---
title: Continuous Charge Fields
module: Continuous Charge Distributions
moduleNumber: 2
lessonNumber: 1
order: 201
summary: >
  A charged rod, ring, or disk is not a point, yet its field is still nothing but
  Coulomb's law added up over the charge it carries. We replace the discrete sum by
  an integral, with $\d q=\lambda\d\ell$, $\sigma\d A$, or $\rho\d V$, so the real
  work becomes geometry: writing the vector from each source element to the field
  point, and letting symmetry cancel the components that must cancel before any
  integral is attempted. We carry the line, ring, and disk fields through in full,
  then check each result against its near field, its far field, and its dimensions.
topics: [Continuous Charge Distributions]
draft: false
sources:
  - book: Tipler & Mosca
    ref: "Ch. 22 — The Electric Field II; §22-1"
---

An extended charge distribution is partitioned into elements $\d q$ whose fields add
vectorially:

$$
\d\vec E=\frac{1}{4\pi\varepsilon_0}\frac{\d q}{r^2}\hat r,
\qquad \vec E=\int \d\vec E.
$$

The density determines $\d q$: $\d q=\lambda\d\ell$ for a line, $\d q=\sigma\d A$ for a
surface, and $\d q=\rho\d V$ for a volume. The vector from a source element to the
field point must be written explicitly. Integrating magnitudes before resolving
directions is generally incorrect.

## Finite line charge

At a point on the perpendicular bisector of a uniformly charged rod, components
parallel to the rod cancel in pairs. For a rod from $-a$ to $a$, field point at
distance $z$, and $r=\sqrt{x^2+z^2}$,

$$
\d E_z=\frac{1}{4\pi\varepsilon_0}\frac{\lambda z\d x}{(x^2+z^2)^{3/2}},
\qquad
E_z=\frac{1}{4\pi\varepsilon_0}\frac{2\lambda a}{z\sqrt{a^2+z^2}}.
$$

For $z\gg a$, this becomes $k(2a\lambda)/z^2$, the field of total charge
$Q=2a\lambda$. For $a\gg z$, it approaches the infinite-line result
$\lambda/(2\pi\varepsilon_0z)$.

$$
% caption: Field of a finite line charge on its perpendicular bisector. A source element $\d q$ at $+x$ and its mirror at $-x$ lie the same distance $r$ from the field point $P$; their components along the rod cancel and the perpendicular components add, leaving a field along $z$.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
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$$

## Ring and disk

Every element of a uniformly charged ring contributes the same axial component;
transverse components cancel. A ring of radius $R$ and total charge $Q$ produces

$$
E_z=\frac{1}{4\pi\varepsilon_0}\frac{Qz}{(z^2+R^2)^{3/2}}.
$$

The field is zero at the centre and falls as $kQ/z^2$ far away. A uniformly charged
disk is built from rings of radius $r$ and charge $\d q=\sigma(2\pi r\d r)$. The
integral yields

$$
E_z=\frac{\sigma}{2\varepsilon_0}\left(1-\frac{z}{\sqrt{z^2+R^2}}\right).
$$

As $R\to\infty$, the field becomes $\sigma/(2\varepsilon_0)$ on either side of
an infinite sheet.

$$
% caption: A charged disk as nested rings. An annulus of radius $r$ and charge $\d q=\sigma\,2\pi r\,\d r$ contributes an axial field at $P$; integrating from the centre to $R$ builds the disk field.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
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$$

> **Worked example (ring on its axis).** A ring of radius $R=5.0\ \mathrm{cm}$ carries
> $Q=10\ \mathrm{nC}$ uniformly. On the axis at $z=10\ \mathrm{cm}$,
>
> $$
> E_z=\frac{1}{4\pi\varepsilon_0}\frac{Qz}{(z^2+R^2)^{3/2}}
> =\frac{(8.99\times10^9)(1.0\times10^{-8})(0.10)}{(0.10^2+0.050^2)^{3/2}}
> =6.4\times10^3\ \mathrm{N/C},
> $$
>
> directed outward along the axis. The field is zero at the centre and peaks at
> $z=R/\sqrt2=3.5\ \mathrm{cm}$, where $E_z=1.4\times10^4\ \mathrm{N/C}$. By
> $z=10\ \mathrm{cm}=2R$ it has dropped to about $70\%$ of the point-charge value
> $kQ/z^2=9.0\times10^3\ \mathrm{N/C}$, the limit it approaches as $z\gg R$.

> **Worked example (disk approaching a sheet).** A disk of radius $R=10\ \mathrm{cm}$
> carries surface density $\sigma=8.0\ \mathrm{nC/m^2}$. On the axis at
> $z=2.0\ \mathrm{cm}$,
>
> $$
> E_z=\frac{\sigma}{2\varepsilon_0}\!\left(1-\frac{z}{\sqrt{z^2+R^2}}\right)
> =(4.5\times10^2\ \mathrm{N/C})\!\left(1-\frac{0.020}{\sqrt{0.020^2+0.10^2}}\right)
> =3.6\times10^2\ \mathrm{N/C}.
> $$
>
> The prefactor $\sigma/2\varepsilon_0=4.5\times10^2\ \mathrm{N/C}$ is the
> infinite-sheet field. One disk radius across and close to the surface, the finite
> disk already reaches about $80\%$ of it. Moving instead to $z\gg R$ sends
> $E_z\to kQ/z^2$ with $Q=\sigma\pi R^2$.

