Theory

Electric flux and Gauss's law

What the flux of the electric field through a surface measures, why the outward flux through a closed surface depends only on the enclosed charge, how Gauss's law gives the field of distributions with spherical, cylindrical and planar symmetry, and what its local form is.

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Electric flux
Flux through a closed surface

Electric flux

Throughout the topic the charges are at rest in an inertial reference frame and are in vacuum.

Let $S$ be an oriented surface, with unit normal vector $\overset{n}{^}$ . The electric flux through $S$ is

Φ_{E} = \iint_{S} E \cdot \overset{n}{^} d A = \iint_{S} E cos θ d A,

where $θ$ is the angle between $E$ and $\overset{n}{^}$ at the point of the surface. The flux is a scalar and is measured in $N \cdot m^{2} /C$ . The product $\overset{n}{^} d A$ is also denoted by $d A$ and is called the vector area element.

If the surface is flat, with area $A$ , and the field is uniform, the integrand is constant and

Φ_{E} = E A cos θ .

The flux is maximum when the field is perpendicular to the surface, that is, parallel to $\overset{n}{^}$ , and it is zero when the field is parallel to the surface. A square of side $0.10 m$ in a uniform field of $2.0 \cdot 1 0^{3} N/C$ making the angle $θ = 3 0^{\circ}$ with the normal is crossed by the flux $2.0 \cdot 1 0^{3} \cdot 0.010 \cdot cos 3 0^{\circ} N \cdot m^{2} /C \approx 17 N \cdot m^{2} /C$ .

Flat surface

S

in a uniform field; the flux

Φ_{E} = E A cos θ

depends on the angle

θ

between

E

and the normal

\overset{n}{^}

Flux and field lines

Field lines are drawn with a density proportional to the magnitude of the field, that is, so that the number of lines crossing a small surface perpendicular to the field is proportional to $E d A$ . With this convention the flux through a surface is proportional to the number of lines crossing it, counted with the plus sign if they pass in the direction of $\overset{n}{^}$ and with the minus sign if they pass in the opposite direction. When a flat surface is tilted with respect to a uniform field, the lines crossing it decrease in proportion to $cos θ$ , in agreement with the formula $E A cos θ$ . For the construction of field lines, see the topic on the electric field.

Flux through a closed surface

Outward flux through a closed surface

A closed surface $S$ , such as a sphere or the surface of a cube, divides space into a bounded interior region and an exterior region. On a closed surface, by convention, the unit normal vector chosen at each point is the one pointing outward from the interior region, and the flux

Φ_{E} = \oint_{S} E \cdot \overset{n}{^} d A

is called the outward flux through $S$ ; the circle on the integral sign indicates that the surface is closed. Where the field leaves the surface the contribution to the flux is positive, where it enters it is negative. A positive outward flux therefore indicates that the field lines leaving $S$ outnumber those entering, and a zero flux that they are equal in number.

Cube in a uniform field

A cube of edge $ℓ$ is in a uniform field $E = E \overset{}{^}$ , with its faces parallel to the coordinate planes. On the face through which the field enters, the outward unit vector is $- \overset{}{^}$ , so the flux is $- E ℓ^{2}$ ; on the opposite face the outward unit vector is $\overset{}{^}$ and the flux is $+ E ℓ^{2}$ . On the other four faces the field is parallel to the surface, therefore the flux is zero. The total outward flux is

Φ_{E} = - E ℓ^{2} + E ℓ^{2} + 0 = 0.

Every field line that enters the cube leaves it, and the two contributions cancel. With $E = 2.0 \cdot 1 0^{3} N/C$ and $ℓ = 0.10 m$ each of the two faces perpendicular to the field is crossed by a flux of magnitude $20 N \cdot m^{2} /C$ , and the total flux is zero.

Cube in a uniform field, seen from the side; the flux is

- E ℓ^{2}

through face 1,

+ E ℓ^{2}

through face 2 and zero through the faces parallel to the field, so the total flux is zero

The theory continues with 6 more sections (gausss law, computing the field with gausss law, distributions with spherical symmetry…), plus all worked examples, the solutions and the PDF.

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