Continuous Charge Distributions

A charged rod holds something like $10^{13}$ extra electrons. Nobody — no computer, no course, no career — is going to sum $10^{13}$ Coulomb fields term by term. Yet by the end of this page you will compute that rod's field exactly, and the trick is the oldest one in calculus: stop counting and start integrating. Chop the source into infinitesimal cells, treat each cell as the point charge it effectively is, let superposition add the contributions — and the impossible sum collapses into an integral you can actually do. This is the article where electrostatics acquires its production machinery. The three charge densities — $\lambda$ along lines, $\sigma$ over surfaces, $\rho$ through volumes — are the vocabulary; the recipe $\mathrm{d}q \to$ point-charge field $\to$ integrate is the grammar; and the classic geometries worked here (the ring, the disk, the infinite line) are sentences you will reuse for the rest of the track. Two of them come with a warning label from the future: the line charge takes us half a page of honest integration now, and Gauss's law will redo it in one line two articles ahead. That contrast is deliberate. You will appreciate the shortcut only after paying full fare once. One habit separates the students who can run this machinery from those who fight it: keeping the field point $\mathbf{r}$ and the source point $\mathbf{r}'$ rigidly distinct. The integral crawls over the source; the field point sits still and watches. Nearly every wrong answer in this subject begins by confusing the two.

Key takeaways

Continuous sources are described by densities: $\lambda = \mathrm{d}q/\mathrm{d}\ell$ (line), $\sigma = \mathrm{d}q/\mathrm{d}a$ (surface), $\rho = \mathrm{d}q/\mathrm{d}\tau$ (volume); total charge is the density integrated over the source.
The field is a superposition integral: $\mathbf{E}(\mathbf{r}) = \dfrac{1}{4\pi\varepsilon_0}\displaystyle\int\dfrac{\hat{\boldsymbol{\eta}}}{\eta^2}\,\mathrm{d}q$ , where $\boldsymbol{\eta}=\mathbf{r}-\mathbf{r}'$ points from each source element to the fixed field point.
The integration variable is the source point $\mathbf{r}'$ ; both $\eta$ and $\hat{\boldsymbol{\eta}}$ change as you crawl over the source — that direction-change is why symmetry is your best friend.
Ring on axis: $E_z = \dfrac{Qz}{4\pi\varepsilon_0(z^2+a^2)^{3/2}}$ ; stack rings into a disk: $E_z=\dfrac{\sigma}{2\varepsilon_0}\left[1-\dfrac{z}{\sqrt{z^2+R^2}}\right]$ ; infinite line: $E = \dfrac{\lambda}{2\pi\varepsilon_0 s}$ .
Every result must pass its limit checks: far away, everything reverts to the point-charge field of the total charge $Q$ ; close up, a disk becomes the infinite sheet $\sigma/2\varepsilon_0$ .
The point charge itself returns as the delta-function limit $\rho = q\,\delta^3(\mathbf{r}'-\mathbf{r}_0)$ — the continuum formalism contains the discrete one.
Key idea: replace $\sum_i q_i$ by $\int\mathrm{d}q$ with $\mathrm{d}q = \lambda\,\mathrm{d}\ell$ , $\sigma\,\mathrm{d}a$ , or $\rho\,\mathrm{d}\tau$ — then every point-charge formula generalizes by superposition.

The core insight

Replace the discrete sum $\sum_i q_i$ with an integral $\int \mathrm{d}q$ over the source, where $\mathrm{d}q = \lambda\,\mathrm{d}\ell$ , $\sigma\,\mathrm{d}a$ , or $\rho\,\mathrm{d}\tau$ ; then every point-charge formula generalizes by superposition.

