College Board AP 2021 (style)-style practice: Section I (multiple choice). A flat square of area A sits in a uniform field E. The field makes an angle of 60^ with the plane of the square. The flux through the square is (A) EA (B) EAcos 60^ (C) EAsin 60^ (D) zero. Justify your reasoning.

A 1-point MCQ on the angle in the flux dot product. The answer is (C). Flux is = E·A = EAcos, where is the angle between E and the area vector A (the normal). If E makes 60^ with the plane, it makes 30^ with the normal, so = EAcos 30^ = EAsin 60^. The trap is using the angle with the plane directly instead of with the normal.

College Board AP 2024 (style)-style practice: Section II (FRQ, quantitative and conceptual). A cube of side s sits in a uniform field E = E_0x. (a) Calculate the flux through the face at x = s (normal +x). (b) Calculate the flux through the face at x = 0 (normal -x). (c) Determine the net flux through the whole cube and explain what it implies about charge inside.

A 5-point FRQ on flux through a closed surface. (a) Right face (1 point): A = s^2x, so = E·A = E_0 s^2 (positive, field exits). (b) Left face (1 point): outward normal is -x, so A = -s^2x and = E·A = -E_0 s^2 (field enters). (c) Net flux and meaning (3 points): the four side faces have EA, contributing zero; the two end faces cancel: _net = E_0 s^2 - E_0 s^2 = 0. By Gauss's law _net = q_enc/_0, so zero net flux means zero net charge enclosed. Markers reward the signed flux on each face, the perpendicular side faces, and linking net flux to enclosed charge.

United StatesPhysics C: Electricity and MagnetismSyllabus dot point

What is electric flux, and how is it computed as a surface integral of the field?

Topic 8.5 Electric Flux: define electric flux as the surface integral of the field and compute it for uniform and non-uniform fields through flat and closed surfaces.

A calculus-based answer to AP Physics C E&M Topic 8.5, covering the area vector, the dot product, the flux surface integral, uniform-field and angle-dependent flux, and the net flux through a closed surface.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

Electric flux measures how much electric field passes through a surface. For a flat area in a uniform field, $\Phi = \vec{E}\cdot\vec{A} = EA\cos\phi$ , where $\vec{A}$ is the area vector (magnitude $A$ , direction along the surface normal) and $\phi$ is the angle between $\vec{E}$ and $\vec{A}$ . In general, $\Phi = \displaystyle\int \vec{E}\cdot d\vec{A}$ , a surface integral over field-area dot products. Flux is maximum when the field is perpendicular to the surface ( $\phi = 0$ ) and zero when parallel ( $\phi = 90^\circ$ ). For a closed surface, the outward normal defines positive flux; the net flux $\oint \vec{E}\cdot d\vec{A}$ counts field exiting minus field entering, and by Gauss's law equals $q_{enc}/\varepsilon_0$ .

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What this topic is asking
The area vector and the dot product
The flux surface integral
Net flux through a closed surface
Try this

What this topic is asking

The College Board (Topic 8.5) wants you to define electric flux as the surface integral of the field through a surface, compute it for uniform and non-uniform fields, and recognize that the net flux through a closed surface measures the charge inside. Flux is the bridge from the field to Gauss's law in the next topic.

The area vector and the dot product

The dot product captures the geometry: only the component of $\vec{E}$ along the normal, $E\cos\phi$ , threads the surface. A field parallel to the surface ( $\phi = 90^\circ$ ) gives zero flux; a field perpendicular to it ( $\phi = 0$ ) gives the maximum $EA$ . Flux is best pictured as the number of field lines crossing the surface: lines that graze along the surface cross nothing, while lines hitting it head-on all pass through. This line-counting picture is exactly why flux measures enclosed charge in Gauss's law, since field lines begin and end only on charge.

The flux surface integral

When the field varies or the surface is curved, divide the surface into patches $d\vec{A}$ , take the dot product on each, and sum:

\Phi = \int_S \vec{E}\cdot d\vec{A}

For a closed surface this becomes $\oint_S \vec{E}\cdot d\vec{A}$ , where the circle on the integral signals a closed surface and $d\vec{A}$ points outward everywhere.

