Why the Field of an Infinite Charged Sheet Does Not Weaken With Distance

A comb run through dry hair picks up bits of paper, and lightning leaps between a cloud and the ground. Both come from electric charges and the fields they create — and one idea, Gauss's law, turns hard field calculations into a few lines.

This lesson covers Coulomb's law and superposition, the fields of point charges and dipoles, and Gauss's law for lines, sheets and shells.

**Coulomb's law states that the force between two point charges is proportional to the product of the charges and inversely proportional to the square of their separation, $F=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}$, and superposition says the net force on a charge is the vector sum of the forces from all the others.

Coulomb's law:**

$$F=\frac{1}{4\pi {\epsilon }_0}\frac{q_1{q}_2}{r^2}\frac{1}{4\pi {\epsilon }_0}=9.0\times 10^9{\text{ N m}}^2{\text{ C}}^{-2}$$

- Like charges repel and unlike charges attract, along the line joining them
- In a medium of dielectric constant K, the force falls to $F/K$
- Charge is conserved and quantised, $q=ne$, with $e=1.6\times 10^{-19}$ C

Superposition. ${\vec{F}}_{\text{net}}={\vec{F}}_{12}+{\vec{F}}_{13}+\dots$, with each pair force calculated as if the other charges were absent.

Worked example 1. Charges of $+2.0\times 10^{-6}$ C and $-3.0\times 10^{-6}$ C are 0.30 m apart:

$$F=\frac{9.0\times 10^9\times 2.0\times 10^{-6}\times 3.0\times 10^{-6}}{0.30^2}=0.60\text{ N, attractive}$$

Worked example 2 — superposition. Charges of $1.0\times 10^{-6}$ C sit at the three corners of an equilateral triangle of side 0.10 m. Each of two charges pushes the third with $\frac{9.0\times 10^9\times (1.0\times 10^{-6}{)}^2}{0.10^2}=0.90$ N, and the two forces are 60° apart, so

$$F_{\text{net}}=2\times 0.90\times \cos 30^{\circ }=1.56\text{ N}$$

An everyday example. Dust clinging to a freshly wiped plastic chair on a dry winter day is held by the attraction between the charged plastic and the dust.

The substance. Electric force dwarfs gravity — between two protons, the electric repulsion is about $10^{36}$ times their gravitational attraction.

**The electric field is the force per unit positive charge, $E=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}$ for a point charge, found by vector addition for a group of charges, and for a dipole it is $\frac{1}{4\pi {\epsilon }_0}\frac{2p}{r^3}$ on the axis and $\frac{1}{4\pi {\epsilon }_0}\frac{p}{r^3}$ on the equatorial line.

Point charges.** $E=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}$, pointing away from a positive charge and towards a negative one, in N C${}^{-1}$; for several charges, $\vec{E}={\vec{E}}_1+{\vec{E}}_2+\dots$

Electric dipole. Charges +q and -q separated by 2a have dipole moment $p=q\times 2a$, pointing from -q to +q. For $r\gg a$:

$$E_{\text{axial}}=\frac{1}{4\pi {\epsilon }_0}\frac{2p}{r^3}\text{ along }\vec{p}{E}_{\text{equatorial}}=\frac{1}{4\pi {\epsilon }_0}\frac{p}{r^3}\text{ opposite to }\vec{p}$$

Worked example 1. At 0.20 m from a charge of $4.0\times 10^{-9}$ C:

$$E=\frac{9.0\times 10^9\times 4.0\times 10^{-9}}{0.20^2}=900{\text{ N C}}^{-1}$$

Worked example 2 — a dipole. Charges of $\pm 1.0\times 10^{-9}$ C, 2.0 mm apart, give $p=2.0\times 10^{-12}$ C m. At 0.10 m:

$$E_{\text{axial}}=\frac{9.0\times 10^9\times 2\times 2.0\times 10^{-12}}{0.10^3}=36{\text{ N C}}^{-1}{E}_{\text{equatorial}}=18{\text{ N C}}^{-1}$$

An everyday example. A charged plastic comb held near a thin stream of tap water bends the stream towards it, because the water molecules behave as tiny electric dipoles.

The substance. **A dipole's field falls as $1/r^3$, faster than a point charge's $1/r^2$** — from far away, its equal and opposite charges nearly cancel.

**Gauss's theorem states that the total electric flux through any closed surface equals the enclosed charge divided by ${\epsilon }_0$, and choosing a surface that matches the symmetry of the charge makes E easy to find.**

$$\varphi =\oint \vec{E}\cdot d\vec{A}=\frac{q_{\text{enc}}}{{\epsilon }_0}$$

Infinite line charge, with charge per unit length $\lambda$: a cylinder of radius r and length l gives $E\times 2\pi rl=\frac{\lambda l}{{\epsilon }_0}$, so $E=\frac{\lambda }{2\pi {\epsilon }_{0}r}$.

Infinite plane sheet, with charge per unit area $\sigma$: a cylinder crossing the sheet with end faces of area A gives $2EA=\frac{\sigma A}{{\epsilon }_0}$, so $E=\frac{\sigma }{2{\epsilon }_0}$ — the same at every distance.

Uniformly charged spherical shell, of charge q and radius R:

- Outside: $E\times 4\pi r^2=\frac{q}{{\epsilon }_0}$, so $E=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}$, as if all the charge were at the centre
- Inside: no charge is enclosed, so $E=0$

Worked example 1 — a line charge. A long wire with $\lambda =2.0\times 10^{-8}$ C m${}^{-1}$, at 0.10 m:

$$E=\frac{2\times 9.0\times 10^9\times 2.0\times 10^{-8}}{0.10}=3.6\times 10^3{\text{ N C}}^{-1}$$

Worked example 2 — a sheet. With $\sigma =1.77\times 10^{-6}$ C m${}^{-2}$ and ${\epsilon }_0=8.85\times 10^{-12}$ C${}^2$ N${}^{-1}$ m${}^{-2}$:

$$E=\frac{1.77\times 10^{-6}}{2\times 8.85\times 10^{-12}}=1.0\times 10^5{\text{ N C}}^{-1}$$

An everyday example. People inside a car struck by lightning are usually safe, because the charge stays on the outside of the metal body, leaving almost no field inside.

The substance. Gauss's theorem always holds, but it only gives E easily when symmetry makes E constant over the surface — for irregular charges, superposition is still needed.