Where this lives today

The chop-and-sum machinery is the beating heart of modern field computation. Boundary-element and finite-element solvers — the software that designs everything from smartphone touchscreens to particle-accelerator injectors — discretize conductors and dielectrics into thousands of little patches, each carrying a $\sigma\,\mathrm{d}a$ , and evaluate exactly this article's superposition integrals with numerical quadrature; the "method of moments" that dominates antenna engineering is equation (3) as a matrix equation. In accelerator physics, "space charge" — the self-field of a bunch of $10^{10}$ protons described by a smooth $\rho(\mathbf{r}')$ — limits the intensity of every high-current machine, and its computation is a disk-and-ring integral running millions of times per simulation. And at the smallest scales the story inverts: instead of computing fields from known densities, physicists infer the density from measured fields. Electron-scattering form factors map the proton's internal charge distribution (r.m.s. radius $0.84$ fm, measured to a precision that fueled a decade-long "proton radius puzzle"), making $\rho(\mathbf{r}')$ itself the experimental frontier.

The physical picture

Zoom into any "continuous" charged object and the continuity is a bluff — the charge sits on discrete electrons and ions. But zoom out to any element $\mathrm{d}\ell$ that is tiny compared with your distance to it yet huge compared with atomic spacing, and the bluff becomes an excellent theory: that element contains thousands of elementary charges, acts like a single point charge $\mathrm{d}q = \lambda\,\mathrm{d}\ell$ , and the graininess washes out below any measurement you will make. This double-zoom — small enough to be a point, large enough to be smooth — is the same continuum move fluid mechanics makes with water, and it is why the densities $\lambda$ , $\sigma$ , $\rho$ deserve to be treated as smooth functions.

Figure

The chop-and-sum machine — each source element dq sends its own Coulomb contribution to the fixed field point P

Once the source is chopped, superposition does the rest: each element contributes the point-charge field it would make alone, aimed along its own separation vector $\boldsymbol{\eta} = \mathbf{r}-\mathbf{r}'$ , and the total field is the vector integral of the contributions. The picture to internalize is a battery of tiny searchlights mounted along the source, every one aimed at $P$ from its own angle. Near the middle of a rod the searchlights on either side aim their transverse components against each other; symmetry cancellations like that are not luck — they are the geometry of the $\hat{\boldsymbol{\eta}}$ 's, and spotting them before integrating is what turns a three-component nightmare into a one-component exercise. Respect the hierarchy of idealizations, too. A "surface charge" is really a thin volume charge viewed from far compared with its thickness; a "line charge" is a thin tube. Each idealization behaves differently up close — the field is finite inside a volume charge, jumps by $\sigma/\varepsilon_0$ across a surface charge, and diverges at a line charge — and knowing which idealization you are holding tells you which of these behaviors to expect at close range.

The machinery

The vocabulary first. Charge distributed along a curve, over a surface, or through a volume is described by

\lambda = \frac{\mathrm{d}q}{\mathrm{d}\ell}, \qquad \sigma = \frac{\mathrm{d}q}{\mathrm{d}a}, \qquad \rho = \frac{\mathrm{d}q}{\mathrm{d}\tau} \tag{1}

where: the units are C/m, C/m², C/m³ respectively, and each density may vary with position over the source. Integrating a density over its home turf returns the total charge, $Q = \int\lambda\,\mathrm{d}\ell = \int\sigma\,\mathrm{d}a = \int\rho\,\mathrm{d}\tau$ — always your first sanity check, because a wrong density normalization corrupts everything downstream. The geometry second, because it is where the errors live. The field point $\mathbf{r}$ is fixed — it is where you want the answer. The source point $\mathbf{r}'$ is the integration variable — it crawls over the distribution. Their difference is the separation vector:

\boldsymbol{\eta} = \mathbf{r}-\mathbf{r}', \qquad \eta = |\mathbf{r}-\mathbf{r}'|, \qquad \hat{\boldsymbol{\eta}} = \boldsymbol{\eta}/\eta \tag{2}

where: both the length $\eta$ and the direction $\hat{\boldsymbol{\eta}}$ change as $\mathbf{r}'$ moves — the searchlights aim from different angles. (Griffiths reserves a special script letter for this vector and calls it the most important geometrical quantity in the book; we use $\boldsymbol{\eta}$ for the same object.) Now the master formula, which is nothing but the point-charge field summed by superposition over the chopped source:

\mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0}\int \frac{\hat{\boldsymbol{\eta}}}{\eta^2}\,\mathrm{d}q = \frac{1}{4\pi\varepsilon_0}\int_{\mathcal{V}} \frac{\rho(\mathbf{r}')\,\hat{\boldsymbol{\eta}}}{\eta^2}\,\mathrm{d}\tau' \tag{3}

where: the second form spells out the volume case ( $\mathrm{d}q=\rho\,\mathrm{d}\tau'$ ); line and surface versions swap in $\lambda\,\mathrm{d}\ell'$ or $\sigma\,\mathrm{d}a'$ . This is a vector integral — three scalar integrals wearing one coat — and the practical strategy is always the same: set up components, use symmetry to identify the components that cancel, and integrate only the survivor. When the dust of this chapter settles, equation (3) will acquire two mightier siblings — the potential integral $V = \frac{1}{4\pi\varepsilon_0}\int\mathrm{d}q/\eta$ (one scalar integral, next tier) and the local law $\nabla\cdot\mathbf{E}=\rho/\varepsilon_0$ — but (3) is the one that works from first principles on any source, symmetric or not.

geometry	result	reach for it when
ring of charge, on axis	$E_z=\dfrac{Qz}{4\pi\varepsilon_0(z^2+a^2)^{3/2}}$	axial symmetry; building block for disks
disk, on axis	$E_z=\dfrac{\sigma}{2\varepsilon_0}\Big[1-\dfrac{z}{\sqrt{z^2+R^2}}\Big]$	capacitor-plate geometry, sheet limit
infinite line	$E=\dfrac{\lambda}{2\pi\varepsilon_0 s}$	long thin wires, $s\ll$ length
infinite sheet	$E=\dfrac{\sigma}{2\varepsilon_0}$ , uniform	close to any large flat charged surface
any source, far away	$E\approx\dfrac{Q_{\text{tot}}}{4\pi\varepsilon_0 r^2}$	the monopole limit — always check it

Derivation AField on the axis of a uniformly charged ring

This is the cleanest nontrivial superposition integral in the subject, and the template for every axial-symmetry calculation you will do. The goal: a ring of radius $a$ carries total charge $Q$ spread uniformly ( $\lambda = Q/2\pi a$ ); find $\mathbf{E}$ at height $z$ on the axis. The strategy: parametrize the ring, write one element's contribution, and let symmetry execute the transverse components before any integral is computed — the surviving axial integral then turns out to have a constant integrand, which is why the answer is exact and tidy. Watch how the geometry factor $\cos\theta = z/\sqrt{z^2+a^2}$ enters once and for all: every element of the ring sits at the same distance from the field point, a luxury unique to this geometry.

Step 1. Chop and label. An element at angle $\phi$ on the ring carries $\mathrm{d}q = \lambda\,a\,\mathrm{d}\phi$ . Every element sits at distance $\eta = \sqrt{z^2+a^2}$ from the axial point $P$ — the same for all $\phi$ . Why is the constant distance such a gift? Because $1/\eta^2$ then pulls out of the integral; only directions remain to be managed.

Step 2. Split each contribution into axial and transverse parts. The element's field $\mathrm{d}E = \frac{1}{4\pi\varepsilon_0}\frac{\mathrm{d}q}{z^2+a^2}$ tilts off-axis; its axial component carries the factor $\cos\theta = z/\sqrt{z^2+a^2}$ , its transverse component the factor $\sin\theta$ .

