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Fields of rings, disks and lines by integrating Coulomb's law, using symmetry to cancel components and checking the far-field limit.
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By the end of this lesson you will be able to set up and evaluate the electric field of continuous charge distributions by integrating Coulomb's law, and check the result in its limits.
You know Coulomb's law for point charges, how to add vectors by components, and from Physics C the fields of symmetric distributions found with Gauss's law. From multivariable calculus you can set up single and double integrals and use cylindrical coordinates. This lesson builds fields directly from Coulomb's law by integration, the method that works even when Gauss's law does not.
| Term | What it means |
|---|---|
| Superposition | The field of many charges is the vector sum of their individual fields. |
| Line charge density | $\lambda$, charge per unit length, in C/m. |
| Surface charge density | $\sigma$, charge per unit area, in C/m². |
| Volume charge density | $\rho$, charge per unit volume, in C/m³. |
| Separation vector | $\vec{\mathscr{r}} = \vec{r} - \vec{r}'$, from the source point to the field point. |
| Coulomb constant | $k = 1/(4\pi\varepsilon_0) = 8.99 \times 10^9$ N m²/C². |
| Point-charge limit | Far from any bounded distribution, its field approaches that of a point charge with the same total. |
The field at $\vec{r}$ of a point charge $q$ at $\vec{r}'$ is
$$\vec{E} = \frac{1}{4\pi\varepsilon_0}\frac{q}{\mathscr{r}^2}\hat{\mathscr{r}}, \qquad \vec{\mathscr{r}} = \vec{r} - \vec{r}'.$$
Fields add as vectors, so for a continuous distribution divide it into pieces $dq = \lambda\,dl$, $\sigma\,da$ or $\rho\,d\tau$ and integrate:
$$\vec{E}(\vec{r}) = \frac{1}{4\pi\varepsilon_0}\int\frac{\hat{\mathscr{r}}}{\mathscr{r}^2}\,dq.$$
The integral is a vector, so it must be done component by component, and symmetry usually kills all but one. On the axis of a ring of radius $R$, every piece is at the same distance $\sqrt{z^2 + R^2}$, the sideways components cancel in opposite pairs, and the axial components add:
$$E_z = \frac{1}{4\pi\varepsilon_0}\frac{Qz}{(z^2 + R^2)^{3/2}}.$$
Stacking rings builds a disk, $E_z = \frac{\sigma}{2\varepsilon_0}\left(1 - \frac{z}{\sqrt{z^2 + R^2}}\right)$, which becomes the infinite-plane field $\sigma/2\varepsilon_0$ as $R \to \infty$. Integrating along an infinite line gives $E = \lambda/(2\pi\varepsilon_0 s)$. Far away, every one of these reduces to $kQ/r^2$.
Another way: picture
Picture standing on the axis of a charged ring. Every bit of charge pushes on you along a line from itself, slanting outward. The slanting parts from opposite sides cancel, leaving only the push along the axis. At the center, the pushes are all sideways and cancel completely; far away, the ring looks like a dot. In between, the field rises from zero, peaks and falls off — a shape no point charge could give.
Another way: steps
Checks. The answer must have units of N/C. Far away, it must reduce to $kQ/r^2$ with $Q$ the total charge. Close to a flat sheet, it must approach $\sigma/2\varepsilon_0$. It must vanish wherever symmetry demands, such as the center of a ring or the midpoint between equal charges. And it must point away from positive charge.
Symmetry is a statement about the source, and it carries over to the field. If rotating a ring about its axis leaves the charge unchanged, then the field on the axis cannot point sideways, because there is no preferred sideways direction for it to choose. The pairwise cancellation in the integral is the calculation's way of saying the same thing.
Using this before integrating saves work and prevents errors. It also tells you when a simple answer is impossible: off the axis of a ring, there is no symmetry to kill the radial component, and the integral gives elliptic functions rather than elementary ones. In such cases the field is computed numerically, as engineers do with finite-element software for real electrode shapes.
The disk result, $\frac{\sigma}{2\varepsilon_0}(1 - z/\sqrt{z^2 + R^2})$, contains two useful limits. Close to the disk, $z \ll R$, the bracket is nearly one and the field is $\sigma/2\varepsilon_0$, independent of distance: near any flat charged surface the surface looks infinite. Far away, $z \gg R$, expand $z/\sqrt{z^2 + R^2} \approx 1 - R^2/2z^2$, and the field becomes $\sigma R^2/(4\varepsilon_0z^2) = kQ/z^2$ with $Q = \sigma\pi R^2$.
