Why the Field Inside a Charged Metal Shell Is Exactly Zero

Adding the fields of countless tiny charges one by one is hopeless for a charged wire or sheet. Gauss's law offers a shortcut: count the flux through a closed surface, and the enclosed charge tells you the field. Along the way we meet the electric dipole, the pair of charges behind much of molecular physics.

This part covers electric flux, the electric dipole, the torque on a dipole, and Gauss's law with its applications.

**Electric flux measures how much electric field passes through a surface; for a uniform field $\vec{E}$ through a flat area $\vec{S}$ it is $\varphi =\vec{E}\cdot \vec{S}=ES\cos \theta$, where $\theta$ is the angle between the field and the surface's normal.

Worked example.** A uniform field of $500$ N C${}^{-1}$ passes through a square of side $0.1$ m, area $0.01$ m${}^2$.

- Field along the normal, $\theta =0^{\circ }$: $\varphi =500\times 0.01=5$ N m${}^2$ C${}^{-1}$
- At $\theta =60^{\circ }$: $\varphi =5\cos 60^{\circ }=2.5$ N m${}^2$ C${}^{-1}$
- Field parallel to the surface, $\theta =90^{\circ }$: $\varphi =0$

Signs for a closed surface. Flux leaving counts as positive and flux entering as negative, so a uniform field gives zero net flux through a closed cube.

An everyday example. A sieve held under a tap catches the most water when it faces the stream squarely and almost none when tilted edge-on — flux depends on $\cos \theta$ in the same way.

The substance. Flux is a scalar, even though it is built from two vectors.

**An electric dipole is a pair of equal and opposite charges $\pm q$ separated by $2a$, with dipole moment $p=q\times 2a$ pointing from $-q$ to $+q$; far away, its field 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.

Axial line**, at distance $r$ from the centre:

$$E_{axial}=\frac{1}{4\pi {\epsilon }_0}\frac{2pr}{(r^2-a^2{)}^2}\approx \frac{1}{4\pi {\epsilon }_0}\frac{2p}{r^3}$$

directed along $\vec{p}$.

Equatorial line, at distance $r$ on the perpendicular bisector:

$$E_{eq}=\frac{1}{4\pi {\epsilon }_0}\frac{p}{(r^2+a^2{)}^{3/2}}\approx \frac{1}{4\pi {\epsilon }_0}\frac{p}{r^3}$$

directed opposite to $\vec{p}$.

Worked example. Charges $\pm 2\times 10^{-9}$ C, $2$ mm apart, give $p=2\times 10^{-9}\times 2\times 10^{-3}=4\times 10^{-12}$ C m. At $r=0.1$ m on the axis:

$$E=\frac{9\times 10^9\times 2\times 4\times 10^{-12}}{(0.1{)}^3}=\frac{0.072}{0.001}=72{\text{ N C}}^{-1}$$

At the same distance on the equatorial line, $E=36$ N C${}^{-1}$.

An everyday example. A water molecule behaves like a tiny dipole, which is why water dissolves common salt so well — its charged ends pull sodium and chloride ions apart.

The substance. **A dipole's field falls as $\frac{1}{r^3}$, faster than a point charge's $\frac{1}{r^2}$**, because its two opposite charges partly cancel.

**In a uniform field the forces $+q\vec{E}$ and $-q\vec{E}$ on the dipole are equal and opposite, so the net force is zero, but they form a couple with torque $\tau =pE\sin \theta$, or $\vec{\tau }=\vec{p}\times \vec{E}$.

Derivation:**

- The force on $+q$ is $q\vec{E}$; the force on $-q$ is $-q\vec{E}$
- The net force is zero, so the dipole's centre does not accelerate
- The two forces act along parallel lines a perpendicular distance $2a\sin \theta$ apart, so

$$\tau =qE\times 2a\sin \theta =pE\sin \theta$$

The torque turns $\vec{p}$ towards the direction of $\vec{E}$.

Worked example. A dipole with $p=4\times 10^{-9}$ C m makes $30^{\circ }$ with a field of $5\times 10^4$ N C${}^{-1}$:

$$\tau =4\times 10^{-9}\times 5\times 10^4\times \sin 30^{\circ }=1\times 10^{-4}\text{ N m}$$

An everyday example. A charged comb attracts neutral bits of paper because it induces tiny dipoles in the paper, and in the comb's non-uniform field those dipoles feel a net pull.

The substance. Zero net force does not mean zero effect — a couple can still rotate a body.

**Gauss's law states that the total electric flux through any closed surface equals $\frac{q_{enc}}{{\epsilon }_0}$; choosing a surface that matches the symmetry gives $\frac{\lambda }{2\pi {\epsilon }_{0}r}$ for a long wire, $\frac{\sigma }{2{\epsilon }_0}$ for an infinite sheet, and $\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}$ outside but zero inside a charged shell.

The law:**

$$\oint \vec{E}\cdot d\vec{S}=\frac{q_{enc}}{{\epsilon }_0}$$

Long straight wire, with charge $\lambda$ per unit length: a cylinder of radius $r$ and length $l$ gives $E\cdot 2\pi rl=\frac{\lambda l}{{\epsilon }_0}$, so

$$E=\frac{\lambda }{2\pi {\epsilon }_{0}r}$$

Infinite plane sheet, with charge $\sigma$ per unit area: a cylinder piercing the sheet with end faces of area $A$ gives $2EA=\frac{\sigma A}{{\epsilon }_0}$, so

$$E=\frac{\sigma }{2{\epsilon }_0}$$

Thin spherical shell of charge $q$ and radius $R$, using a concentric sphere of radius $r$:

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

Worked example. A long wire carries $\lambda =2\times 10^{-8}$ C m${}^{-1}$. At $r=0.1$ m, using $\frac{1}{2\pi {\epsilon }_0}=1.8\times 10^{10}$ N m${}^2$ C${}^{-2}$:

$$E=\frac{1.8\times 10^{10}\times 2\times 10^{-8}}{0.1}=3600{\text{ N C}}^{-1}$$

An everyday example. Sitting inside a car during a thunderstorm is relatively safe, because charge stays on the metal body and the field inside stays close to zero.

The substance. The field of an infinite sheet does not depend on distance — near or far, it is $\frac{\sigma }{2{\epsilon }_0}$.