## Three-dimensional distributions

A volume source has the Coulomb integral

$$
\vec E(\vec r)=\frac{1}{4\pi\varepsilon_0}
\int \rho(\vec r')\frac{\vec r-\vec r'}{|\vec r-\vec r'|^3}\d^3 r'.
$$

The primed coordinate identifies the source location; the unprimed coordinate is
the observation point. This distinction avoids integrating over the field point.
Spherical, cylindrical, or planar symmetry can reduce the integral, but Gauss's law
is usually preferable when the field magnitude is constant over a suitable surface.

$$
% caption: Source and field coordinates. The element $\d q$ sweeps the charged volume at $\vec r'$ while the field point holds fixed at $\vec r$; each contribution uses the separation $\vec R=\vec r-\vec r'$.
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$$

## Establishing the integration geometry

The source variable and field point must be separated before writing an integral.
A charge element at $(x',y',z')$ and a field point at $(x,y,z)$ define

$$
\vec R=(x-x')\hat\imath+(y-y')\hat\jmath+(z-z')\hat k,
\qquad R=\sqrt{(x-x')^2+(y-y')^2+(z-z')^2}.
$$

The corresponding field element is $\d\vec E=k\d q\,\vec R/R^3$. Primes
are not cosmetic: they identify variables that are integrated over. The observation
coordinates remain fixed during the integration. A source at the origin and field
point on an axis can use unprimed shorthand only after that choice is clear.

Symmetry is applied to the vector components, not to the scalar charge element.
A uniformly charged ring has nonzero $\d\vec E$ from every element, while the
transverse components cancel only after matching elements at opposite azimuths. A
single element has no such cancellation. Similarly, a finite rod does not have the
same field as an infinite line merely because its centre is on the same axis.

### The finite rod on its bisector

Take a rod on the $x$ axis from $-a$ to $a$, uniform density $\lambda$, and field
point $(0,z)$. An element has $\d q=\lambda\d x$, distance
$R=\sqrt{x^2+z^2}$, and perpendicular component

$$
\d E_z=k\frac{\lambda\d x}{R^2}\frac{z}{R}
=k\lambda z\frac{\d x}{(x^2+z^2)^{3/2}}.
$$

The antiderivative is

$$
\int\frac{\d x}{(x^2+z^2)^{3/2}}
=\frac{x}{z^2\sqrt{x^2+z^2}}.
$$

Evaluation from $-a$ to $a$ yields the result stated above. The same answer can be
written using the angle $\theta$ subtended by an end of the rod at the field point:

$$
E=\frac{1}{2\pi\varepsilon_0}\frac{\lambda}{z}\sin\theta,
\qquad \tan\theta=\frac az.
$$

The angular form makes the limiting cases immediate. As the rod becomes very long,
$\theta\to\pi/2$ and the infinite-line result follows. As the field point moves
far away, $\sin\theta\approx a/z$ and the total charge $Q=2a\lambda$ appears.

At a point on the rod axis outside an interval from $x_1$ to $x_2$, every field
element is collinear. The signed $x$ component must be retained; an unsigned
integral can incorrectly predict cancellation or the wrong direction. Splitting
the source around a point where the sign changes is often the safest method.

### Ring and disk on the axis

A ring places every source element at distance $R_0=\sqrt{R^2+z^2}$ from an axial
field point. Thus the common axial projection is $z/R_0$ and

$$
E_z=k\int\frac{z\d q}{R_0^3}=\frac{kQz}{(R^2+z^2)^{3/2}}.
$$

The field vanishes at $z=0$ by symmetry. Its derivative at the origin is positive
for a positive ring, so a positive test charge there is pushed away from the centre
along either axial direction. The potential method checks the field: with
$V=kQ/\sqrt{R^2+z^2}$, $E_z=-\d V/\d z$ gives the same expression.

A disk is a continuum of rings. Its element of charge is
$\d q=\sigma(2\pi r\d r)$, not $\sigma\pi r^2\d r$; the latter mistakenly uses the
area enclosed by the ring rather than the area of a thin annulus. Substitution gives

$$
E_z=2\pi k\sigma z\int_0^R\frac{r\d r}{(r^2+z^2)^{3/2}}.
$$

The integral evaluates to the disk expression already given. At a fixed finite
distance, increasing $R$ approaches the infinite-sheet field. At fixed $R$ and
large $z$, it approaches $kQ/z^2$ with $Q=\sigma\pi R^2$.

### Nonuniform density

Charge density need not be constant. A rod with $\lambda(x)=\lambda_0(1+x/a)$
requires $\d q=\lambda(x)\d x$ and no symmetry cancellation about its midpoint. A
surface with radial density $\sigma(r)$ uses $\d q=\sigma(r)r\d r\d\phi$ in polar
coordinates. Coordinate-system Jacobians are part of the area or volume element:

$$
\d V=r\d r\d\phi\d z
\quad\hbox{(cylindrical)},
\qquad
\d V=r^2\sin\vartheta\d r\d\vartheta\d\phi
\quad\hbox{(spherical)}.
$$

Omitting the factor $r$ or $r^2\sin\vartheta$ changes the physical amount of charge
assigned to each coordinate cell. The error cannot be repaired by a later constant.