The constant field near a sheet is why parallel-plate capacitors have uniform fields between them: two sheets of opposite charge give $\sigma/2\varepsilon_0$ each, adding between the plates to $\sigma/\varepsilon_0$ and cancelling outside. Edge effects appear only within about one plate separation of the edges.
An infinite line charge gives $E = 2k\lambda/s$, falling as $1/s$ rather than $1/s^2$. The reason is geometric: as you move away, more of the line contributes at similar angles, partly compensating for each piece's weakening. In general, charge spread in $d$ dimensions produces a field falling as $1/r^{2-d}$: $1/r^2$ for a point, $1/r$ for a line, constant for a plane.
A real wire of length $L$ looks infinite only when $s \ll L$. For a finite segment, the integral gives $E = \frac{k\lambda}{s}(\sin\theta_2 - \sin\theta_1)$, where the angles are measured to the two ends. Power-line engineers use such formulas, with the ground acting as a mirror, to compute the fields under transmission lines, which reach a few kilovolts per meter near the ground under the largest lines.
Gauss's law, in the next lessons, finds fields in a line or two when the charge has spherical, cylindrical or planar symmetry. But most real distributions — a ring, a finite disk, the electrodes of an ion trap — lack the full symmetry Gauss's law needs. Then the Coulomb integral, or its cousin for the potential, is the direct route.
Computers do these integrals routinely. Designers of electron guns, mass spectrometers and the ion optics of electron microscopes divide electrodes into thousands of small pieces and sum their fields, exactly as this lesson does with calculus. The analytic cases here serve as the tests that such programs must pass before anyone trusts them with a real design.
The coulomb is a large charge: two $1$ C charges a meter apart would push with $9 \times 10^9$ N. Everyday static charges are nanocoulombs to microcoulombs. A balloon rubbed on hair carries about $0.1$ μC and makes a field of tens of kilovolts per meter near its surface. Air breaks down at about $3 \times 10^6$ V/m, which limits how much charge a surface in air can hold: $\sigma_{\max} = 2\varepsilon_0E \approx 50$ μC/m² for a single sheet.
Keeping powers of ten straight is most of the work in numerical problems: nC is $10^{-9}$ C, μC is $10^{-6}$ C, cm is $10^{-2}$ m. Writing $k = 8.99 \times 10^9$ N m²/C² and $\varepsilon_0 = 8.85 \times 10^{-12}$ C²/(N m²), and checking that $k\varepsilon_0 = 1/4\pi$, catches many slips before they spread.
Most of the difficulty in a Coulomb integral is bookkeeping, and good coordinates remove most of it. For a ring or a disk, cylindrical coordinates centered on the axis make the distance to every piece depend only on the radius, so the angle integral is trivial. For a sphere, spherical coordinates with the field point on the polar axis make the distance depend only on the polar angle. For a straight segment, measuring position along the segment and converting to the angle seen from the field point turns awkward square roots into sines and cosines.
A useful habit is to write the three ingredients separately before combining them: the charge element $dq$ in the chosen coordinates, the distance $\mathscr{r}$ from that element to the field point, and the component of $\hat{\mathscr{r}}$ that survives. Mistakes usually come from mixing up the source coordinate, which is integrated over, with the field-point coordinate, which is held fixed. Keeping primes on source coordinates, as Griffiths does, makes the difference visible on the page and prevents most of those slips.
Coal and cement plants, steel mills and incinerators remove fine ash from their exhaust with electrostatic precipitators, which the U.S. Environmental Protection Agency credits with capturing over $99$ percent of particulate matter in many installations. The simplest design is a grounded tube with a thin wire along its axis, charged to tens of kilovolts. The wire acts as a line charge, with a field $2k\lambda/s$ that is intense near the wire and weaker toward the tube wall.
Near the wire the field exceeds air's breakdown strength, creating a corona of ions. The ions stick to passing particles, which the field then drives to the tube wall, where they are collected and periodically shaken loose. The $1/s$ law is the key to the design: it makes the field strong enough to ionize the gas only in a thin sheath around the wire, while the rest of the tube carries a gentler field that sweeps charged dust outward without sparking. Frederick Cottrell invented the device at the University of California in 1907.
