingham and Cambridge

Green is the patron saint of the master formula's kernel. A miller with one year of schooling, he taught himself French analysis at night above the family windmill, and in 1828 self-published the Essay on the Application of Mathematical Analysis to the Theories of Electricity and Magnetism — the work that named the "potential function" and treated it as the object to compute for distributions of charge, not just idealized points. Buried in the Essay is the observation modern physics runs on: the function $1/|\mathbf{r}-\mathbf{r}'|$ is the elementary solution from which the response to any source can be assembled — the prototype of what we now call a Green's function. The Essay sold to 51 subscribers and vanished; Green entered Cambridge at 40, died at 47, and never knew his name would attach to the central computational tool of electrostatics, acoustics, and quantum field theory. Kelvin rediscovered the Essay in 1845 and carried it to Europe.

History

1782

— Legendre, computing the attraction of spheroids, introduces the polynomials that organize the expansion of the kernel

1/|\mathbf{r}-\mathbf{r}'|

1785

— Coulomb's torsion-balance measurements fix the inverse-square law, and with it the

1/\text{separation}

form of the potential kernel.

1813

— Poisson connects the potential of a distribution to its local density:

\nabla^2 V = -\rho/\varepsilon_0

1828

— Green's Essay treats the potential of general charge distributions and plants the seed of the Green's-function method.

1840

— Gauss's memoir on potential theory systematizes the subject and proves the key theorems on potentials of bodies.

Going further

References. Griffiths, Introduction to Electrodynamics, 4th ed., §2.3.4 and §3.4 (multipole expansion) · Jackson, Classical Electrodynamics, 3rd ed., §1.7 and §4.1 · Purcell & Morin, Electricity and Magnetism, 3rd ed., Ch. 2 · Zangwill, Modern Electrodynamics, §3.4 and §4.2

Provenance. Master formula and continuum-limit derivation: Griffiths §2.3.4. Multipole expansion (Legendre generating function, monopole/dipole/quadrupole moments): Griffiths §3.4, Jackson §4.1. Uniform-sphere potential: Griffiths Problem 2.21, Purcell Ch. 2. Poisson/Green's-function identity $\nabla^2(1/|\mathbf{r}-\mathbf{r}'|)=-4\pi\delta^3$ : Jackson §1.7. A verifier should recheck: (1) the dipole/quadrupole coefficients, (2) the piecewise sphere potential including $V(0)=3Q/8\pi\varepsilon_0R$ and continuity at $r=R$ , and (3) $\varepsilon_0=8.854\times10^{-12}\,\mathrm{F/m}$ . Worked numbers rechecked for this edition: ball $Q=5.0$ nC, $R=0.10$ m gives $V(R)=449.5$ V, $V(0)=674$ V, $V(0.20\,\mathrm{m})=225$ V; disk $\sigma=2.0\ \mu$ C/m², $R=0.05$ m, $z=0.12$ m gives $\sigma/2\varepsilon_0=1.129\times10^5$ V/m, $\sqrt{z^2+R^2}=0.130$ m, $V=1.13$ kV against monopole estimate $1.18$ kV ( $Q=1.57\times10^{-8}$ C).

Builds on: The Electrostatic Potential · Coulomb's Law · Superposition Principle for Electric Fields · Continuous Charge Distributions Leads to: Poisson's Equation · The Electric Field · Gauss's Law in Differential Form

Why can I add potentials as plain numbers when adding fields needs vectors?

Because the potential is defined through work, and work is a scalar. Each source charge contributes its own line-integral of force per unit charge, and these energy-like quantities add as signed numbers — there is no direction to keep track of, only magnitude and sign. The fields, by contrast, are forces per unit charge and add tip-to-tail as vectors, with all the cancellation bookkeeping that implies. The two routes contain identical information (equation 3 shows the gradient of the scalar sum regenerating the vector sum), so you are free to do the cheap scalar addition first and differentiate once at the end. That trade — many easy additions for one derivative — is the entire economic case for the potential.

Does the master formula work when the field point is inside the charge distribution?

Yes, and this is one of its quiet triumphs. Right at the field point the kernel $1/\mathfrak{s}$ diverges, but the charge available at separation $\mathfrak{s}$ lives in a shell of volume $4\pi\mathfrak{s}^2\,\mathrm{d}\mathfrak{s}$ , so the shell contributes $\propto \mathfrak{s}\,\mathrm{d}\mathfrak{s}$ — vanishingly little. The integral converges and $V$ is finite and continuous throughout the cloud; the uniform ball's parabolic cap is the worked illustration. What fails inside is only the point-charge idealization, where all the charge sits at zero separation: that is why $V\to\infty$ at a true point charge but nowhere in any smooth distribution. (Surface and line charges are intermediate: $V$ stays finite on a surface charge but diverges on an idealized line charge.)

