Many houses in Jaipur now carry solar panels on their roofs. A panel that faces the midday Sun squarely gives its full power. Tilt the same panel away from the Sun and its output falls, although the sunlight is just as bright. Turn it edge-on to the Sun and it gives almost nothing. The panel has not changed; only the way it faces the light has.
A surface faces some way
A flat surface has a size, its area \(A\). It also faces some way. A sheet of paper lying on a desk faces up, and the same sheet pinned to a wall faces sideways. To record both facts in one quantity, draw an arrow at right angles to the surface, with a length equal to its area. This arrow is the area vector \(\vec A\).
The direction at right angles to a surface is its normal. A flat surface has two normals, one on each side, so for an open surface we choose one and say which. For a closed surface, such as a ball or a box, there is no choice. The normal always points outward, away from the space the surface encloses.
The area vector \(\vec A\) of a small flat surface has a size equal to its area and points along its normal. On a closed surface, every area vector points along the outward normal.
The shadow of a tilted surface
Tilt a flat surface so that its normal makes an angle \(\theta\) with some direction. Seen along that direction, the surface looks smaller. Its shadow on a screen at right angles to that direction has area \(A\cos\theta\). This is the component of \(\vec A\) along that direction, found exactly as for a force or a displacement.
A curved surface is cut into many small patches, each nearly flat, each with its own small area vector \(d\vec A\). These add as vectors. On a closed surface they add to zero. The box in Fig 1.1 shows why: opposite faces have equal and opposite area vectors. Any closed shape casts each shadow twice, once from the front and once from the back, with opposite signs.
A closed cubical box has edges of \(2\,\text{m}\). What is the sum of the area vectors of its six faces?
How much field passes through a surface
In the chapter on Fluid Flow, water moving at speed \(v\) carries a volume \(vA\) each second through a ring of area \(A\). The ring is held square to the flow. Tilt the ring and less water passes through, because the ring's shadow across the flow shrinks. An electric field does not flow. But the same product measures how much of the field passes through a surface.
The electric flux \(\Phi\) through a flat surface in a uniform field \(\vec E\) is the dot product of the field and the area vector. Let \(\theta\) be the angle between \(\vec E\) and the normal. Then \(\Phi = EA\cos\theta\): the field times the shadow area \(A\cos\theta\). Its unit is \(\text{N m}^2/\text{C}\), which is also \(\text{V m}\).
The sign follows the angle, as the sign of work followed the angle between a force and a displacement. For \(\theta\) below \(90^\circ\) the field crosses along the normal and the flux is positive. At \(\theta = 90^\circ\) the field runs along the surface and the flux is zero. Above \(90^\circ\) the field crosses against the normal and the flux is negative.
Field lines are drawn with their spacing set by the field, so the number of lines crossing a surface measures the flux. Move the slider. The square has \(A\) \(= 0.2 \times 0.2\) \(= 0.04\,\text{m}^2\), so it carries \(500 \times 0.04\) \(= 20\,\text{N m}^2/\text{C}\) when it faces the field. At \(60^\circ\) the flux is \(20 \times \tfrac12\) \(= 10\,\text{N m}^2/\text{C}\), and 6 of the 12 lines still cross. At \(90^\circ\) no line crosses and the flux is zero.
The solar panel in Jaipur obeys the same rule. Sunlight crossing the panel is set by the panel's shadow across the Sun's rays, \(A\cos\theta\). A panel whose normal is \(60^\circ\) from the Sun has half its area as a shadow, so it catches half the light. Turned edge-on, it casts no shadow across the rays and catches none.
The field in a region is \(\vec E\) \(= (3\,\hat i + 4\,\hat j)\,\text{N/C}\). Find the flux through a flat area of \(2\,\text{m}^2\) whose normal points along (a) \(\hat i\), (b) \(\hat j\), (c) \(\hat k\).
(a) \(\vec A = 2\,\hat i\,\text{m}^2\), so \(\Phi\) \(= \vec E\cdot\vec A\) \(= 3 \times 2\) \(= 6\,\text{N m}^2/\text{C}\). Only the \(x\) component of the field crosses this area.
