EG1216 · Physics 2 — Electricity & Magnetism
Theme 1 · Electrostatics — charge & the electric field

The field of continuous charge distributions

Real charge is spread along rods, around rings, and over sheets. Add up the field of every tiny piece with an integral.

Source: Serway & Jewett, Physics for Scientists and Engineers, 7th ed., §23.5.

Before you start

What you need first

  • Topic 3 — the field — \(E = k|q|/r^2\) for a point charge, and adding fields by superposition.
  • Integration — the idea of summing infinitely many tiny pieces \(\int (\dots)\,dq\) (Physics 1 / Calculus).

What you'll be able to do

  • Describe spread-out charge with \(\lambda\), \(\sigma\), \(\rho\).
  • Set up the field integral \(E = k\!\int dq/r^2\) and use symmetry to drop components.
  • Derive the field on the axis of a charged ring.
  • Use the standard results for a long line and an infinite sheet.

Describing spread-out charge

Charge density: λ, σ, ρ

When charge is spread out, we describe how much charge sits per unit of length, area, or volume — its charge density.

Each is just total charge \(Q\) divided by the size of the thing it is spread over.

rod — λ (C/m) plate — σ (C/m²) ball — ρ (C/m³)
$$ \lambda = \frac{Q}{L} \qquad \sigma = \frac{Q}{A} \qquad \rho = \frac{Q}{V} $$
where:
SymbolMeaningSI unit
λlinear charge density — charge per unit length along a lineC/m
σsurface charge density — charge per unit area on a surfaceC/m²
ρvolume charge density — charge per unit volume in a solidC/m³
Qtotal charge spread over the objectC
L, A, Vthe length, area, or volume it is spread overm, m², m³

The idea

Break it into pieces, then add the fields

A spread-out charge is just many tiny point charges side by side. Each tiny piece \(dq\) makes a tiny field (Topic 3); the total field is the sum of all of them — and "sum of infinitely many tiny pieces" is an integral.

$$ E = k\!\int \frac{dq}{r^{2}} $$
where:
SymbolMeaningSI unit
dqcharge of one tiny piece (e.g. \(dq=\lambda\,dl\) on a line)C
rdistance from that piece to the point where we want the fieldm
kCoulomb's constant \(=8.99\times10^{9}\)N·m²/C²
Symmetry is the trick. The field is a vector, so really each piece adds a little arrow. For a symmetric shape, the sideways parts of those arrows cancel in pairs, and only one direction survives — which turns a hard vector sum into a simple one. We see this next on the ring.

Derivation

The field on the axis of a charged ring

Why we do this one: the ring shows the symmetry trick at its cleanest, and its result is the building block for a disk and a sheet.

Setup & assumptions: a ring of radius \(a\) carries total charge \(Q\) spread evenly. We want the field at a point \(P\) on the axis, a distance \(x\) from the centre. Take \(Q>0\).

axis C a dq x P r θ dE dEₓ dE⊥
A piece \(dq\) at distance \(r\) makes field \(dE\) at \(P\). Split it into axial \(dE_x\) and perpendicular \(dE_\perp\) (measured from the dashed axis).
1Each piece \(dq\) is a point charge a distance \(r=\sqrt{x^2+a^2}\) from \(P\), so its field size is:
$$ dE = k\,\frac{dq}{r^{2}} = k\,\frac{dq}{x^{2}+a^{2}} $$
2Symmetry: for every piece, the one on the opposite side of the ring cancels its perpendicular part \(dE_\perp\). Only the axial part survives, with \(\cos\theta = \dfrac{x}{r} = \dfrac{x}{\sqrt{x^2+a^2}}\):
$$ dE_x = dE\,\cos\theta = k\,\frac{dq}{x^{2}+a^{2}}\cdot\frac{x}{\sqrt{x^{2}+a^{2}}} $$
3Add up all the pieces. Everything except \(dq\) is the same for every piece (all are distance \(r\) away), so it comes out of the integral and \(\int dq = Q\):
$$ E = \int dE_x = \frac{k\,x}{(x^{2}+a^{2})^{3/2}}\int dq $$
4The result — the field on the axis of a ring:
$$ E = \frac{k\,Q\,x}{(x^{2}+a^{2})^{3/2}} $$
where:
SymbolMeaningSI unit
Qtotal charge on the ringC
aradius of the ringm
xdistance from the centre, along the axism
At the centre (\(x=0\)) the result gives \(E=0\): every piece is exactly balanced by the piece across the ring. Far away (\(x \gg a\)) it becomes \(E \approx kQ/x^2\) — the ring looks like a point charge, as it should.
📐 Worked example 1

Field on the axis of a ring

A ring of radius \(a = 0.10\) m carries \(Q = 10\ \mu\text{C}\). Find the field at a point on the axis \(x = 0.20\) m from the centre.