Net flux through a closed surface

For a closed surface, field lines that leave count as positive flux and lines that enter count as negative. Field lines begin and end only on charge, so any line that enters an empty closed surface must also leave: with no charge inside, the net flux is exactly zero. Charge inside is the only way to get a non-zero net flux, which is the content of Gauss's law. A positive enclosed charge is a net source of field lines, giving positive net flux; a negative enclosed charge is a net sink, giving negative net flux. The shape of the surface is irrelevant: stretch or distort it however you like, and as long as it encloses the same charge the net flux is unchanged, because the same number of net lines must escape.

Flux of a uniform field through a tilted disk

A flat disk of radius $0.10$ m sits in a uniform field $E = 500$ N/C. The disk's normal makes a $30^\circ$ angle with the field. Find the flux through the disk.

step 1 Find the area

$A = \pi r^2 = \pi(0.10)^2 = 0.0314$ m squared.

step 2 Identify the angle in the dot product

The angle in $\Phi = EA\cos\phi$ is between the field and the area vector (the normal). Here that angle is given directly as $\phi = 30^\circ$ .

step 3 Compute the flux

$\Phi = EA\cos\phi = (500)(0.0314)\cos 30^\circ = (500)(0.0314)(0.866) = 13.6$ N m squared per C.

step 4 Sanity check the limits

If the disk faced the field ( $\phi = 0$ ), the flux would be $EA = 15.7$ ; if edge-on ( $\phi = 90^\circ$ ), it would be zero. The answer $13.6$ lies sensibly between.

Try this

Q1. A field of $200$ N/C passes perpendicularly through a $0.50$ m squared area. Find the flux. [1 point]

Cue. $\Phi = EA\cos 0 = (200)(0.50) = 100$ N m squared per C.

Q2. State the net electric flux through a closed surface that encloses no charge. [1 point]

Cue. Zero: entering flux equals exiting flux when no charge is inside.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2021 (style)1 marksSection I (multiple choice). A flat square of area

A

sits in a uniform field

E

. The field makes an angle of

60^\circ

with the plane of the square. The flux through the square is (A)

EA

(B)

EA\cos 60^\circ

(C)

EA\sin 60^\circ

(D) zero. Justify your reasoning.

Show worked answer →

A 1-point MCQ on the angle in the flux dot product. The answer is (C).

Flux is $\Phi = \vec{E}\cdot\vec{A} = EA\cos\phi$ , where $\phi$ is the angle between $\vec{E}$ and the area vector $\vec{A}$ (the normal). If $\vec{E}$ makes $60^\circ$ with the plane, it makes $30^\circ$ with the normal, so $\Phi = EA\cos 30^\circ = EA\sin 60^\circ$ . The trap is using the angle with the plane directly instead of with the normal.

AP 2024 (style)5 marksSection II (FRQ, quantitative and conceptual). A cube of side

s

sits in a uniform field

\vec{E} = E_0\hat{x}

. (a) Calculate the flux through the face at

x = s

(normal

+\hat{x}

). (b) Calculate the flux through the face at

x = 0

(normal

-\hat{x}

). (c) Determine the net flux through the whole cube and explain what it implies about charge inside.

Show worked answer →

A 5-point FRQ on flux through a closed surface.

(a) Right face (1 point): $\vec{A} = s^2\hat{x}$ , so $\Phi = \vec{E}\cdot\vec{A} = E_0 s^2$ (positive, field exits).
(b) Left face (1 point): outward normal is $-\hat{x}$ , so $\vec{A} = -s^2\hat{x}$ and $\Phi = \vec{E}\cdot\vec{A} = -E_0 s^2$ (field enters).
(c) Net flux and meaning (3 points): the four side faces have $\vec{E}\perp\vec{A}$ , contributing zero; the two end faces cancel: $\Phi_{net} = E_0 s^2 - E_0 s^2 = 0$ . By Gauss's law $\Phi_{net} = q_{enc}/\varepsilon_0$ , so zero net flux means zero net charge enclosed.

Markers reward the signed flux on each face, the perpendicular side faces, and linking net flux to enclosed charge.

Related dot points

Sources & how we know this

AP Physics C: Electricity and Magnetism Course and Exam Description — College Board (2024)