## A numerical check on the rod field

> **Worked example (finite rod on its bisector).** A rod of length $0.40\ \mathrm m$
> carries $8.0\ \mathrm{nC}$ uniformly, so its half-length is $a=0.20\ \mathrm m$ and
> its line density is $\lambda=Q/(2a)=2.0\times10^{-8}\ \mathrm{C/m}$. Evaluate the
> field at a point on the perpendicular bisector, $z=0.10\ \mathrm m$ from the centre.
> The bisector result gives
>
> $$
> E=\frac{1}{4\pi\varepsilon_0}\frac{2\lambda a}{z\sqrt{a^2+z^2}}
> =\frac{(8.99\times10^9)(2)(2.0\times10^{-8})(0.20)}{(0.10)\sqrt{0.20^2+0.10^2}}
> =3.2\times10^3\ \mathrm{N/C},
> $$
>
> with the unit set by $k\lambda/z$ and the direction pointing away from the positive
> rod. Replacing the rod by a point charge $Q$ at its centre would give
> $kQ/z^2=7.2\times10^3\ \mathrm{N/C}$, larger by the exact factor
> $\sqrt{a^2+z^2}/z=\sqrt5\approx2.2$. Here $z$ is not large compared with the
> half-length $a$, so the point-charge approximation overstates the field; it becomes
> accurate only once $z\gg a$.

## Singularities and physical scope

An ideal line charge has field $E\propto1/r$ and diverges on its axis; an ideal
surface charge has a discontinuity in normal field at the surface. Real sources
have finite microscopic structure. Integrals describe the field at locations where
the continuum approximation is appropriate and exclude self-field questions at an
infinitesimal source element. Conducting surfaces require an additional condition:
electrostatic equilibrium determines their surface-charge distribution together with
the imposed boundary conditions.

## Plane and solid distributions

A uniformly charged rectangular sheet permits direct Coulomb integration
but the limits and vector projections generally require two integrations. At a
point above the centre, pair elements related by reflection cancel their in-plane
components. A circular disk is more convenient because polar coordinates make the
remaining integral one-dimensional. The source geometry, not an aesthetic preference,
determines the efficient coordinate system.

A uniformly charged solid sphere illustrates the distinction between integration
and symmetry. Its external field has the point-charge form because all charge is
contained inside a sphere and the source is spherically symmetric. Its internal
field is not obtained by placing the total charge at the centre. Direct integration
is cumbersome; Gauss's law in the following lesson gives

$$
E(r)=\frac{\rho r}{3\varepsilon_0}
\quad (r<R),
\qquad
E(r)=\frac{\rho R^3}{3\varepsilon_0r^2}
\quad (r>R).
$$

The two expressions agree at the surface. The interior result grows linearly from
zero because the enclosed charge grows as $r^3$ while the area of a spherical
surface grows as $r^2$. A thin spherical shell gives zero interior field, rather
than a field that grows with radius, because it encloses no charge at smaller radii.

### Building the field integral

Construct the field integral in this order.

1. State the observation point and place the source in a coordinate system matched
   to its symmetry.
2. Mark a representative source element and write its charge from the stated
   density and the correct differential length, area, or volume.
3. Write $\vec R=\vec r-\vec r'$ from that element toward the observation
   point.
4. Resolve $\d\vec E$ along axes before integrating. Use symmetry only for
   components that have an explicit partner under a source transformation.
5. Check the result at a symmetry point, far from the distribution, and for units.

These choices determine $\d q$, the source-to-field vector, the surviving components,
and the applicable limits. Coulomb's law then gives the integral.

An integral has a finite answer only when the specified continuous distribution is
physically meaningful at the field point. For example, potential of an infinite
line cannot be assigned a finite value relative to infinity because the integral
contains a logarithmic divergence. Potential differences between two finite radii
are nevertheless well defined. The electric field of that line is finite at every
nonzero radius and is the appropriate quantity for force calculations.

### Continuous charge with point charges

The integral notation does not replace superposition; it is the limiting form of a
sum over many small charges. If a distribution consists of a prescribed continuous
part and several point charges, add their fields:

$$
\vec E=\frac{1}{4\pi\varepsilon_0}\int
\frac{\vec R}{R^3}\d q
+\frac{1}{4\pi\varepsilon_0}\sum_iq_i\frac{\vec R_i}{R_i^3}.
$$

The integral can be evaluated independently for each region when density changes
piecewise. A disk with a central hole is the field of a large disk minus the field
of the missing smaller disk, provided both disks use the same surface density. This
subtraction method is often shorter than changing radial limits and makes the field
direction transparent.

The point-charge limit is recovered only at distances large compared with every
dimension of the source. At those distances, the leading term depends on total
charge. If total charge is zero, the leading term cancels and the dipole term,
which falls as $r^{-3}$, becomes important. The far-field behaviour checks both the
arithmetic and the net charge assigned to the source model. It also tests the source
normalization used in the integral.