1Put the numbers into the ring formula:
$$ E = \frac{(8.99\times10^{9})(10\times10^{-6})(0.20)}{\big((0.20)^{2}+(0.10)^{2}\big)^{3/2}} $$
2Work the bottom: \((0.04+0.01)^{3/2}=(0.05)^{3/2}\approx 0.0112\):
$$ E \approx 1.6\times10^{6}\ \text{N/C} $$
The field points along the axis, away from the ring (the charge is positive).

Two standard results: the long line and the infinite sheet

The same "integrate the pieces" method gives these two results, which we will use rather than re-derive each time:

Long straight line

$$ E = \frac{2k\lambda}{r} $$

Field points straight out from the line. It falls off as \(1/r\) — slower than a point charge — because charge keeps contributing all along the line.

Infinite flat sheet

$$ E = 2\pi k\sigma = \frac{\sigma}{2\varepsilon_0} $$

Field points straight out from the sheet and is uniform — the same size no matter how far away (as long as you stay close compared with the sheet's size).

where (new symbol):
SymbolMeaningSI unit
rdistance from the linem
ε₀permittivity of free space — an alternative to \(k\), with \(k=\tfrac{1}{4\pi\varepsilon_0}\); \(\varepsilon_0 = 8.85\times10^{-12}\)C²/(N·m²)
🔭 Looking ahead: a disk is a stack of rings, and a very large disk becomes the infinite sheet — that uniform field is exactly what sits between two parallel plates. We use it in Topic 5 (a charge moving in a uniform field) and Topic 6 (Gauss's law).
📐 Worked example 2

Field of a long charged wire

A long straight wire has linear charge density \(\lambda = 5\ \mu\text{C/m}\). Find the field \(r = 0.10\) m from the wire.

1Use the long-line result:
$$ E = \frac{2k\lambda}{r} = \frac{2(8.99\times10^{9})(5\times10^{-6})}{0.10} $$
2Work it out:
$$ E \approx 9.0\times10^{5}\ \text{N/C} $$
Move twice as far from the wire and the field halves (it goes as \(1/r\)) — not a quarter, unlike a point charge.

✏️ Try it yourself

A very large flat sheet carries a surface charge density \(\sigma = 5\ \mu\text{C/m}^2\).

(a) Find the electric field just outside the sheet.
(b) What is the field 1 metre from the sheet?

(a) Step 1. \(E = 2\pi k\sigma = 2\pi(8.99\times10^{9})(5\times10^{-6})\) Step 2. \(E \approx 2.8\times10^{5}\ \text{N/C}\). (b) Still \(\approx 2.8\times10^{5}\ \text{N/C}\) — a sheet's field is uniform, so distance does not change it.

Common mistakes

MistakeFix
Using the point-charge \(1/r^2\) for a line or sheet.Pick the result that matches the shape.
Forgetting the perpendicular parts cancel.For a symmetric shape only one component survives — that is what makes the integral easy.
Thinking a sheet's field weakens with distance.An (infinite) sheet's field is uniform.
Leaving \(\mu\text{C}\), cm, or \(\mu\text{C/m}\) unconverted.Convert to C and m first.

Recap — the whole topic on one screen

IdeaWhat you own now
Density\(\lambda=Q/L\), \(\sigma=Q/A\), \(\rho=Q/V\).
Method\(E = k\!\int dq/r^2\); use symmetry to drop the cancelling components.
Ring (axis)\(E = \dfrac{kQx}{(x^2+a^2)^{3/2}}\); zero at the centre, point-like far away.
Long line\(E = \dfrac{2k\lambda}{r}\); falls off as \(1/r\).
Infinite sheet\(E = 2\pi k\sigma = \dfrac{\sigma}{2\varepsilon_0}\); uniform.

Next topic

A charge moving in a uniform field

The sheet (and a pair of parallel plates) gives a uniform field. Next we drop a charge into one and watch it accelerate — straight-line speed-up, or a parabola like a projectile.

→ Topic 5 · Motion of a charge in a uniform field