Electric Flux

Hold a window frame in a steady wind. How much air passes through it depends on three things: how fast the wind blows, how big the frame is, and how the frame is turned. Face-on, the most air gets through; edge-on, none does. Electric flux, written $\Phi_E$, is the same measurement for an electric field: how much field passes through a surface. In the field-line picture, it is proportional to the number of field lines crossing the surface. Gauss's law (topic 8.6) is a statement about flux.

Flux through a flat surface in a uniform field

To describe how a flat surface is turned, give it an area vector $\vec{A}$: its magnitude is the surface's area, and its direction is perpendicular (normal) to the surface. A surface is then fully described by one vector, and its tilt relative to the field is the angle $\theta$ between $\vec{E}$ and $\vec{A}$.

There are two ways to see where the cosine comes from. Split $\vec{E}$ into a part along the normal, $E\cos\theta$, and a part lying in the surface, $E\sin\theta$. Only the normal part carries field through; the part along the surface slides past it. Or keep the whole field and shrink the surface instead: seen from the direction the field comes from, a tilted surface looks like a smaller one, of area $A\cos\theta$, and that "shadow" is what the field lines have to cross.

For example, a square plate $0.40\ \text{m}$ on a side sits in a uniform field of $3.0\times10^{3}\ \text{N/C}$, with its normal at $60^\circ$ to the field. The flux is

Turn the plate face-on ($\theta = 0$) and the flux reaches its maximum, $EA = 480\ \text{N}\cdot\text{m}^2/\text{C}$. Turn it edge-on ($\theta = 90^\circ$, field parallel to the plate) and the flux is zero, even though the field at the plate is just as strong.

The sign of flux, and closed surfaces

A flat surface has two normals, pointing out of its two faces, and the formula needs you to pick one. Flux then has a sign, set by the dot product:

• $\theta < 90^\circ$: positive flux, the field crosses the surface in the direction of $\vec{A}$;
• $\theta = 90^\circ$: zero flux, the field runs along the surface;
• $\theta > 90^\circ$: negative flux, the field crosses against $\vec{A}$.

For an open surface (a flag, a window, a bowl: something with an edge), the choice of normal is yours, and the sign only means something relative to it. Flipping the normal flips the sign of the flux. State the choice and stick with it over the whole surface.

A closed surface (a box, a sphere, a can with both lids: something enclosing a volume with no gaps) removes the choice. By convention every area vector points outward, away from the enclosed volume. Then the sign has a physical meaning: positive flux through a patch means field leaving the volume there, and negative flux means field entering. The net flux through a closed surface, written $\oint\vec{E}\cdot d\vec{A}$, adds the leaving and entering parts with their signs.

In a uniform field, every field line that enters a closed surface also leaves it, so the net flux through any closed surface in a uniform field is zero, whatever its shape. Example 2 checks this face by face. Topic 8.6 shows what a nonzero net flux means: charge inside.

Flux when the field varies: the surface integral

When the field changes in size or direction across the surface, or the surface is curved, $EA\cos\theta$ has no single value to use. Cut the surface into patches $d\vec{A}$ small enough that each is flat and the field is constant across it, find the flux $\vec{E}\cdot d\vec{A}$ through each, and add them:

The skill is choosing patches on which $\vec{E}\cdot d\vec{A}$ is the same everywhere, so the integral has one variable:

• if the field changes along one direction across a flat surface, use thin strips perpendicular to that direction;
• if the field depends on distance from a point on a flat surface (a point charge facing a disk, as in Example 3), use thin rings, $dA = 2\pi s\,ds$;
• if the surface is a sphere or cylinder centered on the charge's symmetry, the field may be the same over the whole surface (topic 8.6 uses exactly this).