### Dimensional and limiting checks

A result for electric field must have units $\mathrm{N/C}$. A line-charge result
often has the scale $k\lambda/r$, a surface-charge result $k\sigma$, and a compact
three-dimensional source far away has scale $kQ/r^2$. These estimates make it
possible to identify a missing power of distance before a detailed derivation is
completed. They do not replace the integral, since they do not determine numerical
coefficients or vector direction.

Limits also identify incorrect expressions. The axial field of a ring must be zero
at its centre, must have a maximum at some nonzero $z$, and must tend to $kQ/z^2$
far away. The field of a positive finite rod on its perpendicular bisector must
point away from the rod and cannot become infinite at nonzero perpendicular distance.
An expression that fails any of these tests has a setup or algebra error.

Density units provide another check: $\lambda$ has $\mathrm{C/m}$, $\sigma$ has
$\mathrm{C/m^2}$, and $\rho$ has $\mathrm{C/m^3}$. Integrating each density over
its associated geometric element must produce coulombs before it is inserted into
Coulomb's law.

Numerical integration is appropriate when a density or boundary has no elementary
antiderivative. The discretization must converge as element size is reduced, and
the same symmetry and unit checks remain necessary. A numerical sum with an
incorrect vector direction converges accurately to the wrong field.

### Line-charge source element

Take a uniformly charged rod on the x axis from $-a$ to $a$, with line density
$\lambda$, and evaluate its field at point $P=(0,z)$. A source element at coordinate
$x$ has charge $\d q=\lambda\d x$. The displacement from that element to P is
$-x\hat\imath+z\hat\jmath$, with magnitude
$R=\sqrt{x^2+z^2}$. Coulomb's law therefore gives a vector element proportional to
the displacement divided by $R^3$. The power of three is required because the
inverse-square magnitude is multiplied by a unit vector containing one further power
of distance.

The contribution from the element at x has horizontal and vertical components. Its
partner at minus x reverses only the horizontal component. Pairing elements before
integration therefore removes the horizontal field exactly and leaves

$$
\d E_z=\frac{1}{4\pi\varepsilon_0}\frac{\lambda z\d x}{(x^2+z^2)^{3/2}}.
$$

Integration from minus a to a gives the perpendicular-bisector field. The symmetry
argument does not state that each element has zero horizontal field; it states that
the integral of paired horizontal components is zero. A nonuniform density or an
observation point displaced from the perpendicular bisector generally destroys this
cancellation.

$$
% caption: Adding the two contributions at $P$. The horizontal parts of $\d\vec E$ from the mirror-image elements point oppositely and cancel; the vertical parts point the same way and sum to the axial field $E_z$.
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\filldraw[draw=black,fill=black!8] (0,0) circle (.08);
\node[below] at (0,-.1) {$P$};
\draw[->,black] (0,0)--(-1.3,1.0);
\node[left] at (-1.3,1.0) {$dE$};
\draw[->,black] (0,0)--(1.3,1.0);
\node[right] at (1.3,1.0) {$dE$};
\draw[dashed,black] (-1.3,1.0)--(1.3,1.0);
\draw[->,acc,very thick] (0,0)--(0,1.55);
\node[right] at (0,1.35) {$E_z$};
\end{tikzpicture}
$$

The setup itself is reusable. State the density, choose a differential element with
the correct units, form the directed source-to-field displacement, and apply symmetry
only after the vector components have been identified. A numerical integration uses
the same construction: replace the integral by a weighted sum of these directed
elements and refine the spacing until the result converges.

### The ring on its axis

Every element of a uniformly charged ring has the same distance to a point on the
ring axis. For ring radius $R$, total charge $Q$, and observation distance $z$ from
the centre, that common distance is $s=\sqrt{R^2+z^2}$. Each element produces a
field magnitude proportional to $\d q/s^2$, but components in the plane of the ring
cancel with components from the diametrically opposite element. Only the axial
component remains after integration. Every element has a continuous set of azimuthal
partners, unlike the single reflected partner available for a finite line.

$$
% caption: A uniform ring on its axis. Every element lies the same distance $s=\sqrt{R^2+z^2}$ from $P$; contributions from diametrically opposite elements cancel transversely and add along the axis.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
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\node[black] at (1.15,1.0) {$s$};
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$$

The axial projection of one element is the field magnitude times $z/s$. Integrating
over the complete ring replaces the sum of charge elements by $Q$ and gives

$$
E_z=\frac{1}{4\pi\varepsilon_0}\frac{Qz}{(R^2+z^2)^{3/2}}.
$$

The expression has several checks. At the ring centre, $z=0$, it gives zero field by
symmetry. Far from the ring, where $z$ is much larger than $R$, it approaches
$kQ/z^2$, the field of total charge concentrated at the origin. The field has a
maximum at a nonzero distance because it begins at zero, rises as the axial
projection grows, then falls by inverse-square distance scaling.