Step 3. Let symmetry kill the transverse sum. The element at $\phi$ and the element at $\phi+\pi$ aim their transverse components in exactly opposite directions with equal magnitudes; summed over the full ring the transverse field vanishes identically. No integral required — this is the searchlight cancellation of the physical picture, made exact. Why insist on arguing this before integrating? Because it converts a three-component vector integral into one scalar integral, and because the argument fails instantly off-axis — which is why the off-axis ring field is genuinely hard (elliptic integrals) while the axial field is one line.

Step 4. Integrate the surviving axial component. The integrand is now constant in $\phi$ :

E_z = \frac{1}{4\pi\varepsilon_0}\,\frac{\cos\theta}{z^2+a^2}\int_0^{2\pi}\lambda a\,\mathrm{d}\phi = \frac{1}{4\pi\varepsilon_0}\,\frac{z}{(z^2+a^2)^{3/2}}\,(2\pi a\lambda) = \frac{1}{4\pi\varepsilon_0}\,\frac{Qz}{(z^2+a^2)^{3/2}}

Why the $3/2$ power? One factor of $\sqrt{z^2+a^2}$ from the inverse-square, half a factor more from the direction cosine — geometry's fingerprint, and a pattern you will see in every axial integral.

Step 5. Audit the limits. Far away ( $z\gg a$ ): $E_z \to Q/4\pi\varepsilon_0 z^2$ — the ring looks like a point of charge $Q$ . ✓ At the centre ( $z=0$ ): $E_z = 0$ — every searchlight's axial component vanishes and the transverse ones cancel; a charge at the exact centre feels nothing. ✓ The field peaks in between, at $z = a/\sqrt{2}$ (set $\mathrm{d}E_z/\mathrm{d}z = 0$ ). The maximum position follows from setting $\mathrm{d}E_z/\mathrm{d}z = 0$ : $(z^2+a^2)^{3/2} - 3z^2(z^2+a^2)^{1/2} = 0 \Rightarrow a^2 = 2z^2$ , i.e. $z = a/\sqrt{2}$ .

The ring is a building block, not just an exercise: a uniformly charged disk is a stack of concentric rings, and integrating the boxed ring result over radii $a' = 0 \to R$ (with $\mathrm{d}q = \sigma\,2\pi a'\,\mathrm{d}a'$ ) gives, on the axis,

E_z = \frac{\sigma}{2\varepsilon_0}\left[1-\frac{z}{\sqrt{z^2+R^2}}\right] \tag{4}

where: the bracket interpolates between two worlds. Close up ( $z\ll R$ ) it tends to $1$ , leaving the infinite-sheet field $\sigma/2\varepsilon_0$ — uniform, distance-independent, the workhorse of capacitor physics. Far away ( $z\gg R$ ), expanding the square root gives $E_z\to Q/4\pi\varepsilon_0 z^2$ with $Q=\sigma\pi R^2$ — the monopole limit again. One formula, both sanity checks built in; worked example 2 puts numbers through it.

Go deeper · Derivation B · 2 minThe infinite line charge, by honest integration

Derivation BThe infinite line charge, by honest integration

We compute the field a perpendicular distance $s$ from an infinitely long straight wire of uniform line charge $\lambda$ — the classic result $E = \lambda/2\pi\varepsilon_0 s$ — by direct superposition. The strategy: exploit the reflection symmetry about the foot of the perpendicular to kill the parallel component, then evaluate the surviving transverse integral with the standard substitution. We do this the hard way on purpose: Gauss's law will reproduce the boxed result in one line two articles from now, and the fair comparison — half a page versus one line — is the best advertisement that chapter will ever get. Note also what the answer says: the field of a line dies as $1/s$ , slower than a point's $1/s^2$ , because receding from a line never makes it look small along one of its dimensions.

Step 1. Set up. Put the wire on the $x$ -axis; seek the field at distance $s$ above the origin. The element at position $x$ carries $\mathrm{d}q = \lambda\,\mathrm{d}x$ at distance $\eta = \sqrt{x^2+s^2}$ from the field point.