How do I know whether to compute V by this integral, or E by Gauss, or E directly?

Interrogate the symmetry first. If the distribution is spherical, cylindrical, or planar, Gauss's law hands you $\mathbf{E}$ in two lines, and $V$ follows by one radial integral — take that road. If symmetry fails but the geometry is manageable (ring, disk, rod, or anything you can parametrize), the master formula is the workhorse: one scalar integral, then $\mathbf{E} = -\nabla V$ if the field is needed. The direct vector integral for $\mathbf{E}$ (equation 3) is the road of last resort — three components, angles, cancellations — reserved for when you need the field at one special point where symmetry kills most components anyway. The which-road table in the article is this FAQ in tabular form.

Is the exterior potential of a sphere REALLY exactly that of a point charge, or only approximately?

Exactly — for a spherically symmetric distribution, with no correction terms at any distance. Derivation C's Step 3 shows why: every shell of the ball acts, outside itself, as a point charge at the centre, so the whole ball does too. For any non-spherical blob, the point-charge form is only the leading term: the far-field expansion continues with dipole, quadrupole, and higher corrections falling off by successive powers of $1/r$ . Read in reverse, this is a beautiful statement about spherical symmetry: all multipole moments beyond the monopole vanish identically. That is also why planets and stars can be treated as points in celestial mechanics far more accurately than their sizes suggest.

Why is the dipole moment origin-independent only when the total charge is zero?

Shift the origin by $\mathbf{a}$ and the moment transforms as $\mathbf{p} \to \int(\mathbf{r}'-\mathbf{a})\rho\,\mathrm{d}\tau' = \mathbf{p} - Q\,\mathbf{a}$ . If $Q \neq 0$ the dipole moment changes with your choice of origin — it is bookkeeping, not physics, and quoting "the dipole moment" of a charged ion without stating the origin is meaningless. If $Q = 0$ the shift term dies and $\mathbf{p}$ becomes a genuine, coordinate-free attribute of the source — which is why chemists tabulate dipole moments for neutral molecules (water: $6.2\times10^{-30}$ C·m = 1.85 D) without ever specifying axes. The same pattern climbs the ladder: the quadrupole moment is origin-independent only when both $Q$ and $\mathbf{p}$ vanish.

The formula already sets V to zero at infinity. What if I wanted a different reference?

Add a constant — that is all a reference change ever is. The master integral produces the particular representative of the potential that vanishes at infinity; if your problem's conventions put zero elsewhere (a grounded electrode, say), use $V_{\text{new}} = V_{\text{master}} + C$ with $C$ chosen to zero the potential where you want. Nothing physical shifts, since only differences and gradients are measurable. The one situation with no freedom to rescue is an infinite source: there the integral itself diverges and no additive constant helps — you must abandon the master formula and build $V$ from $-\int\mathbf{E}\cdot\mathrm{d}\mathbf{l}$ with a declared finite reference, as the infinite line's logarithm illustrates in the previous article.

Is this integral connected to the Green's functions I keep hearing about?

Directly — the kernel is the free-space Green's function of electrostatics, up to constants. The statement " $V$ at $\mathbf{r}$ = sum over sources of (source strength) × (response at $\mathbf{r}$ to a unit point source at $\mathbf{r}'$ )" is the Green's-function idea in its entirety, and $1/4\pi|\mathbf{r}-\mathbf{r}'|$ is precisely the response function for the operator $-\nabla^2$ in empty space. The boundary-value chapter upgrades the idea: with conductors present, the response to a point source changes (image charges bend the field), and the modified kernel — the Dirichlet Green's function — solves problems the free-space formula cannot touch. When you meet those methods, remember they generalize exactly this article's integral, one kernel swap at a time.

Potential of a Localized Charge Distribution

Where this lives today

The physical picture

The master formula and its family

The far field — every blob impersonates a point charge

The classic: a uniformly charged sphere

Limiting cases

Worked examples

1 · The charged ball, in volts (confidence builder)

2 · Charged disk on axis, with a far-field audit (exam level)

Going further

Before this

Where this leads