(b) \(\vec A = 2\,\hat j\,\text{m}^2\), so \(\Phi\) \(= 4 \times 2\) \(= 8\,\text{N m}^2/\text{C}\).
(c) \(\vec E\) has no \(z\) component, so \(\Phi = 0\). The field runs along this area and never crosses it.
Values used: none. Flux needs only the field and the area vector.
A uniform field of \(200\,\text{N/C}\) passes through a flat area of \(0.5\,\text{m}^2\). The field makes \(60^\circ\) with the normal to the area. What is the flux through the area, in N m²/C?
The field through a flat surface makes \(120^\circ\) with the chosen normal. The flux through the surface is
Curved surfaces and changing fields
When the surface curves or the field changes from place to place, cut the surface into small patches. Each patch is nearly flat, with its own area vector \(d\vec A\), and the field over it is nearly uniform. Its flux is \(\vec E\cdot d\vec A\). The flux through the whole surface is the sum of these, an integral over the surface.
A square of side \(2\,\text{m}\) lies in the \(xy\) plane, with one corner at the origin and sides along the axes. The field is \(\vec E = 5x\,\hat k\), in N/C with \(x\) in metres. Take the normal along \(\hat k\). Find the flux through the square.
The field changes with \(x\) but not with \(y\). So cut the square into strips of width \(dx\) and length \(2\,\text{m}\), parallel to the \(y\) axis. Each strip has \(d\vec A = 2\,dx\,\hat k\).
The flux through one strip is \(5x \times 2\,dx = 10x\,dx\).
Add the strips from \(x = 0\) to \(x = 2\): \(\Phi\) \(= \int_0^2 10x\,dx\) \(= 5x^2\,\big|_0^2\) \(= 20\,\text{N m}^2/\text{C}\).
On a closed surface the normals point outward. Lines leaving the surface give positive flux, and lines entering give negative flux. Place a closed cylinder in a uniform field with its axis along the field, as in Fig 1.2. The field enters the left face against its outward normal, \(-EA\), and leaves the right face along it, \(+EA\). Along the curved side the field runs parallel to the surface, so that part carries no flux. The total is zero.
The closed hemisphere of Fig 1.2 works the same way. Its flat face, of radius \(R = 10\,\text{cm}\), stands across a field of \(500\,\text{N/C}\), with its outward normal against the field. So \(\Phi_\text{flat}\) \(= -E\pi R^2\) \(= -500 \times \pi \times 0.01\) \(= -5\pi\) \(\approx -15.7\,\text{N m}^2/\text{C}\). The total must be zero, so the dome carries \(+5\pi\) \(\approx +15.7\,\text{N m}^2/\text{C}\). Every line that enters through the flat face leaves through the dome.
The sign of a flux means nothing until the normal is fixed. For an open surface, say which normal you chose. For a closed surface the normal is always outward, so a negative total means more field enters than leaves.
A hemispherical bowl of radius \(20\,\text{cm}\) stands in a uniform field of \(1000\,\text{N/C}\). Its open circular rim faces the field squarely, so the field enters the open side. What is the flux through the curved surface of the bowl, counted along the field, in N m²/C?
A closed cylinder of cross-section \(A\) lies in a uniform field \(E\), with its axis along the field. Which statements are true?
Michael Faraday had little mathematics, so he pictured forces instead of writing equations for them. From the 1830s he imagined the space around charges and magnets filled with lines of force. He judged the strength of a force by how closely the lines crowded together. Many scientists of his time thought the lines were only a drawing aid. Flux makes Faraday's picture exact: it counts the lines that cross a surface.
- Measuring \(\theta\) from the surface instead of from its normal.In \(\Phi = EA\cos\theta\), \(\theta\) is the angle between the field and the normal. A field along the surface has \(\theta = 90^\circ\) and no flux.
- Drawing an inward normal on a closed surface.On a closed surface every area vector points outward, so entering lines give negative flux.
- Adding the fluxes of the faces of a closed surface as if each were positive.Give each face its sign from its own outward normal before adding.