Step 2. Kill the parallel component by symmetry: elements at $+x$ and $-x$ contribute equal and opposite components along the wire. Only the transverse (radial) component survives, carrying the direction cosine $s/\sqrt{x^2+s^2}$ :

E_s = \frac{\lambda}{4\pi\varepsilon_0}\int_{-\infty}^{\infty}\frac{s\,\mathrm{d}x}{(x^2+s^2)^{3/2}}

Why does the cosine carry $s$ and not $x$ ? The transverse projection of $\hat{\boldsymbol{\eta}}$ is (perpendicular distance)/(hypotenuse) — draw one triangle and it never lies.

Step 3. Evaluate the master integral with $x = s\tan\alpha$ , so $\mathrm{d}x = s\sec^2\alpha\,\mathrm{d}\alpha$ and $(x^2+s^2)^{3/2} = s^3\sec^3\alpha$ :

\int_{-\infty}^{\infty}\frac{\mathrm{d}x}{(x^2+s^2)^{3/2}} = \frac{1}{s^2}\int_{-\pi/2}^{\pi/2}\cos\alpha\,\mathrm{d}\alpha = \frac{2}{s^2}

Why memorize this integral? Because $\int(x^2+s^2)^{-3/2}\mathrm{d}x$ is the single most recurring integral of electrostatics — lines, dipole sheets, Biot–Savart wires all funnel through it.

Step 4. Assemble:

E_s = \frac{\lambda}{4\pi\varepsilon_0}\cdot s\cdot\frac{2}{s^2} = \frac{\lambda}{2\pi\varepsilon_0\,s}

The direction is radially away from the wire (for $\lambda>0$ ), by the same symmetry that killed the parallel part.

Step 5. Read the scaling. $1/s$ , not $1/s^2$ : doubling your distance from a long wire only halves the field. Geometrically, the wire never recedes along its own length — only one of the two "spreading" dimensions is available, so one power of distance is all the dilution you get. (The infinite sheet takes this to its logical end: zero available spreading dimensions, hence no falloff at all.) As a consistency check, the finite-line result $E_s = \frac{\lambda}{4\pi\varepsilon_0 s}\cdot\frac{2L}{\sqrt{s^2+L^2}}$ recovers the boxed infinite-line field for $L\gg s$ , and reduces to the point-charge field $Q/4\pi\varepsilon_0 s^2$ (with $Q=2\lambda L$ ) for $s\gg L$ — both limits confirm the formula at its natural endpoints.

Limiting cases

case	result	meaning
$\rho(\mathbf{r}') = q\,\delta^3(\mathbf{r}'-\mathbf{r}_0)$	(3) collapses to the point-charge field	the continuum formalism contains the discrete case exactly
far from any bounded source	$E \to Q_{\text{tot}}/4\pi\varepsilon_0 r^2$	monopole limit — the universal far-field check
disk, $z\ll R$	$E \to \sigma/2\varepsilon_0$	infinite-sheet limit: uniform, distance-free
disk, $z\gg R$	$E \to Q/4\pi\varepsilon_0z^2$	point limit of the same formula
line vs point vs sheet	$1/s$ vs $1/r^2$ vs const	falloff counts the dimensions available for spreading
element size vs atomic scale	$\ell_{\text{element}}\gg a_{\text{atomic}}$	continuum densities are averages; graininess is invisible below this scale

Common trapr and r′ are different letters because they are different points

The single most common error in all of electrostatics: letting the field point and the source point blur together. The integral in (3) runs over $\mathbf{r}'$ — the source coordinate — while $\mathbf{r}$ is a fixed parameter; the separation $\boldsymbol{\eta} = \mathbf{r}-\mathbf{r}'$ therefore changes length and direction throughout the integration, and $\hat{\boldsymbol{\eta}}$ can never be pulled outside the integral unless symmetry has certified that its direction is effectively constant. Symptoms of the disease: differentiating under the integral with respect to the wrong variable; writing $\int \hat{\mathbf{r}}/r^2\,\mathrm{d}q$ with a single $r$ that silently means two things; getting a nonzero transverse field where symmetry demands zero. The vaccine is notational hygiene — primes on every source quantity ( $\mathrm{d}\ell'$ , $\mathrm{d}a'$ , $\rho(\mathbf{r}')$ ), the separation vector written out, and a drawn triangle before any algebra.

