Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

We now compute the central quantity of this section: the outward flux of $D$ of a single point charge through a closed surface. We start with a sphere centred on the charge, where everything is simple, and then say what happens for any other closed surface around the charge.

A sphere centred on the charge

Let the charge $q$ sit at the origin, and let $S_{R}$ be the sphere of radius $R$ centred on it (Figure 3.2.4a). On $S_{R}$ :

the outward normal is $\overset{n}{^} = a_{r}$ , the same direction as $D$ ,
the magnitude of $D$ is the same at every point, $q / (4 π R^{2})$ , because every point of $S_{R}$ is at the same distance $R$ from the charge, and
the surface element of a sphere of radius $R$ in spherical coordinates (Section 2.3) is the little curved rectangle of sides $R d θ$ (along a meridian) and $R sin θ d ϕ$ (along a circle of latitude, whose radius is $R sin θ$ ), Figure 3.2.4b:

d S_{R} = d S a_{r} = meridian (R d θ) latitude (R sin θ d ϕ) a_{r} = R^{2} sin θ d θ d ϕ a_{r} . (3.2.6)

Since $a_{r} \cdot a_{r} = 1$ ,

D \cdot d S_{R} = \frac{q}{4 π R ^{2}} R^{2} sin θ d θ d ϕ = \frac{q}{4 π} sin θ d θ d ϕ,

and the radius $R$ has cancelled. Integrating over the whole sphere ( $θ$ from $0$ to $π$ , $ϕ$ from $0$ to $2 π$ ),

\oint_{S_{R}} D \cdot d S_{R} = \frac{q}{4 π} \int_{ϕ = 0}^{2 π} \int_{θ = 0}^{π} sin θ d θ d ϕ = \frac{q}{4 π} ϕ integral (2 π) = 1 - (- 1) = 2 [- cos θ]_{0}^{π} = \frac{q}{4 π} (2 π) (2) = q . (3.2.7)

The outward flux through the sphere is exactly $q$ , whatever its radius. This is the precise version of the picture of Figure 3.2.1: the factor $1/ R^{2}$ of Coulomb's law is cancelled by the factor $R^{2}$ of the area.

(a) The charge, the sphere S_R and an arbitrary surface S. — (a) The charge, the sphere $S_{R}$ and an arbitrary surface $S$ .

(b) The surface element dS=R^2\sin\theta\,d\theta\,d\phi. (a) A point charge q at the origin, enclosed by a sphere S_R of radius R centred on the charge and by an arbitrary closed surface S. On S_R the normal is \vec a_r, parallel to \vec D (orange arrows), whereas on S the normal \hat n of an element dS is in general tilted with respect to \vec a_r. The outward flux of \vec D is the same through both surfaces, namely q. (b) The surface element of the sphere is a small curved rectangle of sides R\,d\theta along the meridian and R\sin\theta\,d\phi along the circle of latitude of radius R\sin\theta. — **Figure 3.2.4.** (b) The surface element $d S = R^{2} sin θ d θ d ϕ$ . (a) A point charge $q$ at the origin, enclosed by a sphere $S_{R}$ of radius $R$ centred on the charge and by an arbitrary closed surface $S$ . On $S_{R}$ the normal is $a_{r}$ , parallel to $D$ (orange arrows), whereas on $S$ the normal $\overset{n}{^}$ of an element $d S$ is in general tilted with respect to $a_{r}$ . The outward flux of $D$ is the same through both surfaces, namely $q$ . (b) The surface element of the sphere is a small curved rectangle of sides $R d θ$ along the meridian and $R sin θ d ϕ$ along the circle of latitude of radius $R sin θ$ .

Common mistake · Forgetting the Jacobian

Writing $d S = d θ d ϕ$ on a sphere. Then the flux comes out as $(q /4 π R^{2}) (2 π) (π) = q π / (2 R^{2})$ , which depends on $R$ and even has the wrong unit. The element of a sphere of radius $R$ is $R^{2} sin θ d θ d ϕ$ : both factors $R$ and the factor $sin θ$ (the circles of latitude shrink towards the poles) are essential.

Quick check

What is the flux of $D$ of a point charge $q$ at the origin through the upper hemisphere ( $0 \leq θ \leq π /2$ ) of a sphere centred on it?

$\frac{q}{4 π} (2 π) [- cos θ]_{0}^{π /2} = \frac{q}{4 π} (2 π) (1) = \frac{q}{2}$ , as symmetry demands: half of the lines go up, half go down.

Any closed surface around the charge

The same result holds for an arbitrary closed surface $S$ around the charge (Figure 3.2.4a). The field lines of $q$ begin on the charge and never end in empty space, so the lines that cross $S$ are exactly the lines that cross the small sphere $S_{R}$ . The outward flux through $S$ is therefore the same as through $S_{R}$ , namely $q$ :

\oint_{S} D \cdot d S = \oint_{S_{R}} D \cdot d S_{R} = q .

If the charge lies outside $S$ , every line that enters $S$ also leaves it, and the net flux is zero.

Flux of a point charge

\oint_{S} D \cdot d S = {q, 0, q inside S, q outside S . (3.2.8)

In plain words

Wrap a charge in any closed surface you like, small or large, round or crumpled: exactly $q$ coulombs of flux come out. Wrap a surface beside the charge: whatever flows in, flows out again.

Why the inverse-square law matters

The argument used only two properties of the field (3.2.1): it is radial, and it falls off as $1/ r^{2}$ . The $1/ r^{2}$ is cancelled exactly by the $r^{2}$ growth of the area of a sphere. If the force between charges fell off as $1/ r^{2 + δ}$ with $δ \neq = 0$ , the flux through a sphere would depend on its radius (Figure 3.2.5) and Gauss's law would fail. Gauss's law is therefore not an extra assumption: it is Coulomb's law, rewritten.

A thought experiment: if the field of a point charge were \vec D=q\,r_0^{\delta}\,\vec a_r/(4\pi r^{2+\delta}), the flux through a concentric sphere of radius R would be \Psi=q\,(r_0/R)^{\delta}, different for every sphere. Only the exact inverse-square law (\delta=0) gives the same flux q through every surface. (Precision experiments confirm \delta=0 to extremely high accuracy.) — **Figure 3.2.5.** A thought experiment: if the field of a point charge were $D = q r_{0}^{δ} a_{r} / (4 π r^{2 + δ})$ , the flux through a concentric sphere of radius $R$ would be $Ψ = q (r_{0} / R)^{δ}$ , different for every sphere. Only the exact inverse-square law ( $δ = 0$ ) gives the same flux $q$ through every surface. (Precision experiments confirm $δ = 0$ to extremely high accuracy.)

A charge lying exactly on $S$ is neither inside nor outside, and the surface then catches only a part of its flux, as Exercise 3.2.8 shows for a charge at the corner of a cube. In practice we always choose surfaces that do not pass through point charges.

Quick check

A point charge $q$ sits at the centre of a cube. What is the flux of $D$ through one face?

The flux through the whole cube is $q$ . The six faces are identical as seen from the centre, so each carries $q /6$ (Exercise 3.2.8).

3.2.2 Outward flux of the field of a point charge

A sphere centred on the charge

Any closed surface around the charge