Photocopiers and laser printers work by charging a photoconductive drum uniformly, to a surface density of order $10^{-4}$ C/m², then using light to discharge the areas that should stay white. Near the drum, the field of the remaining charge is the infinite-sheet value $\sigma/2\varepsilon_0$ within a distance comparable to the charged patch's size, and it falls off farther out, just as the disk formula predicts.
Toner particles, charged oppositely, are pulled onto the charged areas. Small features, whose charge patches are narrow, have fields that decay within a short distance, which is why printers must bring toner very close to the drum to reproduce fine lines. Engineers designing these systems use exactly the ring and disk integrals of this lesson, extended to strips and dots, to predict how sharply an image will develop.
Replacing a distribution by a point charge at its center works far away, and exactly for spherically symmetric charge outside it, but nowhere else in general. On the axis of a ring, the point-charge guess $kQ/z^2$ blows up at the center, where the true field is zero. Near a disk, it gives a field growing without limit, when the true field levels off at $\sigma/2\varepsilon_0$.
A second error is adding magnitudes instead of vectors. Every piece of a ring contributes a field of the same size at the center, but they point in all directions and cancel. Always decide which components survive before integrating, and integrate only those.
Write a piece of the ring.
$dq = \lambda R\,d\phi, \qquad \lambda = \dfrac{Q}{2\pi R}$
Uniform charge around the circumference.
Find its distance to the field point.
$\mathscr{r} = \sqrt{z^2 + R^2}$
The same for every piece.
Keep the axial component.
$dE_z = \dfrac{k\,dq}{\mathscr{r}^2}\cdot\dfrac{z}{\mathscr{r}}$
Sideways parts cancel in opposite pairs.
Integrate around the ring.
$E_z = \dfrac{kz}{(z^2 + R^2)^{3/2}}\displaystyle\oint dq = \dfrac{kQz}{(z^2 + R^2)^{3/2}}$
Everything but $dq$ is constant.
Check the far field.
$z \gg R: \quad E_z \to \dfrac{kQ}{z^2}$
A point charge from far away.
Place the line on the $x$ axis and the field point at distance $s$.
$dq = \lambda\,dx, \qquad \mathscr{r} = \sqrt{s^2 + x^2}$
A piece at position $x$ along the line.
Keep the perpendicular component.
$dE_\perp = \dfrac{k\lambda\,dx}{s^2 + x^2}\cdot\dfrac{s}{\sqrt{s^2 + x^2}}$
Components along the line cancel between $\pm x$.
Substitute the angle.
$x = s\tan\theta, \qquad dx = s\sec^2\theta\,d\theta$
Turns the integrand into a cosine.
Simplify the integral.
$E = \dfrac{k\lambda}{s}\displaystyle\int_{-\pi/2}^{\pi/2}\cos\theta\,d\theta$
The powers of $\sec\theta$ cancel.
Evaluate the integral.
$E = \dfrac{k\lambda}{s} \times 2 = \dfrac{2k\lambda}{s}$
$\sin$ from $-1$ to $1$.
Evaluate for $\lambda = 1$ μC/m at $s = 5$ cm.
$E = \dfrac{2 \times 8.99 \times 10^9 \times 10^{-6}}{0.05} = 3.6 \times 10^5\ \text{N/C}$
About a tenth of air's breakdown field.
Divide the disk into rings.
$dq = \sigma\,2\pi r\,dr$
A ring of radius $r$ and width $dr$.
Use the ring result for each.
$dE_z = \dfrac{kz\,\sigma\,2\pi r\,dr}{(z^2 + r^2)^{3/2}}$
All rings' fields point along the axis.
Set up the integral.
$E_z = 2\pi k\sigma z\displaystyle\int_0^R\dfrac{r\,dr}{(z^2 + r^2)^{3/2}}$
Sum over the rings.
Substitute $w = z^2 + r^2$.
$\displaystyle\int\dfrac{r\,dr}{(z^2 + r^2)^{3/2}} = -\dfrac{1}{\sqrt{z^2 + r^2}}$
$dw = 2r\,dr$.
Evaluate between the limits.