$$
% caption: On-axis ring field. $E_z$ is zero at the centre, rises to a maximum at $z=R/\sqrt2$, and falls off as $kQ/z^2$ far away.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->,black] (0,0)--(5,0) node[right] {$z$};
\draw[->,black] (0,0)--(0,2.6) node[above] {$E_z$};
\draw[acc,very thick] plot[domain=0:4.6,samples=140] ({\x},{4.2*\x/pow(1+\x*\x,1.5)});
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\node[below] at (0.707,0) {max};
\node[below left] at (0,0) {$0$};
\end{tikzpicture}
$$

The ring integral also provides a numerical benchmark. A computational sum divides
the ring into equal azimuthal elements, sets each $\d q=Q/N$, evaluates its vector
field, and sums components. Increasing $N$ should converge to the axial expression.
At off-axis points, transverse cancellation is incomplete and the same discrete
vector sum remains valid, though no equally simple scalar reduction is available.

A uniformly charged disk is constructed from concentric rings. A ring of radius r
and thickness dr has area $2\pi r\d r$ and charge $\d q=\sigma2\pi r\d r$ for constant
surface density $\sigma$. The axial field contribution has the same form as the ring
result, with $R$ replaced by the integration radius r. Summing rings from the centre
to disk radius R gives an integral whose numerator contains both the annular radius
and the observation distance. The radius factor comes from the annular area and
distinguishes a disk integral from a line-charge integral.

The integration variable labels source geometry, while z remains fixed as the field
point coordinate. Keep z fixed so the source coordinate alone is integrated. Keep
that distinction visible throughout the algebra. The axial symmetry has already
removed all transverse components before the radial integral is written. At the disk
centre,
the limiting field is finite for a finite surface density. Far from the disk, the
completed integral must approach the field of total charge $Q=\sigma\pi R^2$ at its
centre. Both limits are checks on the annular charge element and integration bounds.

Numerical quadrature replaces a continuous source by a finite collection of charge
elements. A midpoint rule evaluates the integrand at the centre of each source bin
and multiplies by bin width. For a smooth integrand on equal bins, its error generally
falls more rapidly with bin refinement than a left- or right-endpoint sum because
leading local deviations cancel around the midpoint. A trapezoidal rule instead
averages endpoint values and is often convenient when source values are already known
at tabulated positions. Neither method repairs a wrong vector construction: numerical
convergence only shows that the chosen discrete model has approached its own integral.

With source model and observation point held fixed, retain the vector result at each
refinement level and use the successive difference as the numerical diagnostic:

$$
\Delta_N=\left\lVert\vec E_{2N}-\vec E_N\right\rVert.
$$

| check | quantity held fixed | acceptable trend | failure indicated by the result |
| --- | --- | --- | --- |
| charge normalization | total $Q$ and density law | discrete charge approaches the prescribed $Q$ | incorrect element weight or Jacobian |
| vector sum | field point and coordinate basis | each component approaches a stable value | sign or projection error |
| refinement | source partition and quadrature rule | $\Delta_N$ decreases as $N$ increases | unresolved feature or a nonconvergent setup |
| independent limit | source geometry | far field approaches the point-charge result | incorrect total charge or distance dependence |

The comparison uses the same physical source at every $N$. Changing the density law,
source boundary, or observation coordinate during refinement compares different
models rather than estimates quadrature error.

Symmetry should be imposed before numerical work whenever an exact pairing exists.
A uniformly charged rod observed on its perpendicular bisector has equal-charge,
equal-distance bins at plus x and minus x. Their horizontal
components cancel analytically, so a numerical calculation need only sum vertical
components. If a computed horizontal component remains appreciable after pairing,
the source bins, field-point coordinates, or signs have been assigned inconsistently.
The same shortcut applies to a nonuniform density only when that density has the
required reflection symmetry. A density that differs on the two halves of the rod
must retain both components in the quadrature.

Convergence is tested by repeating a calculation with successively smaller bins. A
stable result should approach a limit, while the difference between refinements
provides an estimate of numerical uncertainty. Near a singular source point or a
sharp density discontinuity, uniform bins can converge slowly. Splitting the source
into smooth subintervals, using finer bins near rapid variation, or integrating a
known singular part analytically improves reliability without changing the physical
superposition principle.

Compare results instead of inferring accuracy from a visually smooth sum. Compute the
field with bin width h and again with a reduced width, such as h divided by two. The
difference between the two values estimates the remaining discretization error when
the calculation has entered its convergence regime. If the difference fails to
decrease under refinement, either the source has a feature that is unresolved or the
discretized vector formula is incorrect. Reporting the number of bins without a
refinement comparison gives no direct accuracy measure.

Adaptive element size concentrates source elements where the integrand changes
rapidly. Near the closest point of a line charge to the field point, distance can
change substantially over a short segment, so smaller elements reduce local error.
Farther away, larger segments can give the same accuracy at lower computational cost.
Disk radial bins may need refinement near a radius where an observation point
lies close to the source plane. Adaptive refinement changes numerical sampling; it
does not alter the density definition or replace the required geometric Jacobian in
the source element.

Cancellation can conceal numerical error. Large positive and negative components may
sum to a small net field, so a small absolute residual does not prove that each
component is accurate. Symmetric pair evaluation reduces this loss by combining
analytically cancelling terms before floating-point subtraction. When symmetry is
not exact, retain sufficient precision and compare component sums separately. A
computed zero should be tested against the known symmetry of the physical source,
not accepted merely because rounded values happen to cancel.

