Unit 8 · Topic 8.5

8.5 Electric Flux

Electric flux measures how much electric field passes through a surface. It depends on the field's strength, the surface's area and the angle between them, and it's the quantity that Gauss's law (8.6) is built on.

Key terms

electric flux
area vector
dot product
closed surface
surface integral

Flux through a flat surface

Picture field lines passing through a window. The flux counts how many go through. For a uniform field and a flat surface of area A, $\Phi_E = \vec{E}\cdot\vec{A} = EA\cos\theta$

Flux has units of N·m²/C and is a scalar. The angle θ is between the field and the area vector, not between the field and the surface itself.

The area vector

The area vector $\vec{A}$ has a magnitude equal to the area and points perpendicular to the surface. A flat surface has two possible normal directions; for a closed surface (one with no holes, like a sphere or box), you always choose the outward one.

Orientation	θ	Flux
Field straight through the surface	0°	EA (largest)
Field at 60° to the area vector	60°	½EA
Field parallel to the surface	90°	0
Field straight in through a closed surface	180°	−EA

Nonuniform fields and curved surfaces

If the field changes over the surface, or the surface is curved, break it into tiny patches dA, find $\vec{E}\cdot d\vec{A}$ for each, and add them: $\Phi_E = \displaystyle\int\vec{E}\cdot d\vec{A}$ . For a closed surface you write $\oint$ , and the circle reminds you the surface wraps all the way around.

In practice, you'll pick surfaces where this is easy: patches where the field is parallel to dA (contributing E dA) or perpendicular to it (contributing zero). If E varies along a flat face, like $E = bx$ across a strip, write dA as a thin strip and integrate.

Signs for closed surfaces

With outward area vectors, field leaving a closed surface gives positive flux and field entering gives negative flux. In a uniform field, every line that enters a closed surface also leaves it, so the net flux is zero, whatever the shape.

Now put a point charge q at the center of a sphere of radius r. The field is $\dfrac{kq}{r^2}$ everywhere on the sphere and points straight out, so the flux is $\dfrac{kq}{r^2}\cdot 4\pi r^2 = 4\pi kq = \dfrac{q}{\varepsilon_0}$ . The radius cancels. A bigger sphere has a weaker field over a larger area, and the two effects balance exactly. In field-line terms, the same number of lines crosses every sphere around the charge.

Net flux through a closed surface is only nonzero when field lines start or end inside it, which means there's charge inside. That's the idea Gauss's law makes exact. A handy trick: the flux through any open surface that shares an edge with a flat one (like a hemisphere on a disk) is the same as through the flat one, if no charge sits between them.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Flux through a tilted square
A uniform field of 300 N/C passes through a flat square 0.20 m on a side. The square's area vector makes a 30° angle with the field. Find the flux.
Show the solutionHide the solution
1. Step 1: A = (0.20)² = 0.040 m².
2. Step 2: $\Phi_E = EA\cos\theta = (300)(0.040)\cos 30^\circ \approx 10.4$ N·m²/C.
Answer: About 10 N·m²/C
Example 2Calculator allowed
A cube in a uniform field
A cube with sides of length a sits in a uniform field $\vec{E}$ pointing in the +x direction, with two faces perpendicular to the field. Find the flux through each face and the net flux.
Show the solutionHide the solution
1. Step 1: The face where the field leaves has its outward area vector along +x: flux $+Ea^2$ .
2. Step 2: The face where the field enters has its outward area vector along −x: flux $-Ea^2$ .
3. Step 3: The other four faces are parallel to the field, so θ = 90° and the flux is zero.
4. Step 4: Net flux: $Ea^2 - Ea^2 + 0 = 0$ . There is no charge inside.
Answer: $+Ea^2$ out of one face, $-Ea^2$ into the opposite face, 0 through the others; net 0
Example 3Calculator allowed
A cube in a field that grows with x
The field in a region is $\vec{E} = bx\,\hat{x}$ with b = 200 N/(C·m). A cube 0.10 m on a side has faces at x = 0.10 m and x = 0.20 m. Find the net flux through the cube. Is there charge inside?
Show the solutionHide the solution
1. Step 1: Only the two faces perpendicular to x have flux; the field runs along the other four.
2. Step 2: Right face (x = 0.20 m), field leaving: $+b(0.20)(0.10)^2 = +0.40$ N·m²/C.
3. Step 3: Left face (x = 0.10 m), field entering: $-b(0.10)(0.10)^2 = -0.20$ N·m²/C.
4. Step 4: Net flux: +0.20 N·m²/C. More field leaves than enters, so there must be positive charge inside. Gauss's law (8.6) gives its value: ε₀Φ ≈ 1.8 × 10⁻¹² C.
Answer: +0.20 N·m²/C; yes, about +1.8 × 10⁻¹² C inside

Common mistakes

Measuring θ from the surface instead of from the area vector (the normal). A field skimming along the surface has zero flux, not maximum flux.
Concluding that zero net flux means zero field. A uniform field passes through a closed box with zero net flux.
Pointing area vectors inward on a closed surface. Always use outward normals, so field leaving counts as positive.

On the exam

Expect conceptual flux questions: rank the flux through tilted surfaces, or decide whether the net flux through a closed surface changes when a charge inside moves (it doesn't).
When a field varies over a flat face, set up $\int E\,dA$ with a thin strip of area, and show how dA depends on the variable.

Connected topics

Videos

AP Physics C E&M - Unit 8 - Lesson 5 - Electric Flux
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Electric flux and Gauss's law (part 1) | AP Physics | Khan Academy
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Check yourself

4 questions on 8.5 Electric Flux. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

A uniform electric field of 500 N/C passes through a flat surface of area 0.20 m². The surface's area vector makes a 60° angle with the field. What is the electric flux through the surface?

Question 2 of 4Calculator allowed

A closed cube with no charge inside sits in a uniform electric field that points along +x. What is the net flux through the cube?

Question 3 of 4Calculator allowed

A hemisphere of radius R sits with its flat circular opening perpendicular to a uniform field E. What is the magnitude of the flux through the curved surface?

Question 4 of 4Calculator allowed

The electric field in a region is $\vec{E} = (3.0\,\hat{x} + 4.0\,\hat{y})$ N/C. What is the flux through a flat 2.0 m² square lying in the xz-plane, taking the area vector along +y?

0 of 4 answered