Worked examples

1 · A ring, checked three ways (confidence builder)

A ring of radius $a = 10$ cm carries $Q = 50$ nC uniformly. Find the field on the axis at $z = 20$ cm, and check it against its limits.

Step 1. Plug into the boxed ring formula: $z^2+a^2 = 0.04+0.01 = 0.05\ \mathrm{m^2}$ , so $(z^2+a^2)^{3/2} = 0.0112\ \mathrm{m^3}$ .

Step 2. $E_z = (8.99\times10^9)\,\dfrac{(50\times10^{-9})(0.20)}{0.0112} = 8.0\times10^{3}$ N/C, pointing away from the ring along the axis (positive charge).

Step 3. Upper-bound check: pretending the ring were a point charge at the origin gives $k_eQ/z^2 = 1.12\times10^4$ N/C — larger, as it must be, since the real ring's charge sits off-axis and its contributions arrive tilted, wasting some of their magnitude on cancelled transverse parts.

Step 4. Location check: the field maximum sits at $z = a/\sqrt{2} = 7.1$ cm; our point at 20 cm is well outside it, on the descending, point-charge-approaching tail. At $z = 1$ m the exact and point-charge values agree to within about 1.5% — the monopole limit arriving on schedule.

Answer: $E_z = 8.0\times10^3$ N/C along the axis — below the $1.12\times10^4$ N/C point-charge bound, converging to it at large $z$ .

2 · A disk and its two impostors (exam level)

A disk of radius $R = 5.0$ cm carries $Q = 2.0$ nC uniformly. Find the axial field at $z = 1.0$ cm, and compare with the infinite-sheet and point-charge approximations.

Step 1. Density first: $\sigma = Q/\pi R^2 = (2.0\times10^{-9})/(\pi\times0.0025) = 2.55\times10^{-7}\ \mathrm{C/m^2}$ .

Step 2. The sheet prefactor: $\dfrac{\sigma}{2\varepsilon_0} = \dfrac{2.55\times10^{-7}}{2(8.854\times10^{-12})} = 1.44\times10^4$ N/C.

Step 3. The geometry bracket of equation (4): $\sqrt{z^2+R^2} = \sqrt{0.0001+0.0025} = 0.0510$ m, so the bracket is $1 - 0.010/0.0510 = 1-0.196 = 0.804$ .

Step 4. Exact axial field: $E_z = (1.44\times10^4)(0.804) = 1.16\times10^4$ N/C.

Step 5. Now the impostors. Infinite sheet: $1.44\times10^4$ N/C — 24% high; at $z = R/5$ the disk's edge is already "visible" and the sheet idealization is fraying. Point charge: $k_eQ/z^2 = (8.99\times10^9)(2.0\times10^{-9})/10^{-4} = 1.8\times10^5$ N/C — fifteen times too large, catastrophically wrong, because at $z\ll R$ almost all of the disk's charge is far to the side, not concentrated below the field point. Moral: each idealization has a home regime — sheet for $z\ll R$ , point for $z\gg R$ — and equation (4) is the honest bridge between them.

Answer: $E_z = 1.16\times10^4$ N/C — the sheet approximation runs 24% high at $z=R/5$ , the point approximation fails by a factor of 15.