Special limits provide independent validation. A finite charged rod observed far away
must approach the field of its total charge. A ring field on its axis must vanish at
the centre and have the correct inverse-square far-field behaviour. A large disk
observed near its centre should approach the infinite-sheet result. These checks test
source normalization, distance powers, vector direction, and integration bounds at
once. Failure of any required limit identifies a definite error in the model or
calculation. Agreement in one limit leaves other setup errors possible.

A two-dimensional surface distribution is built from an area element rather than a
line element. In Cartesian coordinates on a planar surface, the source charge is
$\d q=\sigma(x,y)\d x\d y$. The source-to-field displacement must then be written from
each source coordinate pair to the fixed observation point before the electric-field
vector is projected onto chosen coordinate axes. Rectangular boundaries and density functions
given directly in x and y often make Cartesian elements the shortest setup, even
when the later integral requires numerical evaluation.

Polar coordinates are preferable for circular disks, annuli, and radially symmetric
density functions. The area element is $\d A=r\d r\d\phi$, so the charge of a thin
annular sector is $\d q=\sigma(r,\phi)r\d r\d\phi$. The extra factor r is the
Jacobian that accounts for the increasing circumference of annuli. Omitting it gives
equal charge to rings of unequal area and produces an incorrect disk field. The
coordinate choice is therefore part of the physical source model because it fixes
the charge represented by each differential element.

Symmetry can reduce a two-dimensional integral before calculation. At a point on the
axis of a uniformly charged disk, integration over azimuth removes all transverse
components, leaving a one-dimensional radial integral. At an off-axis point, the same
azimuthal cancellation generally fails, and both coordinates must remain. The field
direction should be checked against the source geometry before evaluating the
integral: a positive disk pushes a positive test charge away from its surface, while
reflection symmetries can force components to vanish on selected planes.

Axial symmetry of a finite circular surface is exact only on its axis. Every annular
source element then has a full azimuthal set of partners whose
transverse field components cancel. The remaining axial components have the same
direction for a positive surface charge above the disk. This reduces a two-dimensional
surface integral to a radial integral, but it does not make the finite disk equivalent
to an infinite sheet. The disk radius remains in the distance denominator and sets
the near- and far-field crossover.

Moving P off the axis removes the rotational pairing. Opposite elements of an annulus
are no longer equal distances from the field point, so their transverse contributions
do not cancel exactly. The full surface integral, or a carefully converged numerical
quadrature, is then required to determine both field magnitude and direction.

The axial field of a uniformly charged disk follows by integrating ring contributions
from radius zero to R. The resulting expression contains a constant surface-density
term minus a geometric correction determined by observation distance and disk radius.
Close to the centre of a large disk, the correction is small and the field approaches
the infinite-sheet magnitude on one side. The approximation fails near the edge,
where field lines spread outward and the finite radius cannot be ignored. Near the
disk edge, fringing changes both field magnitude and direction; the infinite-sheet
approximation is invalid there.

Far from the disk, observation distance is large compared with R. The detailed
annular source geometry becomes unresolved, and the leading field must approach that
of total charge $Q=\sigma\pi R^2$ placed at the disk centre. The axial disk result
then reduces to the inverse-square point-charge field. This limit checks both the
area factor in total charge and the distance powers in the integrated expression.
If a numerical disk calculation does not approach the point-charge result as the
field point recedes, its annular weights or radial bounds are incorrect.

The two limits describe different regimes. The near-centre large-disk limit tests
local planar behaviour, while the far-field limit tests total-charge behaviour. A
finite disk calculation must interpolate between them rather than applying either
approximation at every distance.

### Disk-to-sheet limit

The on-axis field of a uniformly charged disk provides a controlled route to the
infinite-sheet result. Hold surface density and observation distance fixed while the
disk radius grows. The outer annuli become increasingly distant from the observation
point, yet their increasing area adds a finite cumulative contribution. In the
limit of radius much larger than the observation distance, the edge is no longer
resolved locally and the axial field approaches the constant field of an infinite
sheet on one side. The direct disk integral therefore gives a validation of the
sheet result without assuming planar symmetry at the outset.

This limit has a clear scope. It applies near the central region of a large disk, not
near its rim. At a fixed finite radius, moving the observation point outward instead
produces the point-charge far-field limit. The two limiting operations change which
geometric scale is dominant, so they cannot be interchanged without stating the
observation regime.