For example, take a square of side $s$ in the $xy$-plane, from $0$ to $s$ in both $x$ and $y$, with normal $+\hat{k}$, in the field $\vec{E} = Cx\,\hat{k} + B\,\hat{\imath}$, where $C$ and $B$ are constants. The $B\,\hat{\imath}$ part lies in the surface and contributes nothing. The $\hat{k}$ part depends only on $x$, so use strips of width $dx$ and length $s$, with $dA = s\,dx$:

With $C = 500\ \text{N}/(\text{C}\cdot\text{m})$ and $s = 0.20\ \text{m}$, $\Phi_E = 2.0\ \text{N}\cdot\text{m}^2/\text{C}$.

Curved surfaces in a uniform field have a shortcut. Cap an open curved surface with a flat one sharing its edge, and the two together form a closed surface, whose net flux in a uniform field is zero. With outward normals on that closed surface, $\Phi_{\text{curved}} + \Phi_{\text{cap, outward}} = 0$, where the cap's outward normal points away from the curved part. Flip the cap's normal to point the same way across as the curved surface's normals (for a dome bulging toward $+z$ with normals pointing away from its center, the cap's flipped normal is $+\hat{k}$), and $\Phi_{\text{curved}} = \Phi_{\text{cap, flipped}}$: both count field crossing from the same side to the other. In a uniform field, the flux through any surface depends only on its edge.

Worked examples

Worked exampleExample 2: every face of a closed wedge

A closed wedge-shaped surface sits in a uniform field $\vec{E} = (500\ \text{N/C})\,\hat{\imath}$. Its bottom is a $0.40\ \text{m}$ by $0.50\ \text{m}$ rectangle on the floor, its back is a vertical $0.30\ \text{m}$ by $0.50\ \text{m}$ rectangle facing $-\hat{\imath}$, its sloped top is a rectangle running from the top of the back down to the far edge of the bottom, and its two ends are right triangles. Find the flux through each face and the net flux.

Show solution

All area vectors point outward, because the surface is closed.

• Back face ($0.30 \times 0.50 = 0.15\ \text{m}^2$): outward normal $-\hat{\imath}$, opposite the field, so $\Phi = -(500)(0.15) = -75\ \text{N}\cdot\text{m}^2/\text{C}$. Field enters here.
• Bottom and both triangular ends: their normals ($-\hat{\jmath}$ and $\pm\hat{k}$) are perpendicular to $\hat{\imath}$, so the field runs along them and each has zero flux.
• Sloped face: its length is $\sqrt{0.30^2 + 0.40^2} = 0.50\ \text{m}$, so its area is $0.50 \times 0.50 = 0.25\ \text{m}^2$. Its outward normal points up and to the right, perpendicular to the slope: $\hat{n} = 0.6\,\hat{\imath} + 0.8\,\hat{\jmath}$ (going down the slope is the direction $0.8\,\hat{\imath} - 0.6\,\hat{\jmath}$, and the unit vector perpendicular to that, pointing out of the wedge, swaps the numbers and fixes the signs). Then $\Phi = (500\,\hat{\imath})\cdot(0.25)(0.6\,\hat{\imath} + 0.8\,\hat{\jmath}) = +75\ \text{N}\cdot\text{m}^2/\text{C}$. Field leaves here.

Net flux: $-75 + 0 + 0 + 0 + 75 = 0$. The sloped face is larger than the back, but it is tilted, and its shadow seen along the field is exactly the back's $0.30\ \text{m}$ by $0.50\ \text{m}$. The same field lines enter through one and leave through the other.

Where students lose points

On the exam

Expect to calculate flux (through a tilted surface, a face of a box, or a surface where the field varies), to compare fluxes by ranking or factor of change as the orientation or area changes, and to evaluate the flux side of Gauss's law, $\oint\vec{E}\cdot d\vec{A}$, which each derivation in topic 8.6 sets equal to $\frac{q_{\text{enc}}}{\varepsilon_0}$. In a derivation, state the patch you chose and why $\vec{E}\cdot d\vec{A}$ is constant on it, and give the sign of a flux with the direction of the area vector you used.