Figure

The disk field and its two impostors — the exact bracket bridges the sheet regime and the point regime

Origins · 1 minSiméon Denis Poisson and how this was discovered

Siméon Denis Poisson

1781–1840 · École Polytechnique, Paris

Poisson never performed an experiment in his life — "life is good for only two things," he said, "doing mathematics and teaching it" — yet he industrialized electrostatics. His 1812 memoir on the distribution of electricity over conductors took Coulomb's pairwise force and Laplace's potential methods and welded them into a machinery for continuous charge: describe the source by a density, integrate, and demand consistency with equilibrium. Out of that program came the equation bearing his name, $\nabla^2 V = -\rho/\varepsilon_0$ — the differential twin of this article's integrals — plus lasting tools across elasticity, magnetism, and probability (the Poisson distribution). A cautionary footnote to his brilliance: reviewing Fresnel's wave theory of light, he derived the "absurd" prediction of a bright spot behind a circular obstacle, intending to kill the theory; Arago promptly measured the spot. The mathematics was, as usual with Poisson, impeccable either way.

History

1785

— Coulomb establishes the pair force law; summing it over extended bodies immediately becomes the practical problem

1812

— Poisson's memoir treats electricity on conductors with continuous surface densities, founding electrostatic potential theory

1813

— Poisson relates the potential's second derivatives to the local charge density: the differential form of this article's integrals

1828

— Green's essay introduces the potential function and boundary methods, systematizing distributed-charge problems

1911

— Rutherford models the nuclear atom with charge distributions, turning scattering into a probe of charge density inside matter

1956

— Hofstadter's electron-scattering experiments map the proton's charge distribution — charge elements resolved at the femtometre scale

Going further

References. Griffiths, Introduction to Electrodynamics, 4th ed., §2.1.4 and §2.3 · Jackson, Classical Electrodynamics, 3rd ed., §1.5–1.7 · Purcell & Morin, Electricity and Magnetism, 3rd ed., Ch. 1–2 · Zangwill, Modern Electrodynamics, §3.2–3.4 Three key integrals underpin the results on this page: the ring axial field $E_z=Qz/4\pi\varepsilon_0(z^2+a^2)^{3/2}$ and its large- $z$ point-charge limit; the shell potential integral $\int_{-1}^{1}(R^2+r^2-2Rru)^{-1/2}\mathrm{d}u=[(R+r)-|R-r|]/Rr$ (relevant to the potential articles of the next tier); and the line-charge master integral $\int_{-\infty}^{\infty}(s^2+x^2)^{-3/2}\mathrm{d}x=2/s^2$ , which gives $\lambda/2\pi\varepsilon_0 s$ . Throughout, $\varepsilon_0=8.854\times10^{-12}\,\mathrm{C^2 N^{-1} m^{-2}}$ . The worked numbers on this page — ring $8.0\times10^3$ N/C vs point bound $1.12\times10^4$ ; disk $1.16\times10^4$ N/C vs sheet $1.44\times10^4$ and point $1.8\times10^5$ ; $\sigma=2.55\times10^{-7}\ \mathrm{C/m^2}$ — follow directly from equations (3)–(4).

Builds on: Coulomb's Law · The Electric Field · Superposition Principle for Electric Fields Leads to: Electrostatic Potential · Gauss's Law in Integral Form · Gauss's Law in Differential Form · Poisson's Equation · Potential of a Localized Distribution

If charge is really made of discrete electrons, how can a "density" be legitimate?

By the double-zoom argument: pick source elements small compared with your distance to them but large compared with atomic spacing. A cube 10 nm on a side in a charged metal contains thousands of atoms, so its charge content is statistically smooth, yet from a millimetre away it is utterly point-like. The densities $\lambda,\sigma,\rho$ are averages over such cells, and every prediction made with them is exact up to corrections of order (atomic spacing/observation distance), invisibly small in any macroscopic problem. The same license underwrites fluid dynamics and elasticity. It fails, appropriately, exactly where you would expect: scanning-probe microscopy at sub-nanometre range sees individual adsorbed charges, not a smooth surface density.

Why do the primes matter so much? It's just notation.