$$
% caption: Disk-to-sheet limit at fixed height. Left: at a fixed axial point, a larger disk radius pushes the edge out of view. Right: the axial field grows with $R$ toward the infinite-sheet value $\sigma/2\varepsilon_0$.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
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\begin{scope}
  \draw[acc,thick,fill=acc!8] (-1.5,0) ellipse (.85 and .24);
  \draw[acc,thick,fill=acc!8] (1.4,0) ellipse (1.5 and .42);
  \filldraw[draw=black,fill=black!8] (0,1.6) circle (.08);
  \node[above] at (0,1.6) {$P$};
  \draw[dashed,black] (0,0)--(0,1.6);
  \node[below] at (-1.5,-.3) {small $R$};
  \node[below] at (1.4,-.5) {large $R$};
\end{scope}
\begin{scope}[shift={(5.6,0)}]
  \draw[->,black] (0,0)--(4.2,0) node[right] {$R$};
  \draw[->,black] (0,0)--(0,2.7) node[above] {$E_z$};
  \draw[acc,very thick] plot[domain=.05:3.8,samples=110] ({\x},{1.95*(1-1/sqrt(1+(\x/.9)^2))});
  \draw[dashed,black] (0,1.95)--(3.85,1.95);
  \node[right,black] at (1.9,2.35) {sheet limit};
\end{scope}
\end{tikzpicture}
$$

Agreement with the sheet limit checks the annular charge element, the radial Jacobian,
and the axial component projection in the disk integration.

### Field and potential for a disk

The axial potential of a charged disk is often easier to integrate than its electric
field because potential contributions are scalars. Concentric rings at radius r have
a common distance to an axial observation point, so their potentials add without
resolving transverse components. After the potential is found, the axial electric
field follows from the negative derivative with respect to observation distance. This
route must reproduce the direct ring-by-ring field integral.

The comparison checks more than algebra. Potential is allowed to have an arbitrary
additive reference, but its derivative is not. A constant offset in numerical
potential does not alter the recovered field, whereas an incorrect radial charge
weight or distance factor changes both the potential curve and its slope. On the
positive side of a positively charged disk, potential decreases with increasing axial
distance, so its negative derivative gives a field directed away from the disk. The
sign is therefore fixed by the potential trend before a numerical value is computed.

For numerical data, a centred finite difference of the potential at nearby axial
points can be compared with the independently summed field. Agreement under grid
and step-size refinement validates the source weights and the directed component
calculation simultaneously.

## Discretization convergence

A numerical line, ring, or disk calculation replaces a continuous density by a
finite number of source elements. The field estimate should approach a fixed value as
the number of elements increases. For a smooth symmetric distribution, doubling the
number of equal bins usually reduces the difference between successive estimates.
The convergence trend matters more than the apparent precision of one calculation:
many displayed digits do not establish accuracy when the source mesh is coarse.

Compare field estimates at $N$, $2N$, and $4N$ elements while holding the field point
and density normalization fixed. A stable sequence supports the quadrature model;
an oscillating or drifting sequence indicates insufficient resolution, a singular
near-source configuration, or a component-sign error. Symmetry-reduced sums should
converge to the same result as full vector sums, providing a second implementation
check.

Convergence cannot validate an incorrect physical source model. It must be combined
with dimensional checks, symmetry tests, and far-field limits.

## Validation and reported assumptions

Every continuous-charge result should be tested against a limiting case that has an
independent physical interpretation. A short rod observed far away becomes a point
charge carrying its total charge. A disk with radius much larger than the observation
distance approaches an infinite sheet near its centre. A ring field vanishes at its
centre and approaches a point-charge field far from the ring. These checks constrain
charge normalization, distance powers, and direction simultaneously. They are more
reliable than a purely algebraic comparison because they test whether the completed
formula represents the intended source geometry.

Experimental reconstruction reverses the calculation: field measurements at many
points can constrain an unknown charge distribution, subject to measurement noise
and incomplete spatial coverage. The inverse problem is generally harder than
computing a field from a known density. Multiple source distributions can produce
similar fields over a limited observation region, so reconstruction requires stated
assumptions about source support, symmetry, smoothness, or total charge. Potential
measurements provide additional scalar information, but their arbitrary reference
must be handled through potential differences or a specified boundary condition.

Reported results should state the adopted source density, coordinate system, field
point, and approximation regime. A point-charge replacement needs a far-distance
justification. A direct integral requires the differential charge element and vector
direction. A numerical result requires element count, refinement evidence, and the
special limits used for validation. Changing the density, coordinates, field point,
or approximation regime changes the modeled field.

Units should be retained until the final result: line, surface, and volume densities
carry distinct dimensions, and their associated differential elements must restore
coulombs before each contribution is inserted into Coulomb's law.

Unit tracking exposes normalization errors during source-element setup.

Gauss's law is valid for every closed surface, but it determines a field magnitude
directly only when source symmetry makes the flux integral simple. A Gaussian
surface permits direct evaluation when the field is constant and normal over a known area or tangent
to the surface so that its flux is zero. Spherical symmetry permits a concentric
sphere; cylindrical symmetry permits a coaxial cylinder; infinite planar symmetry
permits a pillbox. In each case, the selected surface matches the field symmetry, not
merely the shape of an object that encloses charge.