$E_z = 2\pi k\sigma z\left(\dfrac{1}{z} - \dfrac{1}{\sqrt{z^2 + R^2}}\right) = \dfrac{\sigma}{2\varepsilon_0}\left(1 - \dfrac{z}{\sqrt{z^2 + R^2}}\right)$
Using $2\pi k = 1/2\varepsilon_0$.
Check the near limit.
$z \ll R: \quad E_z \to \dfrac{\sigma}{2\varepsilon_0}$
An infinite sheet.
Check the far limit.
$z \gg R: \quad E_z \to \dfrac{\sigma}{2\varepsilon_0}\cdot\dfrac{R^2}{2z^2} = \dfrac{k\sigma\pi R^2}{z^2}$
A point charge $Q = \sigma\pi R^2$.
Evaluate for $\sigma = 2$ μC/m², $R = 4$ cm, $z = 3$ cm.
$E_z = 113\ \text{kN/C} \times \left(1 - \tfrac{3}{5}\right) = 45\ \text{kN/C}$
Forty percent of the infinite-sheet value.
Write the line-charge field.
$E = \dfrac{2k\lambda}{s}$
For an effectively infinite line.
Substitute the values.
$E = \dfrac{2 \times 8.99 \times 10^9 \times 2 \times 10^{-6}}{0.10}$
SI units.
Evaluate the field.
A thin ring carries $23$ nC spread uniformly around it. What is the electric field at the ring's center?
Complete the worked solution: a charge of $6$ nC sits $1$ m to the left of point P and a charge of $2$ nC sits $2$ m to the right. Find each field's magnitude at P and the net field along the line (positive to the right), in N/C, with $k = 8.99 \times 10^9$ N m²/C².
Find the left charge's field at P.
$E_1 = \dfrac{8.99 \times 6}{1^2} =$ p
Points to the right, away from the positive charge.
Find the right charge's field at P.
$E_2 = \dfrac{8.99 \times 2}{2^2} =$ q
Points to the left.
Subtract to find the net field.
$E_{\text{net}} = E_1 - E_2 =$ r
Superposition of vectors along one line.
Match each charge distribution to the magnitude of its electric field.
| $kq/r^2$ | $kQz/(z^2 + R^2)^{3/2}$ | $2k\lambda/s$ | $\sigma/2\varepsilon_0$ | |
|---|---|---|---|---|
| a point charge | ||||
| a ring, on its axis | ||||
| an infinite line | ||||
| an infinite plane |
A ring of radius $R$ has $kQ/R^2 = 432$ N/C. Fill in the field on its axis, in N/C, at $z = 0$, $z = \tfrac{3}{4}R$ and $z = \tfrac{4}{3}R$.
| $E$ (N/C) | |
|---|---|
| $z = 0$ | |
| $z = \tfrac{3}{4}R$ | |
| $z = \tfrac{4}{3}R$ |
A ring of radius $5$ cm carries $6$ nC. What is the field on its axis $12$ cm from its center, in N/C? Use $k = 8.99 \times 10^9$ N m²/C².
Answer: N/C
A disk of radius $R = 4$ cm carries a uniform surface charge $\sigma = 2$ μC/m². What is the electric field on its axis $z = 3$ cm from its center, in kN/C? Use $\varepsilon_0 = 8.85 \times 10^{-12}$ C²/(N m²).
Answer: kN/C
An electrostatic precipitator cleans smokestack gas with a long wire charged to $\lambda = 7$ μC/m running down the stack's axis. Treating the wire as an infinite line charge, what is the field $4$ cm from it, in kN/C?
Answer: kN/C
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A ring of radius $R$ has $kQ/R^2 = 262$ N/C. Fill in the field on its axis, in N/C, at $z = 0$, $z = \tfrac{3}{4}R$ and $z = \tfrac{4}{3}R$.
| $E$ (N/C) | |
|---|---|
| $z = 0$ | |
| $z = \tfrac{3}{4}R$ | |
| $z = \tfrac{4}{3}R$ |
You can compute fields of extended charges. Explain to someone why the field at the center of a charged ring is zero.
18. Your turn: what is the field $10$ cm from a long wire carrying $2$ μC/m?, step 3
$E = 3.6 \times 10^5\ \text{N/C}$
Twice the charge at twice the distance gives the same field as $1$ μC/m at $5$ cm.