Because the notation encodes which point moves. In $\int\rho(\mathbf{r}')\hat{\boldsymbol{\eta}}/\eta^2\,\mathrm{d}\tau'$ , everything primed crawls over the source while $\mathbf{r}$ watches from a fixed seat, and the separation vector built from both changes direction throughout. Blur the prime and you will eventually differentiate with respect to the wrong variable — fatal in the divergence-of-E calculation coming in the Gauss's law articles, where $\nabla$ acts on $\mathbf{r}$ only and must slide past $\rho(\mathbf{r}')$ untouched. Physicists who survived this course keep the habit for life: source coordinates primed, field coordinates bare, triangle drawn before algebra.

How close is "close" for using the infinite-sheet formula, and how far is "far" for the point formula?

Worked example 2 is the calibration: for a disk, the sheet value $\sigma/2\varepsilon_0$ is exact at $z=0$ and runs about 24% high already at $z = R/5$ ; it is within 10% only for $z \lesssim R/10$ . The point formula errs the opposite way and needs roughly $z \gtrsim 3R$ for 10% accuracy ( $z\gtrsim 10R$ for 1%). Between those regimes, use the honest bracket of equation (4). The general habit transfers to every geometry: idealizations are asymptotic statements with a home range, and one evaluation of the exact formula tells you where the range ends.

Is the field infinite right at a charged line or surface? The formulas seem to blow up.

There is a clean hierarchy, worth memorizing. At a volume charge, the field is finite everywhere — the $1/\eta^2$ singularity is tamed by the three powers of $\eta$ in $\mathrm{d}\tau'$ . At a surface charge the field is finite on each side but jumps by $\sigma/\varepsilon_0$ across the sheet (the two-sided average is what acts on the sheet itself). At a line charge the field genuinely diverges as $1/s$ — and at a point charge as $1/r^2$ . The divergences are artifacts of over-idealization: a real "line" is a thin tube of volume charge, and inside the tube the field is finite. The thinner your idealization, the more violent its close-range fiction.

Why is there both a vector integral for E and (soon) a scalar integral for V? Which should I use?

The field integral (3) is three coupled scalar integrals with direction bookkeeping; the potential integral $V=\frac{1}{4\pi\varepsilon_0}\int\mathrm{d}q/\eta$ is a single scalar integral with no directions at all — then one gradient recovers $\mathbf{E}$ . For most extended sources the scalar route is dramatically less work, which is why the next tier builds it with such care. Use the direct field integral when symmetry hands you the direction for free (all the geometries on this page) or when you need only one component; switch to the potential for anything genuinely three-dimensional. Knowing both — and choosing deliberately — is a hallmark of fluency.

Do these integrals ever fail to converge?

For any bounded source with finite total charge, the field integral converges at every point off the source, and even at interior points of a volume charge (the singularity is integrable). Trouble comes from idealized infinite sources: the infinite line and sheet give perfectly finite fields but infinite total charge, so quantities like "the potential at infinity" stop making sense — for the line, the potential must be referenced to a finite radius instead. If an integral you set up diverges for a bounded source, the diagnosis is almost always an error — usually the $\mathbf{r}$ vs $\mathbf{r}'$ disease — not exotic physics.

The line-charge derivation took half a page. Is there really a one-line version coming?

Yes — and you have now earned the right to enjoy it. Gauss's law exploits the cylindrical symmetry you used to cancel components, but before any integration: wrap the wire in an imaginary cylinder of radius $s$ , note the field must be radial with constant magnitude on the cylinder wall, and the flux bookkeeping $E\cdot2\pi s\ell = \lambda\ell/\varepsilon_0$ hands you $E=\lambda/2\pi\varepsilon_0 s$ immediately. The catch: that magic needs enough symmetry to know the field's direction and uniformity in advance — spheres, infinite cylinders, infinite planes, and essentially nothing else. For every other shape, the integrals of this page remain the honest tool. Pay full fare here; ride free there.