A finite rod illustrates a symmetry failure. A sphere can enclose the rod, but points
on the sphere have different distances and directions relative to its charge
elements. The field is therefore neither constant nor everywhere normal on that
surface. Gauss's law fixes total flux, yet the local field magnitude cannot be factored
out of the flux integral. Direct component integration remains the valid method.

$$
% caption: Why a sphere fails for a finite rod. A sphere encloses the rod, but the field varies in magnitude and direction over it, so the flux cannot be reduced to $E$ times one area.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black,dashed,thick] (0,0) circle (1.6);
\draw[acc,very thick] (-.85,0)--(.85,0);
\node[acc,below] at (0,-.12) {rod};
\draw[->,black] (1.6,0)--(2.25,0);
\draw[->,black] (1.13,1.13)--(1.55,1.55);
\draw[->,black] (0,1.6)--(0,2.1);
\draw[->,black] (-1.13,1.13)--(-1.55,1.55);
\draw[->,black] (-1.6,0)--(-2.05,0);
\draw[->,black] (1.13,-1.13)--(1.5,-1.5);
\draw[->,black] (-1.13,-1.13)--(-1.5,-1.5);
\draw[->,black] (0,-1.6)--(0,-2.1);
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$$

Gauss's law still fixes the total flux, but this surface does not permit the
reduction $\Phi_E=EA$. Apply the same test before selecting any Gaussian surface.

Direct Coulomb integration remains necessary when no surface has this property. A
finite uniformly charged rod is symmetric about its midpoint, but its field magnitude
is not constant on any simple closed surface. A sphere around the rod still has a
field that varies from point to point, so the flux integral cannot be reduced to
$E(4\pi r^2)$. The same failure occurs for a finite disk away from its axis. Gauss's
law still constrains net flux, but it does not supply the local field without further
calculation.

Method selection follows the desired quantity. Use direct integration when a source
element and vector geometry can be parameterized cleanly, or when partial symmetry
eliminates components but leaves a manageable integral. Use Gauss's law when full
symmetry makes the field magnitude constant over part of a closed surface. Both
methods arise from the same inverse-square electric field and superposition. The
selected method changes the calculation while retaining the same underlying field
model. Symmetry reduces the computation.

| source and target geometry | primary representation | extracted quantity | independent check |
| --- | --- | --- | --- |
| finite rod or off-axis disk | directed Coulomb element | field components | reflection symmetry and far-field limit |
| spherical, cylindrical, or planar symmetry | closed Gaussian surface | one field magnitude | constant magnitude or zero-flux regions on the surface |
| irregular source with specified density | discrete source panels | vector field and potential | refinement residual and total discrete charge |
| numerical potential map | scalar source sum plus finite difference | $-\nabla V$ | direct vector Coulomb sum at the same point |

The chosen representation must retain the source geometry that sets the symmetry.
A Gaussian surface permits flux reduction only when the field, rather than the
drawing of the surface, has the required symmetry.

A numerical charge model can compute electric potential and electric field from the
same source elements. Potential is a scalar sum of charge divided by source-to-field
distance; field is the corresponding vector sum with an additional directional
factor. Away from source elements, differentiating the numerical potential should
give the negative numerical field. This comparison provides an independent consistency test
because the two calculations fail in different ways: a missing vector direction can
leave a potential result plausible while corrupting field components, whereas a wrong
charge weight changes both quantities.

A discretized distribution is evaluated at several nearby field points to
approximate its spatial derivative with a symmetric finite difference. Compare that
derivative with the field obtained by direct vector summation at the central point.
The finite-difference spacing must be large compared with numerical roundoff but small
compared with the geometric scale over which potential changes. A disagreement that
does not shrink under source-grid refinement or derivative-step refinement indicates
an inconsistency in source coordinates, charge weights, or sign conventions.

This cross-check applies only where the field point does not coincide with a discrete
source element. The ideal point-charge kernel is singular at zero separation, so a
numerical model must exclude self-field evaluations, resolve a finite source shape,
or use an analytic local correction. Treating a singular sampled value as an ordinary
finite bin contribution produces grid-dependent results rather than a physical field.

Numerical validation should use constraints that do not depend on the same summation
routine. Symmetry provides the first check. For a source distribution symmetric under
reflection across a plane, the field component normal to that plane must vanish at
points on the plane when the reflected source charges are equal. A discretized model
that gives a persistent nonzero component after paired refinement has mismatched
source weights, coordinates, or vector signs. Testing individual components is more
informative than testing field magnitude alone because cancellation errors can remain
hidden in a plausible resultant.

The far-field limit provides a second independent check. At observation distances
large compared with every source dimension, a distribution with total charge Q must
produce leading field magnitude proportional to Q divided by distance squared. A
numerical field should therefore approach the point-charge result as the observation
point recedes. If total charge is zero, that leading term cancels and the field must
decrease more rapidly; failure to show that cancellation indicates an error in charge
normalization or source placement. Near-field agreement does not replace this test,
because a local grid can accidentally reproduce one value while carrying the wrong
global charge or symmetry.

Refinement studies and physical limits should agree simultaneously. A result that
converges with bin count but violates a known symmetry or far-field limit has
converged to an incorrectly posed discrete model. Validation requires geometric,
dimensional, and asymptotic checks in sequence. Independent analytic checks should
agree within the stated numerical resolution.

Record the quadrature rule, source partition, and refinement sequence with a
numerical field result. Uniform bins, adaptive bins, and analytic removal of a
near-singular contribution converge differently near sharp edges and close field
points. A reported field should identify the discretization-error estimate and the
physical source model that the discretization approximates.